S Chand ICSE Maths Solutions Class 10 Chapter 6 Ratio Proportion Exercise 6A
Hi students, Welcome to Amans Maths Blogs (AMB). In this post, you will get S Chand ICSE Maths Solutions Class 10 Chapter 6 Ratio Proportion Exercise 6A. This is the chapter of introduction of ‘Ratio and Proportion‘ included in ICSE Class 10 Maths syllabus.
S Chand ICSE Maths Class 10 Chapter 6 Ratio Proportion Exercise 6A Solutions
S Chand ICSE Solutions Class 10 Maths Ratio Proportion Exercise 6A: Ques No 1
Find the value of x in each case:
(i) 8 : 14 :: x : 28
(ii) x : 9 :: 5 : 3
(iii) 12 : x :: 4 : 15
S Chand ICSE Maths Solutions:
(i) Since 8, 14, x, 28 are in proportion, So 8 / 14 = x / 28 ⇒ 8 × 28 = x × 14 ⇒ x = (8 × 28) / 14 = 16
(ii) Since x, 9, 5, 3 are in proportion, So x / 9 = 5 / 3 ⇒ x × 3 = 9 × 5 ⇒ x = (9 × 5) / 3 = 15
(iii) Since 12, x, 4, 15 are in proportion, So 12 / x = 4 / 15 ⇒ 12 × 15 = x × 4 ⇒ x = (12 × 15) / 4 = 45
S Chand ICSE Solutions Class 10 Maths Ratio Proportion Exercise 6A: Ques No 2
Find the fourth proportional to
(i) 25, 15, 40
(ii) 3a2b2, a3, b3
(iii) a2 – 5a + 6, a2 + a – 6, a2 – 9
S Chand ICSE Maths Solutions:
(i) Let x be the fourth proportional to 25, 15, 40.
Then 25 : 15 :: 40 : x
So 25 / 15 = 40 / x ⇒ 25 × x = 15 × 40 ⇒ x = (15 × 40) / 25 = 24
(ii) Let x be the fourth proportional to 3a2b2, a3, b3.
Then 3a2b2 : a3 :: b3 : x
So 3a2b2 / a3 = b3 / x ⇒ 3a2b2 × x = a3 × b3 ⇒ x = ab / 3
(iii) Let x be the fourth proportional to a2 – 5a + 6, a2 + a – 6, a2 – 9.
Then (a2 – 5a + 6) : (a2 + a – 6) :: (a2 – 9) : x
So (a2 – 5a + 6) / (a2 + a – 6) = (a2 – 9) / x
⇒ (a2 – 5a + 6) × x = (a2 + a – 6) × (a2 – 9)
⇒ (a – 3)(a – 2) × x = (a + 3)(a – 2) × (a + 3)(a – 3)
⇒ x = [(a + 3)(a – 2) × (a + 3)(a – 3)] / (a – 3)(a – 2)
⇒ x = (a + 3)2
S Chand ICSE Solutions Class 10 Maths Ratio Proportion Exercise 6A: Ques No 3
Find the third proportional to
(i) 16 and 36
(ii) (x/y + y/x) and x/y
(iii) a2 – b2 and a + b
S Chand ICSE Maths Solutions:
(i) Let x be the third proportional to 16 and 36
Then 16 : 36 :: 36 : x
So 16 / 36 = 36 / x ⇒ 16 × x = 36 × 36 ⇒ x = (36 × 36) / 16 = 81
(ii) Let k be the third proportional to (x/y + y/x) and x/y
Then (x/y + y/x) : (x/y) :: (x/y) :k
So (x/y + y/x) / (x/y) = (x/y) / k
⇒ (x/y + y/x) × k = (x/y) × (x/y)
⇒ k = (x/y)2 / (x/y + y/x)
⇒ k = (x2 / y2) × xy / (x2 + y2)
⇒ k = x3 / y(x2 + y2)
(ii) Let k be the third proportional to a2 – b2 and a + b
Then (a2 – b2) : (a + b) :: (a + b) : k
So (a2 – b2) / (a + b) = (a + b) / k
⇒ (a2 – b2) × k = (a + b) × (a + b)
⇒ (a + b)(a – b) × k = (a + b) × (a + b)
⇒ k = (a + b) / (a – b)
S Chand ICSE Solutions Class 10 Maths Ratio Proportion Exercise 6A: Ques No 4
Find the mean proportional to
(i) 5 and 80
(ii) 360a4 and 250a2b2
(iii) (x – y) and (x3 – x2y)
S Chand ICSE Maths Solutions:
(i) Mean Proportional of 5 and 80 is √(5 × 80) = √400 = 20
(ii) Mean Proportional of 360a4 and 250a2b2 is √(360a4 × 250a2b2) = √(62 × 52 × 102 × a6 × b2) = 6 × 5 × 10 × a3 × b = 300a3b
(iii) Mean Proportional of (x – y) and (x3 – x2y) is √((x – y) × (x3 – x2y)) = √(x2 × (x – y)2) = x(x – y)
S Chand ICSE Solutions Class 10 Maths Ratio Proportion Exercise 6A: Ques No 5
(i) If x, 16, 48, y are in continued proportion, find the value of x and y.
(ii) If x, 9, y, 16 are in continued proportion, find the value of x and y.
S Chand ICSE Maths Solutions:
(i) Given that x, 16, 48, y are in continued proportion.
Then, x / 16 = 16 / 48 = 48 / y
⇒ x / 16 = 16 / 48 and 16 / 48 = 48 / y
⇒ x / 16 = 1 / 3 and 1 / 3 = 48 / y
⇒ x = (1/3) × 16 and y = 3 × 48
⇒ x = 16/3 and y = 144
(ii) Given that x, 9, y, 16 are in continued proportion.
