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Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles (Including Construction of angles)

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 14 Lines and Angles (Including Construction of angles)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

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POINTS TO REMEMBER

1. POINT : A point: is a mark of position; which has no length, no breadth and no thickness. It, in general, is represented by a capital letter as shown alongside.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 1

2. LINE : A line has length, but no breadth or thickness.

“The given figure shows a line AB in which two arrow-heads in opposite directions show that can be extended infinitely in both the directions.
A line may be straight or curved but when we say a line’ it means a straight line only.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 3

3. RAY : It is a straight line which stats from a fixed point and moves in the same direction.

The given figure shows a ray\(\xrightarrow { AB }\) with fixed initial point A ‘and moving in the direction AB.

4. LINE SEGMENT : It is a straight line with its both ends fixed. The given figure shows a line segment, whose both the ends A and B are fixed.

(i) The adjoining figure shows a line AB which can be extended upto infinitey on both the sides of it.

(ii) The adjoining figure shows a ray AB with fixed end as point A and which can be extended upto infinity through point B. It is clear from the figure, that a ray is a part of a line.

(iii) The adjoining figure shows a line-segment AB with fixed ends A and B. It is clear from the figure, that a line-segment is a part of a ray as well as of a line. Also, a line segment is the shortest distance between two fixed points.

5. ANGLE : An angle is formed when two line segments or two rays have a common end-point.
The two line segments, forming an angle, are called the arms of the angle whereas their common end-point is called the vertex of the angle.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 9
The adjacent figure represents an angle ABC or ∠ABC or simply ∠B. AB and BC are the arms of the angle and their common point B is the vertex.

6. MEASUREMENT OF AN ANGLE : The unit of measuring an angle is degree. The symbol for degree is °.
Thus : 60 degree = 60°, 87 degree = 87’ and so on.
If one degree is divided into 60 equal parts, each part is called a minute ( ‘) and if one minute is further divided into 60 equal parts, each part is called a second ( ” ).
Thus, (i) r = 60′ and l’ = 60″
(ii) 9 minutes 45 seconds = 9′ 45″
(iii) 85 degrees 30 minutes 15 seconds = 85° 30′ 15″ and so on.

7. TYPES OF ANGLES :
1. Acute angle :
measures less than 90°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 10
2. Right angle:
measures 90°

3. Obtuse angle :
measures between 90° and 180°

4. Straight angle :
measures 180°

5. Reflex angle :
measures between 180° and 360°

8. MORE ABOUT ANGLES :
(A) Angles about a point: If a number of angles are formed about a point, their sum is always 360°.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 15
In the adjoining figure :
∠AOB + ∠BOC + ∠COD +∠DOE + ∠EOA = 360°
(B) Adjacent angles : Two angles are said to be adjacent angles, if:
(i) they have a common vertex,
(ii) they have a common arm and
(iii) the other arms of the two angles lie on opposite sides of the common arm.
The adjoining figure shows a pair of adjacent angles :
(i) they have a common vertex (O),
(ii) they have a common ann (OB) and
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 16
(iii) the other arms OA and OC of the two angles are on opposite sides of the common arm OB.
(C) Vertically opposite angles : When two straight lines intersect each other four angles are formed.
The pair of angles which lie on the opposite sides of the point of intersection are called vertically opposite angles.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 17
In the adjoining figure, two straight lines AB and CD intersect each other at point 0. Angles AOD and BOC form one pair of vertically opposite angles; whereas angles AOC and BOD form another pair of vertically opposite angles.
Vertically opposite angles are always equal.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 18
i. e. ∠AOD = ∠BOC and ∠AOC = ∠BOD.
Important: In the adjoining figure, rays OX and OY meet a O to form ∠XOY (i.e. ∠a) and reflex ∠XOY (i. e. ∠b). It must be noted that ∠XOY means the smaller angle only unless it is mentioned to take otherwise.

9. COMPLEMENTARY AND SUPPLEMENTARY ANGLES
1. Two angles are called complementary angles, if their sum is one’right angle i.e. 90° Each angle is called the complement of the other.
e.g., 20″ and 70″ are complementary angles, because 20° + 70° = 90°.
Clearly, 20″ is the complement of 70° and 70° is the complement of 20°.
Thus, the complement of angle 53° = 90° – 53° = 37°.
2. Two angles are called supplementary angles, if their sum is two right angles i.e. 180″. Each angle is called the supplement of the other.
e.g., 30″ and 150° are supplementary angles because 30° + 150° = 180°.
Clearly, 30″ is the supplement of 150° and vice-versa.
Thus, the supplement of 105° = 180° – 105° = 75°.

10. Transversal : It is a straight line which cuts two or more given straight lines.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 19
In the adjoining figure, PQ cuts straight lines AB and CD, and so it is a transversal.
When a transversal cuts two given straight lines (refer the adjoining figure), the following pairs of angles are formed.
1. Two pairs of interior alternate angles : Angles marked 1 and 2 form one pair of interior alternate angles, while angles marked 3 and 4 form another pair of interior alternate angles.
2. Two pairs of exterior alternate angles : Angles marked 5 and 8 form one pair, while angles marked 6 and 7 form the other pair of exterior alternate angles.
3. Four pairs of corresponding angles : Angles marked 3 and 6; 1 and 5; 8 and 2; 7 and 4 form the four pairs of corresponding angles.
4. Two pairs of allied or co-interior or conjoined angles : Angles marked 3 and 2 form one pair and angles marked 1 and 4 form another pair of allied angles.

11. PARALLEL LINES : Two straight lines are said to be parallel, if , they do not meet anywhere; no matter how long are they produced in any direction.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 20
The adjacent figure shows two parallel straight lines AB and CD.
When two parallel lines AB and CD are cut by a transversal PQ :
(i) Interior and exterior alternate angles are equal:
i.e. ∠3 = ∠6 and ∠4 = ∠5 [Interior alternate angles]
∠1 = ∠8 and ∠2 = ∠7 [Exterior alternate angles]
(ii) Corresponding angles are equal:
i.e. ∠1 = ∠5;∠2 = ∠6;∠3 = ∠7 and ∠4 = ∠8
(iii) Co-interior or allied angles are supplementary :
i. e. ∠3 + ∠5 = 180° and ∠4 +∠6 = 180°

12. CONDITIONS OF PARALLELISM : If two straight lines are cut by a transversal such that:
(i) a pair of alternate angles are equal, or
(ii) a pair of corresponding angles are equal, or
(iii) the sum of the interior angles on the same side of the transversal is 180°, then the two straight lines are parallel to each other.
Therefore, in order to prove that the given lines are parallel, show either alternate angles are equal or, corresponding angles are equal or, the co-interior angles are supplementary.

Lines and Angles Exercise 14A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
State, true or false :
(i) A line segment 4 cm long can have only 2000 points in it.
(ii) A ray has one end point and a line segment has two end-points.
(iii) A line segment is the shortest distance between any two given points.
(iv) An infinite number of straight lines can be drawn through a given point.
(v) Write the number of end points in
(a) a line segment AB (b) arayAB
(c) alineAB
(vi) Out of \(\overleftrightarrow { AB }\) , \(\overrightarrow { AB }\) , \(\overleftarrow { AB }\) and \(\overline { AB }\) , which one has a fixed length?
(vii) How many rays can be drawn through a fixed point O?
(viii) How many lines can be drawn through three
(a) collinear points?
(b) non-collinear points?
(ix) Is 40° the complement of 60°?
(x) Is 45° the supplement of 45°?
Solution:
(i) False : It has infinite number of points.
(ii) True
(iii) True
(iv) True
(v) (a) 2 (b) 1 (c) 0
(vi) AB
(vii) Infinite
(viii) (a) 1 (b) 3
(ix) False : 40° is the complement of 50° as 40° + 50° = 90°
(x) False : 45° is the supplement of 135° not 45°.

Question 2.
In which of the following figures, are ∠AOB and ∠AOC adjacent angles? Give, in each case, reason for your answer.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 21

Solution:
If ∠AOB and ∠AOC are adjacent angle if they have OA their common arm.
(i) In the figure, OB is their common arm
∴∠AOB and ∠AOC are not adjacent angles.
(ii) In the figure, OC is their common arm
∴∠AOB and ∠AOC also not adjacent angles.
(iii) In this figure, OA is their common arm
∴ ∠AOB and ∠AOC are adjacent angles.
(iv) In this figure, OB is their common arm
∴ ∠AOB and ∠AOC are not adjacent angles.

Question 3.
In the given figure, B AC is a straight line.
Find : (i) x (ii) ∠AOB (iii) ∠BOC
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 23
Solution:

Question 4.
Find yin the given figure.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 25
Solution:

Question 5.
In the given figure, find ∠PQR.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 27
Solution:
SQR is a straight line
∴∠SQT + ∠TQP + ∠PQR = 180°
⇒ x + 70° + 20° – x + ∠PQR = 180°
⇒ 90″ + ∠PQR = 180°
⇒ ∠PQR = 180°-90° = 90°
Hence ∠PQR = 90°

Question 6.
In the given figure. p° = q° = r°, find each.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 28
Solution:

Question 7.
In the given figure, if x = 2y, find x and y
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 30
Solution:

Question 8.
In the adjoining figure, if b° = a° + c°, find b.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 32
Solution:

Question 9.
In the given figure, AB is perpendicular to BC at B.
Find : (i) the value of x.
(ii) the complement of angle x.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 34
Solution:

Question 10.
Write the complement of:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 36

Solution:

Question 11.
Write the supplement of:
(i) 100°
(ii) 0°
(iii) x°
(iv) (x + 35)°
(v) (90 +a + b)° f
(vi) (110 – x – 2y)°
(vii) \(\frac { 1 }{ 5 }\) of a right angle
(viii) 80° 49′ 25″
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 39

Question 12.
Are the following pairs of angles complementary ?
(i) 10° and 80°
(ii) 37° 28′ and 52° 33′
(iii) (x+ 16)°and(74-x)°
(iv) 54° and \(\frac { 2 }{ 5 }\) of a right angle.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 41

Question 13.
Are the following pairs of angles supplementary?
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 42
Solution:

Question 14.
If 3x + 18° and 2x + 25° are supplementary, find the value of x.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 44

Question 15.
If two complementary angles are in the ratio 1:5, find them.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 45

