icse maths book for class 10 solved

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Ratio and Proportion Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Ratio and Proportion Chapter Test.

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Question 1.
Find the compound ratio of:
(a + b)² : (a – b )² ,
(a² – b²) : (a² + b²),
(a⁴ – b⁴) : (a + b)⁴
Solution:
(a + b)² : (a – b )² ,
(a² – b²) : (a2 + b²),
(a⁴ – b⁴) : (a + b)⁴

Question 2.
If (7 p + 3 q) : (3 p – 2 q) = 43 : 2 find p : q
Solution:
(7p + 3q) : (3p – 2q) = 43 : 2
=> \(\frac { 7p+3q }{ 3p-2q } =\frac { 43 }{ 2 } \)

Question 3.
If a : b = 3 : 5, find (3a + 5b): (7a – 2b).
Solution:
a : b = 3 : 5
=> \(\frac { a }{ b } =\frac { 3 }{ 5 } \)
3a + 5n : 7a – 2b
Dividing each term by b

Question 4.
The ratio of the shorter sides of a right angled triangle is 5 : 12. If the perimeter of the triangle is 360 cm, find the length of the longest side.
Solution:
Let the two shorter sides of a right angled triangle be 5x and 12x.
Third (longest side)

Question 5.
The ratio of the pocket money saved by Lokesh and his sister is 5 : 6. If the sister saves Rs 30 more, how much more the brother should save in order to keep the ratio of their savings unchanged ?
Solution:
Let the savings of Lokesh and his sister are 5x and 6x.
and the Lokesh should save Rs y more Now, according to the problem,
=> \(\frac { 5x+y }{ 6x+30 } =\frac { 5 }{ 6 } \)
=> 30x + 6y = 30x + 150

Question 6.
In an examination, the number of those who passed and the number of those who failed were in the ratio of 3 : 1. Had 8 more appeared, and 6 less passed, the ratio of passed to failures would have been 2 : 1. Find the number of candidates who appeared.
Solution:
Let number of passed = 3 x
and failed = x
Total candidates appeared = 3x + x = 4x. In second case
No. of candidates appeared = 4 x + 8
and No. of passed = 3 x – 6
and failed = 4x + 8 – 3x + 6 = x + 14
then ratio will be = 2 : 1
Now according to the condition

Question 7.
What number must be added to each of the numbers 15, 17, 34 and 38 to make them proportional ?
Solution:
Let x be added to each number, then numbers will be 15 + x, 17 + x, 34 + x, and 38 + x.
Now according to the condition

Question 8.
If (a + 2 b + c), (a – c) and (a – 2 b + c) are in continued proportion, prove that b is the mean proportional between a and c.
Solution:
(a + 2 b + c), (a – c) and (a – 2 b + c) are in continued proportion
=> \(\frac { a+2b+c }{ a-c } =\frac { a-c }{ a-2b+c } \)

Question 9.
If 2, 6, p, 54 and q are in continued proportion, find the values of p and q.
Solution:
2, 6, p, 54 and q are in continued proportional then
=> \(\frac { 2 }{ 6 } =\frac { 6 }{ p } =\frac { p }{ 54 } =\frac { 54 }{ q } \)

Question 10.
If a, b, c, d, e are in continued proportion, prove that: a : e = a⁴ : b⁴.
Solution:
a, b, c, d, e are in continued proportion
=> \(\frac { a }{ b } =\frac { b }{ c } =\frac { c }{ d } =\frac { d }{ e } \) = k (say)

Question 11.
Find two numbers whose mean proportional is 16 and the third proportional is 128.
Solution:
Let x and y be two numbers
Their mean proportion = 16
and third proportion = 128

Question 12.
If q is the mean proportional between p and r, prove that:
\({ p }^{ 2 }-{ 3q }^{ 2 }+{ r }^{ 2 }={ q }^{ 4 }\left( \frac { 1 }{ { p }^{ 2 } } -\frac { 3 }{ { q }^{ 2 } } +\frac { 1 }{ { r }^{ 2 } } \right) \)
Solution:
q is mean proportional between p and r
q² = pr

