SSC CGL 25 September 2025 · Shift 3 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 167 of 192
25 September 2025 · Shift 3
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
Mahesh invested an amount of ₹6,000 in a fixed deposit scheme for 3 years at a compound interest rate of 5% p.a. How much will Mahesh get on maturity of the fixed deposit?
Which set contains only non-zero positive numbers?
A.Integers
B.Real Numbers
C.Rational Numbers
D.Natural Numbers
Solution
Natural numbers
Natural numbers are 1, 2, 3, 4, …
They begin at 1, so zero and all negatives are excluded.
Integers, rationals and reals all take in negatives and zero.
Hence option (d).
53
Aman and Ashish travel the same distance at speeds of 15 km/h and 18 km/h respectively. If Aman takes 30 minutes more than Ashish, the distance travelled by each is:
A.42 km
B.45 km
C.40 km
D.36 km
Solution
Time difference
Let the distance be d km; 30 minutes = hour.
− = .
= ⇒ = .
d = 45 km — option (b).
54
If the surface area of a sphere increases by 26.56%, by what percentage does its volume increase?
A.21.23%
B.11.89%
C.33.1%
D.42.4%
Solution
Surface area to volume
Surface area varies as r², so a 26.56% rise is two successive rises of 12.5%: 12.5 + 12.5 + = 26.56 ✓.
So the radius itself grows by 12.5%, i.e. the new radius is times the old.
Volume varies as r³, so it becomes ()³ = times.
Increase = × 100 ≈ 42.4% — option (d).
55
A kitchen wall, measuring 5.6 m in length and 2.4 m in height, needs to be tiled. The tiles are rectangular, 70 cm × 60 cm. If an additional 12% of tiles are required to account for errors and cuts, how many tiles should be purchased?
A.36
B.38
C.40
D.50
Solution
Tiling with extra
Wall area = 5.6 × 2.4 = 13.44 m².
One tile = 0.7 × 0.6 = 0.42 m², so plain requirement = = 32 tiles.
Extra 12% = 32 × = 3.84, i.e. 4 more tiles.
Total = 32 + 4 = 36 — option (a).
56
Three partners A, B and C started a firm with investments in the ratio 5 : 3 : 4. After 8 months, A doubled his capital, B tripled his, and C maintained his investment. If the total profit after one year was ₹2,35,000, what was B's share of profit?
A.₹75,000
B.₹78,000
C.₹77,000
D.₹80,000
Solution
Partnership with changes
Profit share = investment × time, counting the first 8 months and the last 4 separately.
A: 5×8 + 10×4 = 80; B: 3×8 + 9×4 = 60; C: 4×12 = 48.
80 : 60 : 48 = 20 : 15 : 12, a total of 47 units = ₹2,35,000, so 1 unit = ₹5,000.
B has 15 units = ₹75,000 — option (a).
57
An investment is made in a business. In the first year it gains 30%, in the second year it loses 25%, and in the third year it gains 20%. What is the overall gain or loss percentage?
A.20% gain
B.20% loss
C.22% gain
D.17% gain
Solution
Successive change
Take the investment as 100 and multiply the three factors.
100 × × × .
= 100 × 1.3 × 0.75 × 1.2 = 117.
Gain = 17% — option (d).
58
A square and a rectangle have equal areas. The square's side is 22 cm and the rectangle's length is 44 cm. Find its breadth.
A.15 cm
B.16 cm
C.14 cm
D.11 cm
Solution
Equal areas
Area of the square = 22² = 484 cm².
The rectangle has the same area, with length 44 cm.
Breadth = .
= 11 cm — option (d).
59
A composite wall consists of a rectangle 20 m by 8 m with a triangular window of base 15 m and height 8 m cut out. If the wall needs to be painted at ₹111 per m², find the painting cost after deducting the window area.
A.₹12,528
B.₹11,100
C.₹12,000
D.₹11,520
Solution
Area minus a cut-out
Rectangle area = 20 × 8 = 160 m².
Triangular window = × 15 × 8 = 60 m².
Area to paint = 160 − 60 = 100 m².
Cost = 100 × 111 = ₹11,100 — option (b).
60
A regular hexagonal mirror has a side of 20 cm. If 10% of the mirror area is covered by the frame, what is the visible glass area?
A.690.2 cm²
B.710.5 cm²
C.941.15 cm²
D.935.28 cm²
Solution
Hexagon area
Area of a regular hexagon = a².
= × 400 = 600 ≈ 1039.2 cm².
The frame covers 10%, so 90% stays visible.
1039.2 × 0.90 ≈ 935.28 cm² — option (d).
61
Due to an error, a shopkeeper increases the selling price of a ₹400 item by 12.5% instead of decreasing it by 12.5%. What is the percentage increase in the price due to this error?
A man buys 10 shirts for ₹5,000 and sells 8 of them for ₹4,900. If he discards the remaining 2 shirts, what is his overall gain or loss?
