A sum of money invested at compound interest amounts to 3 times itself in 5 years. In how many years will it become 9 times itself?
A.10 years
B.15 years
C.12 years
D.18 years
Solution
CI — multiples
Under compound interest the multiple grows in powers.
9 = 3², so the 3-times growth must happen twice over.
Time = 2 × 5.
= 10 years — option (a).
53
Which of the following numbers belongs to the set of Integers but NOT to the set of Whole Numbers?
A.0
B.
C.−5
D.+5
Solution
Integer not whole
Whole numbers are 0, 1, 2, 3, … — no negatives.
Integers are …, −3, −2, −1, 0, 1, 2, … — negatives included.
−5 is therefore an integer but not a whole number.
Hence option (c).
54
An athlete runs a 350-metre race in 20 seconds. What is his speed in km/h?
A.63 km/h
B.67 km/h
C.64 km/h
D.55 km/h
Solution
Speed conversion
Speed = = 17.5 m/s.
To convert m/s into km/h, multiply by .
17.5 × 3.6.
= 63 km/h — option (a).
55
If the surface area of a sphere is equal to twice its volume, find the radius of the sphere.
A.1 unit
B.3 units
C.2 units
D.1.5 units
Solution
Sphere — area and volume
Surface area = 4πr² and volume = πr³.
Given 4πr² = 2 × πr³.
Cancel 4πr²: 1 = r.
r = 1.5 units — option (d).
56
A runner completes 5.2 km in 38 minutes. What is his speed in km/h?
A.8.21 km/h
B.9.21 km/h
C.11.21 km/h
D.13.42 km/h
Solution
Speed in km/h
Speed = with time in hours.
= × 60.
= .
≈ 8.21 km/h — option (a).
57
Partners A and B entered into a business investing ₹40,000 and ₹90,000 respectively. After 6 months C joins with ₹1,40,000. At year-end the profit was ₹80,000. How much more did C earn than A?
A.₹5,000
B.₹12,000
C.₹10,000
D.₹15,000
Solution
Partnership — C vs A
Capital × time: A = 40,000×12 = 4,80,000; B = 90,000×12 = 10,80,000; C = 1,40,000×6 = 8,40,000.
Ratio = 48 : 108 : 84 = 4 : 9 : 7 — 20 parts in all.
A gets × 80,000 = ₹16,000 and C gets × 80,000 = ₹28,000.
Difference = ₹12,000 — option (b).
58
Three friends share a winning prize in the ratio 3 : 4 : 5. If the total prize is ₹15,000, how much more does the person with the highest share get than the lowest?
A.₹3,500
B.₹3,000
C.₹2,500
D.₹1,250
Solution
Ratio difference
Total parts = 3 + 4 + 5 = 12, so 1 part = = ₹1,250.
Highest = 5 parts and lowest = 3 parts.
Difference = 2 parts = 2 × 1,250.
= ₹2,500 — option (c).
59
A triangular flag has base and height in the ratio 3 : 4. If the area is 216 cm², what are the base and height?
A.16 cm and 12 cm
B.18 cm and 24 cm
C.20 cm and 25 cm
D.18 cm and 20 cm
Solution
Triangle base and height
Let base = 3x and height = 4x.
Area = × 3x × 4x = 6x² = 216.
x² = 36 ⇒ x = 6.
Base = 18 cm and height = 24 cm — option (b).
60
The population of a village increases by 10% annually. If the current population is 72,600, what was it two years ago?
A.60,000
B.59,500
C.62,000
D.61,000
Solution
Population two years ago
Present = past × (1.1)².
72,600 = P × 1.21.
P = .
= 60,000 — option (a).
61
A person invested in two items — one gave an 8% profit and the other a 6% loss. If the total investment was ₹6,000 and the net result was a ₹200 profit, what was the investment in the profit-making item?
A.₹4,000
B.₹3,500
C.₹4,500
D.₹3,000
Solution
Profit and loss mix
Let the profit-making investment be ₹x, so the other is ₹(6,000 − x).
