If x : y = 2 : 5, then find (3x + 2y) : (7x − 2y).
A.3 : 2
B.4 : 1
C.7 : 4
D.9 : 5
Solution
Ratio substitution
Take x = 2, y = 5.
3x + 2y = 6 + 10 = 16.
7x − 2y = 14 − 10 = 4.
16 : 4 = 4 : 1 — option (b).
54
Ram and Mohan invest ₹70,000 and ₹50,000 respectively. After 1 year, the total profit is distributed including simple interest at 10% per annum on the capital of each partner. If the total profit including interest is ₹18,000, what is Ram's share of the profit?
A.₹7,000
B.₹10,500
C.₹8,000
D.₹8,500
Solution
Profit with interest on capital
Interest on capital: Ram 10% of 70,000 = ₹7,000; Mohan 10% of 50,000 = ₹5,000 — total ₹12,000.
Remaining profit = 18,000 − 12,000 = ₹6,000, shared in the capital ratio 7 : 5.
Ram’s share of it = × 6,000 = ₹3,500.
Ram’s total = 7,000 + 3,500 = ₹10,500 — option (b).
55
What is the average of all numbers between 1000 and 1500 that are divisible by 20?
A.1150
B.1250
C.1575
D.1500
Solution
Average of multiples
Multiples of 20 between 1000 and 1500: 1020, 1040, …, 1480.
They form an AP, so average = .
= = 1250.
Hence option (b).
56
There are 2 consecutive even numbers and 3 consecutive odd numbers. The average of the even numbers is 3 more than the average of the odd numbers. If the sum of the even numbers is 2 more than the sum of the odd numbers, find the average of the odd numbers.
A.3
B.4
C.5
D.6
Solution
Consecutive even/odd
Odd: x, x+2, x+4 (sum 3x+6, average x+2). Even: y, y+2 (sum 2y+2, average y+1).
Averages: y + 1 = x + 2 + 3 ⇒ y = x + 4.
Sums: 2y + 2 = 3x + 6 + 2 ⇒ 2(x + 4) + 2 = 3x + 8 ⇒ x = 2.
Average of odd numbers = x + 2 = 4 — option (b).
57
Evaluate: 40% of 175 − 37% of 160 + 10% of 250
A.25
B.50
C.35
D.30
Solution
Percentages
40% of 175 = 70.
37% = , so × 160 = 60.
10% of 250 = 25.
70 − 60 + 25 = 35 — option (c).
58
Aman needed ₹60,000 and split the sum between two lenders: Lender A at 18% p.a. and Lender B at 12% p.a. (simple interest). After 2 years, the total interest paid was ₹19,200. If he had interchanged the principal amounts, his interest would have been ₹2,400 less. How much did Aman borrow from Lender A at 18% p.a.?
A.₹20,000
B.₹16,000
C.₹14,000
D.₹40,000
Solution
SI — two lenders
Interest per year = = ₹9,600 on ₹60,000 ⇒ overall rate 16%.
Alligation between 18% and 12% around 16%: (16 − 12) : (18 − 16) = 4 : 2 = 2 : 1.
Amount at 18% = × 60,000 = ₹40,000 (check: swap gives 8,400/yr, i.e. ₹2,400 less over 2 years ✓).
Hence option (d).
59
A shop sells a shirt for ₹P, incurring a loss of 10%. They mark it up to ₹1.8P and then provide a 20% discount. What is the percentage profit or loss?
A.24.6% Profit
B.28.6% Loss
C.29.6% Profit
D.34.6% Loss
Solution
Loss → markup → discount
Let P = 90 ⇒ CP = = 100.
MP = 1.8 × 90 = 162; SP after 20% off = 162 × 0.8 = 129.6.
Profit = 129.6 − 100 = 29.6 on 100.
29.6% profit — option (c).
60
A shopkeeper marked an item 80% above its cost price. He then offered two successive discounts of 20% and 25%. If he made a profit of ₹160, at what price did he sell the item?
A shopkeeper mixes two types of pulses costing ₹80/kg and ₹100/kg. He sells the mixture at ₹112.5/kg, earning a 25% profit. In what ratio did he mix the two types of pulses?
