Increase Decrease in Ratio Class 10 Concise Exe-7A Selina Solutions Ch-7 Ratio and Proportion (Including Properties and Uses. In this article you will learn how to solve problems on Increase / Decrease in a Ratio. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Increase Decrease in Ratio Class 10 Concise Exe-7A Selina Solutions Ch-7 Ratio and Proportion
| Board | ICSE |
| Publications | Selina |
| Subject | Maths |
| Class | 10th |
| Chapter-7 | Ratio and Proportion (Including Properties and Uses) |
| Writer | R.K. Bansal |
| Exe-7A | Increase/Decrease in a ratio and Composition of Ratios. |
| Edition | 2025-2026 |
Increase Decrease in Ratio
Class 10 Concise Exe-7A Selina Solutions Ch-7 Ratio and Proportion (Including Properties and Uses.
Exercise- 7A
Que-1: If a : b = 5 : 3, find : {(5a-3b)/(5a+3b)}
Sol: Given, a : b = 5 : 3
⇒ 𝑎/𝑏 = 5/3
(5𝑎−3𝑏)/(5𝑎+3𝑏) = {5(𝑎/𝑏)−3}/{5(𝑎/𝑏)+3} …(Dividing each term by b)
= {5(5/3)−3}/{5(5/3)+3}
= {(25/3)−3}/{(25/3)+3}
= (25−9)/(25+9)
= 16/34
= 8/17
Que-2: If x: y = 4: 7, find the value of (3x + 2y): (5x + y).
Sol: Given, x : y = 4 : 7
⇒ 𝑥/𝑦 = 4/7
(3𝑥+2𝑦)/(5𝑥+𝑦) = {3(𝑥/𝑦)+2}/{5(𝑥/𝑦)+1} …(Dividing each term by y)
= {3(4/7)+2}/{5(4/7)+1}
= {(12/7)+2}/{(20/7)+1}
= (12+14)/(20+7)
= 26/27
Que-3: If a : b = 3 : 8, find the value of (4a+3b)/(6a-b)
Sol: Given, a : b = 3 : 8
⇒ 𝑎/𝑏 = 3/8
(4𝑎+3𝑏)/(6𝑎−𝑏) = {4(𝑎/𝑏)+3}/{6(𝑎/𝑏)−1} …(Dividing each term by b)
= {4(3/8)+3}/{6(3/8)−1}
= {(3/2)+3}/{(9/4)−1}
= {(9/2)/(5/4)}
= 18/5
Que-4: If (a – b): (a + b) = 1: 11, find the ratio (5a + 4b + 15): (5a – 4b + 3).
Sol: (a – b) : (a + b) = 1 : 11,
(𝑎−𝑏)/(𝑎+𝑏) = 1/11
11a – 11b = a + b
10a = 12b
𝑎/𝑏 =12/10 = 6/5
So, let a = 6k and b = 5k
(5𝑎+4𝑏+15)/(5𝑎−4𝑏+3) = {5(6𝑘)+4(5𝑘)+15}/{5(6𝑘)−4(5𝑘)+3}
= (30𝑘+20𝑘+15)/(30𝑘−20𝑘+3)
= (50𝑘+15)/(10𝑘+3)
= {5(10𝑘+3)}/(10𝑘+3)
= 5
Hence, (5a + 4b + 15) : (5a – 4b + 3) = 5 : 1
Que-5: Find the number which bears the same ratio to (7/33) and (8/21) does to 4/9.
Sol: Let the required number be 𝑥/𝑦
Now, ratio of 8/21 to 4/9
= {(8/21)/(4/9)}
= (8/21) × (9/4)
= 6/7
Thus, we have
{(𝑥/𝑦)/(7/33)} = 6/7
⇒ 𝑥/𝑦 = {(6/7)/(7/33)}
⇒ 𝑥/𝑦 = (6/7) × (7/33)
⇒ 𝑥/𝑦 = 2/11
Hence, the required number is 2/11.
