ICSE Class 10

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 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 7 Ratio and Proportion Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Chapter Test

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 7 Ratio and Proportion Chapter Test are helpful to complete your math homework.

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

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion MCQS

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 Ratio and Proportion MCQS

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Chapter Test

Choose the correct answer from the given options (1 to 10):

Question 1.
The ratio of 45 minutes to \(5 \frac { 3 }{ 4 } \) hours is
(a) 180:23
(b) 3:23
(c) 23:3
(d) 6:23
Solution:
ratio of 45 minutes to \(5 \frac { 3 }{ 4 } \) hours is
45 minutes to : \(5 \frac { 3 }{ 4 } \) hours

Question 2.
The ratio of 4 litres to 900 mL is
(a) 4 : 9
(b) 40 : 9
(c) 9 : 40
(d) 20 : 9
Solution:
4l : 900 ml
= 4000 ml : 900 ml
= 4000 : 900
= 40 : 9 (b)

Question 3.
When the number 210 is increased in the ratio 5 : 7, the the new number is
(a) 150
(b) 180
(c) 294
(d) 420
Solution:
210 is increased in the ratio 5 : 7, then
New increased number will be
= 210 × \(\\ \frac { 7 }{ 5 } \)
= 294 (c)

Question 4.
Two numbers are in the ratio 7 : 9. If the sum of the numbers is 288, then the smaller number is
(a) 126
(b) 162
(c) 112
(d) 144
Solution:
Ratio in two number = 7 : 9
Sum of numbers = 288
Sum of ratios = 7 + 9
= 16
Smaller number = \(\\ \frac { 288\times 7 }{ 16 } \)
= 126 (a)

Question 5.
A ratio equivalent to the ratio \(\\ \frac { 2 }{ 3 } \) : \(\\ \frac { 5 }{ 7 } \) is
(a) 4:6
(b) 5:7
(c) 15:14
(d) 14:15
Solution:
\(\\ \frac { 2 }{ 3 } \) : \(\\ \frac { 5 }{ 7 } \)
Multiply and divide \(\\ \frac { 2 }{ 3 } \) by 7 and
Multiply and divide \(\\ \frac { 5 }{ 7 } \) by 3

Question 6.
The ratio of number of edges of a cube to the number of its faces is
(a) 2 : 1
(b) 1 : 2
(c) 3 : 8
(d) 8 : 3
Solution:
No. of edges of the cube = 12
No. of faces = 6
Ratio in edges a cube to the number of faces = 12 : 6
= 2 : 1 (a)

Question 7.
If x, 12, 8 and 32 are in proportion, then the value of x is
(a) 6
(b) 4
(c) 3
(d) 2
Solution:
x, 12, 8, 32 are in proportion, then
x × 32 = 12 × 8 (∵ ad = bc)
⇒ x = \(\\ \frac { 12\times 8 }{ 32 } \) = 3
x = 3 (c)

Question 8.
The fourth proportional to 3, 4, 5 is
(a) 6
(b) \(\\ \frac { 20 }{ 3 } \)
(c) \(\\ \frac { 15 }{ 4 } \)
(d) \(\\ \frac { 12 }{ 5 } \)
Solution:
The fourth proportion to 3, 4, 5 will be
= \(\\ \frac { 4\times 5 }{ 3 } \)
= \(\\ \frac { 20 }{ 3 } \) (b)

Question 9.
The third proportional to \(6 \frac { 1 }{ 4 } \) and 5 is
(a) 4
(b) \(8 \frac { 1 }{ 2 } \)
(c) 3
(d) none of these
Solution:
The third proportional to \(6 \frac { 1 }{ 4 } \) and 5 is
⇒ \(6 \frac { 1 }{ 4 } \) : 5 :: 5 : x
⇒ \(\\ \frac { 25 }{ 4 } \) : 5 :: 5 : x
⇒ x = \(\\ \frac { 5\times 5 }{ 25 } \) × 4
⇒ 4 (a)

Question 10.
The mean proportional between \(\\ \frac { 1 }{ 2 } \) and 128 is
(a) 64
(b) 32
(c) 16
(d) 8
Solution:
The mean proportional between \(\\ \frac { 1 }{ 2 } \) and 128 is
= \(\sqrt { \frac { 1 }{ 2 } \times 128 } \)
= √64
= 8 (d)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion MCQS are helpful to complete your math homework.