A is Four times more efficient as a worker than B, which allows A to complete a job in 60 days less than B. If they work together, they can finish the job in:
A.19.2 days
B.22.5 days
C.16 days
D.20 days
Solution
Efficiency 4 : 1
Efficiency A : B = 4 : 1 ⇒ time A : B = 1 : 4.
Difference 3 units = 60 days ⇒ 1 unit = 20 ⇒ A = 20 days, B = 80 days.
Work = 80 units; together 4 + 1 = 5 units/day.
80 ÷ 5 = 16 days — option (c).
63
A cyclist has to cover 160 km. After cycling for 9 hours and 36 minutes, he finds that he has completed th of the total distance. What is his speed in km/h?
A.15.45 km/h
B.16 km/h
C.12.5 km/h
D.10 km/h
Solution
Speed = distance/time
Distance covered = × 160 = 120 km.
Time = 9 h 36 min = 9.6 h.
Speed = = 12.5 km/h.
Hence option (c).
64
Sarah invested in three schemes M, N, and O at SI rates of 5%, 6%, and 10% p.a. Total interest in one year was ₹3664. The amount in O was 90% of M and 125% of N. What was the amount invested in Scheme N?
A.₹14000
B.₹15000
C.₹14400
D.₹16500
Solution
SI — three schemes
O = 0.9M and O = 1.25N ⇒ M : O = 10 : 9, O : N = 5 : 4 ⇒ M : O : N = 50 : 45 : 36.
Interest = 50x(0.05) + 36x(0.06) + 45x(0.10) = 2.5x + 2.16x + 4.5x = 9.16x.
9.16x = 3664 ⇒ x = 400.
N = 36 × 400 = ₹14,400 — option (c).
65
A vertical pole of height H stands on the ground. From point P, the angle of elevation of the top is 45°. From point Q, 20 meters away from P, the angle of elevation is 30°. What is the height of the pole?
A.5( + 1) m
B.10( + 1) m
C.10 m
D.20 m
Solution
Angles of elevation
From P (45°): distance PB = H.
From Q (30°): tan30° = ⇒ 20 + H = H.
H( − 1) = 20 ⇒ H = × .
= 10( + 1) m — option (b).
66
A right circular cone has a radius of 8 cm and a height of 15 cm. A sphere is placed inside the cone touching the base and the slanted surface. Find the radius of this inscribed sphere.
A.5.2 cm
B.4.8 cm
C.3.8 cm
D.6.2 cm
Solution
Sphere inscribed in cone
Slant height l = = 17 cm.
Inscribed sphere radius = inradius of the cross-section triangle = .
= = .
= 4.8 cm — option (b).
67
If the perimeter of a regular octagon is 160 cm, what is the side length?
A.20 cm
B.16 cm
C.24 cm
D.12 cm
Solution
Regular octagon side
A regular octagon has 8 equal sides.
Perimeter = 8 × side.
Side = = 20.
Hence 20 cm — option (a).
68
A circular wheel is divided into 4 equal sectors. If the area of one sector is 90 cm², what is the radius of the wheel?
A.8.6 cm
B.10.7 cm
C.6.5 cm
D.7.5 cm
Solution
Sector → radius
Full circle = 4 × 90 = 360 cm².
πr² = 360 ⇒ r² = ≈ 114.6.
r ≈ 10.7.
Hence 10.7 cm — option (b).
69
If the lateral surface area of a prism is 320 cm² and the height is 16 cm, what is the perimeter of the base?
A.20 cm
B.24 cm
C.36 cm
D.40 cm
Solution
Prism LSA
LSA of a prism = perimeter of base × height.
320 = P × 16.
P = 20.
Hence 20 cm — option (a).
70
Solve system: y = 4x + 2 and y = −2x + 8
A.(1, 6)
B.(2, 5)
C.(3, 7)
D.(0, 1)
Solution
Two lines — intersection
At the intersection both y-values are equal: 4x + 2 = −2x + 8.
6x = 6 ⇒ x = 1.
Substituting back: y = 4(1) + 2 = 6.
The solution is (1, 6) — option (a).
If two circles are touching externally how many common tangents do they have?
A.1
B.2
C.3
D.0
Solution
Tangents — touching circles
Externally touching circles have two direct common tangents.
Plus one transverse tangent at the point of contact.
Total = 3.
Hence option (c).
73
Two equal chords, AB and CD, are at a distance of 9 cm from the center of a circle. If the radius is 41 cm, what is the length of AB?
