SSC CGL 22 September 2025 · Shift 1 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 123 of 192
22 September 2025 · Shift 1
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
Rohan invested a certain sum of money at compound interest. The amount becomes three times the original sum in 3 years. What is the rate of interest per annum?
A.44.2%
B.43.3%
C.50%
D.78%
Solution
CI — find rate
Take P = 1 and A = 3 with t = 3 years.
3 = (1 + R)³, so 1 + R = ∛3.
∛3 ≈ 1.4422, hence R ≈ 0.4422.
≈ 44.2% per annum — option (a).
52
Which of the following numbers has a terminating decimal representation?
A.
B.
C.
D.
Solution
Terminating decimal
A fraction terminates only if the denominator, in lowest terms, has just 2s and 5s as prime factors.
= and 8 = 2³ ⇒ 0.375, which terminates ✓.
= , = and = all carry 3s or 7s and repeat.
Hence option (a).
53
Sanjay and Vijay start from the same point and walk at 6 km/h and 3 km/h respectively, in the same direction. What will be the distance between them after 2 hours?
A.4.5 km
B.3 km
C.6 km
D.4 km
Solution
Relative distance
Walking in the same direction, the gap grows at the difference of speeds.
Relative speed = 6 − 3 = 3 km/h.
Distance = 3 × 2.
= 6 km — option (c).
54
A spherical balloon is inflated, causing its radius to grow by 5%. By what percentage does its surface area increase?
A.10.25%
B.20%
C.21.25%
D.25%
Solution
Surface area change
Surface area = 4πr², so it varies as the SQUARE of the radius.
Radii before and after are in the ratio 20 : 21.
Areas are therefore 400 : 441, a rise of 41 parts on 400.
× 100 = 10.25% — option (a).
55
A water tub can hold 12.5 litres of water. How many such tubs are needed to fill a 1 cubic metre tank?
A.80
B.70
C.65
D.90
Solution
Volume conversion
1 cubic metre = 1000 litres.
Number of tubs = .
= 80.
Hence option (a).
56
P and Q started a business by investing ₹60,000 and ₹30,000 respectively. After 3 months R joined with ₹45,000. At the end of the year, what is R's share in a profit of ₹40,000 (approximately)?
A.₹10,909
B.₹10,109
C.₹15,909
D.₹15,109
Solution
Partnership — R's share
Capital × time: P = 60,000×12, Q = 30,000×12 and R = 45,000×9.
Taking units of 15,000 × month: 48 : 24 : 27, which reduces to 16 : 8 : 9 — 33 parts in all.
R’s share = × 40,000.
≈ ₹10,909 — option (a).
57
Two numbers are in the ratio 3 : 5. If the difference between them is 16, find the numbers.
A.24 and 40
B.18 and 35
C.15 and 35
D.18 and 25
Solution
Ratio and difference
Let the numbers be 3x and 5x.
Their difference is 2x = 16.
x = 8, so the numbers are 24 and 40.
Hence option (a).
58
A rectangular sheet has an area equal to a square plot. If the sheet measures 36 m in length and 25 m in breadth, find the perimeter of the square plot.
A.110 m
B.120 m
C.140 m
D.220 m
Solution
Square from equal area
Area of the rectangle = 36 × 25 = 900 m².
The square has the same area, so s² = 900 ⇒ s = 30 m.
Perimeter = 4s = 4 × 30.
= 120 m — option (b).
59
The area of an equilateral triangle increases by 21%. By what percentage did the side length increase?
A.30%
B.32%
C.27%
D.10%
Solution
Area and side %
Area varies as the square of the side, so a rise of x% in the side gives x + x + percent in area.
Try x = 10: 10 + 10 + = 21% ✓.
That matches the given rise exactly.
So the side rose by 10% — option (d).
60
The difference between 10% of a number and 40% of the same number is 120. What is the number?
Amit buys a set of 3 identical watches at a total cost of ₹3,500. He sells them at ₹1,600, ₹1,700 and ₹1,500 respectively. What is the overall profit or loss?
A.₹1,100 loss
B.₹1,300 profit
C.₹1,600 profit
D.₹1,600 loss
Solution
Overall profit
Total selling price = 1,600 + 1,700 + 1,500 = ₹4,800.
Total cost price = ₹3,500.
Since SP > CP, there is a profit of 4,800 − 3,500.
= ₹1,300 profit — option (b).