Then, x / 9 = 9 / y = y / 16
⇒ x / 9 = 9 / y and 9 / y = y / 16
⇒ x × y = 9 × 9 and y × y = 16 × 9
⇒ xy = 81 and y2 = 144
⇒ xy = 81 and y = √144 = 12
⇒ x × 12 = 81 and y = 12
⇒ x = 81 / 12 and y = 12
⇒ x = 27 / 4 and y = 12
Ques No 6
What number must be added to 3, 5, 7, 10 each in order to get four numbers in proportion?
Solutions:
Let a number x is added to each of 3, 5, 7, 10 in order to get four numbers in proportion.
Then, (3 + x) : (5 + x) :: (7 + x) : (10 + x)
⇒ (3 + x) / (5 + x) = (7 + x) / (10 + x)
⇒ (x + 3)(x + 10) = (x + 5)(x + 7)
⇒ x2 + 13x + 30 = x2 + 12x + 35
⇒ 13x + 30 = 12x + 35
⇒ 13x – 12x = 35 – 30
⇒ x = 5
Ques No 7
What number must be subtracted from each of the numbers 28, 53, 19, 35 so that they are in proportion.
Solutions:
Let a number x is subtracted from each of the numbers 28, 53, 19, 35 in order to get four numbers in proportion.
Then, (28 – x) : (53 – x) :: (19 – x) : (35 – x)
⇒ (28 – x) / (53 – x) = (19 – x) / (35 – x)
⇒ (28 – x)(35 – x) = (19 – x)(53 – x)
⇒ x2 – 63x + 980 = x2 – 72x + 1007
⇒ 72x – 63x = 1007 – 980
⇒ 9x = 27
⇒ x = 3
Ques No 8
(i) Find the two numbers such that the mean proportional between them is 14 and the third proportional to them is 112.
(ii) Find the two numbers such that the mean proportional between them is 18 and the third proportional to them is 144.
Solutions (i):
Let x and y are two numbers.
Then, Mean Proportional of x and y = 14
⇒ √xy = 14
⇒ xy = 196
⇒ x = 196 / y
And, x : y = y : 112
⇒ 112 × x = y2
⇒ 112 × (196 / y) = y2
⇒ y3 = 112 × 196
⇒ y3 = 14 × 8 × 14 × 14
⇒ y = 14 × 2 = 28
and x = 196 / 28 = 7
Thus, required numbers are 7 and 28.
Solutions (ii):
Let x and y are two numbers.
Then, Mean Proportional of x and y = 18
⇒ √xy = 18
⇒ xy = 324
⇒ x = 324 / y
And, x : y = y : 144
⇒ 144 × x = y2
⇒ 144 × (324 / y) = y2
⇒ y3 = 144 × 324
⇒ y3 = 18 × 8 × 18 × 18
⇒ y = 18 × 2 = 36
and x = 324 / 36 = 9
Thus, required numbers are 9 and 36.
Ques No 9
If p + r = 2q and 1/q + 1/s = 2/r, then prove that p : q = r : s
Solutions:
Given that p + r = 2q
⇒ (p/q) + (r/q) = 2… equation (1)
And, 1/q + 1/s = 2/r
⇒ (r/q) + (r/s) = 2… equation (2)
From equation (1) – (2),
⇒ (p/q) – (r/s) = 0
⇒ p/q = r/s
⇒ p : q = r : s
Ques No 10
If b is the mean proportional between a and c, prove that a, c, a2 + b2 and b2 + c2 are proportional.
Solutions:
Given that b is the mean proportional between a and c.
Then, √ac = b
⇒ b2 = ac
Now,
(a2 + b2) / (b2 + c2) = (a2 + ac) / (ac + c2)
⇒ (a2 + b2) / (b2 + c2) = a(a + c) / c(a + c)
⇒ (a2 + b2) / (b2 + c2) = a / c
Thus, a, c, a2 + b2 and b2 + c2 are proportional.
Ques No 11
If (x + 7) is the mean proportional between (x + 3) and (x + 12), find the value of x.
Solutions:
Given that (x + 7) is the mean proportional between (x + 3) and (x + 12).
Then, √((x + 3)(x + 12)) = (x + 7)
⇒ (x + 7)2 = (x + 3)(x + 12)
⇒ x2 + 14x + 49 = x2 + 15x + 36
⇒ 14x – 15x = 36 – 49
⇒ -x = -13
⇒ x = 13
S Chand ICSE Solutions Class 10 Maths Ratio Proportion Exercise 6A: Ques No 12
If (a2 + c2) / (ab + cd) = (ab + cd) / (b2 + d2), then prove that a/b = c/d.
S Chand ICSE Maths Solutions:
Given that (a2 + c2) / (ab + cd) = (ab + cd) / (b2 + d2).
⇒ (a2 + c2) × (b2 + d2) = (ab + cd)2
⇒ a2b2 + a2d2 + b2c2 + c2d2 = a2b2 + c2d2 + 2abcd
⇒ a2d2 + b2c2 = 2abcd
⇒ a2d2 + b2c2 – 2abcd = 0
⇒ (ad – bc)2 = 0
⇒ ad = bc
⇒ a/b = c/d
S Chand ICSE Class 10 Maths Chapterwise Notes & Solutions |
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Chapter 3 : Shares & Dividends Notes & Solutions | Notes | Exercise 3A | Exercise 3B | Revision Exercise |
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Chapter 5 : Quadratic Equations Notes & Solutions | Notes | Exercise 5A | Exercise 5B | Exercise 5C| Exercise 5D| Exercise 5E| Revision Exercise |
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