Q1): Two numbers are in the ratio 3:5. If 12 is subtracted from each number, the new ratio becomes 1:2. What are the two numbers?
A) 36 and 60
B) 30 and 50
C) 48 and 80
D) 24 and 40
Answer: A) 36 and 60
Explanation:
- Let the numbers be 3k and 5k
- After subtracting 12: (3k − 12) and (5k − 12)
- Given new ratio: (3k − 12) : (5k − 12) = 1 : 2
- So, 2(3k − 12) = 1(5k − 12)
- 6k − 24 = 5k − 12
- k = 12
- Numbers = 3k = 36 and 5k = 60
Q2): Three numbers A, B, C are in continued proportion. If A + B + C = 98 and C − A = 42, what is B?
A) 21
B) 28
C) 35
D) 42
Answer: B) 28
Explanation:
- Continued proportion means: A : B = B : C
- So, B² = A × C
- Given: C − A = 42 → C = A + 42
- Given: A + B + C = 98
- Substitute C: A + B + (A + 42) = 98
- 2A + B = 56
- So, B = 56 − 2A
- Now use B² = A × C
- (56 − 2A)² = A(A + 42)
- Solving gives A = 14
- Then B = 56 − 2(14) = 28
Q3): If a:b = 5:7 and b:c = 6:11, what is a:c?
A) 30:77
B) 30:73
C) 33:77
D) 35:77
Answer: A) 30:77
Explanation:
- a:b = 5:7 → a/b = 5/7
- b:c = 6:11 → b/c = 6/11
- a/c = (a/b) × (b/c)
- a/c = (5/7) × (6/11)
- a/c = 30/77
- So, a:c = 30:77
Q4): y varies as x^(3/2). If y = 24 when x = 16, find y when x = 36.
A) 54
B) 72
C) 81
D) 96
Answer: C) 81
Explanation:
- y = k × x^(3/2)
- Given: 24 = k × 16^(3/2)
- 16^(3/2) = (√16)³ = 4³ = 64
- 24 = 64k
- k = 24/64 = 3/8
- Now x = 36
- 36^(3/2) = (√36)³ = 6³ = 216
- y = (3/8) × 216
- y = 81
Q5): x varies inversely as y². If x = 20 when y = 3, find y when x = 5.
A) 4
B) 5
C) 6
D) 9
Answer: C) 6
Explanation:
- x ∝ 1/y²
- So, x × y² = constant
- 20 × 3² = constant
- 20 × 9 = 180
- Now x = 5
- 5 × y² = 180
- y² = 36
- y = 6
Q6): ₹4690 is divided among A, B, C such that A:B = 3:5 and B:C = 4:7. What is C’s share?
A) ₹2100
B) ₹2450
C) ₹2600
D) ₹2800
Answer: B) ₹2450
Explanation:
- A:B = 3:5 → A = 3k, B = 5k
- B:C = 4:7 → B = 4m, C = 7m
- Make B equal: 5k = 4m
- Choose k = 4t and m = 5t
- Then A = 3(4t) = 12t
- B = 5(4t) = 20t
- C = 7(5t) = 35t
- Total ratio = 12 + 20 + 35 = 67
- 67t = 4690 → t = 70
- C = 35t = 35 × 70 = 2450
Q7): Milk:Water = 5:2. If 7 L water is added, ratio becomes 5:3. Find initial mixture quantity.
A) 35 L
B) 42 L
C) 49 L
D) 56 L
Answer: C) 49 L
Explanation:
- Let milk = 5k and water = 2k
- New water = 2k + 7
- New ratio: 5k : (2k + 7) = 5 : 3
- 3(5k) = 5(2k + 7)
- 15k = 10k + 35
- 5k = 35
- k = 7
- Initial mixture = 5k + 2k = 7k = 49 L
Q8): First quantity +20% and second quantity −10%, new ratio = 3:2. Find original ratio.
A) 9:8
B) 8:9
C) 5:4
D) 7:6
Answer: A) 9:8
Explanation:
- Let original ratio be a:b
- After change: 1.2a : 0.9b = 3 : 2
- So, (1.2a)/(0.9b) = 3/2
- a/b = (3/2) × (0.9/1.2)
- 0.9/1.2 = 9/12 = 3/4
- a/b = (3/2) × (3/4) = 9/8
- Original ratio = 9:8
Q9): If A:B = 3:4 and B:C = 5:6, find (A+B):(B+C).
A) 7:10
B) 35:44
C) 5:6
D) 44:35
Answer: B) 35:44
Explanation:
- A:B = 3:4 → A = 3x, B = 4x
- B:C = 5:6 → B = 5y, C = 6y
- Make B equal: 4x = 5y
- Choose x = 5k, y = 4k
- Then A = 15k, B = 20k, C = 24k
- A + B = 15k + 20k = 35k
- B + C = 20k + 24k = 44k
- Ratio = 35k : 44k = 35:44
Q10): a:b = 5:6 and c:d = 7:8. If a+c = 96 and b+d = 112, find a.
A) 32
B) 40
C) 48
D) 56
Answer: B) 40
Explanation:
- a = 5x, b = 6x
- c = 7y, d = 8y
- a + c = 96 → 5x + 7y = 96
- b + d = 112 → 6x + 8y = 112
- Multiply first equation by 6: 30x + 42y = 576
- Multiply second equation by 5: 30x + 40y = 560
- Subtract: (30x + 42y) − (30x + 40y) = 576 − 560
- 2y = 16
- y = 8
- Put y in 5x + 7y = 96
- 5x + 56 = 96
- 5x = 40
- x = 8
- a = 5x = 40