Que-6: If (m+n)/(m+3n) = 2/3, find : 2n²/(3m²+mn)
Sol: Given: (𝑚+𝑛)/(𝑚+3𝑛) = 2/3
⇒ 3m + 3n = 2m + 6n
⇒ 3m – 2m = 6n – 3n
⇒ m = 3n
Now 2𝑛²/(3𝑚²+𝑚𝑛)
= 2𝑛² / {3(3𝑛)²+3𝑛×𝑛}
= 2𝑛²/(27𝑛²+3𝑛²
= 2𝑛²/30𝑛²
= 1/15
Que-7: Find x/y ; when x²+6y² = 5xy.
Sol: x2 + 6y2 = 5xy
Dividing both sides by y2, we get
(𝑥²/𝑦²) + (6𝑦²/𝑦²) = 5𝑥𝑦/𝑦²
(𝑥/𝑦)² + 6 = 5(𝑥/𝑦)
(𝑥/𝑦)² − 5(𝑥/𝑦) + 6 =0
Let 𝑥/𝑦 = 𝑎
∴ a2 – 5a + 6 = 0
⇒ (a – 2)(a – 3) = 0
⇒ a = 2, 3
Hence, 𝑥/𝑦 = 2, 3
Que-8: If the ratio between 8 and 11 is the same as the ratio of 2x-y to x+2y, find the value of 7x/9y.
Sol: (2𝑥−𝑦)/(𝑥+2𝑦) = 8/11
22x – 11y = 8x + 16y
14x = 27y
Given, 𝑥/𝑦 = 27/14
∴ 7𝑥/9𝑦 = {7×27)/(9×14) = 3/2
Que-9: Divide Rs1290 into A,B and C such that A is 2/5 of B and B:C = 4:3
Sol: Given B : C = 4 : 3
⇒ 𝐵/𝐶 = 4/3
And 𝐴 = (2/5)𝐵
⇒ 𝐴/𝐵 = 2/5
Now 𝐴/𝐵 = 2/5 = (2×4)/(5×4) = 8/20
And 𝐵/𝐶 = (4×5)/(3×5) = 20/15
⇒ A : B : C = 8 : 20 : 15
⇒ A = 8x, B = 20x and C = 15x
∴ 8x + 20x + 15x = 1290
⇒ 43x = 1290
⇒ x = 30
A’s share = 8x
= 8 × 30
= Rs. 240
B’s share = 20x
= 20 × 30
= Rs. 600
C’s share = 15x
= 15 × 30
= Rs. 450
Que-10: A school has 630 students. The ratio of the number of boys to the number of girls is 3 : 2. This ratio changes to 7 : 5 after the admission of 90 new students. Find the number of newly admitted boys.
Sol: Let the number of boys be 3x.
Then, the number of girls = 2x
∴ 3x + 2x = 630
⇒ 5x = 630
⇒ x = 126
⇒ Number of boys = 3x = 3 × 126 = 378
And Number of girls = 2x = 2 × 126 = 252
After admission of 90 new students we have total number of student = 630 + 90 = 720
Now, let the number of boys be 7x
Then, number of girls = 5x
∴ 7x + 5x = 720
⇒ 12x = 720
⇒ x = 60
⇒ Number of boys = 7x = 7 × 60 = 420
And Number of girls = 5x = 5 × 60 = 300
∴ Number of newly admitted boys = 420 – 378 = 42
Que-11: What quantity must be subtracted from each term of the ratio 9: 17 to make it equal to 1: 3?
Sol: Let x be subtracted from each term of the ratio 9 : 17.
(9−𝑥)/(17−𝑥) = 1/3
27 – 3x = 17 – x
10 = 2x
x = 5
Thus, the required number which should be subtracted is 5.
Que-12: The monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each saves Rs. 80 every month, find their monthly pocket money.
Sol: Let the monthly pocket money of Ravi and Sanjeev be 5x and 7x respectively.
Also, let their expenditure be 3y and 5y respectively.