A.₹100 loss
B.₹50 loss
C.₹100 profit
D.₹50 profit
Solution
Overall loss
Total cost = ₹5,000 for all ten shirts.
The two discarded shirts fetch nothing, so total revenue = ₹4,900.
5,000 − 4,900 = ₹100.
A loss of ₹100 — option (a).
63
The average of 10 numbers is 64. If two of them are removed and the new average becomes 62, what is the average of the two removed numbers?
A.70
B.71
C.72
D.73
Solution
Removed numbers
Total of 10 numbers = 10 × 64 = 640.
Total of the remaining 8 = 8 × 62 = 496.
The two removed add up to 640 − 496 = 144.
Average = = 72 — option (c).
64
A large storage box in the shape of a right prism with a square base is 2 m tall. If the box holds 720 L of water, what is the side of the base (in cm)?
A.40 cm
B.45 cm
C.60 cm
D.50 cm
Solution
Volume to base side
720 litres = 0.72 m³, since 1000 L = 1 m³.
Volume = base area × height, so base area = = 0.36 m².
Side = = 0.6 m.
= 60 cm — option (c).
65
A departmental store marked up the price of an electronic gadget by 40% above its cost price. During a clearance sale it offered two successive discounts of 20% and 10%. If the final selling price was ₹2,520, what was the cost price of the gadget?
A class of 50 students has an average score of 65. If 10 new students are added and the class average decreases by 1, what is the average score of the 10 new students?
A.59
B.60
C.58
D.62
Solution
Average of new students
Old total = 50 × 65 = 3250.
New average = 64 for 60 students, so new total = 60 × 64 = 3840.
The ten new students together score 3840 − 3250 = 590.
Average = = 59 — option (a).
67
If secA − cosA = 2, then find the value of cosA.
A.−1 +
B.3 +
C. + 2
D.No real value
Solution
Quadratic in cosA
Write secA as and put cosA = c: − c = 2.
Multiply by c: 1 − c² = 2c ⇒ c² + 2c − 1 = 0.
c = = −1 ± .
Taking the value that lies within range, cosA = −1 + — option (a).
68
If cosθ = − and θ lies in the 3rd quadrant, find sinθ.
A.−
B.
C.
D.−
Solution
Sign by quadrant
sin²θ = 1 − cos²θ = 1 − = , so sinθ is in size.
In the third quadrant only tan and cot are positive — sine and cosine are both negative there.
That is why the cosine is given as a negative fraction, and the sine must be negative too.
sinθ = − — option (a).
69
If an angle θ = 3 radians, what percentage of the full circle does it represent?
A.45.74%
B.47.74%
C.50.8%
D.46.83%
Solution
Radians as a fraction
A full circle measures 2π radians.
Fraction = = .
× 100 = 47.746.
≈ 47.74% — option (b).
70
If cosθ = and θ is acute, then find the value of cotθ × tan(90° − θ).
A.1
B.
C.
D.
Solution
cot²θ
tan(90° − θ) = cotθ, so the expression is cotθ × cotθ = cot²θ.
cosθ = gives the (3, 4, 5) triplet, so sinθ = and cotθ = .
cot²θ = ()².
= — option (b).
71
Two circles of radii 5 cm and 3 cm are such that the distance between their centres is 8 cm. How many common tangents can be drawn?
A.1
B.0
C.2
D.3
Solution
Circles touching externally
Here d = 8 and r₁ + r₂ = 5 + 3 = 8, so d = r₁ + r₂.
The circles therefore touch each other externally at exactly one point.
In this case there are two direct common tangents and one at the point of contact.
Total 3 — option (d).
72
A circle has a radius of 12 cm, and a chord makes a 60° angle at the centre. What is the length of that chord?
A.6 cm
B.12 cm
C.12 cm
D.12 cm
Solution
Chord at 60°
The two radii to the chord are equal, so the triangle is isosceles.
With the angle between them 60°, the other two angles are 60° each — the triangle is EQUILATERAL.
So the chord equals the radius.
= 12 cm — option (b).
73
Points P, Q, R and S are located on the circumference of a circle. If ∠RPS = 50° and ∠PQS = 70°, what is the measure of ∠PRS?
A.60°
B.90°
C.50°
D.70°
Solution
Same segment
Angles subtended by the same chord in the same segment are equal.
∠PQS and ∠PRS are both subtended by the chord PS.
So ∠PRS = ∠PQS.
= 70° — option (d).
74
In a circle, points P, Q, R and S lie on the circumference such that chords PR and QS intersect at the centre, and both ∠RPQ and ∠RSQ are subtended by the same arc RQ. If ∠RPQ = 70°, what is the measure of ∠RSQ?
A.80°
B.70°
C.100°
D.110°
Solution
Same arc
Angles subtended by the same arc at the circumference are equal.
Both ∠RPQ and ∠RSQ stand on arc RQ.
So ∠RSQ = ∠RPQ.
= 70° — option (b).
75
A circle with radius r has a tangent line at a point P on the circle. The distance along the tangent from P to the external point Q is 13 cm. The distance from the centre of the circle to point Q is 15 cm. Find the radius of the circle.