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

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3

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 Ratio and Proportion Ex 7.3

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Chapter Test

Question 1.
If a : b : : c : d, prove that
(i) \(\frac { 2a+5b }{ 2a-5b } =\frac { 2c+5d }{ 2c-5d } \)
(ii) \(\frac { 5a+11b }{ 5c+11d } =\frac { 5a-11b }{ 5c-11d } \)
(iii) (2a + 3b)(2c – 3d) = (2a – 3b)(2c + 3d)
(iv) (la + mb) : (lc + mb) :: (la – mb) : (lc – mb)
Solution:
(i) a : b : : c : d
then \(\frac { a }{ b } =\frac { c }{ d } \)
⇒ \(\frac { 2a }{ 5b } =\frac { 2c }{ 5d } \) (multiply by \(\\ \frac { 2 }{ 5 } \) )

Question 2.
(i) If \(\frac { 5x+7y }{ 5u+7v } =\frac { 5x-7y }{ 5u-7v } \) , Show that \(\frac { x }{ y } =\frac { u }{ v } \)
(ii) \(\frac { 8a-5b }{ 8c-5d } =\frac { 8a+5b }{ 8c+5d } \) , prove that \(\frac { a }{ b } =\frac { c }{ d } \)
Solution:
(i) \(\frac { 5x+7y }{ 5u+7v } =\frac { 5x-7y }{ 5u-7v } \)
Applying alternendo \(\frac { 5x+7y }{ 5u+7v } =\frac { 5x-7y }{ 5u-7v } \)
Applying componendo and dividendo

Question 3.
If (4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d), prove that a, b, c, d are in proporton.
Solution:
(4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d)
⇒ \(\frac { 4a+5b }{ 4a-5b } =\frac { 4c+5d }{ 4c-5d } \)
Applying componendo and dividendo

Question 4.
If (pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd) prove that a : b : : c : d
Solution:
(pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd)
⇒ \(\frac { pa+qb }{ pc+qd } =\frac { pq-qb }{ pc-qd } \)
⇒ \(\frac { pa+qb }{ pc-qd } =\frac { pq+qb }{ pc-qd } \)
Applying componendo and dividendo
⇒ \(\frac { pa+qb+pa-qb }{ pa+qb-pa+qb } =\frac { pc+qd+pc-qd }{ pc-qd-pc+qd } \)

Question 5.
If (ma + nb): b :: (mc + nd) : d, prove that a, b, c, d are in proportion.
Solution:
(ma + nb): b :: (mc + nd) : d
⇒ \(\frac { ma+nb }{ b } =\frac { mc+nd }{ d } \)
⇒ mad + nbd = mbc + nbd
⇒ mad = mbc
⇒ ad = bc
⇒ \(\frac { a }{ b } =\frac { c }{ d } \)
Hence a : b :: c : d.

Question 6.
If (11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²), prove that a : b :: c : d.
Solution:
(11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²)
⇒ \(\frac { 11a+{ 13b }^{ 2 } }{ { 11a }^{ 2 }-{ 13b }^{ 2 } } =\frac { { 11c }^{ 2 }+{ 13d }^{ 2 } }{ { 11c }^{ 2 }-{ 13d }^{ 2 } } \)
Applying componendo and dividendo

Question 7.
If (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d), prove that a : b :: c : d.
Solution:
\(\frac { a + 3b + 2c + 6d }{ a – 3b + 2c – 6d } =\frac { a + 3b – 2c – 6d }{ a – 3b – 2c + 6d } \)
⇒ \(\frac { a + 3b + 2c + 6d }{ a + 3b – 2c – 6d } =\frac { a – 3b + 2c – 6d }{ a – 3b – 2c + 6d } \) (by altenendo)
Applying componendo and dividendo

Question 8.
If \(x=\frac { 2ab }{ a+b } \) find the value of \(\frac { x+a }{ x-a } +\frac { x+b }{ x-b } \)
Solution:
\(x=\frac { 2ab }{ a+b } \)
⇒ \(\frac { x }{ a } =\frac { 2b }{ a+b } \)
Applying componendo and dividendo,

