Selina ICSE Mathematics Class 8 Textbook Solutions – Page 2

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 8 Profit, Loss and Discount. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Profit, Loss and Discount Exercise 8A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Megha bought 10 note-books for Rs.40 and sold them at Rs.4.75 per note-book. Find, her gain percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 1

Question 2.
A fruit-seller buys oranges at 4 for Rs.3 and sells them at 3 for Rs.4 Find his profit percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 2

Question 3.
A man buys a certain number of articles at 15 for Rs. 112.50 and sells them at 12 for Rs.108. Find ;
(i) his gain as percent;
(ii) the number of articles sold to make a profit of Rs.75.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 4

Question 4.
A boy buys an old bicycle for Rs. 162 and spends Rs. 18 on its repairs before selling the bicycles for Rs. 207. Find his gain or loss percent.
Solution:
Buying price of the old bicycle = Rs.162
Money spent on repairs = Rs. 18
Real C.P. of the bicycle= 162+18 = Rs.180
S.P. of the bicycle = Rs.207
Profit = S.P. – C.P. = 207 – 162 = Rs. 45
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 6Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 6

Question 5.
An article is bought from Jaipur for Rs. 4,800 and is sold in Delhi for Rs. 5,820. If Rs. 1,200 is spent on its transportations, etc. ; find he loss or the gain as percent.
Solution:
Cost price = Rs. 4,800
Selling Price = Rs. 5,820
Transport etc. charges = Rs. 1,200
Total cost price = Rs, 4,800 + Rs. 1,200 = Rs. 6,000
Loss = Rs. 6000 – Rs. 5820 = Rs. 180
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 7

Question 6.
Mohit sold a T.V. for Rs. 3,600 ; gaining one-sixth of its selling price. Find :
(i) the gain
(ii) the cost price of the T.V.
(iii) the gain percent.
Solution:
S.P. of T.V. = Rs. 3,600
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 8

Question 7.
By selling a certain number of goods for Rs. 5,500; a shopkeeper loses equal to one-tenth of their selling price. Find :
(i) the loss incured
(ii) the cost price of the goods
(iii) the loss as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 9

Question 8.
The selling price of a sofa-set is \(\frac { 4 }{ 5 }\) times of its cost price. Find the gain or the loss as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 10

Question 9.
The cost price of an article is \(\frac { 4 }{ 5 }\) times of its selling price. Find the loss or the gain as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 11

Question 10.
A shopkeeper sells his goods at 80% of their cost price. Find the percent gain or loses ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 12

Question 11.
The cost price of an article is 90% of its selling price. What is the profit or the loss as percent ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 13

Question 12.
The cost price of an article is 30 percent less than its selling price. Find, the profit or loss as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 15

Question 13.
A shop-keeper bought 300 eggs at 80 paisa each. 30 eggs were broken in transaction and then he sold the remaining eggs at one rupee each. Find, his gain or loss as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 16

Question 14.
A man sold his bicycle for Rs.405 losing one-tenth of its cost price, find :
(i) its cc price;
(ii) the loss percent.
Solution:
(i) Let C.P. of the bicycle = Rs. x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 17

Question 15.
A man sold a radio-set for Rs.250 and gained one-ninth of its cost price. Find ;
(i) its cost price;
(ii) the profit percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 18

Profit, Loss and Discount Exercise 8B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the selling price, if:
(i) C.P. = Rs. 950 and profit = 8%
(ii) C.P. = Rs. 1,300 and loss = 13%
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 21

Question 2.
Find the cost price, if :
(i) S.P. = Rs. 1,680 and profit = 12%
(ii) S.P. = Rs. 1,128 and loss = 6%
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 22

Question 3.
By selling an article for Rs.900; a man gains 20%. Find his cost price and the gain.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 24

Question 4.
By selling an article for Rs.704; a person loses 12%. Find his cost price and the loss
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 25

Question 5.
Find the selling price, if :
(i) C.P. = Rs.352; overheads = Rs.28 and profit = 20
(ii) C.P. = Rs.576; overheads = Rs.44 and loss = 16%
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 26

Question 6.
If John sells his bicycle for Rs. 637, he will suffer a loss of 9%. For how much should it be sold, if he desires a profit of 5% ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 27

Question 7.
A man sells a radio-set for Rs.605 and gains 10%. At what price should he sell another radio of the same kind, in order to gain 16% ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 28

Question 8.
By selling a sofa-set for Rs.2,500; the shopkeeper loses 20%. Find his loss percent or profit percent ; if he sells the same sofa-set for Rs.3150.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 29

Question 9.
Mr. Sinha sold two tape-recorders for Rs.990 each; gaining 10% on one and losing 10% on the other. Find his total loss or gain as percent on the whole transaction.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 31

Question 10.
A tape-recorder is sold for Rs. 2,760 at a gain of 15% and a C.D. player is sold for Rs. 3,240 at a loss of 10% Find :
(i) the C.P. of the tape-recorder
(ii) the C.P. of the C.D. player.
(iii) the total C.P. of both.
(iv) the total S.P. of both
(v) the gain % or the loss % on the whole
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 33

Question 11.
Rajesh sold his scooter to Rahim at 8% loss and Rahim, in turn, sold the same scooter to Prem at 5% gain. If Prem paid Rs. 14,490 for the scooter ; find :
(i) the S.P. and the C.P. of the scooter for Rahim
(ii) the S.P. and the C.P. of the scooter for Rajesh
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 34

Question 12.
John sold an article to Peter at 20% profit and Peter sold it to Mohan at 5% loss. If Mohan paid Rs.912 for the article; find how much did John pay for it ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 35

Profit, Loss and Discount Exercise 8C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A stationer buys pens at 5 for Rs.28 and sells them at a profit of 25 %. How much should a customer pay; if he buys
(i) only one pen ;
(ii) three pens ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 37

Question 2.
A fruit-seller sells 4 oranges for Rs. 3, gaining 50%. Find :
(i) C.P. of 4 oranges,
(ii) C.P. of one orange.
(iii) S.P. of one orange.
(iv) profit made by selling one orange.
(v) number of oranges brought and sold in order to gain Rs. 24.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 38

Question 3.
A man sells 12 articles for Rs. 80 gaining 33\(\frac { 1 }{ 3 }\) %. Find the number of articles bought by the man for Rs. 90.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 40

Question 4.
The cost price of 20 articles is same as the selling price of 16 articles. Find the gain percent.
Solution:
C.P. of 20 articles = S.P. of 16 articles.
Let C.P. of 1 article = Re. 1
C.P. of 20 articles = Rs.20
and C.P. of 16 articles = Rs.16
S.P. of 16 articles = Rs.20
[S.P. of 16 articles = C.P. of 20 articles]
Gain = Rs.20 – Rs.16 = Rs.4
Gain% = \(\frac { 4 }{ 16 }\) x 100
= \(\frac { 4 x 100 }{ 16 }\)
= 25%

Question 5.
The selling price of 15 articles is equal to the cost price of 12 articles. Find the gain or loss as percent.
Solution:
S.P. of 15 articles = C.P. of 12 articles
Let C.P. of 1 article = Re.1
C.P. of 12 article = Rs.12
and C.P. of 15 articles = Rs.15
S.P. of 15 articles = Rs.12
[S.P. of 15 articles = C.P. of 12 articles]
Loss = Rs.15 – Rs.12 = Rs.3
Loss% = \(\frac { 3 }{ 15 }\) x 100 = 20%

Question 6.
By selling 8 pens, Shyam loses equal to the cost price of 2 pens. Find his loss percent.
Solution:
Let C.P. of 1 pen = Re.1
C.P. of 2 pens = Rs.2
and C.P. of 8 pens = Rs.8
Loss = Rs.2 ….[Loss = C.P. of 2 Pens]
Loss% = \(\frac { 2 }{ 8 }\) x 100 = 25%

Question 7.
A shop-keeper bought rice worth Rs.4,500. He sold one-third of it at 10% profit.
If he desires a profit of 12% on the whole ; find :
(i) the selling price of the rest of the rice ;
(ii) the percentage profit on the rest of the rice.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 41

Question 8.
Mohan bought a certain number of note-books for Rs.600. He sold \(\frac { 1 }{ 4 }\) of them at 5 percent loss. At what price should he sell the remaining note-books so as to gain 10% on the whole ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 43

Question 9.
Raju sells a watch at 5% profit. Had he sold it for Rs.24 more ; he would have gained 11%. Find the cost price of the watch.
Solution:
Let C.P. of the watch = Rs.100
When profit = 5%; S.P. = Rs.(100+5) = Rs.105
When profit = 11%;
S.P. = Rs.(100 + 11) = Rs .111
Difference of two selling prices = Rs. 111 – Rs. 105 = Rs.6
When watch sold for Rs.6 more; then C.P. of the watch = Rs.100
When watch sold for Re. 1 more; then C.P. of the watch = Rs. \(\frac { 100 }{ 6 }\)
When watch sold for Rs.24 more; then C.P. of the watch = Rs. \(\frac { 100 }{ 6 }\) x 24 = Rs.400

Question 10.
A man sold a bicycle at 5% profit. If the cost had been 30% less and the selling price Rs.63 less, he would have made a profit of 30%. What is the cost price of the bicycle ?
Solution:
Let C.P. of the bicycle = Rs.100
In the I case :
When Profit = 5% ;
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 45

Question 11.
Renu sold an article at a loss of 8 percent. Had she bought it at 10% less and sold for Rs.36 more; she would have gained 20%. Find the cost price of the article.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 46

Profit, Loss and Discount Exercise 8D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
An article is marked for Rs. 1,300 and is sold for Rs. 1,144 ; find the discount percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 48

Question 2.
The marked price of a dinning table is Rs. 23,600 and is available at a discount of 8%. Find its selling price.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 49

Question 3.
A wrist-watch is available at a discount of 9%. If the list-price of the watch is Rs. 1,400 ; find the discount given and the selling price of the watch.
Solution:
List price of the watch = Rs. 1,400
Discount = 9%
Discount = \(\frac { 1400 x 9 }{ 100 }\) = 14 x 9 = Rs. 126
S.P. = (List price – Discount) = Rs. (1400 – 126) = Rs. 1274

Question 4.
A shopkeeper sells an article for Rs. 248.50 after allowing a discount of 10%. Find the list price of the article.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 50

Question 5.
A shop-keeper buys an article for Rs.450. He marks it at 20% above the cost price. Find :
(i) the marked price of the article.
(ii) the selling price, if he sells the articles at 10 percent discount.
(iii) the percentage discount given by him, if he sells the article for Rs.496.80
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 51

Question 6.
The list price of an article is Rs.800 and is available at a discount of 15 percent. Find :
(i) selling price of the article ;
(ii) cost price of the article, if a profit of 13\(\frac { 1 }{ 3 }\) % is made on selling it.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 53

Question 7.
An article is marked at Rs. 2,250. By selling it at a discount of 12%, the dealer makes a profit of 10%. Find :
(i) the selling price of the article.
(ii) the cost price of the article for the dealer.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 55

Question 8.
By selling an article at 20% discount, a shopkeeper gains 25%. If the selling price of the article is Rs. 1,440 ; find :
(i) the marked price of the article.
(ii) the cost price of the article.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 56

Question 9.
A shop-keeper marks his goods at 30 percent above the cost price and then gives a discount of 10 percent. Find his gain percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 57

Question 10.
A ready-made garments shop in Delhi, allows 20 percent discount on its garments and still makes a profit of 20 percent. Find the marked price of a dress which is bought by the shop-keeper for Rs.400.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 58

Question 11.
At 12% discount, the selling price of a pen is Rs. 13.20. Find its marked price. Also, find the new selling price of the pen, if it is sold at 5% discount.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 60

Question 12.
The cost price of an article is Rs. 2,400 and it is marked at 25% above the cost price. Find the profit and the profit percent, if the article is sold at 15% discount.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 61

Question 13.
Thirty articles are bought at Rs. 450 each. If one-third of these articles be sold at 6% loss; at what price must each of the remaining articles be sold in order to make a profit of 10% on the whole?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 63

Question 14.
The cost price of an article is 25% below the marked price. If the article is available at 15% discount and its cost price is Rs. 2,400; find:
(i) Its marked price
(ii) its selling price
(iii) the profit percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 64

Question 15.
Find a single discount (as percent) equivalent to following successive discounts:
(i) 20% and 12%
(ii) 10%, 20% and 20%
(iii) 20%, 10% and 5%
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 65

Question 16.
Find the single discount (as percent) equivalent to successive discounts of:
(i) 80% and 80%
(ii) 60% and 60%
(iii) 60% and 80%
Solution:
(i) Successive discounts = 80% and 80%
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 68

Profit, Loss and Discount Exercise 8E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Rajat purchases a wrist-watch costing Rs. 540. The rate of Sales Tax is 8%. Find the total amount paid by Rajat for the watch.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 70
Total Amount of Watch = ₹ 540 + ₹ 43.20 = ₹ 583.20

Question 2.
Ramesh paid ₹ 345.60 as Sales Tax on a purchase of ₹ 3,840. Find the rate of Sales Tax.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 71

Question 3.
The price of a washing machine, inclusive of sales tax is ₹ 13,530/-. If the Sales Tax is 10%, find its basic (cost) price.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 72

Question 4.
Sarita purchases biscuits costing ₹ 158 on which the rate of Sales Tax is 6%. She also purchases some cosmetic goods costing ₹ 354 on which rate of Sales Tax is 9%. Find the total amount to be paid by Sarita.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 73
Total cost of cosmetic goods = ₹ 354 + ₹ 31.86 = ₹ 385.86
Total amount paid by Sarita = ₹ 167. 48 + 385.86 = ₹ 553. 34

Question 5.
The price of a T.V. set inclusive of sales tax of 9% is ₹ 13,407. Find its marked price. If Sales Tax is increased to 13%, how much more does the customer has to pay for the T.V. ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 74

Question 6.
The price of an article is ₹ 8,250 which includes Sales Tax at 10%. Find how much more or less does a customer pay for the article, if the Sales Tax on the article:
(i) increases to 15%
(ii) decreases to 6%
(iii) increases by 2%
(iv) decreases by 3%
Solution:
Price of an article = ₹ 8,250
Rate of Sales Tax = 10%
Let the list price = x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 75

Question 7.
A bicycle is available for ₹ 1,664 including Sales Tax. If the list price of the bicycle is ₹ 1,600, find :
(i) the rate of Sales Tax
(ii) the price a customer will pay for the bicycle if the Sales Tax is increased by 6%.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 78

Question 8.
When the rate of sale-tax is decreased from 9% to 6% for a coloured T.V. ; Mrs Geeta will save ₹ 780 in buying this T.V. Find the list price of the T.V.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 79

Question 9.
A shopkeeper sells an article for ₹ 21,384 including 10% sales-tax. However, the actual rate of sales-tax is 8%. Find the extra profit made by the dealer.
Solution:
Sale Price of an article including S.T. = ₹ 21384
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 80

Profit, Loss and Discount Exercise 8F – Selina Concise Mathematics Class 8 ICSE Solutions

[In this exercise, all the prices are excluding tax/VAT unless specified]
Question 1.
A shopkeeper buys an article for ₹ 8,000 and sells it for ₹ 10,000. If the rate of tax under VAT is 10%, find :
(i) tax paid by the shopkeeper
(ii) tax charged by the shopkeeper
(iii) VAT paid by the shopkeeper
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 81
(iii) VAT paid by the shopkeeper = ₹ 1000 – ₹ 800 = ₹ 200

Question 2.
A trader buys some goods for ₹ 12,000 and sells them for ₹ 15,000. If the rate of tax under VAT is 12%, find the VAT paid by the trader?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 82

Question 3.
The marked price of an article is ₹ 7,000 and is available at 20% discount. Manoj buys this article and then sold it at its marked price. If the rate of tax at each state is 10%, find the VAT paid by Manoj.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 83

Question 4.
A buys some goods for ₹ 4,000 and sold them to B for ₹ 5,000. B sold these goods to C for ₹ 6,000. If the rate of tax (under VAT) at each stage is 5%, find :
(i) VAT paid by A
(ii) VAT paid by B
Solution:
C.P. of some goods for A = ₹ 4000
C.P. of some goods for B = ₹ 5000
and C.P. for C = ₹ 6000
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 84

Question 5.
A buys an article for ₹ 8,000 and sold it to B at 20% profit. If the rate of tax under VAT is 8%, find :
(i) tax paid by A
(ii) tax charged by A
(iii) VAT paid by A
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 85

Question 6.
A shopkeeper purchases an article for ₹ 12,400 and sells it to a customer for ₹ 17,000. If the tax under VAT is 8%, find the VAT paid by the shopkeeper.
Solution:
C.P. of article = ₹ 12,400
Rate of VAT =8%
Total VAT = ₹ 12,400 x \(\frac { 8 }{ 100 }\) = ₹ 992
S.P. of the article = ₹ 17000
VAT charge 8% = ₹ 17000 x \(\frac { 8 }{ 100 }\) = ₹ 1360
Amount of VAT paid by the shopkeeper = ₹ 1360 – ₹ 992 = ₹ 368

Question 7.
A purchases an article for ₹ 7,200 and sells it to B for ₹ 9,600. B, in turn, sells the article to C for ₹ 11,000. If the tax (under VAT) is 10%, find the VAT paid by A and B.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 86

Question 8.
A manufacturer buys some goods for ₹ 60,000 and pays 5% tax. He sells these goods for ₹ 80,000 and charges tax at the rate of 12%. Find the VAT paid by the manufacturer.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Disco
VAT paid by the manufacturer = ₹ 9600 – ₹ 3000 = ₹ 6600

Question 9.
The cost of an article is ₹ 6,000 to a distributor, he sells it to a trader for ₹ 7,500 and the trader sells it further to a customer for ₹ 8,000. If the rate of tax under VAT is 8%; find the VAT paid by the:
(i) distributor
(ii) trader
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 88

Question 10.
The marked price of an article is ₹ 10,000. A buys it at 30% discount on the marked price and sells it at 10% discount on the marked price. If the rate of tax under VAT is 5%, find the amount of VAT paid by A.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount image - 89

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions (Using ruler and compass only)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions (Using ruler and compass only)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 18 Constructions (Using ruler and compass only). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Constructions Exercise 18A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Given below are the angles x and y.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 1
Without measuring these angles, construct :
(i) ∠ABC = x + y
(ii) ∠ABC = 2x + y
(iii) ∠ABC = x + 2y
Solution:

(i) Steps of Construction :

Draw a line segment BC of any suitable length.
With B as centre, draw an arc of any suitable radius. With the same radius, draw arcs with the vertices of given angles as centres. Let these arcs cut arms of the arc x at points P and Q and arms of angle y at points R and S.
From the arc, with centre B, cut DE = PQ arc of x and EF = RS arc of y
Join BF and produce upto point A.
Thus ∠ABC = x + y