0.08x − 0.06(6,000 − x) = 200.
0.08x − 360 + 0.06x = 200 ⇒ 0.14x = 560.
x = ₹4,000 — option (a).
62
A group of 6 men or 15 women can complete a task in 80 days. Find the time taken by 4 men and 10 women to complete the task.
A.70 days
B.60 days
C.75 days
D.80 days
Solution
Men and women work
6M × 80 = 15W × 80 ⇒ = = , so take M = 5 and W = 2.
Total work = 6 × 5 × 80 = 2400 units.
4 men and 10 women do 4×5 + 10×2 = 40 units a day.
= 60 days — option (b).
63
A machine can fill 110 milk bottles in 5 minutes. How many milk bottles can it fill in 12 minutes?
A.190 bottles
B.250 bottles
C.264 bottles
D.200 bottles
Solution
Unitary method
Rate = = 22 bottles per minute.
In 12 minutes it fills 22 × 12.
= 264.
Hence option (c).
64
The average weight of 12 boxes is 18 kg. If the heaviest box weighs 24 kg and the lightest weighs 12 kg, what is the average weight of the remaining boxes?
A.17 kg
B.18 kg
C.19 kg
D.16 kg
Solution
Average unchanged
Total weight = 12 × 18 = 216 kg.
Removing the two boxes takes away 24 + 12 = 36 kg, leaving 180 kg over 10 boxes.
Average = .
= 18 kg — option (b).
65
A mixture of 200 litres contains acid and water in the ratio 3 : 5. If 40 litres of the mixture is replaced with water, what will be the percentage of acid in the final mixture?
A.29%
B.28%
C.32%
D.30%
Solution
Replacement in mixture
Acid = × 200 = 75 L and water = 125 L.
The 40 L removed carries away × 40 = 15 L of acid, leaving 60 L of acid.
The total stays 200 L after adding water.
× 100 = 30% — option (d).
66
A water tank is shaped like a composite prism: the bottom part is a triangular prism with base area 80 cm² and height 1.5 m, while the top part is a rectangular prism with base 50 cm × 35 cm and height 0.9 m. What is the total volume of the tank in litres? (1 m³ = 1000 L)
A.160.8 L
B.178.5 L
C.169.5 L
D.201.5 L
Solution
Composite prism volume
Triangular part: base area 80 cm² = 0.008 m², so volume = 0.008 × 1.5 = 0.012 m³.
Rectangular part: 0.5 × 0.35 × 0.9 = 0.1575 m³.
Total = 0.012 + 0.1575 = 0.1695 m³.
× 1000 = 169.5 L — option (c).
Find the slope of the line passing through the points (−3, 7) and (5, −9).
A.2
B.−2
C.3
D.−3
Solution
Slope of a line
Slope = .
= .
= .
= −2 — option (b).
69
What is the minimum distance from the origin (0, 0) to a point on the line segment connecting A(6, 0) and B(0, 8)?
A.4.8
B.3.2
C.2.4
D.6.8
Solution
Distance from origin
The line through (6, 0) and (0, 8) is 4x + 3y − 24 = 0.
Perpendicular distance from (0, 0) = .
= .
= 4.8 — option (a).
70
If in △ABC and △DEF, ∠A = ∠D, AB = DE and ∠B = ∠E, which rule proves △ABC ≅ △DEF?
A.SSS
B.SAS
C.ASA
D.AAS
Solution
ASA congruence
Two angles and the side BETWEEN them are given equal.
AB lies between ∠A and ∠B, and DE between ∠D and ∠E.
That is the angle-side-angle condition.
Congruent by ASA — option (c).
71
The value of − is:
A.1 +
B.2
C.−2
D.1 −
Solution
Nested surd
13 − 2 = 7 + 6 − 2 = ( − )², so the root is − .
Its reciprocal, rationalised, is + .
Difference = ( − ) − ( + ).
= −2 — option (c).