So, 5x – 3y = 80 …(i)
And 7x – 5y = 80 …(ii)
Multiplying (i) by 7 and (ii) by 5 and subtracting, we get
35x – 21y = 560
35x – 25y = 400
– + –
4y = 160
⇒ y = 40
From (i),
5x = 80 + 3 × 40 = 200
⇒ x = 40
So, monthly pocket money of Ravi
= ₹ 5 × 40
= ₹ 200
And monthly pocket money of Sanjeev
= ₹ 7 × 40
= ₹ 280
Que-13: The work done by (x – 2) men in (4x + 1) days and the work done by (4x + 1) men in (2x – 3) days are in the ratio 3: 8. Find the value of x.
Sol: Assuming that all the men do the same amount of work in one day and one day work of each man = 1 units, we have,
Amount of work done by (x – 2) men in (4x + 1) days
= Amount of work done by (x – 2)(4x + 1) men in one day
= (x – 2)(4x + 1) units of work
Similarly,
Amount of work done by (4x + 1) men in (2x – 3) days
= (4x + 1)(2x – 3) units of work
According to the given information,
{(𝑥−2)(4𝑥+1)}/{(4𝑥+1)(2𝑥−3)} = 3/8
(𝑥−2)/(2𝑥−3) = 3/8
8x – 16 = 6x – 9
2x = 7
x = 7/2 = 3.5
Que-14: The bus fare between two cities is increased in the ratio 7: 9. Find the increase in the fare, if:
(i) the original fare is Rs 245; (ii) the increased fare is Rs 207.
Sol: According to the given information,
Increased (new) bus fare = 9/7 × original bus fare
(i) We have:
Increased (new) bus fare = 9/7 × Rs. 245 = Rs. 315
∴ Increase in fare = Rs. 315 – Rs. 245 = Rs. 70
(ii) We have:
Rs. 207 = 9/7 × original bus fare
Original bus fare = Rs. 207 × (7/9) = Rs. 161
∴ Increase in fare = Rs. 207 – Rs. 161 = Rs. 46
Que-15: By increasing the cost of entry ticket to a fair in the ratio 10: 13, the number of visitors to the fair has decreased in the ratio 6: 5. In what ratio has the total collection increased or decreased?
Sol: Let the cost of the entry ticket initially and at present be 10 x and 13x respectively.
Let the number of visitors initially and at present be 6y and 5y respectively.
Initially, total collection = 10x × 6y = 60 xy
At present, total collection = 13x × 5y = 65 xy
Ratio of total collection = 60 xy: 65 xy = 12: 13
Thus, the total collection has increased in the ratio 12: 13.
Que-16: In a basket, the ratio between the number of oranges and the number of apples is 7: 13. If 8 oranges and 11 apples are eaten, the ratio between the number of oranges and the number of apples becomes 1: 2. Find the original number of oranges and the original number of apples in the basket.
Sol: Let the original number of oranges and apples be 7x and 13x.
According to the given information,
(7𝑥−8)/(13𝑥−11) = 1/2
⇒ 2(7x – 8) = 1(13x – 11)
⇒ 14x – 16 = 13x – 11
⇒ 14x – 13x = –11 + 16
⇒ x = 5
Thus, the original number of oranges and apples are 7 × 5 = 35 and 13 × 5 = 65 respectively.
Que-17: In a mixture of 126 kg of milk and water, milk and water are in ratio 5 : 2. How much water must be added to the mixture to make this ratio 3 : 2?
Sol: Quantity of milk : Quantity of water = 5 : 2
∴ Quantity of milk = 126 × (5/7) = 90 kg
⇒ Quantity of water = 126 – 90 = 36 kg
New ratio = 3 : 2
Let the quantity of water to be added be x kg
Then, milk : water = 90/(36+𝑥)
90/(36+𝑥) = 3/2
⇒ 180 = 108 + 3x
⇒ 3x = 72
⇒ x = 24
Thus, quantity of water to be added is 24 kg.
