Selina Concise Class 10 Math Chapter 7 Ratio and Proportion Solutions
Exercise – 7(B)
(Q1) Find the fourth proportional to
(i) 1.5, 4.5 and 3.5 (ii) 3a, 6a2 and 2ab2
= Solution:
(i) Let us assume that the fourth proportional to 1.5, 4.5 and be x
∴ 1.5/4.5 = 3.5/x
x = 3.5 × 4.5/1.5
x = 10.5
(ii) 3a, 6a2 and 2ab2 be x
3a/6a2 and 2ab2/x
½ = 2ab2/x
x = 2ab2 × (2a)
x = 4a2b2
(Q2) Find the third proportional to
(i) 2 2/3 and 4 (ii) a – b and a2 – b2
= Solution:
Let us consider the third proportional to 2 2/3 and 4 be x.
∴ 2 2/3, 4, x
(6+2/3)/(4/1) = 4/x
8/3 × 1/4 = 4/x
2/3 = 4/x
x = 4 × 3/2
x = 12/2
x = 6
(ii) a – b, a2 – b2, x
= Solution:
a – b/a2 – b2 = a2 – b2/x
a – b/(a + b) (a-b) = a2 – b2/x (∵ a2 – b2 = (a + b) (a – b)
1/(a + b) = a2 – b2/x
x = (a + b) (a2 – b2)
(Q3) Find the mean proportional between:
(i) 6 + 3√3 and 8 – 4√3
(ii) a – b and a3 – a2b
= Solution:
Let us consider the mean proportional between 6 + 3√3, and 8 – 4√3 be x.
6 + 3√3, x, 8 – 4√3
6 + 3√3/x = x/8 – 4√3
(6 + 3√3) (8 – 4√3) = x2
6 (8 – 4√3) + 3√3 (8 – 4√3) = x2
48 – 24√3 + 24√3 – 12 × (√3×√3) = x2
48 – 24√3 + 24√3 – 12×3 = x2
48 – 24√3 + 24√3 – 36 = x2
48 – 36 = x2
12 = x2
∴ x = √12
x = √4×3
x = 2√3
(ii) a – b, x, a3 – a2b
a-b/x = x/a3 – a2b
(a – b) (a3 – a2b) = x2
(a – b) a2 (a – b) = x2
a2 (a – b)2 = x2
Taking square root on both sides,
√a2 (a – b)2 = x
∴ x = a (a – b)
(Q4) If x + 5 is the mean proportional between x + 2 and x + 9; Find the value of x.
Solution:
Given that x + 5 is the mean proportional between x + 2 and x + 9.
x + 2, x + 5, x + 9
x + 2/x + 5 = x + 5/x + 9
(x + 2) (x + 9) = (x + 5) (x + 5)
x (x + 9) + 2 (x + 9) = x (x + 5) + 5 (x + 5)
x2 + 9x + 2x + 18 = x2 + 5x + 5x + 25
The term x2 are same on both sides so it’s cancel.
9x + 2x + 18 = 5x + 5x + 25
11x + 18 = 10x + 25
11x – 10x = 25 – 18
x = 7
(Q5) If x2, 4 and 9 are in continued proportion, find x.
= Solution:
Given that x2, 4 and 9 are in continued proportion.
x2/4 = 4/9
x2 = 4×4/9
x2 = 16/9
Talking square root on both sides,
√x2 = √16/9
x = 4/3
(Q6) What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?
= Solution:
Let, x be aded to each of the number be
(6 + x), (15 + x), (20 + x) and (43 + x)
6 + x/15 + x = 20 + x/43+x
(6 + x) (43 + x) = (15 + x) (20 + x)
6 (43 + x) + x (43 + x) = 15 (20 + x) + x (20 + x)
258 + 6x + 43x + x2 = 300 + 15x + 20x + x2
258 + 6x + 43 = 300 + 15x + 20x
258 + 49x = 300 + 35x
49x – 35x = 300- 258
14x = 42
x = 42/14
x = 3
∴ 6 + x = 6 + 3 = 9
15 + x = 15 + 3 = 18
20 + x = 20 + 3
43 + x = 43 + 3 = 46
9/18 = 23/46
3/6 = ½
½ = ½
(Q7) (i) If a, b, c are in continued proportion,
a2 + b2/b(a + c) = b(a + c)/b2 + c2 show that
= Solution:
Given that a, b, c are in continued proportion.
a/b = b/c
b2 = ac —– (1)
To show: a2 + b2/b (a + c) = b(a + c)/b2 + c2
∴ L.H.S. = a2 + b2/b (a + c)
From equation (1)
= a2+ ac/b(a + c)
= a (a + c)/b (a + c)
= a/b
R.H.S = b(a + c)/b2 + c2
= b (a + c)/ac + c2
= b (a + c)/c (a + c)
= b/c
∴ L.H.S = R.H.S Hence proved.
(ii) If a, b, c are in continued proportion and a(b –c) = 2b, prove that: a-c = 2(a + b)/a
= Solution:
Given a, b, c are in continued proportion.
∴ a/b = b/c
= b2 = ac ———- (1)
a (b – c) = 2b
a (b – c)/b = 2 ——- (2)
Prove that: a – c = 2(a + b)/a
∴ R.H.S = 2(a + b)/a
= a (b – c)/b × (a + b)/a (From (2))
= (b – c)/b × (a + b)
= ab + b2 – ac – bc/b
R.H.S = ab + ac – ac – bc/b
= ab – bc/b (Using equation (1))
= b (a – c)/b
R.H.S = a – c
∴ L.H.S. = R.H.S.
a – c = 2(a + b)/a Hence proved
(iii) If a/b = c/d, a3c + ac3/b3d + bd3 = (a + c)4/(a + b)4
= Solution:
Let, a/b = c/d = P
∴ a/b = P, c/d = P
a = bp, c = dp
L.H.S = a3c + ac3/b3 d + bd3
= (bp)3 (dp) + (bp) (dp)3/b3d + bd3
= b3p3 dp + bp d3 p3/b3d + bd3
= b3 d p4 + bd3 p4/b3d + bd3
= p4 (b3d + bd3)/(b3d + bd3 )
= P4
R.H.S. = (a + c)4/(b + d)4
= (bp) + (dp)4/(b + d)4
= p4 b4 + d4 p4/(b + d)4
= P4 (b4 + d4)/(b + d)4
R.H.S = p4 (b + d)4/(b + d)4
= p4
∴ L.H.S = R.H.S Hence proved.
Here is your solution of Selina Concise Class 10 Math Chapter 7 Ratio and Proportion Exercise 7B Solutions
Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.