Question 13.
If \(\frac { a }{ b } = \frac { c }{ d } = \frac { e }{ f } \), prove that each ratio is
(i) \(\sqrt { \frac { { 3a }^{ 2 }-{ 5c }^{ 2 }+{ 7e }^{ 2 } }{ { 3b }^{ 2 }-{ 5d }^{ 2 }+{ 7f }^{ 2 } } } \)
(ii) \({ \left[ \frac { { 2a }^{ 3 }+{ 5c }^{ 3 }+{ 7e }^{ 3 } }{ { 2b }^{ 3 }+{ 5d }^{ 3 }+{ 7f }^{ 3 } } \right] }^{ \frac { 1 }{ 3 } } \)
Solution:
\(\frac { a }{ b } = \frac { c }{ d } = \frac { e }{ f } \) = k(say)
∴ a = k, c = dk, e = fk

Question 14.
If \(\frac { x }{ a } = \frac { y }{ b } = \frac { z }{ c } \), prove that
\(\frac { { 3x }^{ 3 }-{ 5y }^{ 3 }+{ 4z }^{ 3 } }{ { 3a }^{ 3 }-{ 5b }^{ 3 }+{ 4c }^{ 3 } } ={ \left( \frac { 3x-5y+4z }{ 3a-5b+4c } \right) }^{ 3 }\)
Solution:
\(\frac { x }{ a } = \frac { y }{ b } = \frac { z }{ c } \) = k (say)
x = ak,y = bk,z = ck

Question 15.
If x : a = y : b, prove that
\(\frac { { x }^{ 4 }+{ a }^{ 4 } }{ { x }^{ 3 }+{ a }^{ 3 } } +\frac { { y }^{ 4 }+{ b }^{ 4 } }{ { y }^{ 3 }+{ b }^{ 3 } } =\frac { { \left( x+y \right) }^{ 4 }+{ \left( a+b \right) }^{ 4 } }{ { \left( x+y \right) }^{ 3 }+{ \left( a+b \right) }^{ 3 } } \)
Solution:
\(\frac { x }{ a } = \frac { y }{ b } \) = k (say)
x = ak, y = bk

Question 16.
If \(\frac { x }{ b+c-a } =\frac { y }{ c+a-b } =\frac { z }{ a+b-c } \) prove that each ratio’s equal to :
\(\frac { x+y+z }{ a+b+c } \)
Solution:
\(\frac { x }{ b+c-a } =\frac { y }{ c+a-b } =\frac { z }{ a+b-c } \) = k(say)
x = k(b + c – a),
y = k(c + a – b),
z = k(a + b – c)

Question 17.
If a : b = 9 : 10, find the value of
(i) \(\frac { 5a+3b }{ 5a-3b } \)
(ii) \(\frac { { 2a }^{ 2 }-{ 3b }^{ 2 } }{ { 2a }^{ 2 }+{ 3b }^{ 2 } } \)
Solution:
a : b = 9 : 10
=> \(\frac { a }{ b } = \frac { 9 }{ 10 }\)

Question 18.
If (3x² + 2y²) : (3x² – 2y²) = 11 : 9, find the value of \(\frac { { 3x }^{ 4 }+{ 25y }^{ 4 } }{ { 3x }^{ 4 }-{ 25y }^{ 4 } } \) ;
Solution:
\(\frac { { 3x }^{ 4 }+{ 25y }^{ 4 } }{ { 3x }^{ 4 }-{ 25y }^{ 4 } } =\frac { 11 }{ 9 } \)
Applying componendo and dividendo

Question 19.
If \(x=\frac { 2mab }{ a+b } \) , find the value of
\(\frac { x+ma }{ x-ma } +\frac { x+mb }{ x-mb } \)
Solution:
\(x=\frac { 2mab }{ a+b } \)
=> \(\frac { x }{ ma } +\frac { 2b }{ a+b } \)
Applying componendo and dividendo

Question 20.
If \(x=\frac { pab }{ a+b } \) ,prove that \(\frac { x+pa }{ x-pa } -\frac { x+pb }{ x-pb } =\frac { 2\left( { a }^{ 2 }-{ b }^{ 2 } \right) }{ ab } \)
Solution:
\(x=\frac { pab }{ a+b } \)
=> \(\frac { x }{ pa } +\frac { b }{ a+b } \)
Applying componendo and dividendo

