Selina Concise Class 10 Math Chapter 7 Ratio and Proportion Solutions
Exercise 7 (C)
(Q1) If a:b = c:d, prove that:
(i) 5a + 7b : 5a – 7b = 5c + 7d : 5c – 7d
(ii) (9a + 13b) (9c – 13d) = (9c + 13d (9a – 13b)
(iii) xa + yb : xc + yd = b:d
= Solution:
Given that a/b = c/d
(i) To Prove: 5a + 7b/5a – 7b = 5c + 7d/5c – 7d
Proof: From given,
a/b = c/d
Multiply by ‘5’ and divide by ‘7’ on both sides,
5a/7b = 5c/7d
Apply componendo and dividendo,
5a + 7b/5a – 7b = 5c + 7d/5c – 7d Hence Proved.
(ii) To prove: (9a + 13b) (9c – 13d) = (9c + 13d) (9a – 13b)
Proof = given a/b = c/d
Multiply on both side by 9/13.
9a/13b = 9C/13d
Apply componendo and dividendo,
9a + 13b/9a – 13b = 9c + 13d/9c- 13d
(9a+ 13b) (9c – 13d) = (9c + 13d) (9a – 13d)
Hence proved
(iii) To prove: xa + yb/xc + yd = b/d
Proof: Given,
a/b = c/d
Multiply on both side by x/y
xa/yb = xc/yd
Apply componendo
xa + yb/yb = xc + yd/yd
xa + yb/xc + yd = yb/yd
xa + yb/xc + yd = b/d (Hence proved)
(Q2) If a:b = c;, prove that
(6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b)
= Solution:
Given that,
a/b = c/d
To prove: (6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b)
Proof: Given that,
a/b = c/d
Multiply on both side by 6/7
6a/7b = 6c/7d
Apply componenedo,
6a + 7b/7b = 6c + 7d/7d
6a + 7b/6c + 7d = 7b/7d
6a + 7b/6c + 7d = b/d —– (i)
Now, a/b = c/d
Multiply on both side by 3/4,
3a/4b = 3c/4d
Apply dividendo,
3a – 4b/4b = 3c – 4d/4d
3a – 4b/3c – 4d = 4b/4d
∴ 3a – 4b/3c – 4d = b/d —– (ii)
From equation (i) and (ii),
We get,
6a + 7b/6c + 7d = 3a – 4b/3c – 4d
Apply Alternnendo,
(6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b) (Hence proved)
(Q3) Given, a/b = c/d, prove that
(3a – 5b)/(3a + 5b) = (3c – 5d)/(3c + 5d)
= Solution:
Given that,
a/b = c/d
Multiply on both side by 3/5
3a/5b = 3c/5d
Apply componendo and dividendo
3a + 5b/3a – 5b = 3c + 5d/3c – 5d
Apply alternendo,
3a – 5b/3a + 5b = 3c – 5d/3c + 5d (Hence proved)
(Q4) If 5x + 6y/5u + 6v = 5x – 6y/5u – 6v
Then prove that x:y = u:v
= Solution:
Given that,
5x + 6y/5u + 6v = 5x – 6y/5u – 6v
To prove: x:y = u:v
Prove: 5x + 6y/5u + 6v = 5x – 6y/5u – 6v
Apply alternendo on both side,
5x + 6y/5x – 6y = 5u + 6v/5u – 6v
Apply componenedo and dividendo
..> 5x + 6y + 5x – 6y/5x + 6y – (5x – 6y) = 5u + 6v + 5u – 6v/5u + 6v – (5u – 6v)
..> 5x + 6y + 5x – 6y/5x + 6y – 5x + 6y = 5u + 6v + 5u – 6v/5u + 6v – 5u + 6v
..> 10x/12y = 10u/12v
x/y = u/v
∴ x:y = u:v (Hence proved)
(Q5) If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d)
= Prove that a:b = c:d