(ii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 3
Proceed in exactly the same way as in part(i)
takes DE = PQ = arc of x.
EF = PQ = arc of x and FG = RS = arc of y.
Join BG and produce it upto A.
Thus ∠ABC = x + x+ y = 2x + y
(iii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 4
Proceed in exactly the same way as in (ii)
taking DE = PQ = arc of x. and EF = RS = arc of y and FG = RS = arc of y.
4. Join BF and produce upto point A.
Thus ∠ABC = x + y + y = x + 2y

Question 2.
Given below are the angles x, y and z.
Without measuring these angles construct :
(i) ∠ABC = x + y + z
(ii) ∠ABC = 2x + y + z
(iii) ∠ABC = x + 2y + z
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 5
Solution:
(i) Steps of Construction :

Draw line segment BC of any suitable length.
With B as centre, draw an arc of any suitable radius. With the same radius, draw arcs with the vertices of given angles as centres. Let these arcs cut arms of the anlge x at the points P and Q and arms of the angle y at points R and S and arms of the angle z at the points L and M.
From the arc, with centre B, cut
DE = PQ = arc of x, EF = RS = arc of y and FG = LM = arc of z.
Join BG and produce it upto A.
Then ∠ABC = x + y + z

(ii) Proceed as in part (i) upto step 2.
3. From the arc, with centre B, cut
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 8
DE = 2PQ = 2 arc of x
EF = RS = arc of y
FG = LM = arc of z
4. Join BG and produce it upto point A
Then ∠ABC = 2x + y + z
(iii) proceed as in (i) upto step 2

3. Here cut arc DE = arc PQ = arc of x arc EF = 2 arc RS = 2 arc of y arc FG = arc LM = arc of z.
4. Join BG and produce it upto A
5. Then ∠ABC = x + 2y + z

Question 3.
Draw a line segment BC = 4 cm. Construct angle ABC = 60°.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 10

Draw a line segment BC = 4 cm
With B as centre, draw an arc of any suitable radius which cuts BC at the point D.
With D as centre, and the same radius as in step 2, draw one more arc which cuts the previous arc at the point E.
Join BE and produce it to the point A.
Thus ∠ABC = 60°

Question 4.
Construct angle ABC = 45° in which BC = 5 cm and AB = 4.6 cm.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 11

Draw a line segment BC = 5 cm
Taking B as centre, draw an arc of any suitable radius, which cuts BC at the point D.
With D as centre and the same radius, as taken in step 2, draw an arc which cuts the previous arc at point E.
With E as centre and the same radius, draw one more arc which cuts the first arc at point F.
With E and F as centres and radii equal to more than half the distance between E at F, draw arc which cut each other at point P.
Join BP to meet EF at L and produce to point O. Then ∠OBC = 90°
Draw BA, the bisector of angle OBC. [With D, L as centres and suitable radius draw two arc meeting each other at Q produced it to R]
=> ∠ABC = 45° [∴ BA is bisector of ∠OBC ∴ ∠ABC = = 45°]
From BR cut arc AB = 4.6 cm

Question 5.
Construct angle ABC = 90°. Draw BP, the bisector of angle ABC. State the measure of angle PBC.
Solution:

Draw ∠ABC = 90° (as in Ques. 4)
Draw bisector of ∠ABC
Then ∠PBC = \(\frac { 1 }{ 2 }\) (90°) = 45°

Question 6.
6. Draw angle ABC of any suitable measure.
(i) Draw BP, the bisector of angle ABC.
(ii) Draw BR, the bisector of angle PBC and draw BQ, the bisector of angle ABP.
(iii) Are the angles ABQ, QBP, PBR and RBC equal?
(iv) Are the angles ABR and QBC equal ?
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 13

Construct any angle ABC
With B as centre, draw an arc EF meeting BC at E and AB at F.
With E, F as centres draw two arc of equal radii meeting each other at the point P.
Join BP. Then BP is the bisector of ∠ABC
∠ABP = ∠PBC = \(\frac { 1 }{ 2 }\) ∠ABC
Similarly draw BR, the bisector of ∠PBC and draw BQ as the bisector of ∠ABP [With the same method as in steps 2, 3]
Then ∠ABQ = ∠QBP = ∠PBR = ∠RBC
∠ABR = \(\frac { 3 }{ 4 }\) ∠ABC and ∠QBC = \(\frac { 3 }{ 4 }\) ∠ABC
∠ABR = ∠QBC.

Constructions Exercise 18B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Draw a line segment AB of length 5.3 cm. Using two different methods bisect AB.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 14

Draw a line segment AB = 5.3 cm
With A as centre and radius equal to more than half of AB, draw arcs on both sides of AB.
With B as centre and with the same radius as taken in step 2, draw arcs on both the sides of AB.
Let the arcs intersect each other at points P and Q.
Join P and Q.
The line PQ cuts the given line segment AB at the point O.
Thus, PQ is a bisector of AB such that
OA = OB = \(\frac { 1 }{ 2 }\) AB

Second Method
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 15
Steps of Construction :

Draw the given line segment AB = 5.3 cm.
At A, construct ∠PAB of any suitable measure. Then ∠PAB = 60° construct ∠QBA = 60°
3.From AP, cut AR of any suitable length and from BQ ; cut BS = AR.
Join R and S
Let RS cut the given line segment AB at the point O.
Thus RS is a bisector of AB such that OA = OB = \(\frac { 1 }{ 2 }\) AB

Question 2.
Draw a line segment PQ = 4.8 cm. Construct the perpendicular bisector of PQ.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 16

Draw a line segment PQ = 4.8 cm.
With P as centre and radius equal than half of PQ, draw arc on both the PQ.
With Q as centre and the same radius as taken in step 2, draw arcs on both sides of PQ.
Let the arcs intersect each other at point A and B
Join A and B.
The line AB cuts the line segment PQ at the point O. Here OP = OQ and ∠AOQ = 90°. Then the line AB is perpendicular bisector of PQ.

Question 3.
In each of the following, draw perpendicular through point P to the line segment AB :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 17
Solution:
(i) Steps of Construction :

With P as centre, draw an arc of a suitable radius which cuts AB at points C and D.
With C and D as centres, draw arcs of equal radii and let these arcs intersect each other at the point Q.
[The radius of these arcs must be more than half of CD and both the arcs must be drawn on the other side]
Join P and Q
Let PQ cut AB at the point O.
Thus, OP is the required perpendicular clearly, ∠AOP = ∠BOP = 90°

(ii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 19

With P as centre, draw an arc of any suitable radius which cuts AB at points C and D.
With C and D as centres, draw arcs of equal radii. Which intersect each other at point A.
[This radius must be more than half of CD and let these arc intersect each other at the point 0]
Join P and O. Then OP is the required perpendicular.
∠OPA = ∠OPB = 90°

(iii) Steps of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 20

With P as centre, draw an arc of any suitable radius which cuts AB at points C and D.
With C and D as centre, draw arcs of equal radii
[The radius of these arcs must be more than half of CD and both the arcs must be drawn on the other side.]
and let these arcs intersect each other at the point Q.
Join Q and P. Let QP cut AB at the point O. Then OP is the required perpendicular.
Clearly, ∠AOP = ∠BOP = 90°

Question 4.
Draw a line segment AB = 5.5 cm. Mark a point P, such that PA = 6 cm and PB = 4.8 cm. From the point P, draw perpendicular to AB.
Solution:
Step of Construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 21

Draw a line segment AB = 5.5 cm
With A as centre and radius = 6 cm, draw an arc.
With B as centre and radius = 4.8 cm draw another arc.
Let these arcs meet each other at the point P.
PA = 6 cm, PB = 4.8
With P as centre and some suitable radius draw an arc meeting AB at the points C and D.
With C as centre and radius more than half of CD, draw an arc.
With D as centre and same radius as in step 6, draw an arc.
Let these arcs meet each other at the point Q.
Join PQ.
The PQ meet AB at point O.
Then PO ⊥ AB i.e; ∠AOP = 90° = ∠POB.

Question 5.
Draw a line segment AB = 6.2 cm. Mark a point P in AB such that BP = 4 cm. Through point P draw perpendicular to AB.
Solution:
Steps of Construction :

Draw a line segment AB = 6.2 cm
Cut off BP = 4 cm
With P as centre and some radius draw arc meeting AB at the points C, D.
With C, D as centres and equal radii [each is more than half of CD] draw two arcs, meeting each other at the point O.
Join OP. Then OP is perpendicular for AB.

Constructions Exercise 18C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Draw a line AB = 6 cm. Mark a point P any where outside the line AB. Through the point P, construct a line parallel to AB.
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 23

Draw a line AB = 6 cm
Take any point Q on the line AB and join it with the given point P.
At point P, construct ∠CPQ = ∠PQB
Produce CP upto any point D.
Thus, CPD is the required parallel line.

Question 2.
Draw a line MN = 5.8 cm. Locate a point A which is 4.5 cm from M and 5 cm from N. Through A draw a line parallel to line MN.
Solution:
Steps of construction :

Draw a line MN = 5.8 cm
With M as centre and radius = 4.5 cm, draw an arc.
With N as centre draw another arc of radius 5 cm. These arcs intersect each other at A.
Join AM and AN.
At point A, draw ∠DAN = ∠ANM
Produce DA to any point C.
Thus CAD is the required parallel line.

Question 3.
Draw a straight line AB = 6.5 cm. Draw another line which is parallel to AB at a distance of 2.8 cm from it.
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 25

Draw a straight line AB = 6.5 cm
Taking point A as centre, draw an arc of radius 2.8 cm.
Taking B as centre, drawn another arc of radius 2.8 cm.
Draw a line CD which touches the two arcs drawn.
Thus CD is the required parallel line.

Question 4.
Construct an angle PQR = 80°. Draw a line parallel to PQ at a distance of 3 cm from it and another line parallel to QR at a distance of 3.5 cm from it. Mark the point of intersection of these parallel lines as A.
Solution:
Steps of construction :

Draw ∠PQR = 80°
With P as centre draw an arc of radius 2 cm.
Again with Q as centre, draw another arc of radius 2 cm. Then BM is a line which touches the two arcs. Then BM is a line parallel to PQ.
With Q as centre, draw an arc of radius 3.5 cm. With R as centre draw another arc of radius 3.5 cm. Draw a line HC which touches these two arcs. Let these two parallel line intersect at A.

Question 5.
Draw an angle ABC = 60°. Draw the bisector of it. Also draw a line parallel to BC a distance of 2.5 cm from it.
Let this parallel line meet AB at point P and angle bisector at point Q. Measure the length of BP and PQ. Is BP = PQ ?
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 27

Draw, ∠ABC = 60°
Draw BD, the bisector of ∠ABC.
Taking B as centre, draw an arc of radius 2.5 cm.
Taking C as centre, draw another arc of radius 2.5 cm.
Draw a line MN which touches these two arcs drawn. Then MN is the required line parallel to BC.
Let this line MN meets AB at P and bisector BD at Q.
Measure BP and PQ.
By measurement we see BP = PQ.

Question 6.
Construct an angle ABC = 90°. Locate a point P which is 2.5 cm from AB and 3.2 cm from BC.
Solution:
Steps of construction :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 28

Draw ∠ABC = 90°
From AB, cut BD = 3.2 cm.
Through point C, draw CH⊥BC. From CH, cut CE = 3.2. Join DE. Now DE is a line parallel to BC and at a distance of 3.2 cm from BC.
From BC cut BM = 2.5 cm.
Through point A, draw AK ⊥ AB. From AK cut AN = 2.5 cm. Join NM. Therefore NM is parallel to AB and at a distance of 2.5 cm from AB.
DE and MN intersect each other at P. Thus P is the required point which is 2.5 cm from AB and 3.2 cm from BC.

Constructions Exercise 18D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Construct a quadrilateral ABCD; if:
(i) AB = 4.3 cm, BC = 5.4, CD = 5 cm, DA = 4.8 cm and angle ABC = 75°.
(ii) AB = 6 cm, CD = 4.5 cm, BC = AD = 5 cm and ∠BCD = 60°.
(iii) AB = 8 cm, BC = 5.4 cm, AD = 6 cm, ∠A = 60° and ∠B = 75°.
(iv) AB = 5 cm, BC = 6.5 cm, CD =4.8 cm, ∠B = 75° and ∠C = 120°.
(v) AB = 6 cm = AC, BC = 4 cm, CD = 5 cm and AD = 4.5 cm.
(vi) AB = AD = 5cm, BD = 7 cm and BC = DC = 5.5 cm
Solution:
(i) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 29
Steps :

Draw AB = 4.3 cm.
At B, draw ∠PBA = 75°
Cut BC = 5.4 cm.
From C & A, draw arcs of radii 5 cm and 4.8 cm respectively which intersect at D.
Join AD and DC.
ABCD is the required quadrilateral.

(ii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 30
Steps :

Draw BC = 5 cm.
Draw ∠PCB = 60° and cut CD = 4.5 cm.
From B and D, draw arcs of radii 6 cm and 5 cm respectively which intersect at A.
Join AB and AD.
Thus ABCD is the required quadrilateral.

(iii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 31
Actual quadrilateral is constructed with the help of above rough figure.

Steps :

Draw AB = 8 cm.
At A, draw ∠PAB = 60° and cut DA = 6 cm.
At B, draw ∠QBA = 75° and cut BC = 5.4 cm.
Join DC.
Thus ABCD is the required quadrilateral.

(iv) Rough figure is as shown below.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 33
Steps :

Draw BC = 6-5 cm.
Draw ∠B = 75° and cut BA = 5 cm.
Draw ∠C = 120° and cut CD = 4.8 cm.
Join AD.
Thus ABCD is the required quadrilateral.

(v) Rough figure is as shown below.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 34
Steps :

Draw AB = 6 cm.
From A and B, draw arcs of radii 6 cm and 4 cm which cut at C.
From A and C, draw arcs of radii 4.5 cm and 5 cm respectively which intersect at D.
Join BC, CD and DA. Thus ABCD is the required quadrilateral.

(vi) Rough figure is as follow :

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 35
Actual construction is as follow (using above rough fig.)

Steps :

Draw AB = 5 cm.
From A & B draw arcs of radii 5 cm and 7cm which intersect at D.
From B & D draw arcs of radii 5.5 cm each which intersect at C.
Join AD, BD, DC and BC.
Thus ABCD is the required quadrilateral.

Question 2.
Construct a parallelogram ABCD, if :
(i) AB = 3.6 cm, BC = 4.5 cm and ∠ABC = 120°.
(ii) BC = 4.5 cm, CD = 5.2 cm and ∠ADC = 75°.
(iii) AD = 4 cm, DC = 5 cm and diagonal BD = 7 cm.
(iv) AB = 5.8 cm, AD = 4.6 cm and diagonal AC = 7.5 cm.
(v) diagonal AC = 6.4 cm, diagonal BD = 5.6 cm and angle between the diagonals is 75°.
(vi) lengths of diagonals AC and BD are 6.3 cm and 7.0 cm respectively, and the angle between them is 45°.
(vii) lengths of diagonals AC and BD are 5.4 cm and 6.7 cm respectively and the angle between them is 60°.
Solution:
(i) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 37
The above rough figure is used to construct the actual ||gm as follow :

Steps :

Draw AB = 3.6 cm.
Draw BP such that ∠B = 120°.
Cut BC = 4.5 cm.
From A, draw arc of radius 4.5 cm.
From C, draw arc of radius 3.6 cm. Which interescts first arc at D.
Join AD and CD.
Hence ABCD is the required ||gm.

(ii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 39
Steps :

Draw CD = 5.2 cm.
Draw ZCDP = 75°
Cut DA = 4.5 cm.
From A draw arc of radius 5.2 cm.
From C, draw arc of radius 4.5 cm which meets first arc at B.
Join AB and CB.
Thus ABCD is the required ||gm.

(iii) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 40
Steps :

Draw AD = 4 cm.
From A, draw an arc of radius 5 cm.
From B, draw an arc of radius 4 cm.
From D, draw an arc of ardius 5 cm which intersect first arc at C.
Join AB, BD, BC and CD.
Thus ABCD is the required || gm.

(iv) Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 41
opposite sides of ||gm are equal
BC = AD = 4.6 cm.
Actual figure is constructed as follow :

Steps:

Draw AB = 5.8 cm.
Draw an arc of radius 4.6 cm with centre B.
Draw an arc of radius 7.5 cm from A which intersects first arc at C.
From A, draw an arc of radius 4.6 cm.
From C, draw an arc of radius 5.8 cm which intersects first arc at D.
Join AD, CD, BC and AC.
Thus ABCD is the required //gm.

(v) Rough figure is as follow.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 43
Steps :

Draw AC = 6.4 cm.
Bisect AC at O.
Draw ∠XOC = 75° and produce XO to Y.
Cut OB = OD = 2 8 cm.
Join AB, BC, AD and CD.
Thus ABCD is the required ||gm.

(vi)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 44
Steps :

Draw AC = 6.3 cm.
Bisect AC at O.
At O, draw ∠XOC = 45° and produce XO to Y.
Cut OB = OD = 3.5 cm (half the diagonal 7 cm.)
Join AB, CB, AD and CD. Thus ABCD is the required || gm.

(vii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 45
Steps :

Draw BD 6.7 cm.
Bisect BD at O.
At O, draw ∠XOD = 60° and produce XO to Y.
Cut OA = OC = 2.7 cm (half the diagonals 5.4 cm)
Join AB, AD, BC and CD.
Thus ABCD is the required ||gm.

Question 3.
Construct a rectangle ABCD ; if :
(i) AB = 4.5 cm and BC = 5.5 cm.
(ii) BC = 61 cm and CD = 6.8 cm.
(iii) AB = 5.0 cm and diagonal AC = 6.7 cm.
(iv) AD = 4.8 cm and diagonal AC = 6.4 cm.
(v) each diagonal is 6 cm and the angle between them is 45°.
(vi) each diagonal is 5.5 cm and the angle between them is 60°.
Solution:
(i)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 46
Steps :

Draw BC = 5.5 cm.
At B, draw ∠XBC = 90°
Cut BA = 4.5 cm.
From A, draw an arc of radius 5.5 cm.
From C, draw an arc of radius 4 5 cm which meets first arc at D.
Join AD and CD.
Thus ABCD is the required rectangle.

(ii)
na Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 47
Steps:

Draw BC = 6.1 cm.
At C, draw ∠PCB = 90°.
Cut CD = 6.8 cm.
Draw an arc of radius 6.8 cm from B.
From D, draw an arc of radius 6.1 cm which meets the first arc at A.
Join AB and AD.
Thus ABCD is the required rectangle.