Question 9.
If \(x=\frac { 8ab }{ a+b } \) find the value of \(\frac { x+4a }{ x-4a } +\frac { x+4b }{ x-4b } \)
Solution:
\(x=\frac { 8ab }{ a+b } \)
⇒ \(\frac { x }{ 4a } =\frac { 2b }{ a+b } \)
Applying componendo and dividendo,

Question 10.
If \(x=\frac { 4\sqrt { 6 } }{ \sqrt { 2 } +\sqrt { 3 } } \) find the value of \(\frac { x+2\sqrt { 2 } }{ x-2\sqrt { 2 } } +\frac { x+2\sqrt { 3 } }{ x-2\sqrt { 3 } } \)
Solution:
\(x=\frac { 4\sqrt { 6 } }{ \sqrt { 2 } +\sqrt { 3 } } \)
⇒ \(\frac { 4\sqrt { 2 } \times \sqrt { 3 } }{ \sqrt { 2 } +\sqrt { 3 } } \)

Question 11.
Solve \(x:\frac { \sqrt { 36x+1 } +6\sqrt { x } }{ \sqrt { 36x+1 } -6\sqrt { x } } =9 \)
Solution:
\(\frac { \sqrt { 36x+1 } +6\sqrt { x } }{ \sqrt { 36x+1 } -6\sqrt { x } } =\frac { 9 }{ 1 } \)
Applying componendo and dividendo,

Question 12.
Find x from the following equations :
(i) \(\frac { \sqrt { 2-x } +\sqrt { 2+x } }{ \sqrt { 2-x } -\sqrt { 2+x } } =3 \)
(ii) \(\frac { \sqrt { x+4 } +\sqrt { x-10 } }{ \sqrt { x+4 } -\sqrt { x-10 } } =\frac { 5 }{ 2 } \)
(iii) \(\frac { \sqrt { 1+x } +\sqrt { 1-x } }{ \sqrt { 1+x } -\sqrt { 1-x } } =\frac { a }{ b } \)
(iv) \(\frac { \sqrt { 12x+1 } +\sqrt { 2x-3 } }{ \sqrt { 12x+1 } -\sqrt { 2x-3 } } =\frac { 3 }{ 2 } \)
(v) \(\frac { 3x+\sqrt { { 9x }^{ 2 }-5 } }{ 3x-\sqrt { { 9x }^{ 2 }-5 } } =5 \)
(vi) \(\frac { \sqrt { a+x } +\sqrt { a-x } }{ \sqrt { a+x } -\sqrt { a-x } } =\frac { c }{ d } \)
Solution:
(i) \(\frac { \sqrt { 2-x } +\sqrt { 2+x } }{ \sqrt { 2-x } -\sqrt { 2+x } } =3 \)
Applying componendo and dividendo,

Question 13.
Solve \(\frac { 1+x+{ x }^{ 2 } }{ 1-x+{ x }^{ 2 } } =\frac { 62\left( 1+x \right) }{ 63\left( 1-x \right) } \)
Solution:
\(\frac { 1+x+{ x }^{ 2 } }{ 1-x+{ x }^{ 2 } } =\frac { 62\left( 1+x \right) }{ 63\left( 1-x \right) } \)
⇒ \(\frac { \left( 1-x \right) \left( 1+x+{ x }^{ 2 } \right) }{ \left( 1+x \right) \left( 1-x+{ x }^{ 2 } \right) } =\frac { 62 }{ 63 } \)
⇒ \(\frac { \left( 1+x \right) \left( 1-x+{ x }^{ 2 } \right) }{ \left( 1-x \right) \left( 1+x+{ x }^{ 2 } \right) } =\frac { 63 }{ 62 } \)

Question 14.
Solve for \(x:16{ \left( \frac { a-x }{ a+x } \right) }^{ 3 }=\frac { a+x }{ a-x } \)
Solution:
\(x:16{ \left( \frac { a-x }{ a+x } \right) }^{ 3 }=\frac { a+x }{ a-x } \)
⇒ \(\left( \frac { a+x }{ a-x } \right) \times { \left( \frac { a+x }{ a-x } \right) }^{ 3 }=16 \)