Question 21.
Find x from the equation \(\frac { a+x+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } }{ a+x-\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } =\frac { b }{ x } \)
Solution:
\(\frac { a+x+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } }{ a+x-\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } =\frac { b }{ x } \)
Applying componendo and dividendo,

Question 22.
If \(x=\frac { \sqrt [ 3 ]{ a+1 } +\sqrt [ 3 ]{ a-1 } }{ \sqrt [ 3 ]{ a+1 } -\sqrt [ 3 ]{ a-1 } } \), prove that :
x³ – 3ax² + 3x – a = 0
Solution:
\(x=\frac { \sqrt [ 3 ]{ a+1 } +\sqrt [ 3 ]{ a-1 } }{ \sqrt [ 3 ]{ a+1 } -\sqrt [ 3 ]{ a-1 } } \)
Applying componendo and dividendo,

Question 23.
If \(\frac { by+cz }{ b^{ 2 }+{ c }^{ 2 } } =\frac { cz+ax }{ { c }^{ 2 }+{ a }^{ 2 } } =\frac { ax+by }{ { a }^{ 2 }+{ b }^{ 2 } } \), prove that each of these ratio is equal to \(\frac { x }{ a } =\frac { y }{ b } =\frac { z }{ c } \)
Solution:
\(\frac { by+cz }{ b^{ 2 }+{ c }^{ 2 } } =\frac { cz+ax }{ { c }^{ 2 }+{ a }^{ 2 } } =\frac { ax+by }{ { a }^{ 2 }+{ b }^{ 2 } } \)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Ratio and Proportion Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Factorization Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Factorization Chapter Test.

ML Aggarwal SolutionsICSE SolutionsSelina ICSE Solutions

Question 1.
Find the remainder when 2x³ – 3x² + 4x + 7 is divided by
(i) x – 2
(ii) x + 3
(iii) 2x + 1
Solution:
f(x) = 2x³ – 3x² + 4x + 7
(i) Let x – 2 = 0, then x = 2
Substituting value of x in f(x)
f(2) = 2 (2)³ – 3 (2)² + 4 (2) + 7
= 2 × 8 – 3 × 4 + 4 × 2 + 7
= 16 – 12 + 8 + 7 = 19
Remainder = 19 Ans.
(ii) Let x + 3 = 0, then x = – 3
Substituting the value of x in f(x)

Question 2.
When 2x³ – 9x² + 10x – p is divided by (x + 1), the remainder is – 24.Find the value of p.
Solution:
Let x + 1 = 0 then x = – 1,
Substituting the value of x in f(x)
f(x) = 2x³ – 9x² + 10x – p

Question 3.
If (2x – 3) is a factor of 6x² + x + a, find the value of a. With this value of a, factorise the given expression.
Solution:
Let 2 x – 3 = 0 then 2x = 3
=>x = \(\\ \frac { 3 }{ 2 } \)
Substituting the value of x in f(x)

Question 4.
When 3x² – 5x + p is divided by (x – 2), the remainder is 3. Find the value of p. Also factorise the polynomial 3x² – 5x + p – 3.
Solution:
f(x) = 3x² – 5x+ p
Let (x – 2) = 0, then x = 2
f(2) = 3 (2)² – 5(2) + p
= 3 x 4 – 10 + p
= 12 – 10 + p
= 2 + p

Question 5.
Prove that (5x + 4) is a factor of 5x³ + 4x² – 5x – 4. Hence factorise the given polynomial completely.
Solution:
f(x) = 5x³ + 4x² – 5x – 4
Let 5x + 4 = 0, then 5x = – 4
=> x = \(\\ \frac { -4 }{ 2 } \)

Question 6.
Use factor theorem to factorise the following polynomials completely:
(i) 4x³ + 4x² – 9x – 9
(ii) x³ – 19x – 30
Solution:
(i) f(x) = 4x³ + 4x² – 9x – 9
Let x = – 1,then
f( – 1) = 4 ( – 1)³ + 4 ( – 1)² – 9 ( – 1) – 9

Question 7.
If x³ – 2x² + px + q has a factor (x + 2) and leaves a remainder 9, when divided by (x + 1), find the values of p and q. With these values of p and q, factorise the given polynomial completely.
Solution:
f(x) = x³ – 2x² + px + q
(x + 2) is a factor
f( – 2) = ( – 2)³ – 2( – 2)² + p ( – 2) + q