(iii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 48
Steps :

Draw AB = 5 cm.
At B, draw ∠XBA = 90°.
From A, draw an arc of radius 6.7 cm which meets XB at C.
From C, draw an arc of a radius 5 cm.
From A, draw an arc of radius equal to BC which meets first arc at D.
Join AD and CD. Thus ABCD is the required rectangle.

(iv)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 49
Steps :

Draw AD = 4.8 cm.
At D, draw ∠XDA = 90°.
From A, draw an arc of radius 6-4 cm which meets DX at C.
From A, draw an arc of radius equal to DC.
From C, draw an arc of radius 4.8 cm which meets first arc at B.
Join AB and CB. Thus ABCD is the required rectangle.

(v)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 50
Steps :

Draw AC = 6 cm.
Bisect AC at O.
At O, draw ∠XOC = 45° and produce XO to Y.
Cut OB = OD = 3 cm (half the diagonal 6 cm)
Join AB, CB, AD and CD.
Thus ABCD is the required rectangle.

(vi)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 51
Steps :

Draw AC = 5.5 cm.
Bisect AC at O.
At O, draw ∠XOC = 60° and produce XO to Y.
Cut OB = OA and OD = OA (half the diagonal AC).
Join AB, BC, AD and CD.
Thus ABCD is the required rectangle.

Question 4.
Construct a rhombus ABCD, if ;
(i) AB = 4 cm and ∠B = 120°.
(ii) BC = 4.7 cm and ∠B = 75°.
(iii) CD = 5 cm and diagonal BD = 8.5 cm.
(iv) BC = 4.8cm, and diagonal AC = 7cm.
(v) diagonal AC = 6 cm and diagonal BD = 5.8 cm.
(vi) diagonal AC = 4.9 cm and diagonal BD = 6 cm.
(vii) diagonal AC = 6.6 cm and diagonal BD = 5.3 cm.
Solution:
(i)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 52
Steps :

Draw AB = 4 cm.
At B, draw ∠XBA = 120°
Cut BC = 4 cm.
Draw arcs of radii 4 cm each from A and C which intersect at D.
Join CD and AD.
Thus ABCD is the required rhombus.

(ii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 53
Steps :

Draw BC = 4.7 cm.
At B, draw ∠XBC = 75°
Cut BA = 4.7 cm.
From A and C, draw arcs of radii 4.7 cm each which intersect at D.
Join AD and CD.
Thus ABCD is the rhombus.

(iii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 54
Steps :

Draw CD = 5 cm.
From C & D draw arcs of radii 5 cm and 8.5 cm respectively which intersect at B.
From B and D, draw arcs of radii 5 cm each which intersect at A.
Join AB and AD.
Thus ABCD is the required rhombus.

(iv)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 55
Steps :

Draw AC = 7 cm.
Draw arcs of radii 4.8 cm each from A and C which intersect at B.
From A & C again draw arcs of radii 4.8 cm each which intersect at D.
Join AB, BC, AD and CD.
Thus ABCD is the required rhombus.

(v)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 56
Steps :

Draw BD = 5.8 cm.
Draw perpendicular bisector XY of BD.
Cut OA = OC = 3 cm (half the diagonal 6 cm)
Join AB, AD, BC and CD.
Thus ABCD is the required rhombus.

(vi)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 57
Steps :

Draw AC = 4.9 cm.
Draw perpendicular bisector XY of AC.
Cut OB = OD = 3 cm (half the diagonal 6 cm)
Join AB, BC, AD and CD.
Thus ABCD is the required rhombus.

(vii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 58
Steps :

Draw BD = 5.3 cm.
Draw perpendicular bisector XY of BD.
Cut OA = OC = 3.3 cm (half the diagonal 6.6 cm)
Join AB, AD, BC and CD.
Thus ABCD is the required rhombus.

Question 5.
Construct a square, if :
(i) its one side is 3.8 cm.
(ii) its each side is 4.3 cm.
(iii) one diagonal is 6.2 cm.
(iv) each diagonal is 5.7 cm.
Solution:
(i)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 59
Steps :

Draw AB = 3.8 cm.
At B, draw ∠PBA = 90°.
Cut BC = 3.8 cm.
From A and C, draw arcs of radii 3.8 cm each which intersect at D.
Join AD and CD.
Thus ABCD is the required square.

(ii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 60
Steps :

Draw AB = 4.3 cm.
Draw ∠PAB = 90° at A.
Cut AD = 4.3 cm.
From B and D, draw arcs of radii 4.3 cm each which intersect at C.
Join AD, BC and CD.
Hence ABCD is the required square.

(iii)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 61
Steps :

Draw BD = 6.2 cm.
Draw perpendicular bisector XY of BD.
Cut OA = OC = 3.1 cm (half the diagonal)
Join AB, AD, BC and CD.
Thus ABCD is the required square.

(iv)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 62
Steps :

Draw BD = 5.7 cm.
Draw perpendicular bisector XY of BD.
From 0, draw arcs of radii equal to OB which cuts XY at A and C.
Join AB, AD, BC and CD.
Thus ABCD is the required square.

Question 6.
Construct a quadrilateral ABCD in which ; ∠A = 120°, ∠B = 60°, AB = 4 cm, BC = 4.5 cm and CD = 5 cm.
Solution:
Rough figure is as follow :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 63
Steps :

Draw AB = 4 cm.
At A, draw ∠PAB = 120°.
At B, draw ∠QBA = 60°.
From BQ, cut BC = 4.5 cm.
From C, draw an arc of radius 5 cm which meets AP at D.
Join CD.
Thus ABCD is the required quadrilateral.

Question 7.
Construct a quadrilateral ABCD, such that AB = BC = CD = 4.4 cm, ∠B = 90° and ∠C = 120°.
Solution:
Rough figure is as follow
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 64
Steps :

Draw BC = 4.4 cm.
At B, draw ∠PBC = 90°.
Cut BA = 4.4 cm.
At C, draw ∠QCB = 120°.
Cut CD = 4.4 cm.
Join AD.
Thus ABCD is the required quadrilateral.

Question 8.
Using ruler and compasses only, construct a parallelogram ABCD, in which : AB = 6 cm, AD = 3 cm and ∠DAB = 60°. In the same figure draw the bisector of angle DAB and let it meet DC at point P. Measure angle APB.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 65
Steps :

Draw AB = 6 cm.
At A draw ∠QAB = 60°.
From AQ cut AD = 3 cm.
From D, draw an arc of radius 6 cm.
From B, draw an arc of radius 3 cm which meets first arc at C.
Join CD and BC.
Thus ABCD is the required ||gm.
Bisect ∠DAB, so that bisector meets CD at P.
Join PB and measure ZAPB.
∴ ∠APB = 90°.

Question 9.
Draw a parallelogram ABCD, with AB = 6 cm, AD = 4.8 cm and ∠DAB = 45°. Draw the perpendicular bisector of side AD and let it meet AD at point P. Also draw the diagonals AC and BD ; and let they intersect at point O. Join O and P. Measure OP.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 66
Steps :

Draw AB = 6 cm.
Draw ∠PAB = 45°.
Cut AD = 4.8 cm.
From D, draw an arc of radius 6 cm.
From B, draw an arc of radius 4.8 cm which meets first arc at C.
Join BC, CD, AD.
Thus ABCD is the required ||gm.
Draw perpendicular bisector XY of AD which cuts AD at P.
Join AC and BD which intersect at O.
Join OP and measure it.
OP = 3 cm.

Question 10.
Using ruler and compasses only, construct a rhombus whose diagonals are 8 cm and 6 cm. Measure the length of its one side.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions image - 67
Steps :

Draw BD = 8 cm.
Draw perpendicular bisector PQ of BD.
Cut OA = OC = 3 cm [half the diagonal 6 cm]
Join AB, AD, BC and CD.
Measure side AB which is 5 cm.
Thus ABCD is the required rhombus.

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 6 Sets. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Sets Exercise 6A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Write the following sets in roster (Tabular) form :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 1
Solution:

Question 2.
Write the following sets in set-builder (Rule Method) form :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 6
Solution:

Question 3.
(i) Is {1, 2, 4, 16, 64} = {x : x is a factor of 32} ? Give reason.
(ii) Is {x : x is a factor of 27} ≠ {3, 9, 27, 54} ? Give reason.
(iii) Write the set of even factors of 124.
(iv) Write the set of odd factors of 72.
(v) Write the set of prime factors of 3234.
(vi) Is {x : x² – 7x + 12 = 0} = {3, 4} ?
(vii) Is {x : x² – 5x – 6 = 0} = {2, 3} ?
Solution:
(i) No, {1, 2, 4, 16, 64} ≠ {x : x is factor of 32}
Because 64 is not a factor of 32
(ii) Yes, {x : x is a factor of 27} + {3, 9, 27, 54}
Because 54 is not a factor of 27
(iii) 1 x 124 = 124
2 x 62 = 124
4 x 31 = 124
Factors of 124 = 1, 2, 4, 31, 62, 124
Set of even factors of 124 = {2, 4, 62, 124}
(iv) 1 x 72 = 72
2 x 36 = 72
3 x 24 = 72
4 x 18 = 72
6 x 12 = 72
8 x 9 = 72
Factors of 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Set of odd factors of 72 = {1, 3, 9}
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 8

Question 4.
Write the following sets in Roster form :
(i) The set of letters in the word ‘MEERUT’.
(ii) The set of letters in the word ‘UNIVERSAL’.
(iii) A = {x : x = y + 3, y ∈N and y > 3}
(iv) B = {p : p ∈ W and p² < 20}
(v) C = {x : x is composite number and 5 < x < 21}
Solution:
(i) Roster form of the set of letters in the word “MEERUT” = {m, e, r, u, t}
(ii) Roster form of the set of letters in the word “UNIVERSAL” = {u, n, i, v, e, r, s, a, l}
(iii) A = {x : x = y + 3, y ∈ N and y > 3}
x = y + 3
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 10

Question 5.
List the elements of the following sets :
(i) {x : x² – 2x – 3 = 0}
(ii) {x : x = 2y + 5; y ∈ N and 2 ≤ y < 6}
(iii) {x : x is a factor of 24}
(iv) {x : x ∈ Z and x² ≤ 4}
(v) {x : 3x – 2 ≤ 10, x ∈ N}
(vi) {x : 4 – 2x > -6, x ∈ Z}
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 12

Sets Exercise 6B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the cardinal number of the following sets :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 16
Solution:

Question 2.
If P = {P : P is a letter in the word “PERMANENT”}. Find n (P).
Solution:
P = (P : P is a letter in the word “PERMANENT”}
or P = {p, e, r, m, a, n, t)
n (P) = 7

Question 3.
State, which of the following sets are finite and which are infinite :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 19
Solution:

Question 4.
Find, which of the following sets are singleton sets :
(i) The set of points of intersection of two non-parallel st. lines in the same plane
(ii) A = {x : 7x – 3 = 11}
(iii) B = {y : 2y + 1 < 3 and y ∈ W}
Note : A set, which has only one element in it, is called a SINGLETON or unit set.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 24

Question 5.
Find, which of the following sets are empty :
(i) The set of points of intersection of two parallel lines.
(ii) A = {x : x ∈ N and 5 < x < 6}
(iii) B = {x : x² + 4 = 0, x ∈ N}
(iv) C = {even numbers between 6 & 10}
(v) D = {prime numbers between 7 & 11}
Note : The set, which has no element in it, is called the empty or null set.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 25

Question 6.
(i) Are the sets A = {4, 5, 6} and B = {x : x² – 5x – 6 = 0} disjoint ?
(ii) Are the sets A = {b, c, d, e} and B = {x : x is a letter in the word ‘MASTER’} joint ?
Note :
(i) Two sets are said to be joint sets, if they have atleast one element in common.
(ii) Two sets are said to be disjoint, if they have no element in common.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 26

Question 7.
State, whether the following pairs of sets are equivalent or not :
(i) A = {x : x ∈ N and 11 ≥ 2x – 1} and B = {y : y ∈ W and 3 ≤ y ≤ 9}
(ii) Set of integers and set of natural numbers.
(iii) Set of whole numbers and set of multiples of 3.
(iv) P = {5, 6, 7, 8} and M = {x : x ∈ W and x < 4}
Note : Two sets are said to be equivalent, if they contain the same number of elements.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 28
B = {3,4,5,6,7,8,9}
n (B) = 7
Cardinal number of set A = 6 and cardinal number of set B = 7
Set A and set B are not equivalent.
(ii) Set of integers has infinite number of elements. Set of natural numbers has infinite number of elements.
Set of integers and set of natural numbers are equivalent because both these sets have infinite number of elements.
(iii) Set of whole numbers, has infinite number of elements. Set of multiples of 3, has infinite number of element.
Set of whole numbers and set of multiples of 3 are equivalent because both these sets have infinite number of elements.
(iv) P = {5,6,7,8}
n (P) = 4
M = {x : x ∈ W and x ≤ 4}
M = {0, 1, 2, 3, 4}
n (M) = 5
Now Cardinal number of set P = 4 and
Cardinal number of set M = 5
These sets are not equivalent.

Question 8.
State, whether the following pairs of sets are equal or not :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 29
Solution:

Question 9.
State whether each of the following sets is a finite set or an infinite set:
(i) The set of multiples of 8.
(ii) The set of integers less than 10.
(iii) The set of whole numbers less than 12.
(iv) {x : x = 3n – 2, n ∈ W, n ≤ 8}
(v) {x : x = 3n – 2,n ∈ Z, n ≤ 8}
(vi) {x : x = \(\frac { n-2 }{ n+1 }\) , n ∈ w)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 33

Question 10.
Answer, whether the following statements are true or false. Give reasons.
(i) The set of even natural numbers less than 21 and the set of odd natural numbers less than 21 are equivalent sets.
(ii) If E = {factors of 16} and F = {factors of 20}, then E = F.
(iii) The set A = {integers less than 20} is a finite set.
(iv) If A = {x : x is an even prime number}, then set A is empty.
(v) The set of odd prime numbers is the empty set.
(vi) The set of squares of integers and the set of whole numbers are equal sets.
(vii) In n(P) = n(M), then P → M.
(viii) If set P = set M, then n(P) = n(M).
(ix) n(A) = n(B) => A = B.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 34

Sets Exercise 6C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find all the subsets of each of the following sets :
(i) A = {5, 7}
(ii) B = {a, b, c}
(iii) C = {x : x ∈ W, x ≤ 2}
(iv) {p : p is a letter in the word ‘poor’}
Solution:
(i) A = {5,7}
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 38

Question 2.
If C is the set of letters in the word “cooler”, find :
(i) Set C
(ii) n(C)
(iii) Number of its subsets
(iv) Number of its proper subsets.
Note : (i) If a set has n elements, the number of its subsets = 2ⁿ
(ii) If a set has n elements, the number of its proper subsets = 2ⁿ – 1
Solution:
(i) C = {c, o, l, e, r}
(ii) n(C) = 5
(iii) Number of its subsets : 2⁵= 2 x 2 x 2 x 2 x 2 = 32
(iv) Number of its proper subsets = 2⁵ – 1 = 32 – 1 = 31

Question 3.
If T = {x : x is a letter in the word ‘TEETH’}, find all its subsets.
Solution:
T = {t,e,h}
Subsets of set T = φ, {r}, {e}, {h}, {t,e}, {t,h}, {e,h}, {t,e,h}

Question 4.
Given the universal set = {-7,-3, -1, 0, 5, 6, 8, 9}, find :
(i) A = {x : x < 2}
(ii) B = {x : -4 < x < 6}
Solution:
Universal set = {-7, -3, -1, 0, 5, 6, 8, 9},
(i) A = {x : x < 2} = {-7, -3, -1, 0}
(ii) B = {x : -4 < x < 6} = {-3, -1, 0, 5}

Question 5.
Given the universal set = {x : x ∈ N and x < 20}, find :
(i) A = {x : x = 3p ; p ∈ N}
(ii) B = {y : y – 2n + 3, n ∈ N}
(iii) C = {x : x is divisible by 4}
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 39

Question 6.
Find the proper subsets of {x : x² – 9x – 10 = 0}
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 40
x = 10
=> x = -1
Given set = {-1, 10}
Proper subsets of this set = φ, {-1}, {10}

Question 7.
Given, A = {Triangles}, B = {Isosceles triangles}, C = {Equilateral triangles}. State whether the following are true or false. Give reasons.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 41
Solution:

Question 8.
Given, A = {Quadrilaterals}, B = {Rectangles}, C = {Squares}, D= {Rhombuses}. State, giving reasons, whether the following are true or false.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 43
Solution:

Question 9.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 46
Solution:

Question 10.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 48
Solution:

Question 11.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 50
Solution:

Question 12.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 52
Solution:

Sets Exercise 6D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 54
Solution:

Question 2.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 56
Solution:

Question 3.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 58

Solution:

Question 4.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 61
Solution:

Question 5.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 64
Solution:

Question 6.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 67
Solution:

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 69
Solution:

Question 8.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 71
Solution:

Question 9.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 74
Solution:

Sets Exercise 6E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 77
Solution:

Question 2.
From the given diagram, find :
(i) A’
(ii) B’
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 79
Solution:

Question 3.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 81
Solution:

Question 4.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 84
Solution:

Question 5.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 86
Solution:

Question 6.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 89
Solution:

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 91
Solution:

Question 8.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 93
Solution:

Question 9.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 96
Solution:

Question 10.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 98
Solution:

Question 11.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 6 Sets image - 100
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 10 Direct and Inverse Variations. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 8 Maths Solutions Physics Chemistry Biology Geography History & Civics

Direct and Inverse Variations Exercise 10A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
In which ofthe following tables, x and y vary directly:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 1
Solution:

Question 2.
If x and y vary directly, find the values of x, y and z:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 4
Solution:

Question 3.
A truck consumes 28 litres of diesel for moving through a distance of 448 km. How much distance will it cover in 64 litres of diesel?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 7

Question 4.
For 100 km, a taxi charges ₹ 1,800. How much will it charge for a journey of 120 km?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 8

Question 5.
If 27 identical articles cost ₹ 1,890, how many articles can be bought for ₹ 1,750?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 9

Question 6.
7 kg of rice costs ₹ 1,120. How much rice can be bought for ₹ 3,680?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 10

Question 7.
6 note-books cost ₹ 156, find the cost of 54 such note-books.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 11

Question 8.
22 men can dig a 27 m long trench in one day. How many men should be employed for digging 135 m long trench of the same type in one day?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 12

Question 9.
If the total weight of 11 identical articles is 77 kg, how many articles of the same type would weigh 224 kg?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 13

Question 10.
A train is moving with uniform speed of 120 km per hour.
(i) How far will it travel in 36 minutes?
(ii) In how much time will it cover 210 km?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 14

Direct and Inverse Variations Exercise 10B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Check whether x and y vary inversely or not.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 15
Solution:
x and y are inversely proportional.
Then xy are equal.
(i) xy = 4 x 6 = 24
xy = 3 x 8 = 24
xy = 12 x 2 = 24
xy = 1 x 24 = 24
xy in each case is equal.
x and y are inversely proportional
(ii) xy = 30 x 60 = 1800
xy= 120 x 30 = 3600
xy = 60 x 30= 1800
xy = 24 x 75 = 1800
xy in each case is not equal.
x and y are not inversely proportional.
(iii) xy = 10 x 90 = 900
= 30 x 30 = 900
= 60 x 20= 1200
= 10 x 90 = 900
xy in each case is not equal.
x and y are not inversely proportional.