Question 15.
If \(x=\frac { \sqrt { a+x } +\sqrt { a-1 } }{ \sqrt { a+1 } -\sqrt { a-1 } } \) , using properties of proportion , show that x² – 2ax + 1 = 0
Solution:
We have \(x=\frac { \sqrt { a+x } +\sqrt { a-1 } }{ \sqrt { a+1 } -\sqrt { a-1 } } \)
⇒ \(\frac { x+1 }{ x-1 } =\frac { 2\sqrt { a+1 } }{ 2\sqrt { a-1 } } \)

Question 16.
Given \(x=\frac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } } +\sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } -\sqrt { { a }^{ 2 }-{ b }^{ 2 } } } \) Use componendo and dividendo to prove that \({ b }^{ 2 }=\frac { { 2a }^{ 2 }x }{ { x }^{ 2 }+1 } \)
Solution:
If \(\frac { x }{ 1 } =\frac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } } +\sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } -\sqrt { { a }^{ 2 }-{ b }^{ 2 } } } \)
Applying componendo and dividendo both sides

Question 17.
Given that \(\frac { { a }^{ 3 }+3{ ab }^{ 2 } }{ { b }^{ 3 }+{ 3a }^{ 2 }b } =\frac { 63 }{ 62 } \). Using componendo and dividendo find a: b. (2009)
Solution:
Given that \(\frac { { a }^{ 3 }+3{ ab }^{ 2 } }{ { b }^{ 3 }+{ 3a }^{ 2 }b } =\frac { 63 }{ 62 } \)
By componendo and dividendo

a : b = 3 : 2

Question 18.
Give \(\frac { { x }^{ 3 }+12x }{ { 6x }^{ 2 }+8 } =\frac { { y }^{ 3 }+27y }{ { 9y }^{ 2 }+27 } \) Using componendo and dividendo find x : y.
Solution:
Give \(\frac { { x }^{ 3 }+12x }{ { 6x }^{ 2 }+8 } =\frac { { y }^{ 3 }+27y }{ { 9y }^{ 2 }+27 } \)
Using componendo-dividendo, we have

Question 19.
Using the properties of proportion, solve the following equation for x; given
\(\frac { x^{ 3 }+3x }{ { 3x }^{ 2 }+1 } =\frac { 341 }{ 91 } \)
Solution:
\(\frac { x^{ 3 }+3x }{ { 3x }^{ 2 }+1 } =\frac { 341 }{ 91 } \)
Applying componendo and dividendo

Question 20.
If \(\frac { x+y }{ ax+by } =\frac { y+z }{ ay+bz } =\frac { z+x }{ az+bx } \) , prove that each of these ratio is equal to \(\\ \frac { 2 }{ a+b } \) unless x + y + z = 0
Solution:
\(\frac { x+y }{ ax+by } =\frac { y+z }{ ay+bz } =\frac { z+x }{ az+bx } \)
= \(\frac { x+y+y+z+z+x }{ ax+by+ay+bz+az+bx } \)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 are helpful to complete your math homework.

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2

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 Ratio and Proportion Ex 7.2

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Chapter Test

Question 1.
Find the value of x in the following proportions :
(i) 10 : 35 = x : 42
(ii) 3 : x = 24 : 2
(iii) 2.5 : 1.5 = x : 3
(iv) x : 50 :: 3 : 2
Solution:
(i) 10 : 35 = x : 42
⇒ 35 × x = 10 × 42

Question 2.
Find the fourth proportional to
(i) 3, 12, 15
(ii) \(\frac { 1 }{ 3 } ,\frac { 1 }{ 4 } ,\frac { 1 }{ 5 } \)
(iii) 1.5, 2.5, 4.5
(iv) 9.6 kg, 7.2 kg, 28.8 kg
Solution:
(i) Let fourth proportional to
3, 12, 15 be x.
then 3 : 12 :: 15 : x

Question 3.
Find the third proportional to
(i) 5, 10
(ii) 0.24, 0.6
(iii) Rs. 3, Rs. 12
(iv) \(5 \frac { 1 }{ 4 } \) and 7.
Solution:
(i) Let x be the third proportional to 5, 10,
then 5 : 10 :: 10 : x