Question 16.
If two supplementary’ angles are in the ratio 2 : 7, find them.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 47

Question 17.
Three angles which add upto 180° are in the ratio 2:3:7. Find them.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 48

Question 18.
20% of an angle is the supplement of 60°. Find the angle.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 49

Question 19.
10% of x° is the complement of 40% of 2x°. Find x
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 50

Question 20.
Use the adjacent figure, to find angle x and its supplement.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 52
Solution:

Question 21.
Find k in each of the given figures.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 54
Solution:

Question 22.
In the given figure, lines PQ, MN and RS intersect at O. If x : y = 1 : 2 and z = 90°, find ∠ROM and ∠POR.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 58
Solution:

Question 23.
In the given figure, find ∠AOB and ∠BOC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 60
Solution:

Question 24.
Find each angle shown in the diagram.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 62
Solution:

Question 25.
AB, CD and EF are three lines intersecting at the same point.
(i) Find x, if y = 45° and z = 90°.
(ii) Find a, if x = 3a, y = 5x and r = 6x.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 65
Solution:

Lines and Angles Exercise 14B – Selina Concise Mathematics Class 7 ICSE Solutions

In questions 1 and 2, given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles :
Question 1.
(i) ∠3 and ∠6
(ii) ∠2 and ∠4
(iii) ∠3 and ∠7
(iv) ∠2 and ∠7
(v) ∠4 and∠6
(vi) ∠1 and ∠8
(vii) ∠1 and ∠5
(viii) ∠1 and ∠4
(ix) ∠5 and ∠7
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 68
Solution:
(i) ∠3 and ∠6 are interior alternate angles.
(ii) ∠2 and ∠4 are adjacent angles.
(iii) ∠3 and ∠7 are corresponding angles.
(iv) ∠2 and ∠7 are exterior alternate angles,
(v) ∠4 and ∠6 are allied or co-interior angles,
(vi) ∠1 and ∠8 are exterior alternate angles.
(vii) ∠1 and ∠5 are corresponding angles.
(viii) ∠1 and ∠4 are vertically opposite angles.
(ix) ∠5 and ∠7 are adjacent angles.

Question 2.
(i) ∠1 and ∠4
(ii) ∠4 and ∠7
(iii) ∠10 and ∠12
(iv) ∠7 and ∠13
(v) ∠6 and ∠8
(vi) ∠11 and ∠8
(vii) ∠7 and ∠9
(viii) ∠4 and ∠5
(ix) ∠4 and ∠6
(x) ∠6 and ∠7
(xi) ∠2 and ∠13
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 69
Solution:
(i) ∠1 and ∠4 are vertically opposite angles.
(ii) ∠4 and ∠7 are alternate angles.
(iii) ∠10 and ∠12 are vertically opposite angles.
(iv) ∠7 and ∠13 are corresponding angles.
(v) ∠6 and ∠8 are vertically opposite angles.
(vi) ∠11 and ∠8 are allied or co-interior angles.
(vii) ∠7 and ∠9 are vertically opposite angles.
(viii) ∠4 and ∠5 are adjacent angles.
(ix) ∠4 and ∠6 are allied or co-interior angles.
(x) ∠6 and ∠7 are adjacent angles.
(xi) ∠2 and ∠13 are allied or co-interior angles.

Question 3.
In the given figures, the arrows indicate parallel lines. State which angles are equal. Give reasons.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 70
Solution:
In the figure (i),
a = b (corresponding angles)
b = c (vertically opposite angles)
a = c (alternate angles)
∴ a = b = c
(ii) In the figure (ii),
x =y (vertically opposite angles)
y=l (alternate angles)
x = I (corresponding angles)
1 = n (vertically opposite angles)
n = r (corresponding angles)
∴ x = y = l = n = r
Again m = k (vertically opposite angles)
k = q (corresponding angles)
∴ m = k = q

Question 4.
In the given figure, find the measure of the unknown angles :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 71
Solution:
a = d (vertically opposite angles)
d=f (corresponding angles)
f= 110° (vertically opposite angles)
∴ a = d = f = 110°
e + 110° = 180° (co-interior angles)
∴ e = 180°- 110° = 70°
b = c (vertically opposite angles)
b = e (corresponding angles)
e = g (vertically opposite angles)
∴ b = c = e = g = 70° ”
Hence a = 110°, b = 70°, e = 70°, d = 110°, e = 70°,f= 110° and g = 70°

Question 5.
Which pair of the dotted line, segments, in the following figures, are parallel. Give reason:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 72

Solution:
(i) In figure (i), If lines are parallel, then 120°+ 50° =180°
But there are co-interior angles
⇒170° = 180°. But it not true Hence, there are not parallel lines
(ii) In figure (ii),
∠1 = 45° (vertically opposite angles)
Lines are parallel if
∠1 + 135° = 180° (co-interior angles)
⇒45°+ 135°= 180°
⇒ 180° = 180° which is true.
Hence, the lines are parallel.
(iiii) In figure (iii),
Lines are parallel if corresponding angles are equal
If 120° =130° which is not correct
∴ Lines are not parallel.
(iv) ∠1 = 110° (vertically opposite angles)
If lines are parallel then
∠1 + 70° = 180° (co-interior angles)
⇒110° + 70°= 180°
⇒180° =180°
Which is correct.
∴ Lines are parallel.
(v) ∠1 + 100°= 180°
⇒∠1 = 180°- 100°= 80 (linear pair)
Lines l1 and l2 will be parallel If ∠1 = 70°
⇒ 80° = 70° which is not true
∴ l1 and 12 are not parallel Again, A, l3and l5 will be parallel
If 80° = 70° (corresponding angle)
Which is not true.
∴l3 and l5 are not parallel
But ∠1 = 80° (alternate angles)
⇒ 80° = 80°
Which is true
∴ l2 and l4 are parallel
(vi) Lines are parallel
If alternate angles are equal
⇒ 50° = 40°
Wliich is not ture lines are not parallel.

Question 6.
In the given figures, the directed lines are parallel to each other. Find the unknown angles.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 74

Solution:

Question 7.
Find x. y and p is the given figures
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 82
Solution:

Question 8.
Find x in the following cases :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 85

Lines and Angles Exercise 14C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Using ruler and compasses, construct the following angles :
(i)30°
(ii)15°
(iii) 75°
(iv) 180°
(v) 165°
(vi) 22.5°
(vii) 37.5°
(viii) 67.5°
Solution:
(i) 30°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P and with same radius cut off the arc at Q.
(iv) Now with centre P and Q draw two arcs intersecting each other at R.
(v) Join BR and produce it to A, forming ZABC
= 30°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 90

(ii) (15°)
Steps of Construction:
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P and with same radius cut off the arc at Q.
(iv) Taking P and Q as curves, draw two arcs intersecting each other at D nnd join BD.
(v) With centre P and R, draw two more arcs intersecting each other at S.
(vi) Join BS and produce it to A.
Then ∠ABC = 15°.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 91

(iii) 75°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius draw an arc and cut off PQ, then QR of the same radius.
(iii) With centre Q and R, draw two arcs intersecting each other at S.
(iv) Join SB.
(v) With centre Q and D draw two arcs intersecting each other at T.
(vi) Join BT and produce it to A.
Then ∠ABC = 75°.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 92

(iv) 180°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius draw arc meeting BC at P.
(iii) With centre P and with same radius cut of arcs PQ, QR and then RS.
(iv) Join BS and produce it to A.
Then ∠ABC = 180°.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 93

(v) 165°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius draw an arc meeting BC at P.
(iii) With centre P and same radius cut off arcs PQ, QR and then RS.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 94
(iv) Join SB.
(v) With centres R and S, draw two arcs intersecting each other at M.
(vi) With centre T and S draw two arcs intersecting each other at L.
(vi) Join BL and produce it to A. Then ∠ABC = 165°

(vi) 22.5°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius, draw an arc meeting BC at P.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 95
(iii) With centre P and some radius, cut off arcs PQ.
(iv) Bisect arc PQ at R and join BR.
(v) Bisect arc QR at S and join BS.
(vii) Now bisect arc PR at T.
(viii) Join BT and produce it to A.
Then ∠ABC = 22 \(\frac { 1 }{ 2 }\) ° or 22.5°.

(vii) 37.5°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius, draw an arc meeting BC at P.
(iii) With centre P and same radius cut off arcs PQ and QR.
(iv) Now bisect arc QR at S and again bisect arc QS at T.
(v) Bisect arc PT at K.
(vi) Join BK and produce it to A.
Then, ∠ABC – 37 \(\frac { 1 }{ 2 }\) °or 37-5.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 96

(viii) 67.5°
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius, draw an arc meeting BC at P.
(iii) With centre P and with same radius, cut arcs PQ and then QR.
(iv) Bisect arc QR at K and again bisect arc QK at S.
(v) Bisect again arc SQ at T.
(vi) Join BT and produce it to A.
Then ∠ABC = 67\(\frac { 1 }{ 2 }\) ° or 67.5°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 97

Question 2.
Draw ∠ABC = 120°. Bisect the angle using ruler and compasses only. Measure each 1 angle so obtained and check whether the angles obtained on bisecting ∠ABC are equal or not.
Solution:
Steps of Construction :
(i) Draw a line segment BC.
(ii) With centre B and some suitable radius, draw an arc meeting BC at P.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 98
(iii) With centre P and with same radius, cut arcs PQ and QR.
(iv) Join BR and produce it to A.
Then ∠ABC = 120°
(v) With centres P and R, draw two arcs intersecting each other at S.
(vi) Join BS and produce it to D. BD is the bisector of ∠ABC.
On measuring each angle, it is of 60° each. Yes, both angles are equal in measure.

Question 3.
Draw a line segment PQ = 6 cm. Mark a point A in PQ so that AP = 2 cm. At point A, construct angle QAR = 60°.
Solution:
Steps of Construction :
(i) Draw a line segment PQ = 6 cm.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 99
(ii) Mark a point A on PQ so that AP = 2 cm.
(iii) With centre A and some suitable radius draw an arc meeting AQ at C.
(iv) With centre C and with same radius, cut arc CB.
(v) Join AB and produce it to R.
Then ∠QAR = 60°

Question 4.
Draw a line segment AB = 8 cm. Mark a point P in AB so that AP = 5 cm. At P, construct angle APQ = 30°.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 8 cm.
(ii) Mark a point P in AB such that AP = 5 cm.
(iii) With centre P and some suitable radius, draw an arc meeting AB in L.
(iv) With centre L and same radius cut arc LM.
(v) Bisect arc LM at N.
(vi) Join PN and produce it to Q.
Then ∠APQ = 30°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 100

Question 5.
Construct an angle of 75° and then bisect it.
Solution:
Steps of Construction :
(i) Draw a line segment BC.
(ii) At B, draw an angle ABC equal to 75°.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 101
(iii) With centres P and T, draw arcs intersecting each other at L.
(iv) Join BL and produce it to D. Then BD bisects ∠ABC.