Question 2.
If x and y vary inversely, find the values of l, m and n :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 16
Solution:

Question 3.
36 men can do a piece of work in 7 days. How many men will do the same work in 42 days?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 19

Question 4.
12 pipes, all of the same size, fill a tank in 42 minutes. How long will it take to fill the same tank, if 21 pipes of the same size are used?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 20

Question 5.
In a fort 150 men had provisions for 45 days. After 10 days, 25 men left the fort. How long would the food last at the same rate?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 21

Question 6.
72 men do a piece of work in 25 days. In how many days will 30 men do the same work?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 22

Question 7.
If 56 workers can build a wall in 180 hours, how many workers will be required to do the same work in 70 hours?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 23

Question 8.
A car takes 6 hours to reach a destination by travelling at the speed of 50 km per hour. How long will it take when the car travels at the speed of 75 km per hour?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 24

Direct and Inverse Variations Exercise 10C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Cost of 24 identical articles is Rs. 108, Find the cost of 40 similar articles.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 25

Question 2.
If 15 men can complete a piece of work in 30 days, in how many days will 18 men complete it?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 26

Question 3.
In order to complete a work in 28 days, 60 men are required. How many men will be required if the same work is to be completed in 40 days ?
Solution:
Let x be number of men required 60 men can do the work in = 28 days
1 man can do the work in = 28 x 60 days
x man can do the work in = \(\frac { 28\times 60 }{ x }\) day
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 27

Question 4.
A fort had provisions for 450 soldiers for 40 days. After 10 days, 90 more soldiers come to the fort. Find in how many days will the remaining provisions last at the same rate ?
Solution:
After 10 days :
For 450 soldiers, provision are sufficient for (40 – 10) days = 30 days
For 1 soldier, provision are sufficient for 30 x 450 days
For 540 soldiers, the provision are sufficient for
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 28

Question 5.
A garrison has sufficient provisions for 480 men for 12 days. If the number of men is reduced by 160; find how long will the provisions last.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 29

Question 6.
\(\frac { 3 }{ 5 }\) quintal of wheat costs Rs.210. Find the cost of :
(i) 1 quintal of wheat
(ii) 0.4 quintal of wheat
Solution:
(i) \(\frac { 3 }{ 5 }\) quintal of wheat costs = Rs.210
1 quintal of wheat costs = 210 x \(\frac { 3 }{ 5 }\) = 70 x 5 = Rs.350
(ii) 1 quintal of wheat costs = Rs.350
0.4 quintal of wheat costs = 350 x 0.4 = Rs. 140.0 = Rs.140

Question 7.
If \(\frac { 2 }{ 9 }\) of a property costs Rs.2,52,000; find the cost of \(\frac { 4 }{ 7 }\) of it.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 30

Question 8.
4 men or 6 women earn Rs. 360 in one day. Find, how much will:
(i) a man earn in one day ?
(ii) a woman earn in one day ?
(iii) 6 men and 4 women earn in one day ?
Solution:
4 men earn Rs. 360 in one day
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 31

Question 9.
16 boys went to canteen to have tea and snacks together. The bill amounted to Rs. 114.40. What will be the contribution of a boy who pays for himself and 5 others ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 32

Question 10.
50 labourers can dig a pond in 16 days. How many labourers will be required to dig an another pond, double in size in 20 days ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 33

Question 11.
If 12 men or 18 women can complete a piece of work in 7 days, in how many days can 4 men and 8 women complete the same work?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 34

Question 12.
If 3 men or 6 boys can finish a work in 20 days, how long will 4 men and 12 boys take to finish the same work ?
Solution:
3 men = 6 boys
4 men = \(\frac { 6 }{ 3 }\) x 4 = 8
Total boys in second case :
= 4 men + 12 boys = 8 + 12 = 20 boys
6 boys can do a piece of work in 20 days
Then let 20 boys will do the same work in x days
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 35

Question 13.
A particular work can be completed by 6 men and 6 women in 24 days; whereas the same work can be completed by 8 men and 12 women in 15 days. Find :
(i) according to the amount of work done, one man is equivalent to how many women.
(ii) the time taken by 4 men and 6 women to complete the same work.
Solution:
6 men + 6 women can finish the work in = 24 days
144 men + 144 women can finish it in = 1 day
8 m + 12 women can finish the work in = 15 days
120 men + 180 women can finish it in = 1 day
(i) 144 men + 144 women = 120 men + 180 women
=> 144 men – 120 men
= 180 – women – 144 women
=> 24 men = 36 women
1 man = \(\frac { 36 }{ 24 }\)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 36

Question 14.
If 12 men and 16 boys can do a piece of work in 5 days and, 13 men and 24 boys can do it in 4 days, how long will 7 men and 10 boys take to do it ?
Solution:
12 men + 16 boys can do a piece of work in = 5 days
60 men + 80 boys can do a piece of work in = 1 day ………. (i)
and 13 men + 24 boys can do the same work in = 4 days
52 men + 96 boys can do the same work in = 1 day ………….(ii)
From (i) and (ii)
60 men + 80 boys = 52 men + 96 boys
=> 60 men – 52 men = 96 boys – 80 boys
=> 8 men = 16 boys
1 men = \(\frac { 16 }{ 8 }\) = 2 boys
Now, in first case,
12 men + 16 boys = 12 x 2 + 16 = 24+16 = 40 boys
In the second case,
7 men + 10 boys = 7 x 2 + 10 = 14 + 10 = 24 boys
Now 40 boys can do a piece of work in = 5 days
1 boy can do the same work in = 5 x 40 days
and 24 boys will do the same work in
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 37

Direct and Inverse Variations Exercise 10D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Eight oranges can be bought for Rs. 10.40. How many more can be bought for Rs. 16.90?
Solution:
Number of oranges bought for Rs. 10.40 = 8
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 38

Question 2.
Fifteen men can build a wall in 60 days. How many more men are required to build another wall of same size in 45 days ?
Solution:
In 60 days a wall can be built by = 15 men
In 1 day a wall can be built by = 15 x 60 men
In 45 days a wall can be built by = \(\frac { 15\times 60 }{ 45 } =\frac { 900 }{ 45 }\) = 20 men
No. of more men required to build the wall in 45 days = 20 – 15 = 5 men

Question 3.
Six taps can fill an empty cistern in 8 hours. How much more time will be taken, if two taps go out of order ? Assume, all the taps supply water at the same rate.
Solution:
Total no. of taps = 6
Out of order taps = 2
Taps in working condition =6 – 2 = 4
6 taps can fill an empty cistern in = 8 hours
1 tap can fill an empty cistern in = 6 x 8 hours
4 taps can fill an empty cistern in = \(\frac { 48 }{ 4 }\) = 12 hours
More time taken when 2 taps are out of order = 12 – 8 = 4 hour

Question 4.
A contractor undertakes to dig a canal, 6 kilometres long, in 35 days and employed 90 men. He finds that after 20 days only 2 km of canal have been completed. How many more men must be employed to finish the work in time ?
Solution:
Length of canal = 6 km
In 20 days canal made = 2 km
Remaining length of canal = 6 – 2 = 4 km
Remaining time = 35 – 20 = 15 days
In 20 days 2 km canal is made by = 90 men
In 1 day 2 km canal is made by = 90 x 20 men
In 15 days 2 km canal is made by = \(\frac { 90\times 20 }{ 15 }\) men
In 15 days 1 km canal is made by = \(\frac { 90\times 20 }{ 15\times 2 }\) men
In 15 days 4 km canal is made by = \(\frac { 90\times 20\times 4 }{ 15\times 2 }\) men = 6 x 10 x 4 men = 240 men
Number of more men to be employed to finish the work in time = 240-90 = 150 men

Question 5.
If 10 horses consume 18 bushels in 36 days. How long will 24 bushels last for 30 horses ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 39

Question 6.
A family of 5 persons can be main¬tained for 20 days with Rs.2,480. Find, how long Rs.6944 maintain a family of 8 persons
Solution:
A family of 5 persons can be maintained with Rs.2480 for = 20 days
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 40

Question 7.
90 men can complete a work in 24 days working 8 hours a day. How many men are required to complete the same work in 18 days working 7\(\frac { 1 }{ 2 }\) hours a day ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 41

Question 8.
Twelve typists, all working with same speed, type a certain number of pages in 18 days working 8 hours a day. Find, how many hours per day must sixteen typists work in order to type the same number of pages in 9 days ?
Solution:
12 typists can type in 18 days with number of working hours in day = 8 hours
1 typist can type in 18 days = 8 x 12 hour
1 typist can type in 9 days = 2 (8 x 12) hour
16 typist can type in a day = \(\frac { 2(8\times 12) }{ 16 }\) = 12 hours

Question 9.
If 25 horses consume 18 quintal in 36 days, how long will 28 quintal last for 30 horses ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 42

Question 10.
If 70 men dig 15,000 sq. m of a field in 5 days, how many men will dig 22,500 sq. m field in 25 days ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 43

Question 11.
A contractor undertakes to build a wall 1000 m long in 50 days. He employs 56 men, but at the end of 27 days, he finds that only 448 m of wall is built. How many extra men must the contractor employ so that the wall is completed in time ?
Solution:
Number of men employed in the beginning = 56
Length of wall = 1000 m No. of days = 50
In the time of 27 days, only 448 m of wall was completed
Remaining period = 50 – 27 = 23 days
and length of wall to be completed = 1000 – 448 = 552
Now in 27 days, 448 m long wall was completed by = 1000 m
in 1 day, 448 m long was completed by = 56 x 27
inf 1 day, 1 m long wall will be completed by
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 44

Question 12.
A group of labourers promises to do a piece of work in 10 days, but five of them become absent. If the remaining labourers complete the work in 12 days, find their original number in the group.
Solution:
Total period = 10 days
But work completed in = 12 days
No. of men were absent = 5
Let the number of men in the beginning = x
Now x men can do a piece work in = 10 days
1 man will do it in = 10x x days
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 45

Question 13.
Ten men, working for 6 days of 10 hours each, finish \(\frac { 5 }{ 21 }\) of a piece of work. How many men working at the same rate and for the same number of hours each day, will be required to complete the remaining work in 8 days ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 46

Direct and Inverse Variations Exercise 10E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A can do a piece of work in 10 days and B in 15 days. How long will they take together to finish it ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 47

Question 2.
A and B together can do a piece of work in 6\(\frac { 2 }{ 3 }\) days ; but B alone can do it in 10 days. How long will A take to do it alone ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 48

Question 3.
A can do a work in 15 days and B in 20 days. If they together work on it for 4 days ; what fraction of the work will be left ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 49

Question 4.
A, B and C can do a piece of work in 6 days, 12 days and 24 days respectively. In what time will they all together do it ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 50

Question 5.
A and B working together can mow a field in 56 days and with the help of C, they could have mowed it in 42 days. How long would C take by himself ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 52

Question 6.
A can do a piece of work in 24 days, A and B can do it in 16 days and A, B and C in 10\(\frac { 2 }{ 3 }\) days. In how many days can A and C do it ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 53

Question 7.
A can do a piece of work in 20 days and B in 15 days. They worked together on it for 6 days and then A left. How long will B take to finish the remaining work ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 55

Question 8.
A can finish a piece of work in 15 days and B can do it in 10 days. They worked together for 2 days and then B goes away. In how many days will A finish the remaining work?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 57

Question 9.
A can do a piece of work in 10 days ; B in 18 days; and A, B and C together in 4 days. In what time would C alone do it ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 58

Question 10.
A can do \(\frac { 1 }{ 4 }\) of a work in 5 days and B can do \(\frac { 1 }{ 3 }\) of the same work in 10 days. Find the number of days in which both working together will complete the work.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 60

Question 11.
One tap can fill a cistern in 3 hours and the waste pipe can empty the full cistern in 5 hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 62

Question 12.
A and B can do a work in 8 days; B and C in 12 days, and A and C in 16 days. In what time could they do it, all working together ?
Solution:
A and B can do a work in = 8 days
B and C can do a work in = 12 days
A and C can do a work in = 16 days
(A+B)’s 1 day work = \(\frac { 1 }{ 8 }\)
(B+C)’s 1 day work = \(\frac { 1 }{ 12 }\)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 63

Question 13.
A and B complete a piece of work in 24 days. B and C do the same work in 36 days ; and A, B and C together finish it in 18 days. In how many days will (i) A alone,
(ii) C alone,
(iii) A and C together, complete the work ?
Solution:
A and B complete a piece of work in = 24 days
B and C complete a piece of work in = 36 days
(A+B+C) complete a piece of work in = 18 days
(A+B)’s 1 day work = \(\frac { 1 }{ 24 }\)
(B+C)’s 1 day work = \(\frac { 1 }{ 36 }\)
(A+B+C)’s 1 day work = \(\frac { 1 }{ 18 }\)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 64

Question 14.
A and B can do a piece of work in 40 days; B and C in 30 days; and C and A in 24 days.
(i) How long will it take them to do the work together ?
(ii) In what time can each finish it working alone ?
Solution:
A and B can do a piece of work in = 40 days
B and C can do a piece of work in = 30 days
C and A can do a piece of work in = 24 days
(A+B)’s 1 day work = \(\frac { 1 }{ 40 }\)
(B+C)’s 1 day work = \(\frac { 1 }{ 30 }\)
(C+A)’s 1 day work = \(\frac { 1 }{ 24 }\)
(i) [(A+B)+(B+C)+(C+A)]’s 1 day work = \(\frac { 1 }{ 40 }\) + \(\frac { 1 }{ 30 }\) + \(\frac { 1 }{ 24 }\)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 65

C can do the work in 40 days
Hence A can do the work in = 60 days
B can do the work in = 120 days
C can do the work in = 40 days

Question 15.
A can do a piece of work in 10 days, B in 12 days and C in 15 days. All begin together but A leaves the work after 2 days and B leaves 3 days before the work is finished. How long did the work last ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 67

Question 16.
Two pipes P and Q would fill an empty cistern in 24 minutes and 32 minutes respectively. Both the pipes being opened together, find when the first pipe must be turned off so that the empty cistern may be just filled in 16 minutes.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 10 Direct and Inverse Variations image - 69

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com Provides Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 5 Playing with APIusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 5 Playing with Number. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Playing with Number Exercise 5A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Write the quotient when the sum of 73 and 37 is divided by
(i) 11
(ii) 10
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -1

Question 2.
Write the quotient when the sum of 94 and 49 is divided by
(i) 11
(ii) 13
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -2

Question 3.
Find the quotient when 73 – 37 is divided by
(i) 9
(ii) 4
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -4

Question 4.
Find the quotient when 94 – 49 is divided by
(i) 9
(ii) 5
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -5

Question 5.
Show that 527 + 752 + 275 is exactly divisible by 14.
Solution:
Property :
abc = 100a + 106 + c ………(i)
bca = 1006 + 10c + a ……..(ii)
and cab = 100c + 10a + b ……….(iii)
Adding, (i), (ii) and (iii), we get abc + bca + cab = 111a + 111b + 111c = 111(a + b + c) = 3 x 37(a + b + c)
Now, let us try this method on
527 + 752 + 275 to check is it exactly divisible by 14
Here, a = 5, 6 = 2, c = 7
527 + 752 + 275 = 3 x 37(5 + 2 + 7) = 3 x 37 x 14
Hence, it shows that 527 + 752 + 275 is exactly divisible by 14

Question 6.
If a = 6, show that abc = bac.
Solution:
Given : a = 6
To show : abc = bac
Proof: abc = 100a + 106 + c …….(i)
(By using property 3)
bac = 1006 + 10a + c —(ii)
(By using property 3)
Since, a = 6
Substitute the value of a = 6 in equation (i) and (ii), we get
abc = 1006 + 106 + c ………(iii)
bac = 1006 + 106 + c ………(iv)
Subtracting (iv) from (iii) abc – bac = 0
abc = bac
Hence proved.

Question 7.
If a > c; show that abc – cba = 99(a – c).
Solution:
Given, a > c
To show : abc – cba = 99(a – c)
Proof: abc = 100a + 10b + c ……….(i)
(By using property 3)
cba = 100c + 10b + a ………..(ii)
(By using property 3)
Subtracting, equation (ii) from (i), we get
abc – cba = 100a + c – 100c – a
abc – cba = 99a – 99c
abc – cba = 99(a – c)
Hence proved.

Question 8.
If c > a; show that cba – abc = 99(c – a).
Solution:
Given : c > a
To show : cba – abc = 99(c – a)
Proof:
cba = 100c + 106 + a ……….(i)
(By using property 3)
abc = 100a + 106 + c ………(ii)
(By using property 3)
Subtracting (ii) from (i)
cba – abc= 100c+ 106 + a- 100a- 106-c
=> cba – abc = 99c – 99a
=> cba – abc = 99(c – a)
Hence proved.

Question 9.
If a = c, show that cba – abc = 0.
Solution:
Given : a = c
To show : cba – abc = 0
Proof:
cba = 100c + 106 + a …………(i)
(By using property 3)
abc = 100a + 106 + c …………(ii)
(By using property 3)
Since, a = c,
Substitute the value of a = c in equation (i) and (ii), we get
cba = 100c + 10b + c ……….(iii)
abc = 100c + 10b + c …………(iv)
Subtracting (iv) from (iii), we get
cba – abc – 100c + 106 + c – 100c – 106 – c
=> cba – abc = 0
=> cba = abc
Hence proved.

Question 10.
Show that 954 – 459 is exactly divisible by 99.
Solution:
To show : 954 – 459 is exactly divisible by 3 99, where a = 9, b = 5, c = 4
abc = 100a + 10b + c
=> 954 = 100 x 9 + 10 x 5 + 4
=> 954 = 900 + 50 + 4 ………(i)
and 459 = 100 x 4+ 10 x 5 + 9
=> 459 = 400 + 50 + 9 ……..(ii)
Subtracting (ii) from (i), we get
954 – 459 = 900 + 50 + 4 – 400 – 50 – 9
=> 954 – 459 = 500 – 5
=> 954 – 459 = 495
=> 954 – 459 = 99 x 5
Hence, 954 – 459 is exactly divisible by 99
Hence proved.