Question 4.
Find the mean proportion of:
(i) 5 and 80
(ii) \(\\ \frac { 1 }{ 12 } \) and \(\\ \frac { 1 }{ 75 } \)
(iii) 8.1 and 2.5
(iv) (a – b) and (a³ – a²b), a> b
Solution:
(i) Let x be the mean proportion of 5 and 80 ,
then 5 : x : : x : 80
x² = 5 x 80
⇒ x = \(\sqrt { 5\times 80 } =\sqrt { 400 } \) = 20
x = 20

Question 5.
If a, 12, 16 and b are in continued proportion find a and b.
Solution:
∵ a, 12, 16, b are in continued proportion, then

Question 6.
What number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion ? (2009)
Solution:
Let x be added to 5, 11, 19 and 37 to make them in proportion.
5 + x : 11 + x : : 19 + x : 37 + x

Question 7.
What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion ? (2004)
Solution:
Let x be subtracted from each term, then
23 – x, 30 – x, 57 – x and 78 – x are proportional
23 – x : 30 – x : : 57 – x : 78 – x

Question 8.
If 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion, find the value of x.
Solution:
∵ 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion.
then (2x – 1) (15x – 9) = (5x – 6) (6x + 2)

Question 9.
If x + 5 is the mean proportion between x + 2 and x + 9, find the value of x.
Solution:
∵ x + 5 is the mean proportion between x + 2 and x + 9, then
(x + 5)² = (x + 2) (x + 9)
⇒ x² + 10x + 25 = x² + 11x + 18
⇒ x² + 10x – x² – 11x = 18 – 25
⇒ – x = – 7
∵ x = 7 Ans.

Question 10.
What number must be added to each of the numbers 16, 26 and 40 so that the resulting numbers may be in continued proportion?
Solution:
Let x be added to each number then
16 + x, 26 + x and 40 + x
are in continued proportion.

Question 11.
Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.
Solution:
Let the two numbers are a and b.
∵ 28 is the mean proportional
∵ a : 28 : : 28 : b

Question 12.
If b is the mean proportional between a and c, prove that a, c, a² + b², and b² + c² are proportional.
Solution:
∵ b is the mean proportional between a and c, then,
b² = a × c ⇒ b² = ac …(i)

Question 13.
If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between (a² + b²) and (b² + c²).
Solution:
b is the mean proportional between a and c then
b² = ac …(i)
Now if (ab + bc) is the mean proportional

Question 14.
If y is mean proportional between x and z, prove that
xyz (x + y + z)³ = (xy + yz + zx)³.
Solution:
∵ y is the mean proportional between
x and z, then
y² = xz …(i)

Question 15.
If a + c = mb and \(\frac { 1 }{ b } +\frac { 1 }{ d } =\frac { m }{ c } \), prove that a, b, c and d are in proportion.
Solution:
a + c = mb and \(\frac { 1 }{ b } +\frac { 1 }{ d } =\frac { m }{ c } \)
a + c = mb

Question 16.
If \(\frac { x }{ a } =\frac { y }{ b } =\frac { z }{ c } \), prove that
(i)\(\frac { { x }^{ 3 } }{ { a }^{ 2 } } +\frac { { y }^{ 3 } }{ { b }^{ 2 } } +\frac { { z }^{ 3 } }{ { c }^{ 2 } } =\frac { { \left( x+y+z \right) }^{ 3 } }{ { \left( a+b+c \right) }^{ 2 } } \)

Solution:
\(\frac { x }{ a } =\frac { y }{ b } =\frac { z }{ c } \)
∴ x = ak, y = bk, z = ck

Question 17.
If \(\frac { a }{ b } =\frac { c }{ d } =\frac { e }{ f } \) prove that :
(i) (b² + d² + f²) (a² + c² + e²) = (ab + cd + ef)²

Solution:
\(\frac { a }{ b } =\frac { c }{ d } =\frac { e }{ f } \) = k(say)
∴ a = bk, c = dk, e =fk