Question 6.
Draw a line segment of length 6 .4 cm. Draw its perpendicular bisector.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 6.4 cm.
(ii) With centres A and B and with some suitable radius, draw arcs intersecting each other at S and R.
(iii) Join SR intersecting AB at Q. Then PQR is the perpendicular bisector of line segment AB
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 102

Question 7.
Draw a line segment AB = 5.8 cm. Mark a point P in AB such that PB = 3.6 cm. At P, draw perpendicular to AB.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) Mark a point P in AB such that PB = 3.6 cm.
(iii) With centre P and some suitable radius draw an arc meeting AB in L.
(iv) With centre L and same radius cut arcs LM and then as N.
(v) Bisect arc MN at S.
(vi) Join PS and produce it to Q. Then PQ is perpendicular to AB at P.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 103

Question 8.
In each case, given below, draw a line through point P and parallel to AB :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 104
Solution:
Steps of construction :
(i) From P. draw a line segment meeting AB at
(ii) With centre Q and some suitable radius draw an arc CD.
(iii) With centre P and same radius draw another arc meeting PQ at E.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 14 Lines and Angles 105
(iv) With centre E and radius equal to CD, cut this arc at F

(v) Join PF and produce it to both sides to L and M. Then line LM is parallel to given line AB.

Selina Concise Mathematics class 7 ICSE Solutions – Congruency: Congruent Triangles

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 19 Congruency: Congruent Triangles

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 19 Congruency: Congruent Triangles

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Selina Class 7 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

POINTS TO REMEMBER

1. Meaning of Congruency : If two geometrical figures coincide exactly, by placing one over the other, the figures are said to be congruent to each other.
1. Two lines AB and CD are said to be congruent if, on placing AB on CD, or CD on AB ; the two AB and CD exactly coincide.
It is possible only when AB and CD are equal in length.Two figures ABCD and PQRS are said to be
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 1
congruent if, on placing ABCD on PQRS or PQRS on ABCD the two figures exactly coincide i.e. A and P coincide. B and Q coincide, C and R coincide and D and S coincide.
It is possible only when :
AB = PQ, BC = QR, CD = RS and AD = PS
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R and ∠D = ∠S.
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 2

2. Congruency in Triangles : Let triangle ABC is placed over triangle DEF ; such that, vertex A falls on vertex D and side AB falls on side DE ; then if the two triangles coincide with each other in such a way that B falls on E ; C falls on F ; side BC coincides with side EF and side AC coincides with side DF, then the two triangles are congruent to each other.
The symbol used for congruency is “ ≡ ” or “ ≅ ”
∴∆ ABC is congruent to ∆ DEF is written as :
∆ ABC = ∆ DEF or ∆ABC = ∆DEF.
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 3

3. Corresponding Sides and Corresponding Angles : In case of congruent triangles ABC and DEF, drawn above ; when A ABC is placed over A DEF to cover it exactly; then the sides of the two triangles, which coincide with each other, are called corresponding sides. ‘ Thus, the side AB and DE are corresponding sides; sides BC and EF are corresponding sides and sides AC and DF are also corresponding sides.
In the same way. the angles of the two triangles which coincide with each other, are called corresponding angles. Thus, three pairs of corresponding angles are ∠A and ∠D ; ∠B and ∠E and also ∠C and ∠F.
Note : The corresponding parts of congruent triangles are always equal (congruent).
∴(i) AB = DE, BC = EF and AC = DF. i.e. corresponding sides are equal.
Also (ii) ∠A = ∠D, ZB = ∠E and ∠C = ∠F
i.e. corresponding angles are equal.
4. Conditions of Congruency:
1. If three sides of one triangle are equal to three sides of the other triangle, each to each, then the two triangles are congruent.
The test is known as : side, side, side and is abbreviated as S.S.S.
In triangle ABC and PQR, given alongside:
AB = PQ ; BC = QR and AC = PR
And, so A ABC is congruent to ∆ PQR e. ∆ ABC = ∆ PQR by S.S.S.
Because in congruent triangles, corresponding sides and corresponding angles are equal.
∴ ∠A = ∠P : ∠B = ∠Q and ∠C =∠R
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 4

2. If two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle, each to each, then the triangles are congruent.
This test is known as : side, angle, side and is abbreviated as S.A.S. In the given triangles, AB = XZ ; BC = XY and ∠ABC = ∠ZXY
∴ ∆ ABC ≅ ∆ ZXY
Note : Triangles will be congruent by S.A.S., only when the angles included by the corresponding equal sides are equal.
The pairs of corresponding sides of these two congruent triangles are : AB and ZX ; BC and XY ; AC and ZY
The pairs of corresponding angles are :
∠B and ∠X ; ∠A and ∠Z : ∠C and ∠Y.
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 5

3. If two angles and the included side of one triangle are equal to the two angles and the included side of the other triangle ; then the triangles are congruent.
This test is known as : angle, side, angle and is abbreviated as A.S.A.
In the given figure :
BC = QR;
∠B = ∠Q and ∠C = ∠R
∴ ∆ABC = ∆ PQR.. (by A.S.A.)
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 6

4. If the hypotenuse and one side of a right angled triangle are equal to the hypotenuse and one side of another right angled triangle, then the two triangles are congruent.
This test is known as : right angle, hypotenuse, side and is abbreviated as R.H.S.
In the given figure :
∠B = ∠E = 90° ; AB = FE
and hypotenuse AC = hypotenuse FD
∴ ∆ ABC = ∆ FED (byRH.S.)
The corresponding angles in this case are :
∠A and ∠F ; ∠B and ∠E ; ∠C and ∠D and the corresponding sides are :
AB and EF ; AC and FD ; BC and ED.
Since the triangles are congruent, therefore all its corresponding sides are equal and corresponding angles are also equal.
∴ BC = ED ; ∠A = ∠F and ∠C = ∠D
Note : If three angles of a triangle are equal to the three angles of the other triangle, then the triangle are not necessarily congruent.
For congruency at least one pair of corresponding sides must be equal.
∴A.A.A. is not a test of congruency.
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 7

Congruency: Congruent Triangles Exercise 19 – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
State, whether the pairs of triangles given in the following figures are congruent or not:
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 8

Solution:
(i) In these triangles, corresponding sides are not equal. Hence these are not congruent triangles.
(ii) In the first A, third angle
= 180°-(40°+ 30°)
= 180° – 70°
= 110°
Now in these two triangles the sides and included angle of the one are equal to the corresponding sides and included angle.
Hence these are congruent triangles
(S.A.S. axiom)
(iii) In these triangles, corresponding two sides are equal but included angles are not-equal. Hence these are not congruent triangles.
(iv) In these triangles, corresponding three sides are equal.
Hence these are congruent triangles.
(S.S.S. Axiom)
(v) In these right triangles, one side and diagonal of the one, are equal to the corresponding side and diagonal are equal. Hence these are congruent triangles. –
(R.H.S. Axiom)
(vi) In these triangles two sides and one angle of the one are equal to the corresponding sides and one angle of the other are equal.
Hence these are congruent triangles.
(S.S. A. Axiom).
(vii) In A ABC. AB = 2 cm, BC = 3.5 cm and ∠C = 80° and in ∆ DEF,
DE = 2 cm, DF = 3.5 cm and ∠D = 80°
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 10
From the figure we see that two corresponding sides are equal but their included angles are not equal.
Hence, these are not congruent triangles

Question 2.
In the given figure, prove that:
∆ABD ≅ ∆ ACD
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 11
Solution:

Question 3.
Prove that:
(i) ∆ABC ≡∆ADC
(ii) ∠B = ∠D
(iii) AC bisects angle DCB
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 13
Solution:

Question 4.
Prove that:
(i) ∆ABD ≡ ∆ACD
(ii) ∠B = ∠C
(iii) ∠ADB = ∠ADC
(iv) ∠ADB = 90°
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 15
Solution:

Question 5.
In the given figure, prove that:
(i) ∆ACB ≅ ∆ECD
(ii) AB = ED
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 17
Solution:

Question 6.
Prove that:
(i) ∆ ABC ≅ ∆ ADC
(ii) ∠B = ∠D
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 19
Solution:

Question 7.
In the given figure, prove that: BD = BC.
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 21
Solution:

Question 8.
In the given figure ;
∠1 = ∠2 and AB = AC. Prove that:
(i) ∠B = ∠C
(ii) BD = DC
(iii) AD is perpendicular to BC.
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 23
Solution:

Question 9.
In the given figure prove tlyat:
(i) PQ = RS
(ii) PS = QR
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 25
Solution:

Question 10.
(i) ∆ XYZ ≅ ∆ XPZ
(ii) YZ = PZ
(iii) ∠YXZ = ∠PXZ
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 27
Solution:

Question 11.
In the given figure, prove that:
(i) ∆ABC ≅ ∆ DCB
(ii) AC=DB
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 29
Solution:

Question 12.
In the given figure, prove that:
(i) ∆ AOD ≅ ∆ BOC
(ii) AD = BC
(iii) ∠ADB = ∠ACB
(iv) ∆ADB ≅ ∆BCA
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 32
Solution:

Question 13.
ABC is an equilateral triangle, AD and BE are perpendiculars to BC and AC respectively. Prove that:
(i) AD = BE
(ii)BD = CE
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 34
Solution:

Question 14.
Use the informations given in the following figure to prove triangles ABD and CBD are congruent.
Also, find the values of x and y.
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 36
Solution:

Question 15.
The given figure shows a triangle ABC in which AD is perpendicular to side BC and BD = CD. Prove that:
(i) ∆ABD ≅ ∆ACD
(ii) AB=AC
(iii) ∠B = ∠C
Selina Concise Mathematics class 7 ICSE Solutions - Congruency Congruent Triangles image - 39
Solution:

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability

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Selina Class 7 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

Probability Exercise 22A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
A coin is tossed once. Find the probability of
(i) getting a head
(ii) not getting a head
Solution:
Total number of possible outcomes are Head (H) and Tail (T) i.e. 2
(i) P (Getting a head) =\(\frac { 1 }{ 2 }\)
(ii) P (Not getting a head) = \(\frac { 1 }{ 2 }\)

Question 2.
A coin is tossed 80 times and the head is obtained 38 times. Now, if a coin tossed once, what will the probability of getting:
(i) a tail
(ii) ahead
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -1

Question 3.
A dice is thrown 20 times and the outcomes are noted as shown below :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -2
Now a dice is thrown at random, find the probability of getting :
Solution:

Question 4.
A survey of 50 boys showed that 21 like tea while 29 dislike it. Out of these boys, one boy is chosen at random. What is the probability that the chosen boy
(i) likes tea
(ii) dislikes tea
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -4

Question 5.
In a cricket match, a batsman hits a boundary 12 times out of 80 balls he plays, further, if he plays one ball more, what will be the probability that:
(i) he hits a boundary
(ii) he does not hit a boundary
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -5

Question 6.
There are 8 marbles in a bag with numbers from 1 to 8 marked on each of them. What is the probability of drawing a marble with number
(i) 3
(ii) 7
Solution:
Total number of marbles = 8
(i) Probability (of getting a marble with number 3) = \(\frac { 1 }{ 8 }\)
(ii) Probability (of getting a marble with number 7) = \(\frac { 1 }{ 8 }\)

Question 7.
Two coins are tossed simultaneously 100 times and the outcomes are as given below:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -6
If the same pair of coins is tossed again at random, find the probability of getting :
(i) two heads
(ii) exactly one head
(iii) no head.
Solution:

Question 8.
A bag contains 4 white and 6 black balls,- all of the same shape and same size. A ball is drawn from the bag without looking into the bag. Find the probability that the ball drawn is :
(i) a black ball
(ii) a white ball
(iii) not a black ball
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -8

Question 9.
In a single throw of a dice, find the probability of getting a number:
(i) 4
(ii) 6
(iii) greater than 4
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -10

Question 10.
Hundred identical cards are numbered from 1 to 100. The cards are well shuffled and then a card is drawn. Find the probability that the number on the card drawn is :
(i) 50
(ii) 80
(iii) 40
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -11

Probability Exercise 22B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Suppose S is the event that will happen tommorow and P(S) = 0.03.
(i) State in words, the complementary event S’.
(ii) Find P(S’)
Solution:
Given, P(S) = 0.03
(i) The event will not happen tommorow.
(ii) P(S’) = 1 – P(S)
P(S’) = 1 – 0.03 [∵ P(S) + P(S’) = 1]
P(S’) = 0.97

Question 2.
Five Students A, B, C, D and E are competing in a long distance race. Each student’s probability of winning the race is given below:
A → 20 %, B → 22 %, C → 7 %, D → 15% and E → 36 %
(i) Who is most likely to win the race ?
(ii) Who is least likely to win the race ?
(iii) Find the sum of probabilities given.
(iv) Find the probability that either A or D will win the race.
(v) Let S be the event that B will win the race.
(a) Find P(S)
(b) State, in words, the complementary event S’.
(c) Find P(S’)
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -13

Question 3.
A Ticket is randomly selected from a basket containing3 green, 4 yellow and 5 blue tickets. Determine the probability of getting:
(i) a green ticket
(ii) a green or yellow ticket.
(iii) an orange ticket.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -14

Question 4.
Ten cards with numbers 1 to 10 written on them are placed in a bag. A card is chosen from the bag at random. Determine the probability of choosing:
(i) 7
(ii) 9 or 10
(iii) a number greater than 4
(iv) a number less than 6
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -16

Question 5.
A carton contains eight brown and four white eggs. Find the probability that an egg selected at random is :
(i) brown
(ii) white
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -17

Question 6.
A box contains 3 yellow, 4 green and 8 blue tickets. A ticket is chosen at random. Find the probability that the ticket is :
(i) yellow
(ii) green
(iii) blue
(iv) red
(v) not yellow
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -18

Question 7.
The following table shows number of males and number of females of a small locality in different age groups.

If one of the persons, from this locality, is picked at random, what is the probability that
(a) the person picked is a male ?
(b) the person picked is a female ?
(c) the person picked is a female aged 21-50 ?
(d) the person is a male with age upto 50 years?
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 22 Probability image -19

Selina Concise Mathematics class 7 ICSE Solutions – Ratio and Proportion (Including Sharing in a Ratio)

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics class 7 ICSE Solutions – Ratio and Proportion (Including Sharing in a Ratio)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 7 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

POINTS TO REMEMBER

Ratio
A ratio is a method to compare two quantities of the same kind with same unit; by dividing the first quantity by the second. The symbol (:) is used for ratio between two quantities e.g. a : b.
Note:
(i) A ratio is a pure number and has no unit.
(ii) A ratio must always be expressed in its lowest terms in simplest form.
(iii) If each term of a ratio is multiplied or divided by the same number or quantity, the ratio remains the same.
Proportion :
Proportion is equality of two ratios : e.g. a : b = c : d
i.e. Ratio between first and second is equal to ratio between third and fourth term.
(ii) a and d are called extreme terms and b and c are called mean terms
and a x d = b x c
(iii) Fourth term is called fourth proportional.
Continued Proportion
Three quantities are called in continued proportion if the ratio between first and second is equal to the ratio between second and third i. e.
a, b, c are in continued proportion if a : b = b : c
b the middle term is called the mean proportional between a and c and c, the third term is called the third proportional to a and b.

EXERCISE 6 (A)

Question 1.
Express each of the given ratio in its simplest form :
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 1

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 2

Question 2.
Divide 64 cm long string into two parts in the ratio 5 : 3.

Answer:
Sum of ratios = 5 + 3 = 8
∴ first part = \(\frac { 5 }{ 8 }\) of 64 cm = 40 cm
Second part = \(\frac { 3 }{ 8 }\) of 64 cm = 24 cm

Question 3.
Rs. 720 is divided between x and y in the ratio 4:5. How many rupees will each get?

Answer:
Sol. Total amount = Rs. 720 Ratio between x, y = 4 : 5
Sum of ratios = 4 + 5 = 9
x’s share = \(\frac { 4 }{ 9 }\) of Rs. 720 = Rs. 320
y’s share =\(\frac { 5 }{ 9 }\) of Rs. 720 = Rs. 400

Question 4.
The angles of a triangle are in the ratio 3 :2 : 7. Find each angle.

Answer:
Ratio in angles of a triangle = 3:2:7
Sum of ratios = 3 + 2 + 7=12
Sum of angles of a triangle = 180°
∴ First angle = \(\frac { 3 }{ 12 }\)x 180°= 45°
Second angle = \(\frac { 2 }{ 12 }\) x 180°= 30°
Third angle = \(\frac { 7 }{ 12 }\) x 180°= 105°

Question 5.
A rectangular field is 100 m by 80 m. Find the ratio of
(i) length to its breadth
(ii) breadth to its perimeter.

Answer:
Length of field (l) = 100 m
Breadth (b) = 80 m
∴Perimeter = 2 (l + b) = 2 (100 + 80) m = 2 x 180 = 360 m
(i) Ratio between length and breadth
= 100 : 80 = 5 : 4
(Dividing by 20, the HCF of 100 and 80)

(ii) Ratio between breadth and its perimeter
= 80 : 360 = 2 : 9
(Dividing by 40, the HCF of 80 and 360)

Question 6.
The sum of three numbers, whose ratios are 3 \(\frac { 1 }{ 3 }\) : 4 \(\frac { 1 }{ 5 }\) : 6 \(\frac { 1 }{ 8 }\) is 4917.Find the numbers.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 5

Question 7.
The ratio between two quantities is 3 : the first is Rs. 810, find the second.

Answer:
Ratio between two quantities = 3 : 4
Sum of ratio = 3+4 = 7
∴ Second quantity = Rs. \(\frac { 810 x 4 }{ 3 }\)
= Rs. 270 x 4 = Rs. 1080

Question 8.
Two numbers are in the ratio 5 : 7. Their difference is 10. Find the numbers.

Answer:
Ratio between two numbers = 5:7
Difference = 7-5 = 2
If difference is 2, then first number = 5
and if difference is 10, then first number
= \(\frac { 5 }{ 2 }\) x 10=25
and second number = \(\frac { 7 }{ 2 }\) x 10 = 35

Question 9.
Two numbers are in the ratio 10 : 11. Their sum is 168. Find the numbers.

Answer:
Ratio between two numbers = 10 : 11
Sum of ratios = 10 + 11=21
Total sum = 168
∴first number = \(\frac { 168 }{ 21 }\)x 10 =80
Second number = \(\frac { 168 }{ 21 }\)x 11 =88 Ans.

Question 10.
A line is divided in two parts in the ratio 2.5 : 1.3. If the smaller one is 35T cm, find the length of the line.

Answer:
Ratio between two parts of a line
= 2-5 : 1-3 =25 : 13
Sum of ratios = 25 + 13 = 38
Length of smaller part = 35.1 cm 38
Now length of line = \(\frac { 38 }{ 13 }\) x 35.1 cm
= 38 x 2.7 cm = 102.6 cm

Question 11.
In a class, the ratio of boys to the girls is 7:8. What part of the whole class are girls.

Answer:
Ratio between boys and girls = 7:8
Sum of ratios = 7 + 8 = 15
∴ Girls are \(\frac { 8 }{ 15 }\) of the whole class.

Question 12.
The population of a town is ’ 50,000, out of which males are \(\frac { 1 }{ 3 }\) of the whole population. Find the number of females. Also, find the ratio of the number of females to the whole population.

Answer:
Total population = 180,000
Population of males = \(\frac { 1 }{ 3 }\) of 180,000 = 60,000
∴ Population of females = 180,000 – 60,000 = 120,000
Ratio of females to whole population
= 120,000 : 180,000 = 2:3

Question 13.
Ten gram of an alloy of metals A and B contains 7.5 gm of metal A and the rest is metal B. Find the ratio between :
(i) the weights of metals A and B in the alloy.
(ii) the weight of metal B and the weight of the alloy.