A.24 cm
B.48 cm
C.72 cm
D.80 cm
Solution
Chord length
Perpendicular from centre bisects the chord: half-chord² = r² − d².
= 41² − 9² = 1681 − 81 = 1600 ⇒ half-chord = 40.
AB = 2 × 40 = 80.
Hence 80 cm — option (d).
74
A sector of a circle with radius 8 cm has a central angle of 45°. What is the area of the corresponding segment?
Take x = 2, y = 5.
3x + 2y = 6 + 10 = 16.
7x − 2y = 14 − 10 = 4.
16 : 4 = 4 : 1 — option (b).
Interest on capital: Ram 10% of 70,000 = ₹7,000; Mohan 10% of 50,000 = ₹5,000 — total ₹12,000.
Remaining profit = 18,000 − 12,000 = ₹6,000, shared in the capital ratio 7 : 5.
Ram’s share of it = × 6,000 = ₹3,500.
Ram’s total = 7,000 + 3,500 = ₹10,500 — option (b).
Multiples of 20 between 1000 and 1500: 1020, 1040, …, 1480.
They form an AP, so average = .
= = 1250.
Hence option (b).
Interest per year = = ₹9,600 on ₹60,000 ⇒ overall rate 16%.
Alligation between 18% and 12% around 16%: (16 − 12) : (18 − 16) = 4 : 2 = 2 : 1.
Amount at 18% = × 60,000 = ₹40,000 (check: swap gives 8,400/yr, i.e. ₹2,400 less over 2 years ✓).
Hence option (d).
Let P = 90 ⇒ CP = = 100.
MP = 1.8 × 90 = 162; SP after 20% off = 162 × 0.8 = 129.6.
Profit = 129.6 − 100 = 29.6 on 100.
29.6% profit — option (c).
CP = 100 ⇒ MP = 180.
SP = 180 × 0.8 × 0.75 = 108 ⇒ profit 8% = ₹160.
1% = 20 ⇒ SP = 108 × 20.
= ₹2,160 — option (c).
CP of mixture = = ₹90/kg.
Alligation: (100 − 90) : (90 − 80).
= 10 : 10 = 1 : 1.
Hence option (b).
Efficiency A : B = 4 : 1 ⇒ time A : B = 1 : 4.
Difference 3 units = 60 days ⇒ 1 unit = 20 ⇒ A = 20 days, B = 80 days.
Work = 80 units; together 4 + 1 = 5 units/day.
80 ÷ 5 = 16 days — option (c).
Distance covered = × 160 = 120 km.
Time = 9 h 36 min = 9.6 h.
Speed = = 12.5 km/h.
Hence option (c).
O = 0.9M and O = 1.25N ⇒ M : O = 10 : 9, O : N = 5 : 4 ⇒ M : O : N = 50 : 45 : 36.
Interest = 50x(0.05) + 36x(0.06) + 45x(0.10) = 2.5x + 2.16x + 4.5x = 9.16x.
9.16x = 3664 ⇒ x = 400.
N = 36 × 400 = ₹14,400 — option (c).
From P (45°): distance PB = H.
From Q (30°): tan30° = ⇒ 20 + H = H.
H( − 1) = 20 ⇒ H = × .
= 10( + 1) m — option (b).
Slant height l = = 17 cm.
Inscribed sphere radius = inradius of the cross-section triangle = .
= = .
= 4.8 cm — option (b).
Full circle = 4 × 90 = 360 cm².
πr² = 360 ⇒ r² = ≈ 114.6.
r ≈ 10.7.
Hence 10.7 cm — option (b).
At the intersection both y-values are equal: 4x + 2 = −2x + 8.
6x = 6 ⇒ x = 1.
Substituting back: y = 4(1) + 2 = 6.
The solution is (1, 6) — option (a).
Externally touching circles have two direct common tangents.
Plus one transverse tangent at the point of contact.
Total = 3.
Hence option (c).
Perpendicular from centre bisects the chord: half-chord² = r² − d².
= 41² − 9² = 1681 − 81 = 1600 ⇒ half-chord = 40.
AB = 2 × 40 = 80.
Hence 80 cm — option (d).
Sector = × π × 64 = 8π.
Triangle = r² sin45° = × 64 × = 16.
Segment = sector − triangle.
= 8π − 16 — option (c).