Playing with Number Exercise 5B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -6
Solution:

Question 2.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -8
Solution:

Question 3.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -10
Solution:

Question 4.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -12
Solution:

Question 5.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -14
Solution:

Question 6.

Solution:
As we need A at unit place and 9 at ten’s place,
A = 6 as 6 x 6 = 36
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -17

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -18
Solution:

Question 8.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -20
Solution:

Question 9.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -22
Solution:
As we need B at unit place and A at ten’s place,
B = 0 as 5 x 0 = 0
Now we want to find A, 5 x A = A (at unit’s place)
A = 5, as 5 x 5 = 25
C = 2

Question 10.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -24
Solution:

Question 11.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 5 Playing with Number image -26
Solution:

Playing with Number Exercise 5C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find which of the following nutpbers are divisible by 2:
(i) 192
(ii) 1660
(iii) 1101
(iv) 2079
Solution:
A number having its unit digit 2,4,6,8 or 0 is divisible by 2,
So, Number 192, 1660 are divisible by 2.

Question 2.
Find which of the following numbers are divisible by 3:
(i) 261
(ii) 111
(iii) 6657
(iv) 2574
Solution:
A number is divisible by 3 if the sum of its digits is divisible by 3,
So, 261, 111 are divisible by 3.

Question 3.
Find which of the following numbers are divisible by 4:
(i) 360
(ii) 3180
(iii) 5348
(iv) 7756
Solution:
A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
So, Number 360, 5348, 7756 are divisible by 4.

Question 4.
Find which of the following numbers are divisible by 5 :
(i) 3250
(ii) 5557
(iii) 39255
(iv) 8204
Solution:
A number having its unit digit 5 or 0, is divisible by 5.
So, 3250, 39255 are all divisible by 5.

Question 5.
Find which of the following numbers are divisible by 10:
(i) 5100
(ii) 4612
(iii) 3400
(iv) 8399
Solution:
A number having its unit digit 0, is divisible by 10.
So, 5100, 3400 are all divisible by 10.

Question 6.
Which of the following numbers are divisible by 11 :
(i) 2563
(ii) 8307
(iii) 95635
Solution:
A number is divisible by 11 if the difference of the sum of digits at the odd places and sum of the digits at even places is zero or divisible by 11.
So, 2563 is divisible by 11.

Playing with Number Exercise 5D – Selina Concise Mathematics Class 8 ICSE Solutions

For what value of digit x, is :
Question 1.
1×5 divisible by 3 ?
Solution:
1×5 is divisible by 3
=> 1 + x + 5 is a multiple of 3
=> 6 + x = 0, 3, 6, 9,
=> x = -6, -3, 0, 3, 6, 9
Since, x is a digit
x = 0, 3, 6 or 9

Question 2.
31×5 divisible by 3 ?
Solution:
31×5 is divisible by 3
=> 3 + 1 + x + 5 is a multiple of 3
=> 9 + x = 0, 3, 6, 9,
=> x = -9, -6, -3, 0, 3, 6, 9,
Since, x is a digit
x = 0, 3, 6 or 9

Question 3.
28×6 a multiple of 3 ?
Solution:
28×6 is a multiple of 3
2 + 8+ x + 6 is a multiple of 3
=> 16 + x = 0, 3, 6, 9, 12, 15, 18
=> x = -18, -5, -2, 0, 2, 5, 8
Since, x is a digit = 2, 5, 8

Question 4.
24x divisible by 6 ?
Solution:
24x is divisible by 6
=> 2 + 4+ x is a multiple of 6
=> 6 + x = 0, 6, 12
=> x = -6, 0, 6
Since, x is a digit
x = 0, 6

Question 5.
3×26 a multiple of 6 ?
Solution:
3×26 is a multiple of 6
3 + x + 2 + 6 is a multiple of 3
=> 11 + x = 0, 3, 6, 9, 12, 15, 18,21,
=> x = -11, -8, -5, -2, 1, 4, 7, 10, ….
Since, x is a digit
x = 1, 4 or 7

Question 6.
42×8 divisible by 4 ?
Solution:
42×8 is divisible by 4
=> 4 + 2 + x + 8 is a multiple of 2
=> 14 + x = 0, 2, 4, 6, 8,
=> x = -8, -6, -4, -2, 2, 4, 6, 8,
Since, x is a digit 2, 4, 6, 8

Question 7.
9142x a multiple of 4 ?
Solution:
9142x is multiple of 4
=> 9 + 1 + 4 + 2 + x is a multiple of 4
=> 16 + x = 0, 4, 8, ………
x = -8, -4, 0, 4, 8
Since, x is a digit
4, 8

Question 8.
7×34 divisible by 9 ?
Solution:
7×34 is multiple of 9
=> 7 + x + 3+ 4 is a multiple of 9
=> 14 + x = 0, 9, 18, 27,
=> x = -1, 4, 13,
Since, x is a digit
x = 4

Question 9.
5×555 a multiple of 9 ?
Solution:
Sum of the digits of 5×555
=5 + x + 5 + 5 + 5 = 20 + x
It is multiple by 9
The sum should be divisible by 9
Value of x will be 7

Question 10.
3×2 divisible by 11 ?
Solution:
Sum of the digit in even place = x
and sum of the digits in odd place = 3 + 2 = 5
Difference of the sum of the digits in even places and in odd places = x – 5
3×2 is a multiple of 11
=> x – 5 = 0, 11, 22,
=> x = 5, 16, 27,
Since, x is a digit x = 5

Question 11.
5×2 a multiple of 11 ?
Solution:
Sum of a digit in even place = x
and sum of the digits in odd place = 5 + 2 = 7
Difference of the sum of the digits in even places and in odd places = x – 7
5×2 is a multiple of 11
=> x – 7 = 0, 11, 22,
=> x = 7, 18, 29,
Since, x is a digit
x = 7

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations (Including Number Lines)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations (Including Number Lines)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 15 Linear Inequations (Including Number Lines). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Linear Inequations Exercise 15A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
If the replacement set is the set of natural numbers, solve.
(i) x – 5 < 0
(ii) x + 1 < 7
(iii) 3x – 4 > 6
(iv) 4x + 1 > 17
Solution:
(i) x – 5 < 0
x – 5 + 5 <0 + 5 ………(Adding 5)
=> x < 5
Required answer = {1, 2, 3, 4}
(ii) x + 1 ≤ 7 => x + 1 – 1 ≤ 7 – 1 (Subtracting 1)
=> x ≤ 6
Required answer = {1, 2, 3, 4, 5, 6}
(iii) 3x – 4 > 6
3x – 4 + 4 > 6 + 4 (Adding 4)
=> 3x > 10
\(\frac { 3x }{ 3 }\) > \(\frac { 10 }{ 3 }\) …(Dividing by 3)
=> x > \(\frac { 10 }{ 3 }\)
=> x > \(3\frac { 1 }{ 3 }\)
Required answer = { 4, 5, 6, …}
(iv) 4x + 1 ≥ 17
=> 4x + 1 – 1 ≥ 17 – 1 (Subtracting)
=> 4x ≥ 16
=> \(\frac { 4x }{ 4 }\) ≥ \(\frac { 16 }{ 4 }\) (Dividing by 4)
=> x ≥ 4
Required answer = {4, 5, 6, …}

Question 2.
If the replacement set = {-6, -3, 0, 3, 6, 9}; find the truth set of the following:
(i) 2x – 1 > 9
(ii) 3x + 7 < 1
Solution:
(i) 2x – 1 > 9
⇒ 2x – 1 + 1 > 9 + 1 (Adding 1)
⇒ 2x > 10
⇒ x > 5 (Dividing by 2)
⇒ x > 5
Required answer = {6, 9}
(ii) 3x + 7 ≤ 1
⇒ 3x + 7 – 7 ≤ 1 – 7 (Subtracting 7)
⇒ 3x ≤ – 6
⇒ x ≤ – 2
Required Answer = {-6, -3}

Question 3.
Solve 7 > 3x – 8; x ∈ N
Solution:
7 > 3x – 8
=> 7 – 3x > 3x – 3x – 8 (Subtracting 3x)
=> 7 – 7 – 3x > 3x – 3x – 8 – 7 (Subtracting 7)
=> -3x > -15
=> x < 5 (Dividing by -3)
Required Answer = {1, 2, 3, 4}
Note : Division by negative number reverses the inequality.

Question 4.
-17 < 9y – 8 ; y ∈ Z
Solution:
-17 < 9y – 8
=> -17 + 8 < 9y – 8 + 8 (Adding 8)
=> -9 < 9y
=> -1 < y (Dividing by 9)
Required number = {0, 1, 2, 3, 4, …}

Question 5.
Solve 9x – 7 ≤ 28 + 4x; x ∈ W
Solution:
9x – 1 ≤ 28 + 4x
=> 9x – 4x – 7 ≤ 28 + 4x – 4x (Subtracting 4x)
=> 5x – 7 ≤ 28
=> 5x – 7 + 7 ≤ 28 + 7 (Adding 7)
=> 5x ≤ 35
=> x ≤ 7 (Dividing by 5)
Required answer = {0, 1, 2, 3, 4, 5, 6, 7}

Question 6.
Solve : \(\frac { 2 }{ 3 }\)x + 8 < 12 ; x ∈ W
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -1

Question 7.
Solve -5 (x + 4) > 30 ; x ∈ Z
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -2

Question 8.
Solve the inquation 8 – 2x > x – 5 ; x ∈ N.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -3
x = 1, 2, 3, 4 (x ∈ N)
Solution set = {1, 2, 3, 4}

Question 9.
Solve the inequality 18 – 3 (2x – 5) > 12; x ∈ W.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -4

Question 10.
Solve : \(\frac { 2x+1 }{ 3 }\) + 15 < 17; x ∈ W.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -5

Question 11.
Solve : -3 + x < 2, x ∈ N
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -6

Question 12.
Solve : 4x – 5 > 10 – x, x ∈ {0, 1, 2, 3, 4, 5, 6, 7}
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -7
Solution set = {4, 5, 6, 7}
Question 13.
Solve : 15 – 2(2x – 1) < 15, x ∈ Z.
Solution:

Question 14.
Solve : \(\frac { 2x+3 }{ 5 }\) > \(\frac { 4x-1 }{ 2 }\) , x ∈ W.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -9

Linear Inequations Exercise 15B – Selina Concise Mathematics Class 8 ICSE Solutions

Solve and graph the solution set on a number line :
Question 1.
x – 5 < -2 ; x ∈ N
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -10

Question 2.
3x – 1 > 5 ; x ∈ W
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -11

Question 3.
-3x + 12 < -15 ; x ∈ R.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -12

Question 4.
7 > 3x – 8 ; x ∈ W
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -13

Question 5.
8x – 8 < – 24 ; x ∈ Z
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -14

Question 6.
8x – 9 > 35 – 3x ; x ∈ N
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -16

Question 7.
5x + 4 > 8x – 11 ; x ∈ Z
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -17

Question 8.
\(\frac { 2x }{ 5 }\) + 1 < -3 ; x ∈ R
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -18

Question 9.
\(\frac { x }{ 2 }\) > -1 + \(\frac { 3x }{ 4 }\) ; x ∈ N
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -20

Question 10.
\(\frac { 2 }{ 3 }\) x + 5 ≤ \(\frac { 1 }{ 2 }\) x + 6 ; x ∈ W
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -21

Question 11.
Solve the inequation 5(x – 2) > 4 (x + 3) – 24 and represent its solution on a number line.
Given the replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -23

Question 12.
Solve \(\frac { 2 }{ 3 }\) (x – 1) + 4 < 10 and represent its solution on a number line.
Given replacement set is {-8, -6, -4, 3, 6, 8, 12}.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -24

Question 13.
For each inequation, given below, represent the solution on a number line :
(i) \(\frac { 5 }{ 2 }\) – 2x ≥ \(\frac { 1 }{ 2 }\) ; x ∈ W
(ii) 3(2x – 1) ≥ 2(2x + 3), x ∈ Z
(iii) 2(4 – 3x) ≤ 4(x – 5), x ∈ W
(iv) 4(3x + 1) > 2(4x – 1), x is a negative integer
(v) \(\frac { 4 – x }{ 2 }\) < 3, x ∈ R
(vi) -2(x + 8) ≤ 8, x ∈ R
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 15 Linear Inequations image -26

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers

August 11, 2022January 17, 2019 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 1 Rational Numbers. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Rational Numbers Exercise 1A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Add, each pair of rational numbers, given below, and show that their addition (sum) is also a rational number:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 1
Solution:

Question 2.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 8
Solution:

Question 3.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 15
Solution:

Question 4.
For each pair of rational numbers, verify commutative property of addition of rational numbers:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 23
Solution:

This verifies the commutative property for the addition of rational numbers.

This verifies the commutative property for the addition of rational numbers.

This verifies the commutative property for the addition of rational numbers.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 29
This verifies the commutative property for the addition of rational numbers.

This verifies the commutative property for the addition of rational numbers.

This verifies the commutative property for the addition of rational numbers.

Question 5.
For each set of rational numbers, given below, verify the associative property of addition of rational numbers:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 33
Solution:

This verifies associative property of the addition of rational numbers.

This verifies associative property of the addition of rational numbers.

This verifies associative property of the addition of rational numbers.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 41

Question 6.
Write the additive inverse (negative) of:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 43
Solution:

Question 7.
Fill in the blanks:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 45
Solution:

Question 8.
State, true or false:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 48
Solution:
(i) False
(ii) False
(iii) True
(iv) True
(v) False
(vi) False

Rational Numbers Exercise 1B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 49
Solution:

Question 2.
Subtract:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 54
Solution:

Question 3.
The sum of two rational numbers is \(\frac { 9 }{ 20 }\). If one of them is \(\frac { 2 }{ 5 }\), find the other.
Solution:
The sum of two rational numbers = \(\frac { 9 }{ 20 }\)
And, one of the numbers = \(\frac { 2 }{ 5 }\)
The other rational number
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 59

Question 4.
The sum of the two rational numbers is \(\frac { -2 }{ 3 }\). If one of them is \(\frac { -8 }{ 5 }\), find the other.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 60

Question 5.
The sum of the two rational numbers is -6. If one of them is \(\frac { -8 }{ 5 }\), find the other.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 62

Question 6.
Which rational number should be added to \(\frac { -7 }{ 8 }\) to get \(\frac { 5 }{ 9 }\) ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 63

Question 7.
Which rational number should be added to \(\frac { -5 }{ 9 }\) to get \(\frac { -2 }{ 3 }\) ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 64

Question 8.
Which rational number should be subtracted from \(\frac { -5 }{ 6 }\) to get \(\frac { 4 }{ 9 }\) ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 65

Question 9.
(i) What should be subtracted from -2 to get \(\frac { 3 }{ 8 }\)
(ii) What should be added to -2 to get \(\frac { 3 }{ 8 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 66

Question 10.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 68

Solution:

Rational Numbers Exercise 1C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 73
Solution:

Question 2.
Multiply:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 76
Solution:

Question 3.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 79
Solution:

Question 4.
Multiply each rational number, given below, by one (1):
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 84
Solution:

Question 5.
For each pair of rational numbers, given below, verify that the multiplication is commutative:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 87
Solution:

Question 6.
Write the reciprocal (multiplicative inverse) of each rational number, given below :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 90
Solution:

Question 7.
Find the reciprocal (multiplicative inverse) of:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 92

Solution:

Question 8.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 95
Solution:

Question 9.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 98
Solution:

Question 10.
Name the multiplication property of rational numbers shown below :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 101
Solution:
(i) Commutativity property.
(ii) Associativity property.
(iii) Distributivity property.
(iv) Existence of inverse.
(v) Existence of identity.
(vi) Existence of inverse.

Question 11.
Fill in the blanks:
(i) The product of two positive rational numbers is always ……………
(ii) The product of two negative rational numbers is always ……………
(iii) If two rational numbers have opposite signs then their product is always …………..
(iv) The reciprocal of a positive rational number is ………. and the reciprocal of a negative raitonal number is ……………
(v) Rational number 0 has ………….. reciprocal.
(vi) The product of a rational number and its reciprocal is ………..
(vii) The numbers ……….. and ……….. are their own reciprocals.
(viii) If m is reciprocal of n, then the reciprocal of n is ………….
Solution:
(i) The product of two positive rational numbers is always positive.
(ii) The product of two negative rational numbers is always positive.
(iii) If two rational numbers have opposite signs then their product is always negative.
(iv) The reciprocal of a positive rational number is positive and the reciprocal of a negative raitonal number is negative.
(v) Rational number 0 has no reciprocal.
(vi) The product of a rational number and its reciprocal is 1.
(vii) The numbers 1 and -1 are their own reciprocals.
(viii)If m is reciprocal of n, then the reciprocal of n is m.