Answer:
Total weight of A and B metals = 10 gm A’s weight = 7.5 gm B’s weight = 10 – 7.5 = 2.5 gm

(i) Ratio between A and B = 7.5 : 2.5
= \(\frac { 75 }{ 10 }\) : \(\frac { 25 }{ 10 }\) =3:1

(ii) Ratio between B and total alloy
= 2.5 : 10 = \(\frac { 25 }{ 10 }\) : 10
⇒ 25 : 100 = 1 : 4

Question 14.
The ages of two boys A and B are 6 years 8 months and 7 years 4 months respectively. Divide Rs. 3,150 in the ratio of their ages.

Answer:
A’s age = 6 years 8 months
= 6 x 12 + 8 = 72 + 8 = 80 months
B’s age = 7 years 4 months = 7 x 12 + 4 = 84 + 4 = 88 months
∴ Ratio between them = 80 : 88 = 10 : 11
Amount = Rs. 3150
Sum of ratios = 10 + 11 =21
∴A’s share = \(\frac { 3150 x 10 }{ 21 }\) = 1500 = Rs. 1500

B’s share = \(\frac { 3150 x 11 }{ 21 }\) = 1650 = Rs. 1650

Question 15.
Three persons start a business and spend Rs. 25,000; Rs. 15,000 atid Rs. 40,000 respectively. Find the share of each out of a profit of Rs. 14,400 in a year.

Answer:
A’s investment = Rs. 25000
B’s investment = Rs. 15000
C’s investment = Rs. 40000
∴ Ratio between their investment
= 25000 : 15000 : 40000
=5 : 3 : 8
Sum of ratios = 5 + 3 + 8=16 Total profit = ₹ 14400
∴ A’s share = \(\frac { 14400 }{ 16 }\) x 5 = ₹ 4500
B’s share = \(\frac { 14400 }{ 16 }\) x 3 = ₹ 2700
C’s share = \(\frac { 14400 }{ 16 }\) x 8 = ₹ 7200

Question 16.
A plot of land, 600 sq m in area, is divided between two persons such that the first person gets three-fifth of what the second gets. Find the share of each.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 6

Question 17.
Two poles of different heights are standing vertically on a horizontal field. At a particular time, the ratio between the lengths of their shadows is 2 :3. If the height of the smaller pole is 7.5 m, find the height of the other pole.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 7

Question 18.
Two numbers are in the ratio 4 : 7. If their L.C.M. is 168, find the numbers.

Answer:
Given, Ratio in two numbers = 4:7
and their L.C.M. = 168
Let first number = 4x
and second number = 7x
Now, L.C.M. of 4x and 7x
= 4 x 7 x x = 28x
∴ 28x = 168
x = \(\frac { 168 }{ 28 }\)
x = 6
∴ Required numbers = 4x and 7x = 4 x 6 = 24 and 7 x 6 = 42

Question 19.
is divided between A and B in such a way that A gets half of B. Find :
(i) the ratio between the shares of A and B.
(ii) the share of A and the share of B.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 8

Question 20.
The ratio between two numbers is 5 : 9. Find the numbers, if their H.C.F. is 16.

Answer:
Let the first number be 5x and second number be 9x
H.C.F. of 5x and 9x = Largest number common to 5x and 9x = x
Given H.C.F. = 16 ⇒ x = 16
∴Required numbers = 5x and 9x = 5×16 and 9×16 = 80 and 144

Question 21.
A bag contains ₹ 1,600 in the form of ₹10 and ₹20 notes. If the ratio between the numbers of ₹10 and ₹20 notes is 2 : 3; find the total number of notes in all.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 9

Question 22.
The ratio between the prices of a scooter and a refrigerator is 4 : 1. If the scooter costs ₹45,000 more than the refrigerator, find the price of the refrigerator.

Answer:
Ratio between the prices of scooter and a refrigerator = 4:1
Cost price of scooter = ₹45,000
Let the cost of scooter = 4x
Cost of refrigerator = 1x
According to condition,
Cost of scooter > Cost of refrigerator
⇒ 4x- 1x = 45000
⇒ 3x = 45000
x = \(\frac { 45000 }{ 3}\)
⇒ x = ₹15000
.’. Price of refrigerator = ₹15000

EXERCISE 6 (B)

Question 1.
Check whether the following quantities form a proportion or not ?
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 10

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 11

Question 2.
Find the fourth proportional of
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 13

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 14

Question 3.
Find the third proportional of
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 16

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 17

Question 4.
Find the mean proportional between
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 19

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 20

Question 5.

Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 22
Answer:

Question 6.
If x: y – 5 :4 and 2 : x = 3 :8, find the value of y.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 27

Question 7.
Find the value of x, when 2.5 : 4 = x : 7.5.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 28

Question 8.
Show that 2, 12 and 72 are in continued proportion.

Answer:
Selina Concise Mathematics class 7 ICSE Solutions - Ratio and Proportion (Including Sharing in a Ratio) image - 29

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 7 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

EXERCISE 2 (A)

Question 1.
Write down a rational number whose numerator is the largest number of two digits and denominator is the smallest number of four digits.

Solution:
Largest two digit = 99
Smallest, number of four digit = 1000 Now numerator = 99 and denominator = 1000
∴ Rational number = \(\frac { 99 }{ 1000 }\)

Question 2.
Write the numerator of each of the following rational numbers:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 1

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 2

Question 3.
Write the denominator of each of the following rational numbers:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 4

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 5

Question 4.
Write down a rational number numerator (-5) x (-4) and denominator (28 – 27) x (8 – 5).

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 6

Question 5.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 7

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 8

Question 6.
Separate positive and negative rational numbers from the following :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 9
Solution:

Question 7.
Find three rational numbers equivalent to
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 12

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 13

Question 8.
Which of the following are not rational numbers :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 15

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 16

Question 9.
Express each of the following integers as a rational number with denominator 7 :
(i) 5
(ii) -8
(iii) 0
(iv) -16
(v) 7

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 17

Question 10.
Express \(\frac { 3 }{ 5 }\) as a rational number with denominator:

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 19

Question 11.
Express \(\frac { 4 }{ 7 }\) as a rational number with numerator :

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 20

Question 12.
Find x, such that:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 21

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 22

Question 13.
Express each of the following rational numbers to the lowest terms :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 26

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 27

Question 14.
Express each of the following rational numbers in the standard form.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 29

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 30

EXERCISE 2 (B)

Question 1.
Mark the following pairs of rational numbers on the separate number lines :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 32

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 33

Question 2.
Compare:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 35

Solution:

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 37

Question 3.
Compare:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 40

Solution:

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 42

Question 4.
Arrange the given rational numbers in ascending order :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 45

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 46

Question 5.
Arrange the given rational numbers in descending order:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 48

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 49

Question 6.
Fill in the blanks :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 51

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 52

EXERCISE 2 (C)

Question 1.
Add:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 54

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 55

Question 2.
Add:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 57

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 58

Question 3.
Evaluate:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 64

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 65

Question 4.
Evaluate:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 68

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 69

Question 5.
Subtract :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 72

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 73

Question 6.
Subtract :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 75

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 76

Question 7.
The sum of two rational numbers is \(\frac { 11 }{ 24 }\). If one of them is \(\frac { 3 }{ 8 }\), find the other.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 80

Question 8.
The sum of two rational numbers is \(\frac { -7 }{ 11 }\). If one of them is \(\frac { 13 }{ 24 }\), find the other.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 81

Question 9.
The sum of two rational numbers is -4. If one of them is \(-\frac { 13 }{ 12 }\) , find the other.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 82

Question 10.
What should be added to \(-\frac { 3 }{ 6 }\) to get \(-\frac { 11 }{ 24 }\) ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 83

Question 11.
What should be added to \(\frac { -3 }{ 5 }\) to get 2?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 85

Question 12.
What should be subtracted from \(\frac { -4 }{ 5 }\) to get 1?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 86

Question 13.
The sum of two numbers is \(-\frac { 6 }{ 5 }\). If one of them is -2, find the other.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 88

Question 14.
What should be added to \(\frac { -7 }{ 12 }\) to get \(\frac { 3 }{ 8 }\)?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 89

Question 15.
What should be subtracted from \(\frac { 5 }{ 9 }\)to get \(\frac { 9 }{ 5 }\) ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 90

EXERCISE 2 (D)

Question 1.
Evaluate:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 91

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 92

Question 2.
Multiply:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 95

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 96

Question 3.
Evaluate:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 100

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 102

Question 4.
Find the cost of 3 \(\frac { 1 }{ 2 }\) m cloth, if one metre cloth costs ₹325 \(\frac { 1 }{ 2 }\).

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 107

Question 5.
A bus is moving with a speed of 65 \(\frac { 1 }{ 2 }\) km per hour. How much distance will it cover in 1 \(\frac { 1 }{ 3 }\) hours.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 108

Question 6.
Divide:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 109

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 110

Question 7.
Evaluate:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 112

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 113

Question 8.
The product of two numbers is 14. If one of the numbers is \(\frac { -8 }{ 7 }\), find the other.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 116

Question 9.
The cost of 11 pens is ₹3 \(\frac { 2 }{ 3 }\). Find the cost of one pen.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 117

Question 10.
If 6 identical articles can be bought for ₹2 \(\frac { 6 }{ 17 }\). Find the cost of each article.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 118

Question 11.
By what number should \(\frac { -3 }{ 8 }\) be multiplied so that the product is \(\frac { -9 }{ 16 }\) ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 119

Question 12.
By what number should \(\frac { -5 }{ 7 }\) be divided so -15 that the result is\(\frac { -15 }{ 28 }\) ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 120

Question 13.
Evaluate :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 121

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 122

Question 14.
Seven equal piece are made out of a rope 5 of 21 \(\frac { 5 }{ 7 }\) m. Find the length of each piece.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 123

EXERCISE 2 (E)

Question 1.
Evaluate:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 124

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 125

Question 2.
The sum of two rational numbers is \(\frac { -3 }{ 8 }\). If one of them is \(\frac { 3 }{ 16 }\), find the other,

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 134

Question 3.
The sum of two rational numbers is -5. If one of them is \(\frac { -52 }{ 25 }\) , find the other.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 135

Question 4.
What rational number should be added to \(-\frac { 3 }{ 16 }\) to get \(\frac { 11 }{ 24 }\)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 136

Question 5.
What rational number should be added to \(-\frac { 3 }{ 5 }\) to get 2?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 138

Question 6.
What rational number should be subtracted from \(-\frac { 5 }{ 12 }\)to get \(\frac { 5 }{ 24 }\)

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 139

Question 7.
What rational number should be subtracted from \(\frac { 5 }{ 8 }\) to get \(\frac { 8 }{ 5 }\) ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 140

Question 8.
Evaluate:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 141

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 142

Question 9.
The product of two rational numbers is 24. If one of them is \(-\frac { 36}{ 11 }\), find the other.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 147

Question 10.
By what rational number should we multiply \(\frac { 20 }{ -9 }\) ,so that the product may be \(\frac { -5 }{ 9 }\) ?