Que-18: (a) If A: B = 3: 4 and B: C = 6: 7, find:
(i) A: B: C (ii) A: C
(b) If A : B = 2 : 5 and A : C = 3 : 4, find
(i) A : B : C
Sol: (a)
(i) A : B : C
𝐴/𝐵 = 3/4
= (3/4) × (3/4)
= 9/12
𝐵/𝐶 = 6/7
= (6/7) × (2/2)
= 12/14
A : B : C = 9 : 12 : 14
(ii) A : C
𝐴/𝐵 = 3/4
𝐵/𝐶 = 6/7
∴ 𝐴/𝐶 = (𝐴/𝐵)/(𝐶/𝐵)
= (3/4)/(7/6)
= (3/4) × (6/7)
= 9/14
∴ A : C = 9 : 14
(b) To compare 3 ratios, the consequent of the first ratio and the antecedent of the 2nd ratio must be made equal.
Given that A : B = 2 : 5 and A : C = 3 : 4
Interchanging the first ratio, we have,
B : A = 5 : 2 and A : C = 3 : 4
L.C.M of 2 and 3 is 6
⇒ B : A = 5 × 3 : 2 × 3 and A : C = 3 × 2 : 4 × 2
⇒ B : A = 15 : 6 and A : C = 6 : 8
⇒ B : A : C = 15 : 6 : 8
⇒ A : B : C = 6 : 15 : 8
Que-19: (i) If 3A = 4B = 6C; find A: B: C.
(ii) If 2a = 3b and 4b = 5c, find: a : c.
Sol: (i) 3A = 4B = 6C
3𝐴 = 4𝐵 ⇒ 𝐴/𝐵 = 4/3
This means, A : B = 4 : 3
4𝐵 = 6𝐶 ⇒ 𝐵/𝐶 = 6/4 = 3/2
⇒ B : C = 3 : 2
The value of B in 4 : 3 is equal to the value of B in 3 : 2
Hence, A : B : C = 4 : 3 : 2
(ii) We have,
2a = 3b ⇒ a/b = 3/2
And 4b = 5c ⇒ bc = 5/4
Now, a/b = 3/2 = (3×5)/(2×5) = 15/10 and b/c = 5/4 = (5×2)/(4×2) = 10/8
⇒ a : b : c = 15 : 10 : 8
⇒ a : c = 15 : 8
Que-20: Find the compound ratio of :
(i) 2 : 3, 9 : 14 and 14 : 27
(ii) 2a : 3b, mn : x² and x : n
(iii) √2 : 1, 3 : √5 and √20 : 9
Sol: (i) required compound ration = 2 × 9 × 14 : 3 × 14 × 27
= (2×9×14)/(3×14×27)
= 2/9
= 2 : 9
(ii) Required compound ratio = 2a × mn × x : 3b × x2 × n
= (2𝑎×𝑚𝑛×𝑥)/(3𝑏×𝑥²×𝑛)
= 2𝑎𝑚/3𝑏𝑥
= 2am : 3bx
(iii) Required compound ratio = √2 ×3 ×√20 : 1 × √5 × 9
= (√2×3×√20)/(1×√5×9)
= (√2×√4)/3
= 2√2/3
= 2√2 : 3
Que-21: Find duplicate ratio of:
(i) 3: 4 (ii) 3√3 : 2√5
Sol: (i) Duplicate ratio of 3 : 4 = 32 : 42 = 9 : 16
(ii) Duplicate ratio of 3√3 : 2√5 = (3√3)² : (2√5)² = 27 : 20
Que-22: Find triplicate ratio of :
(i) 1 : 3 (ii) m/2 : n/3
Sol: (i) Triplicate ratio of 1 : 3
= 13 : 33
= 1 : 27
(ii) Triplicate ratio of 𝑚/2 : 𝑛/3
= (𝑚/2)³ : (𝑛/3)³
= 𝑚³/8 : 𝑛³/27
= (𝑚³/8) / (𝑛³/27)
= 27m³ : 8n³
Que-23: Find sub-duplicate ratio of:
(i) 9: 16 (ii) (x – y)4: (x + y)6
Sol: (i) Sub-duplicate ratio of 9 : 16
= √9 :√16
= 3 : 4
(ii) Sub-duplicate ratio of (x – y)4 : {(x + y)6}
√(x – y)4 : √(x + y)6
= (x-y)² : (x+y)³
Que-24: Find the sub-triplicate ratio of:
(i) 64: 27 (ii) x3: 125y3
Sol: (i) Sub-triplicate ratio of 64 : 27 = ∛64 : ∛27 = 4 : 3
(ii) Sub-triplicate ratio of x³ : 125y³ = ∛x³ : ∛125y³ = x : 5y
Que-25: Find the reciprocal ratio of :
(i) 5 : 8 (ii) x/3 : y/7
Sol: (i) The reciprocal ratio of 5 : 8
= 1/5 : 1/8 = 8 : 5
(ii) The reciprocal ratio of x/3 : y/7
1/(x/3) : 1/(y/7) = 3/x : 7/y
= (3/x)/(7/y) = 3y/7x = 3y : 7x
Que-26: If (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.