Question 18.
If ax = by = cz; prove that
\(\frac { { x }^{ 2 } }{ yz } +\frac { { y }^{ 2 } }{ zx } +\frac { { z }^{ 2 } }{ xy } \) = \(\frac { bc }{ { a }^{ 2 } } +\frac { ca }{ { b }^{ 2 } } +\frac { ab }{ { c }^{ 2 } } \)
Solution:
Let ax = by = cz = k

Question 19.
If a, b, c and d are in proportion, prove that:
(i) (5a + 7b) (2c – 3d) = (5c + 7d) (2a – 3b)
(ii) (ma + nb) : b = (mc + nd) : d
(iii) (a⁴ + c⁴) : (b⁴ + d⁴) = a² c² : b² d².

Solution:
∵ a, b, c, d are in proportion
\(\\ \frac { a }{ b } \) = \(\\ \frac { c }{ d } \) = k(say)

Question 20.
If x, y, z are in continued proportion, prove that:\(\frac { { \left( x+y \right) }^{ 2 } }{ { \left( y+z \right) }^{ 2 } } =\frac { x }{ z } \). (2010)
Solution:
x, y, z are in continued proportion
Let \(\\ \frac { x }{ y } \) = \(\\ \frac { y }{ z } \) = k

Question 21.
If a, b, c are in continued proportion, prove that:
\(\frac { { pa }^{ 2 }+qab+{ rb }^{ 2 } }{ { pb }^{ 2 }+qbc+{ rc }^{ 2 } } =\frac { a }{ c } \)
Solution:
Given a, b, c are in continued proportion
\(\frac { { pa }^{ 2 }+qab+{ rb }^{ 2 } }{ { pb }^{ 2 }+qbc+{ rc }^{ 2 } } =\frac { a }{ c } \)
Let \(\\ \frac { a }{ b } \) = \(\\ \frac { b }{ c } \) = k

Question 22.
If a, b, c are in continued proportion, prove that:
(i) \(\frac { a+b }{ b+c } =\frac { { a }^{ 2 }(b-c) }{ { b }^{ 2 }(a-b) } \)

Solution:
As a, b, c, are in continued proportion
Let \(\\ \frac { a }{ b } \) = \(\\ \frac { b }{ c } \) = k

Question 23.
If a, b, c, d are in continued proportion, prove that:
(i) \(\frac { { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } }{ { b }^{ 3 }+{ c }^{ 3 }+{ d }^{ 3 } } =\frac { a }{ d } \)

Solution:
a, b, c, d are in continued proportion
∴ \(\frac { a }{ b } =\frac { b }{ c } =\frac { c }{ d } =k(say)\)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2 are helpful to complete your math homework.

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

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1

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 Ratio and Proportion Ex 7.1

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Chapter Test

Question 1.
An alloy consists of \(27 \frac { 1 }{ 2 } \) kg of copper and \(2 \frac { 3 }{ 4 } \) kg of tin. Find the ratio by weight of tin to the alloy
Solution:
Copper = \(27 \frac { 1 }{ 2 } \) kg = \(\\ \frac { 55 }{ 2 } \) kg,
Tin = \(2 \frac { 3 }{ 4 } \) kg = \(\\ \frac { 11 }{ 4 } \) kg

Question 2.
Find the compounded ratio of:
(i) 2 : 3 and 4 : 9
(ii) 4 : 5, 5 : 7 and 9 : 11
(iii) (a – b) : (a + b), (a + b)² : (a² + b²) and (a⁴ – b⁴) : (a² – b²)²
Solution:
(i) 2 : 3 and 4 : 9
Compound ratio = \(\\ \frac { 2 }{ 3 } \) x \(\\ \frac { 4 }{ 9 } \)
= \(\\ \frac { 8 }{ 27 } \) or 8 : 27
(ii) 4 : 5, 5 : 7 and 9 : 11

Question 3.
Find the duplicate ratio of
(i) 2 : 3
(ii) √5 : 7
(iii) 5a : 6b
Solution:
(i) Duplicate ratio of 2 : 3 = (2)² : (3)² = 4 : 9
(ii) Duplicate ratio of √5 : 7 = (√5)² : (7)² = 5 : 49
(iii) Duplicate ratio of 5a : 6b = (5a)² : (6b)² = 25a² : 36b²

