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

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 \(\\ \frac { 2 }{ 5 } \) )
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q1.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q1.3

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
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q2.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q2.2

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
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q3.1

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
⇒ \(\frac { pa+qb+pa-qb }{ pa+qb-pa+qb } =\frac { pc+qd+pc-qd }{ pc-qd-pc+qd } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q4.1

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
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q6.1

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
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q7.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q7.2

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,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q8.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q8.2

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,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q9.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q9.2

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:
\(x=\frac { 4\sqrt { 6 } }{ \sqrt { 2 } +\sqrt { 3 } } \)
⇒ \(\frac { 4\sqrt { 2 } \times \sqrt { 3 } }{ \sqrt { 2 } +\sqrt { 3 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q10.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q10.2

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,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q11.1

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,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q12.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q12.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q12.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q12.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q12.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q12.6
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q12.7
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q12.8
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q12.9

Question 13.
Solve \(\frac { 1+x+{ x }^{ 2 } }{ 1-x+{ x }^{ 2 } } =\frac { 62\left( 1+x \right) }{ 63\left( 1-x \right) } \)
Solution:
\(\frac { 1+x+{ x }^{ 2 } }{ 1-x+{ x }^{ 2 } } =\frac { 62\left( 1+x \right) }{ 63\left( 1-x \right) } \)
⇒ \(\frac { \left( 1-x \right) \left( 1+x+{ x }^{ 2 } \right) }{ \left( 1+x \right) \left( 1-x+{ x }^{ 2 } \right) } =\frac { 62 }{ 63 } \)
⇒ \(\frac { \left( 1+x \right) \left( 1-x+{ x }^{ 2 } \right) }{ \left( 1-x \right) \left( 1+x+{ x }^{ 2 } \right) } =\frac { 63 }{ 62 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q13.1

Question 14.
Solve for \(x:16{ \left( \frac { a-x }{ a+x } \right) }^{ 3 }=\frac { a+x }{ a-x } \)
Solution:
\(x:16{ \left( \frac { a-x }{ a+x } \right) }^{ 3 }=\frac { a+x }{ a-x } \)
⇒ \(\left( \frac { a+x }{ a-x } \right) \times { \left( \frac { a+x }{ a-x } \right) }^{ 3 }=16 \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q14.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q14.2

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 \(x=\frac { \sqrt { a+x } +\sqrt { a-1 } }{ \sqrt { a+1 } -\sqrt { a-1 } } \)
⇒ \(\frac { x+1 }{ x-1 } =\frac { 2\sqrt { a+1 } }{ 2\sqrt { a-1 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q15.1

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
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q16.1

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
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q17.1
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
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q18.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q18.2

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
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q19.1

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:
\(\frac { x+y }{ ax+by } =\frac { y+z }{ ay+bz } =\frac { z+x }{ az+bx } \)
= \(\frac { x+y+y+z+z+x }{ ax+by+ay+bz+az+bx } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 Q20.1

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