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 2 Rational Numbers image - 148

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions (Decimals)

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions(Decimals)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 7 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

POINTS TO REMEMBER

Decimal fraction (or a decimal number)
A decimal fraction is a fraction whose denominator is 10 or a higher power of 10.
In order to express a given decimal fraction in shorter form, the denominator is not written but its absence is shown by a dot which is called a decimal point inserted in a proper place.
Note :
(i) When there is no number is the left of the decimal point, generally, a zero is written.
(ii) Generally, a decimal fraction has two parts, the first part which is on the right ox the decimal point is called decimal part and the part on the left side of the decimal point, is called integeral part.
(iii) The decimal part is always less than 1.
Reading Decimal Numbers
The integral part is read according to its value and the decimal part is read by naming each digit, in order, separately
Converting a decimal number into a vulgar fraction :
Remove the decimal point from the decimal number and write in its denominator with as many zeros as the number of digits are in the decimal parts to the right of 1.
In the decimal number, the name of each place is given as under is the place value chart:

Thousands Hundreds Tens Units • Tenths Hundredths Thousandth

Decimal point and so on
Converting a given fraction in to a decimal fraction :
(a) When the denominator in given fraction is 10. 100, 1000 etc., then count from extreme right to left, mark decimal point after as many digits of the numerator as there are zeroes in the denominator.
(b) When the denominator of the given number is other then 10 or higher power of 10, then divide in an ordinary way and mark the decimal point in the quotient just after the division of unit digit is completed. After this, any number of zeroes can be borrowed to.complete the division.
Note : The number of figures that follow the decimal part is called the number of decimal places.
Addition and Subtraction of decimal numbers.
(a) Addition : Write the given decimal numbers in such a way, that the decimal points of all the numbers fall in the same vertical line. Digits with the same place value are placed one below the other units are written below units, tens below tens and so on.
Addition is started from the right side, as done in the usual addition (empty places may be filled up by zeroes). In the result (total), the decimal point is placed under decimal points of the numbers added.
(b) Subtraction : In subtraction also, the numbers are written in such a way that their decimals are in ihe same vertical line. Digits with the same place value are placed one below the other (empty places may be filled by zeroes).
Subtraction is started from the right side, as in the case of normal subtraction. In the result, decimal point is placed just under the other decimal points.
Multiplication of decimal numbers :
(1) Multiplication by 10,100, 1000 etc. shift the decimal point, in the multiplicand, to the right by as many digits as there are zeroes in the multiplier.
(2) Multiplication by a whole number : Multiply in an ordinary way, without considering the decimal point. In the product, the decimal point should be fixed by counting as many digits from the right as there are decimal places in the multiplicand.
Thus, (i) 0.3 x 6 = 1.8 (ii) 0.26 x 18 = 4.68 and so on.
(3) Multiplication of a decimal number by another decimal number :
Multiply the two numbers in a normal way, ignoring their decimals. In the product, decimal point is fixed counting from right, the digits equal to the sum of decimal places in the multiplicand and the multiplier.
Thus, 32.5 x ().()7 = 2.275
Since, the multiplicand (32.5) has one decimal place and multiplier (0.07) has two decimal places, their product will have 1+2 = 3 decimal places.
Division of decimal numbers :
(1) Division by 10, 100, 1000 etc : Shift the decimal points to the left as many digits as there are
zeroes in the divisor. .
(2) Division by a whole number : Divide in the normal manner, ignoring the decimal, and mark tire decimal point; in the quotient, while just crossing over the decimal point in the dividend.
Recurring Decimals :
On performing a division, sometimes we find that the same remainder is left, no matter how long we continue the division process. For this reason, the same digit appeares again and again in the quotient. This fact is shown by puting a dot as a bar over the repeating digits.
Rounding off of decimal numbers :
(i) If the answer required is correct to two decimal places, we retain digits upto three decimal places.
(ii) If the digit in the third decimal place is five or more than five, then the digit in the second decimal place is increased by one and, if the digit in the third decimal place is less than five, then the digit in the second decimal place is not altered.
(iii)The third digit which was retained is now omitted.
Thus, for getting 8.4813 correct to two decimal places.
Write the given number upto three decimal places i.e. 8.481.
Since, the digit in the third decimal place is 1 which is less than 5.
∴The digit in the second decimal place is not altered.
And, so 8.4813 = 8.48, correct to two decimal places.
In the same way, to get 3.946824 correct to nearest thousandths i.e., correct to three decimal places, first write it as 3.9468.
Then according to the rule, the digit in the third place changes from 6 to 7.
3.9468 = 3.947, correct to three decimal places.

Decimal Fractions Exercise 4A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Convert the following into fractions in their lowest terms :
(i) 3.75
(ii) 0.5
(iii) 2.04
(iv) 0.65
(v) 2.405
(vi) 0.085
(vii) 8.025

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 1

Question 2.
Convert into decimal fractions
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 2

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 4
Question 3.
Write the number of decimal places in :
(i) 0.4762
(ii) 7.00349
(iii) 8235.403
(iv) 35.4
(v) 2.608
(vi) 0.000879

Answer:
(i) In 0.4762, there are four places.
(ii) In 7.00349, there are five places.
(iii) In 8235.403, there are three places.
(iv) In 35.4, there is one place.
(v) In 2.608, there are three places.
(vi) In 0.000879, there are six places.

Question 4.
Write the following decimals as word statements :
(i) 0.4,0.9,0.1
(ii) 1.9, 4.4, 7.5
(iii) 0.02, 0.56, 13.06
(iv) 0.005,0.207, 111.519
(v) 0.8, 0.08, 0.008, 0.0008
(vi)256.1, 10.22, 0.634

Answer:
(i) 0.4 = zero point four, 0.9 = zero point nine, 0.1 = zero point one.
(ii) 1 .9 = one point nine, 4.4 = four point four, 7.5 = seven point five.
(iii) 0.02 = zero point zero two, 0.56 = zero point five six, 13.06 = thirteen point zero six.
(iv) 0.005 = zero point zero zero five, 0.207 = zero point two zero seven, 111.519 = one hundred eleven point five one nine.
(v) 0.8 = zero point eight, 0.08 = zero point zero eight, 0.008 = zero point zero zero eight, 0.0008 = zero point zero zero zero eight
(vi) 256.1 = Two hundred fifty six point one, 10.22 = Ten point two two, 0.634 = zero point six three four.

Question 5.
Convert the given fractions into like fractions:
(i) 0.5,3.62,43.987 and 232.0037
(ii) 215.78, 33.0006, 530.3 and 0.03569

Answer:
(i) 0.5, 3.62, 43.987 and 232.0037
In these decimals, the greatest places of decimal is 4
∴0.5 = 0.5000
3.62 = 3.6200
43.987 = 43.9870
232.0037 = 232.0037

(ii) 215.78, 33.0006, 530.3 and 0.03569
In these decimals, the greatest places of decimal is 5
∴215.78 = 215.78000
33.0006 = 33.00060
530.3 = 530.30000
0.03569 = 0.03569

Decimal Fractions Exercise 4B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Add:
(i) 0.5 and 0.37 (ii) 3.8 and 8.7
(iii) 0.02, 0.008 and 0.309
(iv) 0. 4136, 0. 3195 and 0.52
(v) 9.25, 3.4 and 6.666
(vi) 3.007, 0.587 and 18.341
(vii) 0.2, 0.02 and 2.0002
(viii) 6. 08, 60.8, 0.608 and 0.0608
(ix) 29.03, 0.0003, 0.3 and 7.2
(x) 3.4, 2.025, 9.36 and 3.6221

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 5

Question 2.
Subtract the first! number from the second :
(i) 5.4, 9.8
(ii) 0.16, 4.3
(iii) 0.82, 8.6
(v) 2.237, 9.425
(vi) 41 .03, 59.46
(vii) 3.92. 26.86
(viii) 4.73, 8.5
(ix) 12.63, 36.2
(x) 0.845, 3.71

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 8

Question 3.
Simplify :
(i) 28.796 -13.42 – 2.555
(ii) 93.354 – 62.82 – 13.045
(iii) 36 – 18.59 – 3.2
(iv) 86 + 16.95 – 3.0042
(v) 32.8 – 13 – 10.725 +3.517
(vi) 4000 – 30.51 – 753.101 – 69.43
(vii) 0.1835 + 163.2005 – 25.9 – 100
(viii) 38.00 – 30 + 200.200 – 0.230
(ix) 555.555 + 55.555 – 5.55 – 0.555

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 10

Question 4.
Find the difference between 6.85 and 0.685.
Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 14

Question 5.
Take out the sum of 19.38 and 56.025 then subtract it from 200. 111.
Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 15

Question 6.
Add 13.95 and 1.003 ; and from the result, subtract the sum of 2.794 and 6.2.
Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 16

Question 7.
What should be added to 39.587 to give 80.375 ?

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 18

Question 8.
What should be subtracted from 100 to give 19.29?
Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 19

Question 9.
What is the excess of 584.29 over 213.95 ?
Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 20

Question 10.
Evaluate:
(i) (5.4 – 0.8) + (2.97 -1.462)
(ii) (6.25 + 0.36) -(17.2 – 8.97)
(iii) 9.004 + (3 -2.462)
(iv) 879.4 – (87.94 – 8 .794)

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 21

Question 11.
What is the excess of 75 over 48.29?
Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 22

Question 12.
If A = 237.98 and B = 83.47.
Find :
(i) A – B
(ii) B – A.
Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 23

Question 13.
The cost of one kg of sugar increases from ?28.47 to T32.65. Find the increase in cost.