Question 4.
Find the triplicate ratio of
(i) 3 : 4
(ii) \(\\ \frac { 1 }{ 2 } \) : \(\\ \frac { 1 }{ 3 } \)
(iii) 1³ : 2³
Solution:
(i) Triplicate ratio of 3 : 4
= (3)³ : (4)³
= 27 : 64

Question 5.
Find the sub-duplicate ratio of
(i) 9 : 16
(ii) \(\\ \frac { 1 }{ 4 } \) : \(\\ \frac { 1 }{ 9 } \),
(iii) 9a² : 49b²
Solution:
(i) Sub-duplicate ratio of 9 : 16
= √9 : √16
= 3 : 4

Question 6.
Find the sub-triplicate ratio of
(i) 1 : 216
(ii) \(\\ \frac { 1 }{ 8 } \) : \(\\ \frac { 1 }{ 125 } \)
(iii) 27a³ : 64b³
Solution:
(i) Sub-triplicate ratio of 1 : 216
= \(\sqrt [ 3 ]{ 1 } :\sqrt [ 3 ]{ 216 } \)

Question 7.
Find the reciprocal ratio of
(i) 4 : 7
(ii) 3² : 4²
(iii) \(\frac { 1 }{ 9 } :2 \)
Solution:
(i) Reciprocal ratio of 4 : 7 = 7 : 4
(ii) Reciprocal ratio of 3² : 4² = 4² : 3² = 16 : 9
(iii) Reciprocal ratio of \(\frac { 1 }{ 9 } :2 \) = \(2:\frac { 1 }{ 9 } \) = 18 : 1

Question 8.
Arrange the following ratios in ascending order of magnitude:
2 : 3, 17 : 21, 11 : 14 and 5 : 7
Solution:
Writing the given ratios in fraction
\(\frac { 2 }{ 3 } ,\frac { 17 }{ 21 } ,\frac { 11 }{ 14 } ,\frac { 5 }{ 7 } \)

Question 9.
(i) If A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7, find A : D
(ii) If x : y = 2 : 3, and y : z = 4 : 7, find x : y : z
Solution:
Let A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7
\(\frac { A }{ B } =\frac { 2 }{ 3 } ,\frac { B }{ C } =\frac { 4 }{ 5 } ,\frac { C }{ D } =\frac { 6 }{ 7 } \)

Question 10.
(i) If A: B = \(\frac { 1 }{ 4 } :\frac { 1 }{ 5 } \) and B : C = \(\frac { 1 }{ 7 } :\frac { 1 }{ 6 } \), find A : B : C.
(ii) If 3A = 4B = 6C, find A : B : C
Solution:
A : B = \(\frac { 1 }{ 4 } \times \frac { 5 }{ 1 } =\frac { 5 }{ 4 } \)
B : C = \(\frac { 1 }{ 7 } \times \frac { 6 }{ 1 } =\frac { 6 }{ 7 } \)

Question 11.
(i) If \(\frac { 3x+5y }{ 3x-5y } =\frac { 7 }{ 3 } \) , Find x : y
(ii) ) If a : b = 3 : 11, find (15a – 3b) : (9a + 5b). a
Solution:
(i) \(\frac { 3x+5y }{ 3x-5y } =\frac { 7 }{ 3 } \)
⇒ 9x + 15y = 21x – 35y [By cross multiplication]
⇒ 21x – 9x = 15y + 35y

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Question 12.
(i) If (4x² + xy) : (3xy – y²) = 12 : 5, find (x + 2y) : (2x + y).
(ii) If y (3x – y) : x (4x + y) = 5 : 12. Find (x² + y²) : (x + y)².
Solution:
(4x² + xy) : (3xy – y²) = 12 : 5
⇒ \(\frac { { 4x }^{ 2 }+xy }{ 3xy-{ y }^{ 2 } } =\frac { 12 }{ 5 } \)
⇒ 20x² + 5xy = 36xy – 12y²

