Ratio and Proportion Class 10 OP Malhotra Exe-6A ICSE Maths Solutions

Ratio and Proportion Class 10 OP Malhotra Exe-6A ICSE Maths Solutions Ch-6. Step by step solutions of questions as latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Ratio and Proportion Class 10 OP Malhotra Exe-6A ICSE Maths Solutions Ch-6

Board	ICSE
Publications	S Chand
Subject	Maths
Class	10th
Chapter-6	Ratio and Proportion
Writer	OP Malhotra
Exe-6A	Mean, third , fourth , continue proportion
Edition	2024-2025

Exercise- 6A

Ratio and Proportion Class 10 OP Malhotra ICSE Maths Solutions Ch-6

Que-1: Find the value of x in each case:
(i) 8 : 14 :: x : 28
(ii) x : 9 :: 5 : 3
(iii) 12 : x :: 4 : 15

Sol: (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

Que-2: Find the fourth proportional to
(i) 25, 15, 40
(ii) 3a²b², a³, b³
(iii) a² – 5a + 6, a² + a – 6, a² – 9

Sol: (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

Que-3: Find the third proportional to
(i) 16 and 36
(ii) (x/y + y/x) and x/y
(iii) a² – b² and a + b

Sol:  (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)

Que-4: Find the mean proportional to
(i) 5 and 80
(ii) 360a⁴ and 250a²b²
(iii) (x – y) and (x³ – x²y)

Sol: (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)

Que-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.

Sol:  (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

Que-6: What number must be added to 3, 5, 7, 10 each in order to get four numbers in proportion?

Sol:  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

Que-7: What number must be subtracted from each of the numbers 28, 53, 19, 35 so that they are in proportion.

Sol:  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

Que-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. 

Sol:  (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.

(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.

Que-9: If p + r = 2q and (1/q) + (1/s) = (2/r), then prove that p : q = r : s

Sol:  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

Que-10: If b is the mean proportional between a and c, prove that a, c, a² + b² and b² + c² are proportional.

Sol: 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.

Que-11: If (x + 7) is the mean proportional between (x + 3) and (x + 12), find the value of x.

Sol: 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

Que-12:  If (a² + c²) / (ab + cd) = (ab + cd) / (b² + d²), then prove that a/b = c/d.

Sol: 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

–: End of  Ratio and Proportion Class 10 OP Malhotra Exe-6A ICSE Maths Solutions Ch-6 :–

Return to :  OP Malhotra S Chand Solutions for ICSE Class-10 Maths

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