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 24

Decimal Fractions Exercise 4C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Multiply:
(i) 0.87 by 10
(ii) 2.948 by 100
(iii) 6.4 by 1000
(iv) 5.8 by 4
(v) 16.32 by 28
(vi) 5. 037 by 8
(vi) 4.6 by 2.1
(viii) 0.568 by 6.4

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 25

Question 2.
Multiply each number by 10, 100, 1000 :
(i) 0.5
(ii) 0.112
(iii) 4.8
(iv) 0.0359
(v) 16.27
(vi) 234.8

Answer:
(i) 0.5 x 10 = 5,0.5 x 100 = 50,
0.5 x 1000 = 500
(ii) 0.112 x 10= 1.12,0.112 x 100
= 11.2, 0.112 x 1000= 112
(iii) 4.8 x 10 = 48, 4.8 x 100 = 480,
4.8 x 1000 = 4800
(iv) 0.0359 x 10 = 0.359,0.0359 x 100 = 3.59, 0.0359 x 1000 = 35-9
(v) 16.27 x 10 = 162.7, 16.27 x 100 = 1627, 16.27 x 1000= 16270
(vi) 234.8 x 10 = 2348, 234.8 x 100 = 23480, 234.8 x 1000 = 234800

Question 3.
Evaluate:
(i) 5.897 x 2.3
(ii) 0.894 x 87
(iii) 0.01 x 0.001
(iv) 0.84 x 2.2 x 4
(v) 4.75 x 0.08 x 3
(vi) 2.4 x 3.5 x 4.8
(vii) 0.8 x 1.2 x 0.25
(viii) 0.3 x 0.03 x 0.003
(ix) 12.003 x (0.2)5

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 26

Question 4.
Divide :
(i) 54.9 by 10
(ii) 7.8 by 100
(iii) 324.76 by 1000
(iv) 12.8 by 4
(v) 27.918 by 9
(vi) 4.672 by 8
(vii) 4.32 by 1.2
(viii) 7.644 by 1.4
(ix) 4.8432 by 0.08

Answer:

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 29

Question 5.
Divide each of the given numbers by 10, 100, 1000 and 10000
(i) 2.1
(ii) 8.64
(iii) 5-01
(iv) 0.0906
(v) 0.125
(vi) 111.11
(vii) 0.848 x 3
(viii)4.906 x (0.2) ²
(ix) (1.2)² x(0.9)²

Answer:

(i) 2.1 ÷ 10 = 0.21, 2.1 ÷ 100 = 0.021,
2.1 ÷ 1000 = 0.0021
and 2.1 ÷10000 = 0.00021

(ii) 8.64 ÷ 10 = 0-864, 8.64 ÷ 100 = 0-0864,
8.64 ÷ 1000 = 0-00864
and 8.64 ÷ 10000 = 0.000864

(iii)5.01 ÷ 10 = 0.501,
5.01 ÷ 100 = 0.0501,
5.01 ÷1000 = 0.00501,
5.01 ÷ 10000 = 0.000501

(iv) 0.0906 ÷ 10 = 0.00906,
0.0906 ÷ 100 = 0.000906,
0.0906 ÷ 1000 = 0.0090906,
0.0906 ÷ 10000 = 0.00000906

(v) 0.125
Now 0.125 + 10 = 0.0125,
0.125 ÷ 100 = 0-00125,
0.125 ÷1000 = 0.000125,
0.125 ÷ 10000 = 0.0000125

(vi) 111.11÷ 10= 11.111,
111.11÷ 100 = 1.1111,
111.11 ÷ 1000 = 0.11111,
111.11 ÷ 10000 = 0.011111

(vii) 0.848 x 3 = 2.544 ,
Now 2.544 ÷ 10 = 0.2544,
2.544 ÷ 100 = 0-02544,
2-544 ÷ 1000 = 0-002544,
2-544 ÷ 10000 = 0-0002544

(viii) 4.906 x (0.2)² = 4.906 x 0.2 x 0.2
= 4.906 x 0.04 = 0.19624
Now 0.19624 + 10 = 0.019624,
0.19624 + 100 = 0.0019624,
0.19624 + 1000 = 0.00019624,
0.19624 + 10000 = 0.000019624

(ix) (1.2)² x (0.9)² = 1.2 x 1.2 x 0.9 x 0.9 = 1.44 x 0.81 = 1.1664
Now 1 .1664 + 10 = 0.11664,
1.1664 ÷ 100 = 0.011664,
1.1664 ÷ 1000 = 0.0011664,
1.1664 ÷ 10000 = 0.00011664

Question 6.
Evaluate :
(i) 9.75 + 5
(ii) 4.4064 + 4
(iii) 27.69 + 30
(iv) 19.25 + 25
(v) 20.64+ 16
(vi) 3.204 + 9
(vii) 0.125 + 25
(viii) 0.14616 + 72
(ix) 0.6227+ 1300
(x) 257.894+ 0-169
(xi) 6.3 + (0.3)²

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 31

Question 7.
Evaluate:
(i) 4.3 x 0.52 x 0.3
(ii) 3.2 x 2.5 x 0.7
(iii) 0.8 x 1.5 x 0.6
(iv) 0.3 x 0.3 x 0.3
(v) 1.2 x 1.2 x 0.4
(vi) 0.4 x 0.04 x 0.004
(vii) 0.5 x 0.6 x 0.7
(Viii) 0.5 x 0.06 x 0.007

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 35

Question 8.
Evaluate:
(i) (0.9)²
(ii) (0.6)² x 0.5
(iii) 0.3 x (0.5)²
(iv) (0.4)³
(v) (0.2)³ x 5
(vi) (0.2)³ x 0.05

Answer:

(i) (0.9)²
⇒ 0.9 x 0.9 = 0.81
(Sum of decimal places 1 + 1=2)

(ii) (0.6)² x 0.5
⇒ 0.6 x 0.6 x 0.5
⇒ 0.36 x 0.5 = 0.180 or 0.18
(Sum of decimal places = 1 + 1 + 1 = 3)

(iii) 0.3 x (0.5)²
⇒ 0.3 x 0.5 x 0.5
⇒ 0.3 x 0.25 = 0.075
(Sum of decimal places 1 + 1 + 1 = 3)
(iv) (0.4)³
⇒ 0.4 x 0.4 x 0.4
⇒ 0.16 x 0.4 = 0.064
(Sum of decimal places 1 + 1 + 1 = 3)

(v) (0.2)³ x 5
⇒ 0.2 x 0.2 x 0.2 x 5
⇒ 0.08 x 5 = 0.40 or 0.4
(Sum of decimal places 1 + 1 + 1 = 3)

(vi) (0.2)³ x 0.05
⇒ 0.2 x 0.2 x 0.2 x 0.05
⇒ 0.008 x 0.05 = 0.00040
(Sum of decimal places = 5)

Question 9.
Find the cost of 36.75 kg wheat at the rate of ₹12.80 per kg.
Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 41

Question 10.
The cost of a pen is ₹56.15. Find the cost of 16 such pens.
Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 42

Question 11.
Evaluate:
(i) 0.0072 ÷ 0.06
(ii) 0.621 ÷ 0.3
(iii) 0.0532 ÷ 0.005
(iv) 0.01162 ÷ 0.14
(v) (7.5 x 40.4) ÷ 25
(vi) 2.1 ÷ (0.1 x 0.1)

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 43

Question 12.
Fifteen indentical articles weigh 31.50 kg. Find the weight of each article.
Answer:
Weight of 15 articles = 31.50 kg
∴ Weight of one article
= 31.50- 15 = 2.1 kg

Question 13.
The product of two numbers is 211.2. If one of these two numbers is 16.5, find the other number.
Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 45

Question 14.
One dozen identical articles cost ₹45.96. Find the cost of each article.
Answer:
∴ Weight of one dozen articles = ₹45.96
One dozen = 12
∴ Cost of one article = 45.96 + 12 = ₹3.83

Decimal Fractions Exercise 4D – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Find whether the given division forms a terminating decimal or a non-terminating decimal:
(i) 3 ÷ 8
(ii)8 ÷ 3
(iii) 6÷ 5
(iv) 5 ÷ 6
(v) 12.5 ÷ 4
(vi) 23 ÷ 0.7
(vii) 42 ÷ 9
(viii) 0.56÷ 0.11

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 46

Question 2.
Express as recurring decimals :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 51

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 52

Question 3.
Convert into vulgar fraction :
(i) 0.\(\bar { 3 }\)
(ii) 0.\(\bar { 8 }\)
(iii) 4.\(\bar { 4 }\)
(iv) 23.\(\bar { 7 }\)

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 58

Question 4.
Convert into vulgar fraction :
(i) 0.\(\bar { 35 }\)
(ii) 2.\(\bar { 23 }\)
(iii) 1.\(\bar { 28 }\)
(iv) 5.\(\bar { 234 }\)

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 60

Question 5.
Convert into vulgar fraction :
(i) 0 0.3\(\bar { 7 }\)
(ii) 0.2\(\bar { 45 }\)
(iii) 0.68\(\bar { 5 }\)
(iv) 0.4\(\bar { 42 }\)

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 61

Decimal Fractions Exercise 4E – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Round off:
(i) 0 .07, 0.112, 3.59, 9.489 to the nearest tenths.
(ii) 0.627, 100.479, 0 065 and 0.024 to the nearest hundredths.
(iii) 4.83,0.86,451 .943 and 9.08 to the nearest whole number.

Answer:
(i) 0.07 = 0.1,
0.112 = 0.1
3 . 59 = 3.6, 9.489 = 9.5
(ii) 0.627 = 0.63,
100.479 = 100.48
0.065 = 0.07,
0.024 = 0.02
(iii) 4.83 = 5,
0.86= 1,
451.943 = 452
9.08 = 9

Question 2.
Simplify, and write your answers correct to the nearest hundredths :
(i) 18 .35 x 1.2
(ii) 62.89 x 0.02

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 63

Question 3.
Write the number of significant figures (digits) in:
(i) 35.06
(ii) 0.35
(iii) 7.0068
(iv) 19 .0
(v) 0.0062
(vi) 0 4.2 x 0.6
(vii) 0.08 x 25
(viii) 3.6 ÷ 0.12 .