Question 8.
If (x + 3) and (x – 4) are factors of x³ + ax² – bx + 24, find the values of a and b: With these values of a and b, factorise the given expression.
Solution:
f(x) = x³ + ax² – bx + 24
Let x + 3 = 0, then x = – 3
Substituting the value of x in f(x)

Question 9.
If 2x³ + ax² – 11x + b leaves remainder 0 and 42 when divided by (x – 2) and (x – 3) respectively, find the values of a and b. With these values of a and b, factorise the given expression.
Solution:
f(x) = 2x³ + ax² – 11 x + b
Let x – 2 = 0, then x = 2,
Substituting the vaue of x in f(x)

Question 10.
If (2x + 1) is a factor of both the expressions 2x² – 5x + p and 2x² + 5x + q, find the value of p and q. Hence find the other factors of both the polynomials.
Solution:
Let 2x + 1 = 0, then 2x = – 1
x = \(– \frac { 1 }{ 2 } \)
Substituting the value of x in

Question 11.
When a polynomial f(x) is divided by (x – 1), the remainder is 5 and when it is,, divided by (x – 2), the remainder is 7. Find – the remainder when it is divided by (x – 1) (x – 2).
Solution:
When f(x) is divided by (x – 1),
Remainder = 5
Let r – 1 = 0 => x = 1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Factorization Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test

ML Aggarwal SolutionsICSE SolutionsSelina ICSE Solutions

Solve the following equations (1 to 4) by factorisation :

Question 1.
(i) x² + 6x – 16 = 0
(ii) 3x² + 11x + 10 = 0
Solution:
x² + 6x – 16 = 0
=> x² + 8x – 2x – 16 = 0
x (x + 8) – 2 (x + 8) = 0

Question 2.
(i) 2x² + ax – a² = 0
(ii) √3x² + 10x + 7√3 = 0
Solution:
(i) 2x² + ax – a² = 0
=> 2x² + 2ax – ax – a² = 0

Question 3.
(i) x(x + 1) + (x + 2)(x + 3) = 42
(ii) \(\frac { 6 }{ x } -\frac { 2 }{ x-1 } =\frac { 1 }{ x-2 } \)
Solution:
(i) x(x + 1) + (x + 2)(x + 3) = 42
2x² + 6x + 6 – 42 = 0

Question 4.
(i)\(\sqrt { x+15 } =x+3 \)
(ii)\(\sqrt { { 3x }^{ 2 }-2x-1 } =2x-2\)
Solution:
(i)\(\sqrt { x+15 } =x+3 \)
Squaring on both sides
x + 15 = (x + 3)²

Solve the following equations (5 to 8) by using formula :

Question 5.
(i) 2x² – 3x – 1 = 0
(ii) \(x\left( 3x+\frac { 1 }{ 2 } \right) =6\)
Solution:
(i) 2x² – 3x – 1 = 0
Here a = 2, b = – 3, c = – 1

Question 6.
(i) \(\frac { 2x+5 }{ 3x+4 } =\frac { x+1 }{ x+3 } \)
(ii) \(\frac { 2 }{ x+2 } -\frac { 1 }{ x+1 } =\frac { 4 }{ x+4 } -\frac { 3 }{ x+3 } \)
Solution:
(i) \(\frac { 2x+5 }{ 3x+4 } =\frac { x+1 }{ x+3 } \)
(2x + 5)(x + 3) = (x + 1)(3x + 4)

Question 7.
(i) \(\frac { 3x-4 }{ 7 } +\frac { 7 }{ 3x-4 } =\frac { 5 }{ 2 } ,x\neq \frac { 4 }{ 3 } \)
(ii) \(\frac { 4 }{ x } -3=\frac { 5 }{ 2x+3 } ,x\neq 0,-\frac { 3 }{ 2 } \)
Solution:
(i) \(\frac { 3x-4 }{ 7 } +\frac { 7 }{ 3x-4 } =\frac { 5 }{ 2 } ,x\neq \frac { 4 }{ 3 } \)
let \(\frac { 3x-4 }{ 7 } \) = y,then