Rational Numbers Exercise 1D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 102

Solution:

Question 2.
Divide:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 108
Solution:

Question 3.
The product of two rational numbers is -2. If one of them is \(\frac { 4 }{ 7 }\), find the other.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 111

Question 4.
The product of two numbers is \(\frac { -4 }{ 9 }\). If one of them is \(\frac { -2 }{ 27 }\), find the other.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 112

Question 5.
m and n are two rational numbers such that
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 113
Solution:

Question 6.
By what number must \(\frac { -3 }{ 4 }\) be multiplied so that the product is \(\frac { -9 }{ 16 }\) ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 115

Question 7.
By what number should \(\frac { -8 }{ 13 }\) be multiplied to get 16?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 116

Question 8.
If 3\(\frac { 1 }{ 2 }\) litres of milk costs ₹49, find the cost of one litre of milk?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 117

Question 9.
Cost of 3\(\frac { 2 }{ 5 }\) metre of cloth is ₹88\(\frac { 1 }{ 2 }\). What is the cost of 1 metre of cloth?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 118

Question 10.
Divide the sum of \(\frac { 3 }{ 7 }\) and \(\frac { -5 }{ 14 }\) by \(\frac { -1 }{ 2 }\).
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 119

Question 11.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 121
Solution:

Question 12.
The product of two rational numbers is -5. If one of these numbers is \(\frac { -7 }{ 15 }\), find the other.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 125

Question 13.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 127
Solution:

Rational Numbers Exercise 1E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 130
Solution:

Question 2.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 132
Solution:

Question 3.
Insert one rational number between (0 7 and 8 (ii) 3.5 and 5
(i) 2 and 3.2
(ii) 3.5 and 5
(iii) 2 and 3.2
(iv) 4.2 and 3.6
(v) \(\frac { 1 }{ 2 }\) and 2
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 134

Question 4.
Insert two rational numbers between
(i) 6 and 7
(ii) 4.8 and 6
(iii) 2.7 and 6.3
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 136

Question 5.
Insert three rational numbers between
(i) 3 and 4
(ii) 10 and 12
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 139

Question 6.
Insert five rational numbers between \(\frac { 3 }{ 5 }\) and \(\frac { 2 }{ 5 }\)
Solution:
LCM of denominators 5 and 3 is 15
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 140

Question 7.
Insert six rational numbers between \(\frac { 5 }{ 6 }\) and \(\frac { 8 }{ 9 }\)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 141

Question 8.
Insert seven rational numbers between 2 and 3.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 1 Rational Numbers image - 143

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon

August 11, 2022May 14, 2018 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 20 Area of Trapezium and a Polygon. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Area of Trapezium and a Polygon Exercise 20A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the area of a triangle, whose sides are :
(i) 10 cm, 24 cm and 26 cm
(ii) 18 mm, 24 mm and 30 mm
(iii) 21 m, 28 m and 35 m
Solution:
(i) Sides of ∆ are
a = 10 cm
b = 24 cm
c = 26 cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 1
(ii) Sides of ∆ are
a = 18 mm
b = 24 mm
c = 30 mm

(iii) Sides of ∆ are
a = 21 m
b = 28 m
c = 35 m

Question 2.
Two sides of a triangle are 6 cm and 8 cm. If height of the triangle corresponding to 6 cm side is 4 cm ; find :
(i) area of the triangle
(ii) height of the triangle corresponding to 8 cm side.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 5

Question 3.
The sides of a triangle are 16 cm, 12 cm and 20 cm. Find :
(i) area of the triangle ;
(ii) height of the triangle, corresponding to the largest side ;
(iii) height of the triangle, corresponding to the smallest side.
Solution:
Sides of ∆ are
a = 20 cm
b = 12 cm
c = 16 cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 6

Question 4.
Two sides of a triangle are 6.4 m and 4.8 m. If height of the triangle corresponding to 4.8 m side is 6 m; find :
(i) area of the triangle ;
(ii) height of the triangle corresponding to 6.4 m side.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 8
= 9/2 = 4.5 m
Hence (i) 14.4 m² (ii) 4.5 m

Question 5.
The base and the height of a triangle are in the ratio 4 : 5. If the area of the triangle is 40 m²; find its base and height.
Solution:
Let base of ∆ = 4x m
and height of ∆ = 5x m
area of ∆ =40 m²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 9

Question 6.
The base and the height of a triangle are in the ratio 5 : 3. If the area of the triangle is 67.5 m²; find its base and height.
Solution:
Let base = 5x m
height = 3x m
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 10
base = 5x = 5 x 3 = 15 m
height = 3x = 3 x 3 = 9 m

Question 7.
The area of an equilateral triangle is 144√3 cm²; find its perimeter.
Solution:
Let each side of an equilateral triangle = x cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 11

Question 8.
The area of an equilateral triangle is numerically equal to its perimeter. Find its perimeter correct to 2 decimal places.
Solution:
Let each side of the equilateral traingle = x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 12

Question 9.
A field is in the shape of a quadrilateral ABCD in which side AB = 18 m, side AD = 24 m, side BC = 40m, DC = 50 m and angle A = 90°. Find the area of the field.
Solution:
Since ∠A = 90°
By Pythagorus Theorem,
In ∆ABD,
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 13

Question 10.
The lengths of the sides of a triangle are in the ratio 4 : 5 : 3 and its perimeter is 96 cm. Find its area.
Solution:
Let the sides of the triangle ABC be 4x, 5x and 3x
Let AB = 4x, AC = 5x and BC = 3x
Perimeter = 4x + 5x + 3x = 96
=> 12x = 96
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 14

Question 11.
One of the equal sides of an isosceles triangle is 13 cm and its perimeter is 50 cm. Find the area of the triangle.
Solution:
In isosceles ∆ABC
AB = AC = 13 cm But perimeter = 50 cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 15

Question 12.
The altitude and the base of a triangular field are in the ratio 6 : 5. If its cost is ₹ 49,57,200 at the rate of ₹ 36,720 per hectare and 1 hectare = 10,000 sq. m, find (in metre) dimensions of the field,
Solution:
Total cost = ₹ 49,57,200
Rate = ₹ 36,720 per hectare
Total area of the triangular field
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 16

Question 13.
Find the area of the right-angled triangle with hypotenuse 40 cm and one of the other two sides 24 cm.
Solution:
In right angled triangle ABC Hypotenuse AC = 40 cm
One side AB = 24 cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 17

Question 14.
Use the information given in the adjoining figure to find :
(i) the length of AC.
(ii) the area of a ∆ABC
(iii) the length of BD, correct to one decimal place.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 18
Solution:

Area of Trapezium and a Polygon Exercise 20B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the length and perimeter of a rectangle, whose area = 120 cm² and breadth = 8 cm
Solution:
area of rectangle = 120 cm²
breadth, b = 8 cm
Area = l x b
l x 8 = 120
l = 120/8 = 15 cm
Perimeter = 2 (l+b) = 2(15+8) = 2 x 23 = 46 cm
Length = 15 cm; Perimeter = 46 cm

Question 2.
The perimeter of a rectangle is 46 m and its length is 15 m. Find its :
(i) breadth
(ii) area
(iii) diagonal.
Solution:
(i) Perimeter of rectangle = 46 m
length, l = 15 m
2 (l+b) = 46
2(15 + b) = 46
15+b = 46/2 = 23
b = 23 – 15
b = 8 m
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 20

Question 3.
The diagonal of a rectangle is 34 cm. If its breadth is 16 cm; find its :
(i) length
(ii) area
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 21
AC² = AB²+BC² (By Pythagoras theorem)
(34)² = l² + (16)²
1156 = l² + 256
l² = 1156 – 256
l² = 900
l = √900 = 30 cm
area = l x b = 30 x 16 = 480 cm²
(i) 30 cm (ii) 480 cm²

Question 4.
The area of a small rectangular plot is 84 m². If the difference between its length and the breadth is 5 m; find its perimeter.
Solution:
Area of a rectangular plot = 84 m²
Let breadth = x m
Then length = (x + 5) m
Area = l x b
x(x + 5) = 84
x² + 5x – 84 = 0
=> x²+ 12x – 7x – 84 = 0
=> x(x + 12) – 7(x + 12) = 0
=> (x + 12) (x – 7) = 0
Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 7 = 0, then x = 7
Length = x + 5 = 7 + 5 = 12m
and breadth = x = 7 m
Perimeter = 2(l + b) = 2(12+ 7) = 2 x 19 m = 38 m

Question 5.
The perimeter of a square is 36 cm; find its area
Solution:
Perimeter of Square = 36 cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 22

Question 6.
Find the perimeter of a square; whose area is : 1.69 m²
Solution:
Area of square= 1.69 m²
Side = √area = √1.69 = 1.3 m
Perimeter = 4 x side = 4 x 1.3 = 5.2 m

Question 7.
The diagonal of a square is 12 cm long; find its area and length of one side.
Solution:
Let side of square = a cm
diagonal = 12 cm
By Pythagoras Theorem, a² + a² = (12)²
2a² = 144
a² = 72
Area of square = a² = 72 cm²
a² = 72
a = √72 = 8.49 cm

Question 8.
The diagonal of a square is 15 m; find the length of its one side and perimeter.
Solution:
Diagonal of square = 15 m
Let side of square = a
a² + a² = (15)² = 225
a² = 225/2 = 112.50
a = √112.50 = 10.6 m
Perimeter = 4 x a = 10.6 x 4 = 42.4 m

Question 9.
The area of a square is 169 cm². Find its:
(i) one side
(ii) perimeter
Solution:
Let each side of the square be x cm.
Its area = x² = 169 (given)
x = √169
x = 13 cm
(i) Thus, side of the square = 13 cm
(ii) Again perimeter = 4 (side) = 4 x 13 = 52 cm

Question 10.
The length of a rectangle is 16 cm and its perimeter is equal to the perimeter of a square with side 12.5 cm. Find the area of the rectangle.
Solution:
Length of the rectangle = 16 cm
Let its breadth be x cm
Perimeter = 2 (16 + x) = 32 + 2x
Also perimeter = 4(12.5) = 50 cm.
According to statement,
32 + 2x = 50
=> 2x = 50 – 32 = 18
=> x = 9
Breadth of the rectangle = 9 cm.
Area of the rectangle (l x b)= 16 x 9 = 144 cm²

Question 11.
The perimeter of a square is numerically equal to its area. Find its area.
Solution:
Let each side of the square be x cm.
Its perimeter = 4x,
Area =x²
By the given condition 4x = x²
=> x² – 4x = 0
=> x (x – 4) = 0
=> x = 4 [x ≠ 0]
Area = x² = (4)² = 4 x 4 = 16 sq.units.

Question 12.
Each side of a rectangle is doubled. Find the ratio between :
(i) perimeters of the original rectangle and the resulting rectangle.
(ii) areas of the original rectangle and the resulting rectangle.
Solution:
Let length of the rectangle = x
and breadth of the rectangle = y
(i) Perimeter P = 2(x + y)
Again, new length = 2x
New breadth = 2y
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 23

Question 13.
In each of the following cases ABCD is a square and PQRS is a rectangle. Find, in each case, the area of the shaded portion.
(All measurements are in metre).
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 24
Solution:
(i) Area of the shaded portion
= Area of the rectangle PQRS – Area of square ABCD
= 3.2 x 1.8 – (1.4)² (∵ PQ = 3.2 and PS = 1.8) Side of square AB = 1.4
= 5.76 – 1.96 = 3.80 = 3.8 m²
(ii) Area of the shaded portion = Area of square ABCD – Area of rectangle PQRS
= 6 x 6 – (3.6) (4.8) = 36 – 17.28 = 18.72 m²

Question 14.
A path of uniform width, 3 m, runs around the outside of a square field of side 21 m. Find the area of the path.
Solution:
According to the given information the figure will be as shown alongside.
Clearly, length of the square field excluding path = 21 m.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 25
Area of the square side excluding the path = 21 x 21 = 441 m²
Again, length of the square field including the path = 21 + 3 + 3 = 27 m
Area of the square field including the path = 27 x 27 = 729 m²
Area of the path = 729 – 441 = 288 m²

Question 15.
A path of uniform width, 2.5 m, runs around the inside of a rectangular field 30 m by 27 m. Find the area of the path.
Solution:
According to the given statement the figure will be as shown alongside.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 26
Clearly, the length of the rectangular field including the path = 30 m.
Breadth = 27 m.
Its Area = 30 x 27 = 810 m²
Width of the path = 2.5 m
Length of the rectangular field including the path = 30 – 2.5 – 2.5 = 25 m.
Breadth = 27 – 2.5 – 2.5 = 22m
Area of the rectangular field including the path = 25 x 22 = 550 m²
Hence, area of the path = 810 – 550 = 260 m².

Question 16.
The length of a hall is 18 m and its width is 13.5 m. Find the least number of square tiles, each of side 25 cm, required to cover the floor of the hall,
(i) without leaving any margin.
(ii) leaving a margin of width 1.5 m all around. In each case, find the cost of the tiles required at the rate of Rs. 6 per tile
Solution:
(i) Length of hall (l) = 18 m and breadth (b) = 13.5 m
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 27

Question 17.
A rectangular field is 30 m in length and 22m in width. Two mutually perpendicular roads, each 2.5 m wide, are drawn inside the field so that one road is parallel to the length of the field and the other road is parallel to its width. Calculate the area of the crossroads.
Solution:
Length of rectangular field (l) = 30 m and breadth (b) = 22m
width of parallel roads perpendicular to each other inside the field =2.5m
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 29
Area of cross roads = width of roads (Length + breadth) – area of middle square
= 2.5 (30 + 22) – (2.5)²
= 2.5 x 52 – 6.25 m²
= (130 – 6.25) m = 123.75 m²

Question 18.
The length and the breadth of a rectangular field are in the ratio 5 : 4 and its area is 3380 m². Find the cost of fencing it at the rate of ₹75 per m.
Solution:
Ratio in length and breadth = 5 : 4
Area of rectangular field = 3380 m²
Let length = 5x and breadth = 4x
5x x 4x = 3380
=> 20x²= 3380
x² = 3380/20 = 169 = (13)2
x = 13
Length = 13 x 5 = 65 m
Breadth =13 x 4 = 52 m
Perimeter = (l + b) = 2 x (65 + 52) m = 2 x 117 = 234 m
Rate of fencing = ₹ 75 per m
Total cost = 234 x 75 = ₹ 17550

Question 19.
The length and the breadth of a conference hall are in the ratio 7 : 4 and its perimeter is 110 m. Find:
(i) area of the floor of the hall.
(ii) number of tiles, each a rectangle of size 25 cm x 20 cm, required for flooring of the hall.
(iii) the cost of the tiles at the rate of ₹ 1,400 per hundred tiles.
Solution:
Ratio in length and breadth = 7 : 4
Perimeter = 110 m
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 30

Area of Trapezium and a Polygon Exercise 20C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
The following figure shows the cross-section ABCD of a swimming pool which is trapezium in shape.
If the width DC, of the swimming pool is 6.4cm, depth (AD) at the shallow end is 80 cm and depth (BC) at deepest end is 2.4m, find Its area of the cross-section.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 32
Solution:
Area of the cross-section = Area of trapezium ABCD

= 1024 cm² or = 10.24 sq.m.

Question 2.
The parallel sides of a trapezium are in the ratio 3 : 4. If the distance between the parallel sides is 9 dm and its area is 126 dm² ; find the lengths of its parallel sides.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 34
Let parallel sides of trapezium be
a = 3x
b = 4x
Distance between parallel sides, h = 9 dm
area of trapezium = 126 dm²

Question 3.
The two parallel sides and the distance between them are in the ratio 3 : 4 : 2. If the area of the trapezium is 175 cm², find its height.
Solution:
Let the two parallel sides and the distance between them be 3x, 4x, 2x cm respectively
Area = \(\frac { 1 }{ 2 }\) (sum of parallel sides) x (distance between parallel sides)
= \(\frac { 1 }{ 2 }\) (3x + 4x) x 2x = 175 (given)
=> 7x x x = 175
=> 7x² = 175
=> x² = 25
=> x = 5
Height i.e. distance between parallel sides = 2x = 2 x 5 = 10 cm

Question 4.
A parallelogram has sides of 15 cm and 12 cm; if the distance between the 15 cm sides is 6 cm; find the distance between 12 cm sides.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 36
BQ = \(\frac { 15 }{ 2 }\) = 7.5 cm

Question 5.
A parallelogram has sides of 20 cm and 30 cm. If the distance between its shorter sides is 15 cm; find the distance between the longer sides.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 37

Question 6.
The adjacent sides of a parallelogram are 21 cm and 28 cm. If its one diagonal is 35 cm; find the area of the parallelogram.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 38

Question 7.
The diagonals of a rhombus are 18 cm and 24 cm. Find:
(i) its area ;
(ii) length of its sides.
(iii) its perimeter;
Solution:
Diagonal of rhombus are 18 cm and 24 cm.
area of rhombus = \(\frac { 1 }{ 2 }\) x Product of diagonals
= \(\frac { 1 }{ 2 }\) x 18 x 24
= 216 cm²
(ii) Diagonals of rhombus bisect each other at right angles.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 40

Question 8.
The perimeter of a rhombus is 40 cm. If one diagonal is 16 cm; find :
(i) its another diagonal
(ii) area
Solution:
(i) Perimeter of rhombus = 40 cm
side = \(\frac { 1 }{ 4 }\) x 40 = 10 cm
One diagonal = 16 cm
Diagonals of rhombus bisect each other at right angles.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 41

Question 9.
Each side of a rhombus is 18 cm. If the distance between two parallel sides is 12 cm, find its area.
Solution:
Each side of the rhombus = 18 cm
base of the rhombus = 18 cm
Distance between two parallel sides = 12 cm
Height = 12 cm
Area of the rhombus = base x height = 18 x 12 = 216 cm²

Question 10.
The length of the diagonals of a rhombus is in the ratio 4 : 3. If its area is 384 cm², find its side.
Solution:
Let the lengths of the diagonals of rhombus are 4x, 3x.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 42

Question 11.
A thin metal iron-sheet is rhombus in shape, with each side 10 m. If one of its diagonals is 16 m, find the cost of painting its both sides at the rate of ₹ 6 per m².
Also, find the distance between the opposite sides of this rhombus.
Solution:
Side of rhombus shaped iron sheet = 10 m and one diagonals (AC) = 16 m
Join BD diagonal which bisects AC at O
The diagonals of a rhombus bisect each other at right angle
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 43

Question 12.
The area of a trapezium is 279 sq.cm and the distance between its two parallel sides is 18 cm. If one of its parallel sides is longer than the other side by 5 cm, find the lengths of its parallel sides.
Solution:
Area of trapezium = 279 sq.cm
Distance between two parallel lines (h) = 18 cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 45

Question 13.
The area of a rhombus is equal to the area of a triangle. If base of ∆ is 24 cm, its corresponding altitude is 16 cm and one of the diagonals of the rhombus is 19.2 cm. Find its other diagonal.
Solution:
Area of a rhombus = Area of a triangle Base of triangle = 24 cm
and altitude = 16 cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 47

Question 14.
Find the area of the trapezium ABCD in which AB//DC, AB = 18 cm, ∠B = ∠C = 90°, CD = 12 cm and AD = 10 cm.
Solution:
In trapezium ABCD,
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 48

Area of Trapezium and a Polygon Exercise 20D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the radius and area of a circle, whose circumference is :
(i) 132 cm
(ii) 22 m
Solution:
(i) Circumference of circle = 132 cm
2πr = 132
2 x 22/7 x r = 132
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 49

Question 2.
Find the radius and circumference of a circle, whose area is :
(i) 154 cm²
(ii) 6.16 m²
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 50

Question 3.
The circumference of a circular table is 88 m. Find its area.
Solution:
Circumference of circle = 88 m
2πr = 88 m
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 52

Question 4.
The area of a circle is 1386 sq.cm ; find its circumference.
Solution:
Area of circle = 1386 cm²
πr² = 1386
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 53

Question 5.
Find the area of a flat circular ring formed by two concentric circles (circles with same centre) whose radii are 9 cm and 5 cm.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 54

Question 6.
Find the area of the shaded portion in each of the following diagrams :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 55
Solution:

Question 7.
The radii of the inner and outer circumferences of a circular running track are 63 m and 70 m respectively. Find :
(i) the area of the track ;
(it) the difference between the lengths of the two circumferences of the track.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 59
= 2x 22/7 x 63 = 396 m
Difference between lengths of two circumferences = 440 – 396 = 44 m
Hence (i) 2926 m² (ii) 44 m

Question 8.
A circular field cf radius 105 m has a circular path of uniform width of 5 m along and inside its boundary. Find the area of the path.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 60

Question 9.
There is a path of uniform width 7 m round and outside a circular garden of diameter 210 m. Find the area of the path.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 61