Sol: If (x + 3) : (4x + 1) is duplicate ratio of 3 : 5
find the value of x
We have
(𝑥+3)/(4𝑥+1) = 3²/5²
=> 25x + 75 = 36x + 9
=> 11x = 66
=> x = 6
Que-27: If m: n is the duplicate ratio of m + x: n + x; show that x2 = mn.
Sol: 𝑚/𝑛 = (𝑚+𝑥)²/(𝑛+𝑥)²
𝑚/𝑛 = (𝑚²+𝑥²+2𝑚𝑥) / (𝑛²+𝑥²+2𝑛𝑥)
mn2 + mx2 + 2mnx = m2n + nx2 + 2mnx
mn2 – m2n + mx2 – nx2 = 0
mn(m – n) – x2(m – n) = 0
x2(m – n) = mn(m – n)
x2 = mn
Que-28: If (3x – 9) : (5x + 4) is the triplicate ratio of 3 : 4, find the value of x.
Sol: We have,
(3𝑥−9)/(5𝑥+4) = 3³/4³
⇒ (3𝑥−9)/(5𝑥+4) = 27/64
⇒ {3(𝑥−3)}/(5𝑥+4) = 27/64
⇒ (𝑥−3)/(5𝑥+4) = 9/64
⇒ 64x – 192 = 45x + 36
⇒ 19x = 228
⇒ x = 12
Que-29: Find the ratio compounded of the reciprocal ratio of 15: 28, the sub-duplicate ratio of 36: 49 and the triplicate ratio of 5: 4.
Sol: Reciprocal ratio of 15 : 28 = 28 : 15
Sub-duplicate ratio of 36 : 49
= √36 : √49
= 6 : 7
Triplicate ratio of 5 : 4
= 53 : 43
= 125 : 64
Required compounded ratio
= (28×6×125)/(15×7×64)
= 25/8
= 25 : 8
Que-30: (a) If r2 =pq, show that p : q is the duplicate ratio of (p + r) : (q + r).
(b) If (p-x) : (q-x) be the duplicate ratio of p:q then show that : (1/p) + (1/q) = 1/x
Sol: (a) Given, r2 = pq
Duplicate ratio of (p + r) : (q + r)
= (p + r)2 : (q + r)2
= (p2 + r2 + 2pr) : (q2 + r2 + 2qr)
= (p2 + pq + 2pr) : (q2 + pq + 2qr)
= p(p + q + 2r) : q(q + p + 2r)
= p : q
Thus, p : q is the duplicate ratio of (p + r) : (q + r).
(b) We have,
(𝑝−𝑥)/(𝑞−𝑥) = 𝑝²/𝑞²
⇒ q2(p – x) = p2(q – x)
⇒ pq2 – q2x = p2q – p2x
⇒ p2x – q2x = p2q – pq2
⇒ x(p2 – q2) = pq(p – q)
⇒ x(p – q)(p + q) = pq(p – q)
⇒𝑥 = 𝑝𝑞/(𝑝+𝑞)
⇒ (𝑝+𝑞)/𝑝𝑞 = 1/𝑥
⇒ (𝑝/𝑝𝑞) + (𝑞/𝑝𝑞) = 1/𝑥
⇒ (1/𝑝) + (1/𝑞) = 1/𝑥
— : End of Increase Decrease in Ratio Class 10 Concise Exe-7A Selina Solutions. :–
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