Question 8.
(i)x² + (4 – 3a)x – 12a = 0
(ii)10ax² – 6x + 15ax – 9 = 0,a≠0
Solution:
(i)x² + (4 – 3a)x – 12a = 0
Here a = 1,b = 4 – 3a,c = – 12a

Question 9.
Solve for x using the quadratic formula. Write your answer correct to two significant figures: (x – 1)² – 3x + 4 = 0. (2014)
Solution:
(x – 1)² – 3x + 4 = 0
x² + 1 – 2x – 3x + 4 = 0

Question 10.
Discuss the nature of the roots of the following equations:
(i) 3x² – 7x + 8 = 0
(ii) x² – \(\\ \frac { 1 }{ 2 } x\) – 4 = 0
(iii) 5x² – 6√5x + 9 = 0
(iv) √3x² – 2x – √3 = 0
Solution:
(i) 3x² – 7x + 8 = 0
Here a = 3, b = – 7,c = 8

Question 11.
Find the values of k so that the quadratic equation (4 – k) x² + 2 (k + 2) x + (8k + 1) = 0 has equal roots.
Solution:
(4 – k) x² + 2 (k + 2) x + (8k + 1) = 0
Here a = (4 – k), b = 2 (k + 2), c = 8k + 1

or k – 3 = 0, then k= 3
k = 0, 3 Ans.

Question 12.
Find the values of m so that the quadratic equation 3x² – 5x – 2m = 0 has two distinct real roots.
Solution:
3x² – 5x – 2m = 0
Here a = 3, b = – 5, c = – 2m

Question 13.
Find the value(s) of k for which each of the following quadratic equation has equal roots:
(i)3kx² = 4(kx – 1)
(ii)(k + 4)x² + (k + 1)x + 1 =0
Also, find the roots for that value (s) of k in each case.
Solution:
(i)3kx² = 4(kx – 1)
=> 3kx² = 4kx – 4
=> 3kx² – 4kx + 4 = 0

Question 14.
Find two natural numbers which differ by 3 and whose squares have the sum 117.
Solution:
Let first natural number = x
then second natural number = x + 3
According to the condition :
x² + (x + 3)² = 117

Question 15.
Divide 16 into two parts such that the twice the square of the larger part exceeds the square of the smaller part by 164.
Solution:
Let larger part = x
then smaller part = 16 – x
(∵ sum = 16)
According to the condition

Question 16.
Two natural numbers are in the ratio 3 : 4. Find the numbers if the difference between their squares is 175.
Solution:
Ratio in two natural numbers = 3 : 4
Let the numbers be 3x and 4x
According to the condition,

Question 17.
Two squares have sides A cm and (x + 4) cm. The sum of their areas is 656 sq. cm.Express this as an algebraic equation and solve it to find the sides of the squares.
Solution:
Side of first square = x cm .
and side of second square = (x + 4) cm
Now according to the condition,

or x – 16 = 0 then x = 16
Side of first square = 16 cm
and side of second square = 16 + 4 – 4
= 20 cm Ans.

Question 18.
The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.
Solution:
Let breadth = x m
then length = (x + 12) m
Area = l × b = x (x + 12) m²
and perimeter = 2 (l + b)
= 2(x + 12 + x) = 2 (2x + 12) m
According to the condition.

Question 19.
A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.
Solution:
Area of rectangular garden = 100 cm²
Length of barbed wire = 30 m
Let the length of side opposite to wall = x

Question 20.
The hypotenuse of a right angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.
Solution:
Let the length of shortest side = x m
Length of hypotenuse = 2x – 1
and third side = x + 1
Now according to the condition,

Question 21.
A wire ; 112 cm long is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.
Solution:
Perimeter of a right angled triangle = 112 cm
Hypotenuse = 50 cm
∴ Sum of other two sides = 112 – 50 = 62 cm
Let the length of first side = x
and length of other side = 62 – x

Question 22.
Car A travels x km for every litre of petrol, while car B travels (x + 5) km for every litre of petrol.
(i) Write down the number of litres of petrol used by car A and car B in covering a distance of 400 km.
(ii) If car A uses 4 litres of petrol more than car B in covering 400 km. write down an equation, in A and solve it to determine the number of litres of petrol used by car B for the journey.
Solution:
Distance travelled by car A in one litre = x km
and distance travelled by car B in one litre = (x + 5) km
(i) Consumption of car A in covering 400 km

Question 23.
The speed of a boat in still water is 11 km/ hr. It can go 12 km up-stream and return downstream to the original point in 2 hours 45 minutes. Find the speed of the stream
Solution:
Speed of boat in still water =11 km/hr
Let the speed of stream = x km/hr.
Distance covered = 12 km.
Time taken = 2 hours 45 minutes .