Question 13.
(i) If (x – 9) : (3x + 6) is the duplicate ratio of 4 : 9, find the value of x.
(ii) If (3x + 1) : (5x + 3) is the triplicate ratio of 3 : 4, find the value of x.
(iii) If (x + 2y) : (2x – y) is equal to the duplicate ratio of 3 : 2, find x : y.
Solution:
(i) \(\frac { x-9 }{ 3x+6 } ={ \left( \frac { 4 }{ 9 } \right) }^{ 2 }\)
⇒ \(\frac { x-9 }{ 3x+6 } =\frac { 16 }{ 81 } \)

Question 14.
(i) Find two numbers in the ratio of 8 : 7 such that when each is decreased by \(12 \frac { 1 }{ 2 } \), they are in the ratio 11 : 9.
(ii) The income of a man is increased in the ratio of 10 : 11. If the increase in his income is Rs 600 per month, find his new income.
Solution:
(i) The ratio is 8 : 7
Let the numbers be 8x and 7x,
According to condition,

Question 15.
(i) A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 91 kg.
(ii) A school collected Rs 2100 for charity. It was decided to divide the money between an orphanage and a blind school in the ratio of 3 : 4. How much money did each receive?
Solution:
(i) Ratio between the original weight and reduced weight = 7 : 5
Let original weight = 7x
then reduced weight = 5x
If original weight = 91 kg.

Question 16.
(i) The sides of a triangle are in the ratio 7 : 5 : 3 and its perimeter is 30 cm. Find the lengths of sides.
(ii) If the angles of a triangle are in the ratio 2 : 3 : 4, find the angles.
Solution:
(i) Perimeter of a triangle = 30 cm.
Ratio among sides = 7 : 5 : 3
Sum of ratios 7 + 5 + 3 = 15

Question 17.
Three numbers are in the ratio \(\frac { 1 }{ 2 } :\frac { 1 }{ 3 } :\frac { 1 }{ 4 } \) If the sum of their squares is 244, find the numbers.
Solution:
The ratio of three numbers \(\frac { 1 }{ 2 } :\frac { 1 }{ 3 } :\frac { 1 }{ 4 } \)
= \(\frac { 6:4:3 }{ 12 } \)
= 6 : 4 : 3
Let first number 6x, second 4x and third 3x
.’. According to the condition
(6x)² + (4x)² + (3x)² = 244
⇒ 36x² + 16x² + 9x² = 244

Question 18.
(i) A certain sum was divided among A, B and C in the ratio 7 : 5 : 4. If B got Rs 500 more than C, find the total sum divided.
(ii) In a business, A invests Rs 50000 for 6 months, B Rs 60000 for 4 months and C, Rs 80000 for 5 months. If they together earn Rs 18800 find the share of each.
Solution:
(i) Ratio between A, B and C = 7 : 5 : 4
Let A’s share = 7x
B’s share = 5x
and C’s share = 4x
Total sum = 7x + 5x + 4x = 16x

Question 19.
(i) In a mixture of 45 litres, the ratio of milk to water is 13 : 2. How much water must be added to this mixture to make the ratio of milk to water as 3 : 1 ?
(ii) The ratio of the number of boys to the number of girls in a school of 560 pupils is 5 : 3. If 10 new boys are admitted, find how many new girls may be admitted so that the ratio of the number of boys to the number of girls may change to 3 : 2.
Solution:
(i) Mixture of milk and water = 45 litres
Ratio of milk and water =13 : 2
Sum of ratio = 13 + 2 = 15

Question 20.
(i) The monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each saves Rs 80 every month, find their monthly pocket money.
(ii) In class X of a school, the ratio of the number of boys to that of the girls is 4 : 3. If there were 20 more boys and 12 less girls, then the ratio would have been 2 : 1, How many students were there in the class?
Solution:
(i) Let the monthly pocket money of Ravi and Sanjeev be 5x and 7x respectively.
Also, let their expenditure be 3y and 5y respectively.

Question 21.
In an examination, the ratio of passes to failures was 4 : 1. If 30 less had appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. How many students appeared for the examination
Solution:
Let the number of passes = 4x
and number of failures = x
The total number of students appeared = 4x + x = 5x
In the second case, the number of students appeared = 5x – 30
and number of passes = 4x – 20

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1 are helpful to complete your math homework.

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