SSC CGL 21 September 2025 · Shift 1 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 111 of 192
21 September 2025 · Shift 1
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
What is the smallest number that must be added to 5600 to make it a perfect square?
A.23
B.24
C.25
D.26
Solution
Nearest perfect square
Find the squares on either side of 5600.
74² = 5476 and 75² = 5625.
5625 − 5600 = 25.
So 25 must be added — option (c).
52
A company produced 6000 chocolates at a total cost of ₹1,80,000. They distributed 600 chocolates free as samples. For the remaining chocolates they offered a 10% discount on the market price of ₹40 per chocolate, and additionally gave 1 chocolate free with every 19 sold. What is the profit or loss percentage?
A.5.6% profit
B.2.6% profit
C.2.6% loss
D.5.6% loss
Solution
Chocolates — profit %
Chocolates left after the free samples = 6000 − 600 = 5400.
Price after 10% discount = 40 × = ₹36; with 1 free in every 20 handed over, the effective price = 36 × = ₹34.20.
Revenue = 5400 × 34.20 = ₹1,84,680 against a cost of ₹1,80,000.
Profit% = × 100 = 2.6% profit — option (b).
53
What is the compound interest on ₹12,000 for 2 years at 10% per annum?
A.₹2,425
B.₹2,520
C.₹2,540
D.₹2,625
Solution
CI for 2 years
Amount = 12,000(1 + )².
= 12,000 × = ₹14,520.
CI = 14,520 − 12,000.
= ₹2,520 — option (b).
54
A certain sum becomes 6 times itself in 7 years at simple interest. In how many years will it become 14 times itself at the same rate?
A.18 years
B.20 years
C.22 years
D.16 years
Solution
SI — multiples
Under simple interest, = .
= ⇒ = .
t₂ = .
= 18 years — option (a).
55
Which of the following is a perfect square and an integer?
A.Square root of 7
B.Square root of 16
C.4.3
D.5.4
Solution
Perfect square integer
= 4, a whole number.
is an irrational number, roughly 2.646.
4.3 and 5.4 are decimals, not integers.
Hence option (b).
56
The speed of a car is 90 km/h. Find the distance covered by the car in 80 seconds.
If the ratio of the capital of A and B is 4 : 2 and the time for which they invested is in the ratio 2 : 4, then the profit ratio is:
A.1 : 1
B.4 : 1
C.5 : 2
D.3 : 1
Solution
Capital × time
Profit share ∝ capital × time.
A: 4 × 2 = 8 and B: 2 × 4 = 8.
Ratio = 8 : 8.
= 1 : 1 — option (a).
58
A company makes a gift box in the shape of a right pyramid with a square base. The base side is 8 cm and its height is 15 cm. Due to safety limits the box can be filled only up to 80% of its total volume. If 1 cm³ of material costs ₹0.60, what will be the cost to fill one such box?
A.₹153.6
B.₹163.6
C.₹182.4
D.₹192.3
Solution
Pyramid volume + cost
Base area = 8² = 64 cm².
Volume of a pyramid = × base area × height = × 64 × 15 = 320 cm³.
Usable volume = 80% of 320 = 256 cm³.
Cost = 256 × 0.60 = ₹153.6 — option (a).
59
Consider a right trapezoidal field. Its two parallel sides are 40 metres and 60 metres long. One of its non-parallel sides, which acts as the height, measures 20 metres. What is the area of this field?
A.900 sq. m
B.1000 sq. m
C.1220 sq. m
D.1100 sq. m
Solution
Trapezium area
Area of a trapezium = (sum of parallel sides) × height.
= (40 + 60) × 20.
= × 100 × 20.
= 1000 sq. m — option (b).
60
A regular hexagon is drawn inside a circle of radius 14 cm. Find the approximate area of the hexagon.
A.500 cm²
B.509 cm²
C.520 cm²
D.530 cm²
Solution
Hexagon in a circle
In a regular hexagon inscribed in a circle, the side equals the radius.
Area = a² = × 14².
= × 196.
≈ 509 cm² — option (b).
61
In a school the number of boys and girls is in the ratio 5 : 4. After one year the number of boys increases by 20% and the number of girls by 10%. What will be the new ratio of boys to girls?
A.11 : 9
B.10 : 8
C.25 : 22
D.15 : 11
Solution
Ratio after % change
Take boys = 5x and girls = 4x.
New boys = 5x × 1.20 = 6x; new girls = 4x × 1.10 = 4.4x.
Ratio = 6 : 4.4 = 60 : 44.
= 15 : 11 — option (d).
62
A triangle has sides 4 cm and 8 cm, and the angle between them is 120°. What is its area?
A.10 cm²
B.8 cm²
C.6 cm²
D.12 cm²
Solution
Area with included angle
Area = ab sinθ.
= × 4 × 8 × sin120°.
sin120° = , so area = 16 × .
= 8 cm² — option (b).
63
A man spends 60% of his income. His income increases by 15% but his expenditure remains the same. By what percent have his savings increased?
A.37.5%
B.36.5%
C.35.25%
D.38.5%
Solution
Savings increase
Take income = 100, so expenditure = 60 and savings = 40.
New income = 115 with the same expenditure 60, so new savings = 55.
Increase = 55 − 40 = 15.
× 100 = 37.5% — option (a).
64
Sameer can complete a work in 18 days. In how many days will the work be completed by Ajay, if his efficiency is 80% more than that of Sameer?
A.14
B.10
C.12
D.20
Solution
Efficiency and time
Efficiency ratio Ajay : Sameer = 180 : 100 = 9 : 5.
Total work = 18 × 5 = 90 units (taking Sameer as 5 units/day).