62
25 men can do a piece of work in 5 days. Find the time required by 10 men to do two times the work.
A.20 days
B.25 days
C.30 days
D.36 days
Solution
Men, days and work
M₁D₁ × 2 = M₂D₂ because the work is doubled.
25 × 5 × 2 = 10 × D.
250 = 10D.
D = 25 days — option (b).
63
Saveena and Tejasvini together can do a piece of work in 15 days. Tejasvini alone can do it in 20 days. In how many days can Saveena alone do it?
A.60
B.62
C.40
D.50
Solution
Work — one alone
Take the work as LCM(15, 20) = 60 units.
Together they do = 4 units a day; Tejasvini alone does = 3 units.
So Saveena does 4 − 3 = 1 unit a day.
Time = = 60 days — option (a).
64
The average salary of 20 workers in a factory is ₹48,000. If the salary of a supervisor is added, the average rises to ₹49,500. What is the supervisor's salary?
A.₹64,000
B.₹74,000
C.₹79,500
D.₹89,500
Solution
Supervisor's salary
Total of 20 workers = 48,000 × 20 = ₹9,60,000.
With the supervisor there are 21 people: 49,500 × 21 = ₹10,39,500.
Supervisor’s salary = 10,39,500 − 9,60,000.
= ₹79,500 — option (c).
65
Two right prisms have the same height of 12 cm. One has a regular hexagonal base with side 5 cm and the other a square base with side 5 cm. What is the ratio of their volumes?
A. : 1
B.3 : 2
C.3 : 5
D.3 : 2
Solution
Prism volumes ratio
Volume = base area × height, and the heights are equal, so only the bases matter.
Hexagon = a² and square = a².
Ratio = : 1.
= 3 : 2 — option (b).
66
A 32-litre mixture of milk and water contains milk and water in the ratio 3 : 5. How much water must be added to make the ratio 3 : 6?
A.3 L
B.4 L
C.5 L
D.7.5 L
Solution
Add water to a mixture
Milk stays fixed, so keep the milk part at 3 units in both ratios.
Initially 3 + 5 = 8 units = 32 L, so 1 unit = 4 L.
Water rises from 5 units to 6 units — a gain of 1 unit.
1 unit = 4 L of water must be added — option (b).
67
If sinA = and A lies in the 2nd quadrant, then find the value of sinA × secA.
A.
B.−
C.
D.−
Solution
Second quadrant
(9, 40, 41) is a Pythagorean triplet, so the base is 40.
In the second quadrant cosine is negative, hence secA = −.
sinA × secA = × (−).
= − — option (b).
68
What is the length of the segment joining (2, 4) and (8, 12)?
A.10
B.12
C.14
D.13
Solution
Distance formula
d = .
= .
= = .
= 10 — option (a).
69
In a triangle ABC, side BC is extended to a point D. The exterior angle at C is 112°. The interior angles at A and B are in the ratio 3 : 4. What is the measure of the largest interior angle of the triangle?
A.60°
B.68°
C.105°
D.15°
Solution
Exterior angle
The exterior angle equals the sum of the two opposite interior angles: 3x + 4x = 112°.
7x = 112° ⇒ x = 16°, so ∠A = 48° and ∠B = 64°.
∠C = 180° − 112° = 68°.
The largest is 68° — option (b).
70
Two triangles △ABC and △DEF are congruent, written as △ABC ≅ △DEF. If ∠A = 40°, what is the measure of ∠D?
A.40°
B.50°
C.60°
D.130°
Solution
Congruent triangles
In congruent triangles the corresponding parts are equal.
The order of letters shows that ∠A corresponds to ∠D.
So ∠D = ∠A.
= 40° — option (a).
71
Given x + = 9, determine the value of .
A.
B.
C.
D.
Solution
x − 1/x from x + 1/x
(x − )² = (x + )² − 4 = 81 − 4 = 77, so x − = .
Divide the required expression by x: = .
= .
Rationalising gives — option (a).
72
A chord of length 8 cm is at a distance of 6 cm away from the centre of a circle. What is the approximate radius of the circle?
A.10 cm
B.14 cm
C.7 cm
D.25 cm
Solution
Chord and radius
The perpendicular from the centre bisects the chord, so half-chord = 4 cm.
r = = .
= ≈ 7.2.
≈ 7 cm — option (c).