A. cm
B. cm
C. cm
D. cm
Solution
Tangent and radius
The radius meets the tangent at right angles, so OPQ is a right triangle with OQ as hypotenuse.
r² + 13² = 15².
r² = 225 − 169 = 56.
r = cm — option (a).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Amount = P(1 + )ᵗ.
= 6,000 × (1.05)³.
= 6,000 × 1.157625.
= ₹6,945.75 — option (a).
Let the distance be d km; 30 minutes = hour.
− = .
= ⇒ = .
d = 45 km — option (b).
Surface area varies as r², so a 26.56% rise is two successive rises of 12.5%: 12.5 + 12.5 + = 26.56 ✓.
So the radius itself grows by 12.5%, i.e. the new radius is times the old.
Volume varies as r³, so it becomes ()³ = times.
Increase = × 100 ≈ 42.4% — option (d).
Wall area = 5.6 × 2.4 = 13.44 m².
One tile = 0.7 × 0.6 = 0.42 m², so plain requirement = = 32 tiles.
Extra 12% = 32 × = 3.84, i.e. 4 more tiles.
Total = 32 + 4 = 36 — option (a).
Profit share = investment × time, counting the first 8 months and the last 4 separately.
A: 5×8 + 10×4 = 80; B: 3×8 + 9×4 = 60; C: 4×12 = 48.
80 : 60 : 48 = 20 : 15 : 12, a total of 47 units = ₹2,35,000, so 1 unit = ₹5,000.
B has 15 units = ₹75,000 — option (a).
Take the investment as 100 and multiply the three factors.
100 × × × .
= 100 × 1.3 × 0.75 × 1.2 = 117.
Gain = 17% — option (d).
Area of the square = 22² = 484 cm².
The rectangle has the same area, with length 44 cm.
Breadth = .
= 11 cm — option (d).
Rectangle area = 20 × 8 = 160 m².
Triangular window = × 15 × 8 = 60 m².
Area to paint = 160 − 60 = 100 m².
Cost = 100 × 111 = ₹11,100 — option (b).
Area of a regular hexagon = a².
= × 400 = 600 ≈ 1039.2 cm².
The frame covers 10%, so 90% stays visible.
1039.2 × 0.90 ≈ 935.28 cm² — option (d).
12.5% = . Correct price = 400 × = ₹350.
Wrong price = 400 × = ₹450.
Excess = 450 − 350 = ₹100, on the correct price of ₹350.
× 100 ≈ 28.57% — option (d).
Total of 10 numbers = 10 × 64 = 640.
Total of the remaining 8 = 8 × 62 = 496.
The two removed add up to 640 − 496 = 144.
Average = = 72 — option (c).
720 litres = 0.72 m³, since 1000 L = 1 m³.
Volume = base area × height, so base area = = 0.36 m².
Side = = 0.6 m.
= 60 cm — option (c).
SP = CP × × × .
= CP × 1.4 × 0.8 × 0.9 = 1.008 CP.
1.008 CP = 2,520 ⇒ CP = .
= ₹2,500 — option (d).
Old total = 50 × 65 = 3250.
New average = 64 for 60 students, so new total = 60 × 64 = 3840.
The ten new students together score 3840 − 3250 = 590.
Average = = 59 — option (a).
Write secA as and put cosA = c: − c = 2.
Multiply by c: 1 − c² = 2c ⇒ c² + 2c − 1 = 0.
c = = −1 ± .
Taking the value that lies within range, cosA = −1 + — option (a).
sin²θ = 1 − cos²θ = 1 − = , so sinθ is in size.
In the third quadrant only tan and cot are positive — sine and cosine are both negative there.
That is why the cosine is given as a negative fraction, and the sine must be negative too.
sinθ = − — option (a).
A full circle measures 2π radians.
Fraction = = .
× 100 = 47.746.
≈ 47.74% — option (b).
tan(90° − θ) = cotθ, so the expression is cotθ × cotθ = cot²θ.
cosθ = gives the (3, 4, 5) triplet, so sinθ = and cotθ = .
cot²θ = ()².
= — option (b).
Here d = 8 and r₁ + r₂ = 5 + 3 = 8, so d = r₁ + r₂.
The circles therefore touch each other externally at exactly one point.
In this case there are two direct common tangents and one at the point of contact.
Total 3 — option (d).
The two radii to the chord are equal, so the triangle is isosceles.
With the angle between them 60°, the other two angles are 60° each — the triangle is EQUILATERAL.
So the chord equals the radius.
= 12 cm — option (b).
Angles subtended by the same chord in the same segment are equal.
∠PQS and ∠PRS are both subtended by the chord PS.
So ∠PRS = ∠PQS.
= 70° — option (d).
The radius meets the tangent at right angles, so OPQ is a right triangle with OQ as hypotenuse.
r² + 13² = 15².
r² = 225 − 169 = 56.
r = cm — option (a).