72
The line segment from the centre of a circle to the midpoint of a chord is 8 cm long. If the radius is 17 cm, what is the length of the chord?
A.12 cm
B.24 cm
C.36 cm
D.30 cm
Solution
Chord length
That segment is perpendicular to the chord, giving a right triangle.
Half-chord = = .
= = 15 cm.
Chord = 2 × 15 = 30 cm — option (d).
73
A chord of a circle subtends an angle of 80° at the centre. What is the angle subtended at a point on the circle in the same segment?
A.60°
B.40°
C.30°
D.90°
Solution
Angle at circumference
The angle at the centre is twice the angle at any point on the remaining circumference.
So the angle at the circle = .
= 40°.
Hence option (b).
Denominator = 529 + 36 − 64.
= 501.
= 0.00005.
= 5 × 10⁻⁵ — option (a).
Under compound interest the multiple grows in powers.
9 = 3², so the 3-times growth must happen twice over.
Time = 2 × 5.
= 10 years — option (a).
Speed = = 17.5 m/s.
To convert m/s into km/h, multiply by .
17.5 × 3.6.
= 63 km/h — option (a).
Surface area = 4πr² and volume = πr³.
Given 4πr² = 2 × πr³.
Cancel 4πr²: 1 = r.
r = 1.5 units — option (d).
Capital × time: A = 40,000×12 = 4,80,000; B = 90,000×12 = 10,80,000; C = 1,40,000×6 = 8,40,000.
Ratio = 48 : 108 : 84 = 4 : 9 : 7 — 20 parts in all.
A gets × 80,000 = ₹16,000 and C gets × 80,000 = ₹28,000.
Difference = ₹12,000 — option (b).
Total parts = 3 + 4 + 5 = 12, so 1 part = = ₹1,250.
Highest = 5 parts and lowest = 3 parts.
Difference = 2 parts = 2 × 1,250.
= ₹2,500 — option (c).
Let base = 3x and height = 4x.
Area = × 3x × 4x = 6x² = 216.
x² = 36 ⇒ x = 6.
Base = 18 cm and height = 24 cm — option (b).
Present = past × (1.1)².
72,600 = P × 1.21.
P = .
= 60,000 — option (a).
Let the profit-making investment be ₹x, so the other is ₹(6,000 − x).
0.08x − 0.06(6,000 − x) = 200.
0.08x − 360 + 0.06x = 200 ⇒ 0.14x = 560.
x = ₹4,000 — option (a).
6M × 80 = 15W × 80 ⇒ = = , so take M = 5 and W = 2.
Total work = 6 × 5 × 80 = 2400 units.
4 men and 10 women do 4×5 + 10×2 = 40 units a day.
= 60 days — option (b).
Acid = × 200 = 75 L and water = 125 L.
The 40 L removed carries away × 40 = 15 L of acid, leaving 60 L of acid.
The total stays 200 L after adding water.
× 100 = 30% — option (d).
Triangular part: base area 80 cm² = 0.008 m², so volume = 0.008 × 1.5 = 0.012 m³.
Rectangular part: 0.5 × 0.35 × 0.9 = 0.1575 m³.
Total = 0.012 + 0.1575 = 0.1695 m³.
× 1000 = 169.5 L — option (c).
The line through (6, 0) and (0, 8) is 4x + 3y − 24 = 0.
Perpendicular distance from (0, 0) = .
= .
= 4.8 — option (a).
Two angles and the side BETWEEN them are given equal.
AB lies between ∠A and ∠B, and DE between ∠D and ∠E.
That is the angle-side-angle condition.
Congruent by ASA — option (c).
That segment is perpendicular to the chord, giving a right triangle.
Half-chord = = .
= = 15 cm.
Chord = 2 × 15 = 30 cm — option (d).
The angle at the centre is twice the angle at any point on the remaining circumference.
So the angle at the circle = .
= 40°.
Hence option (b).
Hours = 4.2 = = .
Wage = = 84 ÷ .
= 84 × = .
= — option (a).