Question 10.
A wire, when bent in the form of a square encloses an area of 484 cm². Find :
(i) one side of the square ;
(ii) length of the wire ;
(iii) the largest area enclosed; if the same wire is bent to form a circle.
Solution:
(i) Area of Square = 484 cm²
Side of Square = √Area = √484 = 22 cm
(ii) Perimeter, i.e. length of wire = 4 x 22 = 88 cm
(iii) Circumference of circle = 88 cm
2πr = 88
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 62

Question 11.
A wire, when bent in the form of a square; encloses an area of 196 cm². If the same wire is bent to form a circle; find the area of the circle.
Solution:
Area of Square = 196 cm²
Side of Square = √Area = √196 = 14 cm
Perimeter of Square = 4 x 14 cm
i.e. length of wire = 56 cm
Circumference of circle = 56 cm
2πr = 56
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 63
= 2744/11
249.45 cm²

Question 12.
The radius of a circular wheel is 42 cm. Find the distance travelled by it in :
(i) 1 revolution ;
(ii) 50 revolutions ;
(iii) 200 revolutions ;
Solution:
(i) Radius of wheel, r = 42 cm
Circumference i.e. distance travelled in 1 revolution = 2πr = 2 x 22/7 x 42 = 264 cm
(ii) Distance travelled in 50 revolutions = 264 x 50 = 13200 cm = 132 m
(iii) Distance travelled in 200 revolutions = 264 x 200 = 52800 cm = 528 m
Hence (i) 264 cm (ii) 132 m (iii) 528 m

Question 13.
The diameter of a wheel is 0.70 m. Find the distance covered by it in 500 revolutions. If the wheel takes 5 minutes to make 500 revolutions; find its speed in :
(i) m/s
(ii) km/hr.
Solution:
Diameter = 0.70 m
Radius, r = 0.35 m
Distance covered in 1 revolution, i.e. circumference = 2πr = 2 x 22/7 x 0.35 = 2.20 m
Distance covered in 500 revolutions = 2.20 x 500 = 1100 m
Time taken = 5 minutes = 5 x 60 = 300 sec.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 64

Question 14.
A bicycle wheel, diameter 56 cm, is making 45 revolutions in every 10 seconds. At what speed in kilometre per hour is the bicycle travelling ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 65

Question 15.
A roller has a diameter of 1.4 m. Find :
(i) its circumference ;
(ii) the number of revolutions it makes while travelling 61.6 m.
Solution:
Diameter = 1.4 m
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 67

Question 16.
Find the area of the circle, length of whose circumference is equal to the sum of the lengths of the circumferences with radii 15 cm and 13 cm.
Solution:
In a circle
Circumference = Sum of circumferences of two circle of radii 15 cm and 13 cm
Now circumference of first smaller circle = 2πr
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 68

Question 17.
A piece of wire of length 108 cm is bent to form a semicircular arc bounded by its diameter. Find its radius and area enclosed.
Solution:
Length of wire = 108 cm
Let r be the radius of the semicircle
πr+ 2r = 108
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 69

Question 18.
In the following figure, a rectangle ABCD enclosed three circles. If BC = 14 cm, find the area of the shaded portion (Take π = 22/7)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon image - 70
Solution:
In rectangle ABCD, BC = 14 cm

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacity

August 11, 2022May 12, 2018 by Prasanna

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacity (Cuboid, Cube and Cylinder)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacity (Cuboid, Cube and Cylinder)

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Surface Area, Volume and Capacity Exercise 21A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the volume and the total surface area of a cuboid, whose :
(i) length = 15 cm, breadth = 10 cm and height = 8 cm.
(ii) l = 3.5 m, b = 2.6 m and h = 90 cm,
Solution:
(i) Length =15 cm, Breadth = 10 cm, Height = 8 cm.
Volume of a cuboid = Length x Breadth x Height = 15 x 10 x 8 =1200 cm³.
Total surface area of a cuboid 2 (l x b + b x h + h x l) = 2 (15 x 10 + 10 x 8 + 8 x 15) = 2(150 + 80 +120) = 2 x 350 = 700 cm²
(ii) Length = 3.5 m Breadth = 2.6 m, Height = 90 cm = \(\frac { 90 }{ 100 }\) m = 0.9 m.
Volume of a cuboid = l x b x h = 3.5 x 2.6 x 0.9 = 8.19 m³
Total surface area of a cuboid = 2(l x b + b x h + h x l)
= 2 (3.5 x 2.6 + 2.6 x 0.9 + 0.9 x 3.5) = 2 (910 + 2.34 + 3.15) = 2(14.59)= 29.18 m²

Question 2.
(i) The volume of a cuboid is 3456 cm³. If its length = 24 cm and breadth = 18 cm ; find its height.
(ii) The volume of a cuboid is 7.68 m³. If its length = 3.2 m and height = 1.0 m; find its breadth.
(iii) The breadth and height of a rectangular solid are 1.20 m and 80 cm respectively. If the volume of the cuboid is 1.92 m³; find its length.
Solution:
(i) Volume of the given cuboid = 3456 cm³.
Length of the given cuboid = 24 cm.
Breadth of the given cuboid = 18 cm
We know,
Length x Breadth x Height = Volume of a cuboid
⇒ 24 x 18 x Height = 3456
⇒ Height = \(\frac { 3456 }{ { 24 }\times { 18 } }\)
⇒ Height = \(\frac { 3456 }{ 432 }\)
⇒ Height = 8 cm
(ii) Volume of a cuboid = 7.68 m³
Length of a cuboid = 3.2 m
Height of a cuboid = 10m
We know
Length x Breadth x Height = Volume of a cuboid
3.2 x Breadth x 1.0 = 7.68
⇒ Breadth = \(\frac { 7.68 }{ { 3.2 }\times { 1.0 } }\)
⇒ Breadth = \(\frac { 7.68 }{ 3.2 }\)
⇒ Breadth = 2.4 m
(iii) Volume of a rectangular solid = 1.92 m³
Breadth of a rectangular solid = 1.20 m
Height of a rectangular solid = 80 cm = 0.8 m
We know
Length x Breadth x Height = Volume of a rectangular solid (cubical)
Length x 1.20 x 0.8 = 1.92
Length x 0.96 = 1.92
⇒ Length = \(\frac { 1.92 }{ 0.96 }\)
⇒ Length = \(\frac { 192 }{ 96 }\)
⇒ Length = 2 m

Question 3.
The length, breadth and height of a cuboid are in the ratio 5 : 3 : 2. If its volume is 240 cm³; find its dimensions. (Dimensions means : its length, breadth and height). Also find the total surface area of the cuboid.
Solution:
Let length of the given cuboid = 5x
Breadth of the given cuboid = 3x
Height of the given cuboid = 2x
Volume of the given cuboid = Length x Breadth x Height
= 5x x 3x x 2x = 30x³
But we are given volume = 240 cm³
30x³ = 240 cm³
⇒ x³ = \(\frac { 240 }{ 30 }\)
⇒ x³ = 8
⇒ x = \({ 8 }^{ \frac { 1 }{ 3 } }\)
⇒ x = \(\left( { 2 }\times { 2 }\times { 2 } \right) ^{ \frac { 1 }{ 3 } }\)
⇒ x = 2 cm
Length of the given cube = 5x = 5 x 2 = 10 cm
Breadth of the given cube = 3x = 3 x 2 = 6 cm
Height of the given cube = 2x = 2 x 2 = 4cm
Total surface area of the given cuboid = 2(l x b + b x h + h x l)
= 2(10 x 6 + 6 x 4 + 4 x 10) = 2(60 + 24 + 40) = 2 x 124 = 248 cm²

Question 4.
The length, breadth and height of a cuboid are in the ratio 6 : 5 : 3. If its total surface area is 504 cm2; find its dimensions. Also, find the volume of the cuboid.
Solution:
Let length of the cuboid = 6x
Breadth of the cuboid = 5x
Height of the cuboid = 3x
Total surface area of the given cuboid = 2 (I x b + b x h + h x l)
= 2(6x x 5x + 5x x 3x + 3x x 6x) = 2(30×2 + 15×2 + 18×2)
= 2 x 63×2 = 126x²
But we are given total surface area of the given cuboid = 504 cm²
126x² = 504 cm²
=> x² = \(\frac { 504 }{ 126 }\)
=> x² = 4
=> x = √4
=> x = 2 cm.
Length of the cuboid = 6x = 6 x 2 = 12 cm
Breadth of the cuboid = 5x = 5 x 2 = 10cm
Height of the cuboid = 3x = 3 x 2 = 6 cm
Volume of the cuboid = l x b x h = 12 x 10 x 6 = 720 cm³

Question 5.
Find the volume and total surface area of a cube whose each edge is :
(i) 8 cm
(ii) 2 m 40 cm.
Solution:
(i) Edge of the given cube = 8 cm
Volume of the given cube = (Edge)³ = (8)³ = 8 x 8 x 8 = 512 cm³
Total surface area of a cube = 6(Edge)² = 6 x (8)² = 384 cm²
(ii) Edge of the given cube = 2 m 40 cm = 2.40 m
Volume of a cube = (Edge)³
Volume of the given cube = (2.40)³ = 2.40 x 2.40 x 2.40 = 13.824 m²
Total surface area of the given cube = 6 x 2.4 x 2.4 = 34.56 m²

Question 6.
Find the length of each edge of a cube, if its volume is :
(i) 216 cm³
(ii) 1.728 m³
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -1

Question 7.
The total surface area of a cube is 216 cm2. Find its volume.
Solution:
6(Edge)² = Total surface area of a cube
6(Edge)² = 216 cm²
=> (Edge)² = \(\frac { 216 }{ 6 }\)
=> (Edge)² = 36
=> Edge = √36
=> Edge = 6 cm
Volume of the given cube = (Edge)³ = (6)³ = 6 x 6 x 6 = 216 cm³

Question 8.
A solid cuboid of metal has dimensions 24 cm, 18 cm and 4 cm. Find its volume.
Solution:
Length of the cuboid = 24 cm
Breadth of the cuboid = 18 cm
Height of the cuboid = 4 cm
Volume of the cuboid = l x b x h = 24 x 18 x 4 = 1728 cm³

Question 9.
A wall 9 m long, 6 m high and 20 cm thick, is to be constructed using bricks of dimensions 30 cm, 15 cm and 10 cm. How many bricks will be required.
Solution:
Length of the wall = 9 m = 9 x 100 cm = 900 cm
Height of the wall = 6 m = 6 x 100 cm = 600 cm
Breadth of the wall = 20 cm
Volume of the wall = 900 x 600 x 20 cm3 = 10800000 cm³
Volume of one Brick = 30 x 15 x 10 cm³ = 4500 cm³
Number of bricks required to construct the wall = \(\frac { Volume\quad of\quad wall }{ Volume\quad \quad of\quad one\quad brick }\)
= \(\frac { 10800000 }{ 4500 }\)
= 2400

Question 10.
A solid cube of edge 14 cm is melted down and recasted into smaller and equal cubes each of edge 2 cm; find the number of smaller cubes obtained.
Solution:
Edge of the big solid cube = 14 cm
Volume of the big solid cube = 14 x 14 x 14 cm3 = 2744 cm³
Edge of the small cube = 2 cm
Volume of one small cube = 2 x 2 x 2 cm³ = 8 cm³
Number of smaller cubes obtained = \(\frac { Volume\quad of\quad big\quad cube }{ Volume\quad of\quad one\quad small\quad cube }\)
= \(\frac { 2744 }{ 8 }\) = 343

Question 11.
A closed box is cuboid in shape with length = 40 cm, breadth = 30 cm and height = 50 cm. It is made of thin metal sheet. Find the cost of metal sheet required to make 20 such boxes, if 1 m² of metal sheet costs Rs. 45.
Solution:
Length of closed box (l) = 40 cm
Breadth (b) = 30 cm
and height (h) = 50 cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -2
Total surface area = 2 (lb + bh + hl)
= 2 (40 x 30 + 30 x 50 + 50 x 40) cm²
= 2 (1200 + 1500 + 2000) cm²
= 2 x 4700 = 9400 cm²
Surface area of sheet used for 20 such boxes = 9400 x 20 = 188000 cm²
Cost of 1 m² sheet = Rs. 45
Total cost = \(\frac { { 18000 }\times { 45 } }{ { 100 }\times { 100 }\times { 100 } }\)
= Rs.846

Question 12.
Four cubes, each of edge 9 cm, are joined as shown below :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -3
Write the dimensions of the resulting cuboid obtained. Also, find the total surface area and the volume of the resulting cuboid.
Solution:
Edge of each cube = 9 cm
(i) Length of the cuboid fonned by 4 cubes (l) = 9 x 4 = 36 cm
Breadth (b) = 9 cm and height (h) = 9 cm
(ii) Total surface area of the cuboid = 2(lb + bh + hl)
= 2 (36 x 9 + 9 x 9 + 9 x 36) cm²
= 2 (324 + 81 + 324) cm²
= 2 x 729 cm²
= 1458 cm²
(iii) Volume = l x b x h = 36 x 9 x 9 cm² = 2916 cm³

Surface Area, Volume and Capacity Exercise 21B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
How many persons can be accommodated in a big-hall of dimensions 40 m, 25 m and 15 m ; assuming that each person requires 5 m³ of air?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -4

Question 2.
The dimension of a class-room are; length = 15 m, breadth = 12 m and height = 7.5 m. Find, how many children can be accommodated in this class-room ; assuming 3.6 m³ of air is needed for each child.
Solution:
Length of the room = 15 m
Breadth of the room = 12 m
Height of the room = 7.5 m
Volume of the room = L x B x H = 15 x 12 x 7.5 m3 = 1350 m³
Volume of air required for each child = 3.6 m³
No. of children who can be accommodated in the class room.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -5

Question 3.
The length, breadth and height of a room are 6 m, 5.4 m and 4 m respectively. Find the area of :
(i) its four-walls
(ii) its roof.
Solution:
Length of the room = 6 m
Breadth of the room = 5.4 m
Height of the room = 4 m
(i) Area of four walls = 2(L+B) x H
= 2(6 + 5.4) x 4 = 2 x 11.4 x 4 = 91.2 m²
(ii) Area of the roof = L x B = 6 x 5.4 = 32.4 m²

Question 4.
A room 5 m long, 4.5 m wide and 3.6 m high has one door 1.5 m by 2.4 m and two windows, each 1 m by 0.75 m. Find :
(i) the area of its walls, excluding door and windows ;
(ii) the cost of distempering its walls at the rate of Rs.4.50 per m².
(iii) the cost of painting its roof at the rate of Rs.9 per m².
Solution:
Length of the room = 5 m
Breadth of the room = 4.5 m
Height of the room = 3.6 m
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -6

Question 5.
The dining-hall of a hotel is 75 m long ; 60 m broad and 16 m high. It has five – doors 4 m by 3 m each and four windows 3 m by 1.6 m each. Find the cost of :
(i) papering its walls at the rate of Rs.12 per m²;
(ii) carpetting its floor at the rate of Rs.25 per m².
Solution:
Length of the dining hall of a hotel = 75 m
Breadth of the dining hall of a hotel = 60 m
Height of the dining hall of a hotel = 16 m
(i) Area of four walls of the dining hall = 2[L+B) x H = 2(75 + 60) x 16
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -7

Question 6.
Find the volume of wood required to make a closed box of external dimensions 80 cm, 75 cm and 60 cm, the thickness of walls of the box being 2 cm throughout.
Solution:
External length of the closed box = 80 cm
External Breadth of the closed box = 75 cm
External Height of the closed box = 60 cm
External volume of the closed box = 80 x 75 x 60 = 360000 cm³
Internal length of the closed box = 80 – 4 = 76 cm
Internal Breadth of the closed box = 75 – 4 = 71 cm
Internal Height of the closed box = 60 – 4=56 cm
Internal volume of the closed box = 76 x 71 x 56 cm = 302176 cm³
Volume of wood required to make the closed box = 360000 – 302176 = 57824 cm³

Question 7.
A closed box measures 66 cm, 36 cm and 21 cm from outside. If its walls are made of metal-sheet, 0.5 cm thick ; find :
(i) the capacity of the box ;
(ii) volume of metal-sheet and
(iii) weight of the box, if 1 cm³ of metal weights 3.6 gm.
Solution:
External length of the closed box = 66cm.
External breadth of the closed box = 36 cm
External height of the closed box =21 cm
External volume of the closed box= 66 x 36 x 21 = 49896 cm³
Internal length of the box =(66 – 2 x 0.5) = 66 – 1 = 65 cm
Internal breadth of the box =(36 – 2 x 0.5) = 36 – 1 = 35 cm
Internal height of the box = (21 – 2 x 0.5) = 21 – 1 = 20 cm
Internal Volume of the box = 65 x 35 x 20 = 45500 cm³
(i) Capacity of the box = 45500 cm³
(ii) Volume of metal sheet of the box = External volume – Internal volume
= 49896 – 45500 = 4396 cm³
(iii) 1 cm3 of metal weigh 3.6 grams.
Weight of the box = 4396 x 3.6 gm = 15825.6 gm

Question 8.
The internal length, breadth and height of a closed box are 1 m, 80 cm and 25 cm. respectively. If its sides are made of 2.5 cm thick wood ; find :
(i) the capacity of the box
(ii) the volume of wood used to make the box.
Solution:
Internal length of the closed box = 1m = 100 cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -8

Question 9.
Find the area of metal-sheet required to make an open tank of length = 10 m, breadth = 7.5 m and depth = 3.8 m.
Solution:
Length of the tank = 10 m
Breadth of the tank = 7.5 m
Depth of the tank = 3.8 m
Area of four walls = 2[L+B] x H = 2(10 + 7.5) x 3.8
= 2 x 17.5 x 3.8 = 35 x 3.8 = 133 m²
Area of the floor = L x B = 10 x 7.5 = 75 m
Area of metal sheet required to make the tank =Area of four walls + Area of floor = 133 m² + 75 m² = 208 m²

Question 10.
A tank 30 m long, 24 m wide and 4.5 m deep is to be made. It is open from the top. Find the cost of iron-sheet required, at the rate of ₹ 65 per m², to make the tank.
Solution:
Length of the tank = 30 m
Width of the tank = 24 m
Depth of the tank = 4.5 m
Area of four walls of the tank = 2[L+B] x H = 2(30 + 24) x 4.5 = 2 x 54 x 4.5 m² = 486 m²
Area of the floor of the tank = L x B = 30 x 24 = 720 m²
Area of Iron sheet required to make the tank = Area of four walls + Area of floor = 486 + 720 = 1206 m²
Cost of iron sheet required @ ₹ 65 per m² = 1206 x 65 = ₹ 78390