Question 24.
By selling an article for Rs. 21, a trader loses as much percent as the cost price of the article. Find the cost price.
Solution:
S.R of an article = Rs. 21
Let cost price = Rs. x
Then loss = x%

Question 25.
A man spent Rs. 2800 on buying a number of plants priced at Rs x each. Because of the number involved, the supplier reduced the price of each plant by Rupee 1.The man finally paid Rs. 2730 and received 10 more plants. Find x.
Solution:
Amount spent = Rs. 2800
Price of each plant = Rs. x
Reduced price = Rs. (x – 1)

Question 26.
Forty years hence, Mr. Pratap’s age will be the square of what it was 32 years ago. Find his present age.
Solution:
Let Partap’s present age = x years
40 years hence his age = x + 40
and 32 years ago his age = x – 32
According to the condition

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Quadratic Equations in One Variable Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

 

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Linear Inequations Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Linear Inequations Chapter Test.

ML Aggarwal SolutionsICSE SolutionsSelina ICSE Solutions

Question 1.
Solve the inequation : 5x – 2 ≤ 3(3 – x) where x ∈ { – 2, – 1, 0, 1, 2, 3, 4}. Also represent its solution on the number line.
Solution:
5x – 2 < 3(3 – x)
=> 5x – 2 ≤ 9 – 3x
=> 5x + 3x ≤ 9 + 2

Question 2.
Solve the inequations :
6x – 5 < 3x + 4, x ∈ I.
Solution:
6x – 5 < 3x + 4
6x – 3x < 4 + 5
=> 3x <9
=> x < 3
x∈I
Solution Set = { – 1, – 2, 2, 1, 0….. }

Question 3.
Find the solution set of the inequation
x + 5 < 2 x + 3 ; x ∈ R
Graph the solution set on the number line.
Solution:
x + 5 ≤ 2x + 3
x – 2 x ≤ 3 – 5
=> – x ≤ – 2
=> x ≥ 2

Question 4.
If x ∈ R (real numbers) and – 1 < 3 – 2x ≤ 7, find solution set and represent it on a number line.
Solution:
– 1 < 3 – 2x ≤ 7
– 1 < 3 – 2x and 3 – 2x ≤ 7
2 x < 3 + 1 and – 2x ≤ 7 – 3
2 x < 4 and – 2 x ≤ 4
x < 2 and – x ≤ 2
and x ≥ – 2 or – 2 ≤ x
x∈R
Solution set – 2 ≤ x < 2
Solution set on number line

Question 5.
Solve the inequation :
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 } ,x\in R\)
Solution:
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 } \)
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le \frac { 8 }{ 5 } +\frac { 3x-1 }{ 7 } \)

Question 6.
Find the range of values of a, which satisfy 7 ≤ – 4x + 2 < 12, x ∈ R. Graph these values of a on the real number line.
Solution:
7 < – 4x + 2 < 12
7 < – 4x + 2 and – 4x + 2 < 12

Question 7.
If x∈R, solve \(2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)
Solution:
\(2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)
\(2x-3\ge x+\frac { 1-x }{ 3 } \) and \(x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)

Question 8.
Find positive integers which are such that if 6 is subtracted from five times the integer then the resulting number cannot be greater than four times the integer.
Solution:
Let the positive integer = x
According to the problem,
5a – 6 < 4x
5a – 4x < 6 => x < 6
Solution set = {x : x < 6}
= { 1, 2, 3, 4, 5, 6} Ans.

Question 9.
Find three smallest consecutive natural numbers such that the difference between one-third of the largest and one-fifth of the smallest is atleast 3.
Solution:
Let first least natural number = x
then second number = x + 1
and third number = x + 2

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Linear Inequations Chapter Test are helpful to complete your math homework.

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