73
A chord AB of a circle has a length of 20 cm. The angle subtended by the chord at a point C on the major arc is 15°. Find the radius of the circle.
A.17 cm
B.20 cm
C.12 cm
D.18 cm
Solution
Chord and angle
The angle at the centre is twice the angle at the circumference: ∠AOB = 2 × 15° = 30°.
In △AOB, OA = OB = r and the angle between them is 30°.
Working with this isosceles triangle and the chord of 20 cm gives r = 20 cm.
Hence option (b).
74
If tanA = and A ∈ (0, ), then what is sinA in terms of x and y?
A.
B.
C.
D.
Solution
sinA from tanA
tanA = means perpendicular = x and base = y.
Hypotenuse = .
sinA = .
= — option (c).
75
You earn ₹157.5 in 10.5 hours. Find your hourly wage as a fraction.
A.
B.
C.
D.
Solution
Hourly wage
Hourly wage = .
= .
Multiply both by 10: = 15.
= — option (a).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Take P = 1 and A = 3 with t = 3 years.
3 = (1 + R)³, so 1 + R = ∛3.
∛3 ≈ 1.4422, hence R ≈ 0.4422.
≈ 44.2% per annum — option (a).
A fraction terminates only if the denominator, in lowest terms, has just 2s and 5s as prime factors.
= and 8 = 2³ ⇒ 0.375, which terminates ✓.
= , = and = all carry 3s or 7s and repeat.
Hence option (a).
Walking in the same direction, the gap grows at the difference of speeds.
Relative speed = 6 − 3 = 3 km/h.
Distance = 3 × 2.
= 6 km — option (c).
Surface area = 4πr², so it varies as the SQUARE of the radius.
Radii before and after are in the ratio 20 : 21.
Areas are therefore 400 : 441, a rise of 41 parts on 400.
× 100 = 10.25% — option (a).
Capital × time: P = 60,000×12, Q = 30,000×12 and R = 45,000×9.
Taking units of 15,000 × month: 48 : 24 : 27, which reduces to 16 : 8 : 9 — 33 parts in all.
R’s share = × 40,000.
≈ ₹10,909 — option (a).
Area varies as the square of the side, so a rise of x% in the side gives x + x + percent in area.
Try x = 10: 10 + 10 + = 21% ✓.
That matches the given rise exactly.
So the side rose by 10% — option (d).
M₁D₁ × 2 = M₂D₂ because the work is doubled.
25 × 5 × 2 = 10 × D.
250 = 10D.
D = 25 days — option (b).
Take the work as LCM(15, 20) = 60 units.
Together they do = 4 units a day; Tejasvini alone does = 3 units.
So Saveena does 4 − 3 = 1 unit a day.
Time = = 60 days — option (a).
Total of 20 workers = 48,000 × 20 = ₹9,60,000.
With the supervisor there are 21 people: 49,500 × 21 = ₹10,39,500.
Supervisor’s salary = 10,39,500 − 9,60,000.
= ₹79,500 — option (c).
Milk stays fixed, so keep the milk part at 3 units in both ratios.
Initially 3 + 5 = 8 units = 32 L, so 1 unit = 4 L.
Water rises from 5 units to 6 units — a gain of 1 unit.
1 unit = 4 L of water must be added — option (b).
(9, 40, 41) is a Pythagorean triplet, so the base is 40.
In the second quadrant cosine is negative, hence secA = −.
sinA × secA = × (−).
= − — option (b).
d = .
= .
= = .
= 10 — option (a).
The exterior angle equals the sum of the two opposite interior angles: 3x + 4x = 112°.
7x = 112° ⇒ x = 16°, so ∠A = 48° and ∠B = 64°.
∠C = 180° − 112° = 68°.
The largest is 68° — option (b).
(x − )² = (x + )² − 4 = 81 − 4 = 77, so x − = .
Divide the required expression by x: = .
= .
Rationalising gives — option (a).
The perpendicular from the centre bisects the chord, so half-chord = 4 cm.
r = = .
= ≈ 7.2.
≈ 7 cm — option (c).
The angle at the centre is twice the angle at the circumference: ∠AOB = 2 × 15° = 30°.
In △AOB, OA = OB = r and the angle between them is 30°.
Working with this isosceles triangle and the chord of 20 cm gives r = 20 cm.
Hence option (b).
tanA = means perpendicular = x and base = y.
Hypotenuse = .
sinA = .
= — option (c).