Answer:
(i) 35.06 : In this significant figures i.e. digits are 4
(ii) In 0.35, significant figures are 2
(iii) In 7.0068, significant figures are 5
(iv) In 19.0, significant figures are 3
(v) In 0.0062, significant figures are 2
(vi) In 4.2 x 0.6 = 2.52, significant figure are 3
(vii) In 008 x 25 = 2.00 = 2 significant figure is 1
(viii) In 3.6 ÷0 .12 or 360 ÷ 12 = 30, significant figure are 2.

Question 4.
Write :
(i) 35.869,0 008426,4.952 and 382.7, correct lo three significant figures.
(ii) 60.974. 2.8753, 0.001789 and 400.04, correct to four significant figures.
(iii) 14.29462, 19.2, 46356.82 and 69, correct to five significant figures.

Answer:
(i) Correct to three significant figures are
35.869 → 35.9
0.008426 →0.00843
4. 952→ 4.95
382.7 →383
(ii) Correct to four significant figures
60.974 →60.97
2. 8753 → 2.875
0.001789 → 0.001789
400.04 → 400.0
(iii) Correct to five significant figures
14.29462→ 14.295
19.2 → 19.200
46356.82 →46357
69 → 69.000

Decimal Fractions Exercise 4F – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
The weight of an object is 3 .06 kg. Find the total weight of 48 similar objects.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 64

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 65

Question 2.
Find die cost of 17.5 m cloth at the rate of Rs. 112.50 per metre.

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 66

Question 3.
One kilogramme of oil costs Rs. 73.40. Find the cost of 9.75 kilogramme of the oil.

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 67

Question 4.
Total weight of 8 identical objects is 51.2 kg. Find the weight of each object.
Answer:
Weight of 8 objects = 51-2 kg
∴ Weight of 1 object = 51 -2 + 8 kg = 6-4 kg Ans.

Question 5.
18.5 m of cloth costs Rs. 666. Find the cost of 3.8 m cloth.

Answer:
Cost of 18.5 m cloth = Rs. 666
Cost of 1 m cloth = Rs. 666 ÷18.5 and cost of 3.8 m cloth
= Rs. (666 ÷18.5) x 3-8 = Rs. (6660 ÷ 185) x 3.8 = Rs. 36 x 3.8 = Rs. 136.80
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 68

Question 6.
Find die value of:
(i) 0.5 of Rs. 7.60 + 1.62 of Rs. 30
(ii) 2.3 of 7.3 kg + 0.9 of 0.48 kg
(iii) 6.25 of 8.4 – 4.7 of 3.24
(iv) 0.98 of 235 – 0 .09 of 3.2

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 69

Question 7.
Evaluate:
(i) 5.6 – 1 .5 of 3.4
(ii) 4.8 ÷ 0.04 of 5
(iii) 0.72 of 80 + 0.2
(iv) 0.72 ÷ 80 of 0.2
(v) 6.45 + (3.9 – 1.75)
(vi) 0.12 of(0.104 – 0.02)+ 0.36 x 0.5

Answer:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 4 Decimal Fractions image - 70

Selina Concise Chemistry Class 7 ICSE Solutions – Language of Chemistry

August 11, 2022January 27, 2019 by Prasanna

Selina Concise Chemistry Class 7 ICSE Solutions – Language of Chemistry

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Chemistry. You can download the Selina Concise Chemistry ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Chemistry for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 7 Chemistry Chapter 5 Language of Chemistry

Points to Remember :

A chemical reaction involves the transformation of original substance into an altogether new substance(s).
A chemical reaction can be represented with the help of the symbols or the formulae of the elements and the compounds taking part in that reaction. This gives a chemical equation.
Certain necessary conditions for a chemical reaction to happen are close contact, solution form, heat, light and catalyst.
Characteristics of chemical reactions are change of colour, evolution of a gas, formation of a precipitate, change of state, change of smell and evolution/absorption of heat.
A complete chemical equation symbolically represents the reactants, products and their physical states.
The substances that react with each other are called reactants and they are represented on the left hand side of the equation. The substances that are formed as a result of the reaction are called products. They are represented on the right hand side of the equation.
A chemical equation needs to be balanced to make it follow the law of the conservation of mass.
The law of conservation of mass states that mass can be neither created nor destroyed, it can only be transformed from one form to another.
A chemical equation gives both qualitative and quantitative information about the reactants and products.

EXERCISE

Question 1.
(a) Define chemical reaction.
(b) What is a chemical equation?
(c) Why do we need to balance chemical equations?
Answer:
(a) Chemical reaction : Any chemical change in matter which involves its transformation into one or more new substances is called a chemical reaction.
(b) Chemical equation : A chemical equation is the symbolic representation of a chemical reaction using the symbols and the formula of the substances involved in the reaction.
(c) A chemical equation needs to be balanced so as to make the number of the atoms of the reactants equal to the number of the atoms of the products.

Question 2.
State four conditions necessary for chemical reactions to take place.
Answer:
Conditions necessary for chemical reactions :

Close contact
Solution form
Heat
Light
Catalyst

3. Differentiate between :
(a) Reactants and products.

Reactants	Products
The substances that react with one another are called reactants. Reactants are written on the left hand side of equation.	The new substances formed are called products. Products are written on the right hand side of equation.

(b) Chemical reaction and chemical equation.

Chemical reaction	Chemical Equation
Any chemical change in matter which involves its transformation into one or more new substances is called a chemical reaction.	A chemical equation is the symbolic representation of a chemical reaction using the symbols and the formula of the substances involved in the reaction.

(c) A balanced and a skeletal chemical equation.

Balanced Equation	Skeletal Equation
A balanced chemical equation is one in which the number of atoms each element on the reactant side is equal to the number of atoms of that element on the product side.	In a skeletal equation the number of atoms on reactant side are not equal to number of atoms of product side.

Question 4.
Write word equations for the following skeletal equations:
(a) KClO₃ → KCl + O₂(b) Zn + HCl → ZnCl₂ + H₂(c)FeCl₂ + Cl₂ → FeCl₃(d) CO + O₂ → CO₂(e) Ca + O₂ → CaO
(f) Na + O_{2 →}Na₂O
(g) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
(h) AgBr → Ag + Br₂(i) KNO₂ → KNO₂ + O₂

Answer:
(a) 2KClO₃ → 2KCl+ 3O₂(b) Zn + 2HCl → ZnCl₂ + H₂(c) 2FeCl₂ + Cl₂ → 2FeCl₃(d) 2CO + O₂ →2CO₂(e) 2Ca + O₂ →2CaO
(f) 4Na + O₂ → 2Na₂O
(g) 2NaOH + H₂SO₄→ Na₂SO₄ + 2H₂O
(h) 2AgBr → 2Ag + Br₂(i) 2KNO₃→ 2KN0₂ + O₂

Question 5.
Balance the following chemical equations :
(a) FeS + HCl → FeCl₂ + H₂S
(b) Na₂CO₃ + HCl → NaCI + H₂O + CO₂
(c) H₂ + O₂ → H₂O
(d) Na₂0 + H₂0 → NaOH
Answer:
(a) FeS + 2HCl → FeCl₂ + H₂S
(b) Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
(c) 2H₂+ O₂ → 2H₂O
(d) Na₂O + H₂O → 2NaOH

Question 6.
What information do you get from the equation H₂+ Cl₂ → 2HCl ?
Answer:
(a)Hydrogen and chlorine molecules are the reactants.
(b)They are in gaseous form.
(c)The product is hydrogen chloride gas.
(d)Two molecules of hydrogen chloride are formed.

Question 7.
Write your observations for the following chemical reactions and name the products formed :
(a) When sugar is heated.
(b) When manganese dioxide is added to potassium chlorate and heated.
(c) When dilute acetic acid is poured on baking soda.
(d) When an aqueous solution of sodium chloride is mixed with an aqueous solution of silver nitrate.
(e) When ammonium chloride is heated with sodium hydroxide.
(f) When water is added to quick lime?

Answer:
(a) Black solid mass (charcoal) is formed along with water vapours.
(b) Manganese dioxide acts as a catalyst for the decomposition of potassium chlorate into potassium chloride and oxygen at a lower temperature.
(c) Sodium acetate, CO₂and water is formed.
(d) A white insoluble solid precipitate of silver chloride is formed along with Sodium nitrate.
(e) When solid ammonium chloride is heated with sodium hydroxide solution, a gas ammonia is evolved which is recognised by its strong pungent smell.
Selina Concise Chemistry Class 7 ICSE Solutions - Language of Chemistry-7
(f) When water is added to quick lime, a large amount of heat energy is evolved.

Question 8.
Write symbolic representation for the following word equations and balance them :
(a) Calcium carbonate → Calcium oxide + Carbon dioxide
(b) Carbon + Oxygen → Carbon dioxide
(c) Calcium oxide + Water → Calcium hydroxide
(d) Aluminium + Chlorine → Aluminium chloride
(e) Iron + Sulphur → Iron sulphide

Answer:

Selina Concise Chemistry Class 7 ICSE Solutions - Language of Chemistry-8

OBJECTIVE TYPE QUESTIONS

1. Fill in the blanks:
(a) The substances which undergo chemical change are called reactants.
(b) The substances formed as a result of a chemical reaction are called products.
(c) During a chemical reaction transfer of energy takes place.
(d) The basic conditions necessary for a chemical reaction is close contact.
(e) In some chemical reactions an insoluble precipitate is formed when two solutions are mixed.

2. Write ‘true’ or ‘false’ for the following statements :
(a) No new substance is formed during a chemical reaction : True
(b) Hydrogen sulphide has rotten egg smell : True
(c) When potassium iodide solution is added to lead acetate solution a red precipitate is formed : False
(d) A black residue is formed when sugar is heated : True
(e) When iron and sulphur are heated together a grey mass is formed which is attracted by a magnet : False
(f) A chemical equation gives only qualitative information of a chemical reaction : False

MULTIPLE CHOICE QUESTIONS

1. A chemical equation is a statement that describes a chemical change in terms of
(a) symbols and formulae
(b) energy
(c) number of atoms
(d) colours

2. Balancing a chemical equation is based on
(a) Law of conservation of mass
(b) Mass of reactants and products
(c) Symbols and formulae
(d) None of the above

3. Copper carbonate when heated, it turns :
(a) Blue
(b) Green
(c) Black
(d) Yellow

Thousands	Hundreds	Tens	Units	•	Tenths	Hundredths	Thousandth
				Decimal point			and so on