Surface Area, Volume and Capacity Exercise 21C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
The edges of three solid cubes are 6 cm, 8 cm and 10 cm. These cubes are melted and recast into a single cube. Find the edge of the resulting cube.
Solution:
Edge of first solid cube = 6 cm
Volume = (6)³ = 216 cm³
Edge of second cube = 8 cm
Volume = (8)³ = 512 cm³
Edge of third cube = 10 cm
Volume = (10)³= 1000 cm³
Sum of volumes of three cubes = 216 + 512 + 1000= 1728 cm³
Let a be the edge of so formed cube volume = a³
a³ = 1728 = (12)³
a = 12 cm

Question 2.
Three solid cubes of edges 6 cm, 10 cm and x cm are melted to form a single cube of edge 12 cm, find the value of x.
Solution:
Edge of first cube = 6 cm
Volume = (6)³ = 216 cm³
Edge of second cube = 10 cm
Volume = (10)³ = 1000 cm³
Edge of third cube = x
Volume =x³
Edge of resulting cube = 12 cm
Volume = (12)³ = 1728 cm³
216 + 1000 + x³ = 1728
x³ = 1728 – 216 – 1000 = 512 = (8)³
x = 8
Edge of third cube = 8 cm

Question 3.
The length of the diagonals of a cube is 8√3 cm.
Find its:
(i) edge
(ii) total surface area
(iii) Volume
Solution:
(i) Length of diagonal of a cube = 8√3 cm
Length of edge = \(\frac { 8\surd 3 }{ \surd 3 }\) = 8 cm
(ii) Total surface area = 6a² = 6 x 8² = 6 x 64 cm² = 384 cm²
(iii) Volume = a³ = (8)³= 512 cm³

Question 4.
A cube of edge 6 cm and a cuboid with dimensions 4 cm x x cm x 15 cm are equal in volume. Find:
(i) the value of x.
(ii) total surface area of the cuboid.
(iii) total surface area of the cube.
(iv) which of these two has greater surface and by how much?
Solution:
Edge of a cube = 6 cm
Volume = a³ = (6)³ = 216 cm³
Dimensions of a cuboid = 4 cm x x cm x 15 cm
Volume = 60x cm³
Volume of both is equal
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -9

Question 5.
The capacity of a rectangular tank is 5.2 m³ and the area of its base is 2.6 x 104 cm²; find its height (depth).
Solution:
Capacity of a tank = 5.2 m³
and area of its base = 2.6 x 10⁴ cm²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -10

Question 6.
The height of a rectangular solid is 5 times its width and its length is 8 times its height. If the volume of the wall is 102.4 cm³, find its length.
Solution:
Height of rectangular solid = 5 x width
and length = 8 x height = 8 x 5 x width = 40 x width
Volume = 102.4 cm³
Let width = w
Then height = 40w
and height = 5w
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -11

Question 7.
The ratio between the lengths of the edges of two cubes are in the ratio 3 : 2. Find the ratio between their:
(i) total surface area
(ii) volume.
Solution:
Ratio in edges of two cubes = 3:2
Let edge of first cube = 3x
Then edge of second cube = 2x
(i) Now total surface area of first cube = 6 x (3x)² = 6 x 9x² = 54x²
and of surface area of second cube = 6 x (2x)² = 6 x 4x² = 24x²
Ratio = 54x²: 24x² = 9:4
(ii) Volume of first cube = (3x)³ = 27x³
and second cube = (2x)³ = 8x³
Ratio = 27x³: 8x³ = 27 :8

Question 8.
The length, breadth and height of a cuboid (rectangular solid) are 4 : 3 : 2.
(i) If its surface are is 2548 cm², find its volume.
(ii) If its volume is 3000 m³, find its surface area.
Solution:
Surface area of cuboid = 2548 cm²
Ratio in length, breadth and height of a cuboid = 4 : 3 : 2
Let length = 4x, Breadth = 3x and height = 2x
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -12

Surface Area, Volume and Capacity Exercise 21D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
The height of a circular cylinder is 20 cm and the diameter of its base is 14 cm. Find:
(i) the volume
(ii) the total surface area.
Solution:
Height of cylinder (h) = 20 cm
and diameter of its base (d)= 14 cm
and radius of its base (r)= 14/2 = 7 cm
(i) Volume = πr²h
= 22/7 x 7 x 7 x 20 cm³ = 3080 cm³
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -13
(ii) Total surface area = 2πr(h + r)
= 2 x 22/7 x 7 (20 + 7) cm² = 44 x 27 = 1188 cm²

Question 2.
Find the curved surface area and the total surface area of a right circular cylinder whose height is 15 cm and the diameter of the cross-section is 14 cm.
Solution:
Diameter of the base of cylinder = 14 cm
Radius (r) = 14/2 cm = 7 cm
Height (h) = 15 cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -14
Curved surface area = 2πrh
= 2 x 22/7 x 7 x 15 = 660 cm²
Total surface area = 2πr (h + r)
= 2 x 22/7 x 7(15 + 7)
= 2 x 22/7 x 7 x 22 = 968 cm²

Question 3.
Find the height of the cylinder whose radius is 7 cm and the total surface area is 1100 cm².
Solution:
Total surface area =1100 cm²
Radius = 7 cm
Let height of the cylinder = h
Then, total surface area = 2πr(h + r)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -15

Question 4.
The curved surface area of a cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.
Solution:
Height (h) = 14 cm
Curved surface area (2πrh) = 88 cm²
Then, 2πrh = 88 cm²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -16

Question 5.
The ratio between the curved surface area and the total surface area of a cylinder is 1 : 2. Find the ratio between the height and the radius of the cylinder.
Solution:
Let r be the radius and h be the height of a right circular cylinder, then Curved surface area = 2πrh
and total surface area = 2πrh x 2πr² = 2πr(h + r)
But their ratio is 1 : 2
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -17

Question 6.
Find the capacity of a cylindrical container with internal diameter 28 cm and height 20 cm.
Solution:
Diameter = 28 cm
Radius = 28/2 cm = 14 cm
Height = 20 cm
Volume = πr²h = 22/7 x 14 x 14 x 20
Volume = 12320 cm³

Question 7.
The total surface area of a cylinder is 6512 cm² and the circumference of its bases is 88 cm. Find:
(i) its radius
(ii) its volume
Solution:
Let r be the radius and h be the height of the given cylinder.
Circumference = 2πr = 88 cm (Given)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -18

Question 8.
The sum of the radius and the height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm². Find the height and the volume of the cylinder.
Solution:
Let r and h be the radius and height of the solid cylinder respectively.
Given, r + h = 37 cm
Total surface area of the cylinder = 1628 cm² (Given)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -20

Question 9.
A cylindrical pillar has radius 21 cm and height 4 m. Find :
(i) the curved surface area of the pillar
(ii) cost of polishing 36 such cylindrical pillars at the rate of ₹ 12 per m².
Solution:
Radius of the cylinder = 21 cm
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -21

Question 10.
If radii of two cylinders are in the ratio 4 : 3 and their heights are in the ratio 5 : 6, find the ratio of their curved surfaces.
Solution:
Ratio in radii of two cylinders = 4 : 3
and ratio in their heights = 5 : 6
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -22

Surface Area, Volume and Capacity Exercise 21E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A cuboid is 8 m long, 12 m broad and 3.5 high, Find its
(i) total surface area
(ii) lateral surface area
Solution:
Length of a cuboid = 8 m
Breadth of a cuboid = 12 m
Height of a cuboid = 3.5 m
(i) Total surface area = 2(lb + bh + hl)
= 2(8 x 12 + 12 x 3.5 + 3.5 x 8)
= 2(96 + 42 + 28)
= 2 x 166 = 332 m²
(ii) Lateral surface area = 2h(l + b)
= 2 x 3.5(8 + 12) = 7 x 20= 140 m²

Question 2.
How many bricks will be required for constructing a wall which is 16 m long, 3 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm ?
Solution:
Length of the wall = 16 m = 16 x 100 cm = 1600 cm
Height of the wall = 3 m = 3 x 100 cm = 300 cm
Breadth of the wall = 22.5 cm
Volume of the wall = 1600 x 300 x 22.5 cm3 = 1,08,00,000 cm³
Volume of one brick = 25 x 11.25 x 6 cm3 = 1687.5 cm³
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -23

Question 3.
The length, breadth and height of cuboid are in the ratio 6 : 5 : 3. If its total surface area is 504 cm², find its volume.
Solution:
Let length of the given cuboid = 6x
Breadth of the given cuboid = 5x
Height of the given cuboid = 3x
Total surface area of the given cuboid
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -24

Question 4.
The external dimensions of an open wooden box are 65 cm, 34 cm and 25 cm. If the box is made up of wood 2 cm thick, find the capacity of the box and the volume of wood used to make it.
Solution:
External length of the open box = 65 cm
External breadth of the open box = 34 cm
External height of the open box = 25 cm
External volume of the open box = 65 x 34 x 25 cm³ = 55250 cm³
Internal length of open box = 65 – (2 x 2) cm = 61 cm
Internal breadth of a open box = 34 – (2 x 2) cm = 30 cm
Internal height of open box = 25 – 2 cm = 23 cm
Internal volume of open box or capacity of the box = 61 x 30 x 23 cm³ = 42090 cm³
Volume of wood required to make the closed box = 55250 – 42090 cm³ = 13160 cm³

Question 5.
The curved surface area and the volume of a toy, cylindrical in shape, are 132 cm² and 462 cm³ respectively. Find, its diameter and its length.
Solution:
Let the radius of a toy = r and
height of the toy = h
Curved surface area of a toy = 132 cm²
=> 2πrh = 132 cm²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -25

Question 6.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹ 10 per m² is ₹ 15,000, find the height of the hall.
Solution:
Let length, breadth and height of the rectangular hall be l m, b m and h m respectively.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -27
Area of four walls = Area of cuboid – Area of floor – Area of top
= 2 (lb + bh + hl) – (l x b) – (l x b)
= 2(lb) + 2 (bh) + 2(hl) – 2lb = 2 lh + 2 bh
= 2h(l + b)
= 2h x 125 [From (i)]
= 250h m²
Area of four walls = 250h m²
Cost of painting 1 m² area = ₹ 10
Cost of painting 250h m² area = ₹ 10 x 250h = 2500h
15000 = 2500h
h = 15000/2500
The height of the hall is 6 m.

Question 7.
The length of a hall is double its breadth. Its height is 3 m. The area of its four walls (including doors and windows) is 108 m², find its volume.
Solution:
Let the breadth be x
and the length be 2x
Height = 3 m
Area of four walls = 108 m²
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -28

Question 8.
A solid cube of side 12 cm is cut into 8 identical cubes. What will be the side of the new cube? Also, find the ratio between the surface area of the original cube and the total surface area of all the small cubes formed.
Solution:
Here, cube of side 12 cm is divided into 8 cubes of side 9 cm.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacityb image -29
Given that,
Their volumes are equal.
Volume of big cube of 12 cm = Volume of 8 cubes of side a cm
(Side of big cube)³ = 8 x (Side of small cube)³
(12)³ = 8 x a³

Question 9.
The diameter of a garden roller is 1.4 m and it 2 m long. Find the maximum area covered by it 50 revolutions?
Solution:
Diameter of the roller = 1.4 m
Radius (r) = 1.4/2 = 0.7 m
and length (h) = 2m
Curved surface area = 2πrh = 2 x 22/7 x 0.7 x 2 cm² = 8.8 m²
Area covered in 50 complete revolutions = 8.8 x 50 m² = 440 m²
Area of the playground = 440 m²

Question 10.
In a building, there are 24 cylindrical pillars. For each pillar, radius is 28 m and height is 4 m. Find the total cost of painting the curved surface area of the pillars at the rate of ₹ 8 per m².
Solution:
Radius (r) of each pillar = 28 m
Height (h) = 4 m
Curved surface area of each pillar = 2πrh
= 2 x 22/7 x 28 x 4 m² = 704 m²
Surface area of 24 pillars = 704 x 24 m² = 16,896 m²
Cost of cleaning = ₹ 8 per m²
Total cost = ₹ 16,896 x 8 = ₹ 1, 35, 168

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling

August 11, 2022May 12, 2018 by Veerendra

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 22 Data Handling. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Data Handling Exercise 22A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Arrange the following data as an array (in ascending order):
(i) 7, 5, 15, 12, 10, 11, 16
(ii) 6.3, 5.9, 9.8, 12.3, 5.6, 4.7
Solution:
(i) Ascending order = 5, 7, 10, 11, 12, 15, 16
(ii) Ascending order = 4.7, 5.6, 5.9, 6.3, 9.8, 12.3

Question 2.
Arrange the following data as an array (descending order):
(i) 0 2, 0, 3, 4, 1, 2, 3, 5
(ii) 9.1, 3.7, 5.6, 8.3, 11.5, 10.6
Solution:
(i) Descending order = 5, 4, 3, 3, 2, 2, 1, 0
(ii) Descending order = 11.5, 10.6, 9.1, 8.3, 5.6, 3.7

Question 3.
Construct a frequency table for the following data:
(i) 6, 7, 5, 6, 8, 9, 5, 5, 6, 7, 8, 9, 8, 10, 10, 9, 8, 10, 5, 7, 6, 8.
(ii) 3, 2, 1, 5, 4, 3, 2, 5, 5, 4, 2, 2, 2, 1, 4, 1, 5, 4.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 1

Question 4.
Following are the marks obtained by 30 students in an examinations.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 2
Taking class intervals 0-10, 10-20, ……… 40-50 ; construct a frequency table.
Solution:

Question 5.
Construct a frequency distribution table for the following data ; taking class-intervals 4-6, 6-8, ……… 14-16.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 4
Solution:

Question 6.
Fill in the blanks:
(i) Lower class limit of 15-18 is ………
(ii) Upper class limit of 24-30 is ……..
(iii) Upper limit of 5-12.5 is ………
(iv) If the upper and the lower limits of a class interval are 16 and 10 ; the class-interval is ……..
(v) If the lower and the upper limits of a class interval are 7.5 and 12.5 ; the class interval is ……..
Solution:
(i) Lower class limit of 15 – 18 is 15.
(ii) Upper class limit of 24 – 30 is 30.
(iii) Upper limit of 5 – 12.5 is 12.5
(iv) If the upper and lower limits of a class interval are 16 and 10 ; the class interval is 10 – 16
(v) If the lower and upper limits of a class interval are 7.5 and 12.5 ; the class interval is 7.5 – 12.5

Data Handling Exercise 22B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Hundred students from a certain locality use different modes of travelling to school as given below. Draw a bar graph.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 7
Solution:

Question 2.
Mr. Mirza’s monthly income is Rs. 7,200. He spends Rs. 1,800 on rent, Rs. 2,700 on food, Rs. 900 on education of his children ; Rs. 1,200 on Other things and saves the rest.
Draw a pie-chart to represent it.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 9

Question 3.
The percentage of marks obtained, in different subjects by Ashok Sharma (in an examination) are given below. Draw a bar graph to represent it.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 10
Solution:

Question 4.
The following table shows the market position of different brand of tea-leaves.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 12
Draw it-pie-chart to represent the above information.
Solution:

Question 5.
Students of a small school use different modes of travel to school as shown below:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 14
Draw a suitable bar graph.
Solution:

Question 6.
For the following table, draw a bar-graph
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 16
Solution:

Question 7.
Manoj appeared for ICSE examination 2018 and secured percentage of marks as shown in the following table:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 18
Represent the above data by drawing a suitable bar graph.
Solution:

Question 8.
For the data given above in question number 7, draw a suitable pie-graph.
Solution:
∵ 60 + 45 + 42 + 48 + 75 = 270
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 20

Question 9.
Mr. Kapoor compares the prices (in Rs.) of different items at two different shops A and B. Examine the following table carefully and represent the data by a double bar graph.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 21
Solution:

Question 10.
The following tables shows the mode of transport used by boys and girls for going to the same school.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling image - 23
Draw a double bar graph representing the above data.
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability

August 11, 2022May 12, 2018 by Veerendra

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 23 Probability. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Probability Exercise 23 – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A die is thrown, find the probability of getting:
(i) a prime number
(ii) a number greater than 4
(iii) a number not greater than 4.
Solution:
A die has six numbers : 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 1

Question 2.
A coin is tossed. What is the probability of getting:
(i) a tail? (ii) ahead?
Solution:
On tossing a coin once,
Number of possible outcome = 2
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 2

Question 3.
A coin is tossed twice. Find the probability of getting:
(i) exactly one head (ii) exactly one tail
(iii) two tails (iv) two heads
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 3

Question 4.
A letter is chosen from the word ‘PENCIL’ what is the probability that the letter chosen is a consonant?
Solution:
Total no. of letters in the word ‘PENCIL = 6
Total Number of Consonant = ‘PNCL’ i.e. 4
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 5

Question 5.
A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. A ball is drawn from the bag without looking into it. What is the probability that the ball drawn is:
(i) a red ball
(ii) not a red ball
(iii) a white ball.
Solution:
Total number of possible outcomes = 3
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 6

Question 6.
6. In a single throw of a die, find the probability of getting a number
(i) greater than 2
(ii) less than or equal to 2
(iii) not greater than 2.
Solution:
A die has six numbers = 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 7

Question 7.
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size.
A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 8

Question 8.
In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will be an odd number.
(iii) will not be an even number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 9

Question 9.
In a single throw of a die, find the probability of getting :
(i) 8
(ii) a number greater than 8
(iii) a number less than 8
Solution:
On a die the numbers are 1, 2, 3, 4, 5, 6 i.e. six.
∴ Number of possible outcome = 6
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 10

Question 10.
Which of the following can not be the probability of an event?
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 11
Solution:
The probability of an event cannot be
(ii) 3.8 i.e. the probability of an even cannot exceed 1.
(iv) i.e. -0.8 and
(vi) -2/5, This is because probability of an even can never be less than 1.

Question 11.
A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball,
(ii) a black ball
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 12

Question 12.
Three identical coins are tossed together. What is the probability of obtaining:
all heads?
exactly two heads?
exactly one head?
no head?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 13

Question 13.
A book contains 92 pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is 9?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 14

Question 14.
Two coins are tossed together. What is the probability of getting:
(i) at least one head
(ii) both heads or both tails.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 15

Question 15.
From 10 identical cards, numbered 1, 2, 3, …… , 10, one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 2 (ii) 3
(iii) 2 and 3 (iv) 2 or 3
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 16

Question 16.
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 0
(ii) 12
(iii) less than 12
(iv) less than or equal to 12
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 17

Question 17.
A die is thrown once. Find the probability of getting:
(i) a prime number
(ii) a number greater than 3
(iii) a number other than 3 and 5
(iv) a number less than 6
(v) a number greater than 6.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 18

Question 18.
Two coins are tossed together. Find the probability of getting:
(i) exactly one tail
(ii) at least one head
(iii) no head
(iv) at most one head
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability image - 19