= Solution:
Given that,
(7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d)
To prove: a:b = c:d
Proof: (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d)
Apply invertendo on both sides,
7a + 8b/7a – 8b = 7c + 8d/7c – 8d
Apply componendo and dividendo,
7a + 8b + 7a – 8b/7a + 8b – (7a – 8b) = 7a + 8b + 7c – 8d/7c + 8d – (7c – 8d)
14a/7a + 8b – 7a + 8b = 14c/7c + 8d – 7c + 8d
14a/16b = 14c/16d
a/b = c/d
∴ a:b = c:d (Hence proved)
(Q6) (i) If x = 6ab/(a + b), Find the value of:
x + 3a/x – 3a + x + 3b/x – 3b
= Solution:
Given that,
x = 6ab/(a + b)
Find: x+3a/x–3a + x+3b/x–3b
x = 6ab/(a+b)
= (3×2)ab/a+b
x = 3a×2b/a+b
x/3a = 2b/a+b
Apply componenedo and dividendo
x+3a/x-3a = 2b+a+b/2b – (a+b)
x + 3a/x – 3a = 2b + a + b/2b – a – b
x + 3a/x – 3a = 3b + a/b + a
x + 3a/x – 3a = a + 3b/- (a – b) —— (i)
x = 6ab/a+b
x = 3b×2a/a+b
x/3b = 2a/a + b
Apply componendo and dividendo,
x + 3b/x – 3b = 2a + a + b/2a – (a + b)
x + 3b/x – 3b = 2a + a + b/2a – a – b
x + 3b/x – 3b = 3a + b/a – b —– (ii)
Now, x + 3a/x – 3a + x + 3b/x – 3b
From equation (i) and (ii)
a + 3b/- (a – b) + 3a + b/a – b
Take L.C.M
– a – 3b + 3a + b/(a – b) = 2a – 2b/(a – b)
= 2 (a – b)/(a – b)
= 2
(ii) Solution:
Given:
a = 4√6/√2 + √3
a = 2×2×√2×√3/√2 +√3
a/2√2 = 2√3/√2 + √3
Apply componendo and dividendo,
a + 2√2/a – 2√2 = 2√3+√2+√3/2√3-√2-√3
a+2√2/a-2√2 = 3√3+√2/√3 – √2 —— (i)
a = 4√6/√2 + √3
a = 2×2×√2×√3/√2 + √3
a/2√3 = 2√2/√2 + √3
Apply componendo and divideno,
a+2√3/a-2√3= 2√2+√2+√3/2√2-√2-√3
a+2√3/a-2√3 = 3√2+√3/√2 – √2 —— (ii)
Now, a+2√2/a-2√2 + a+2√3/a-2√3
3√3+√2/√3-√2 + 3√2+√3/-(√3-√2)
3√3+√2/√3-√2 – (3√2+√3)/(√3-√2)
3√3+√2-3√2-√3/(√3-√2)
..> +2√3 – 2√2/√3 – √2 = 2(√3 – √2)/(√3 – √2)
= 2
(Q7) If (a+b+c+d) (a-b-c+d) = (a+b-c-d) (a-b+c-d), prove that, a:b = c:d
Solution:
Given that,
(a+b+c+d) (a-b-c+d) = (a+b-c-d) (a-b+c-d)
Prove that: a:b = c:d
Proof: (a+b+c+d) (a-b-c+d) = (a+b-c-d) (a-b+c-d)
(a+b+c+d)/(a+b-c-d) = (a-b+c-d)/(a-b-c+d)
Apply componendo and = dividendo,
a+b+c+d+a+b-c-d/a+b+c+d-(a+b-c-d) = a-b+c-d+a-b-c+d/a-b+c-d-(a-b-c+d)
= a+b+c+d+a+b-c-d/a+b+c+d-a-b+c+d = a-b+c-d+a-b-c+d/a-b+c-d-a+b+c-d
2a+2b/2c+2d = 2a-2b/2c-2d
2(a+b)/2(c+d) = 2(a-b)/2(c-d)
(a+b)/(c+d) = (a-b)/(c-d)
Apply invertendo,
a+b/a-b = c+d/c-b
Apply componendo and dividendo,
a+b+a-b/a+b-a+b = c+d+c-d/c+d-c+d
2a/2b = 2c/2d
a/b = c/d
a:b = c:d (Hence proved)
Here is your solution of Selina Concise Class 10 Math Chapter 7 Ratio and Proportion Exercise 7C Solutions
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