Ajay does 9 units a day.
Time = = 10 days — option (b).
65
A right prism has a base in the shape of a trapezium with parallel sides 8 cm and 10 cm, and height 6 cm. If the prism height is 18 cm, what is the volume?
A.972 cm³
B.776 cm³
C.820 cm³
D.962 cm³
Solution
Trapezoidal prism
Base area = (8 + 10) × 6 = × 18 × 6 = 54 cm².
Volume of a prism = base area × prism height.
= 54 × 18.
= 972 cm³ — option (a).
66
The average of a group of 9 numbers is 51. If the first four numbers have an average of 47 and the last four numbers have an average of 55, what is the fifth number?
A.52
B.51
C.44
D.50
Solution
Averages overlap
Total of all nine = 9 × 51 = 459.
First four = 4 × 47 = 188 and last four = 4 × 55 = 220.
Those eight add up to 408.
Fifth number = 459 − 408 = 51 — option (b).
67
A chemist prepares 80 litres of antiseptic solution by mixing two liquids 'A' and 'B' in the ratio 3 : 5. If the cost of 'A' is ₹55 per litre and 'B' is ₹40 per litre, what is the approximate average cost per litre of the final solution?
A.₹46
B.₹44
C.₹47
D.₹49
Solution
Average cost of mixture
A = × 80 = 30 L and B = × 80 = 50 L.
Total cost = 30 × 55 + 50 × 40 = 1650 + 2000 = ₹3,650.
Average = = 45.625.
≈ ₹46 — option (a).
68
If sinA + cosA = , then find the value of sin2A.
A.1
B.0
C.−1
D.
Solution
sin2A from sinA + cosA
Square both sides: (sinA + cosA)² = 2.
sin²A + cos²A + 2 sinA cosA = 2.
Since sin²A + cos²A = 1, we get 2 sinA cosA = 1.
sin2A = 2 sinA cosA = 1 — option (a).
Line through a point
Substitute the point: 7 = m(3) + 1.
3m = 7 − 1 = 6.
m = .
= 2 — option (a).
71
If cosA = and A is an acute angle, find tan(90° − A).
A.
B.
C.
D.
Solution
tan(90° − A)
cosA = gives the triplet (3, 4, 5), so sinA = and tanA = .
tan(90° − A) = cotA.
cotA = = .
Hence option (a).
72
If the measure of an exterior angle of a regular polygon is 30°, determine the total number of its sides.
A.10
B.12
C.8
D.6
Solution
Exterior angle
The exterior angles of any polygon add up to 360°.
Number of sides = .
= .
= 12 — option (b).
73
In a triangle with vertices A(x, y), B(6, 2) and C(3, 7), the centroid is at G(4, 4). Find the coordinates of vertex A.
A.(3, 3)
B.(4, 2)
C.(3, 4)
D.(2, 3)
Solution
Vertex from centroid
Centroid = (, ).
x: 4 = ⇒ 12 = x + 9 ⇒ x = 3.
y: 4 = ⇒ 12 = y + 9 ⇒ y = 3.
A = (3, 3) — option (a).
74
If x = 3 + , find the value of x² − 6x + 7.
A.3 + 2
B.4 +
C.0
D.2
Solution
Surd substitution
x − 3 = , so squaring gives x² − 6x + 9 = 2.
That is x² − 6x = 2 − 9 = −7.
Now x² − 6x + 7 = −7 + 7.
= 0 — option (c).
75
In △PQR a median PS is drawn to the side QR. What is the ratio of the area of △PQS to the area of △PRS?
A.1 : 1
B.3 : 2
C.2 : 1
D.1 : 2
Solution
Median divides area
A median bisects the opposite side, so QS = SR.
Both smaller triangles have the same height from P.
Equal bases with the same height give equal areas.
Ratio = 1 : 1 — option (a).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Find the squares on either side of 5600.
74² = 5476 and 75² = 5625.
5625 − 5600 = 25.
So 25 must be added — option (c).
Chocolates left after the free samples = 6000 − 600 = 5400.
Price after 10% discount = 40 × = ₹36; with 1 free in every 20 handed over, the effective price = 36 × = ₹34.20.
Revenue = 5400 × 34.20 = ₹1,84,680 against a cost of ₹1,80,000.
Profit% = × 100 = 2.6% profit — option (b).
Amount = 12,000(1 + )².
= 12,000 × = ₹14,520.
CI = 14,520 − 12,000.
= ₹2,520 — option (b).
Under simple interest, = .
= ⇒ = .
t₂ = .
= 18 years — option (a).
Convert to m/s: 90 × = 25 m/s.
Distance = speed × time.
= 25 × 80.
= 2000 m — option (a).
Profit share ∝ capital × time.
A: 4 × 2 = 8 and B: 2 × 4 = 8.
Ratio = 8 : 8.
= 1 : 1 — option (a).
Base area = 8² = 64 cm².
Volume of a pyramid = × base area × height = × 64 × 15 = 320 cm³.
Usable volume = 80% of 320 = 256 cm³.
Cost = 256 × 0.60 = ₹153.6 — option (a).
Area of a trapezium = (sum of parallel sides) × height.
= (40 + 60) × 20.
= × 100 × 20.
= 1000 sq. m — option (b).
In a regular hexagon inscribed in a circle, the side equals the radius.
Area = a² = × 14².
= × 196.
≈ 509 cm² — option (b).
Take boys = 5x and girls = 4x.
New boys = 5x × 1.20 = 6x; new girls = 4x × 1.10 = 4.4x.
Ratio = 6 : 4.4 = 60 : 44.
= 15 : 11 — option (d).
Area = ab sinθ.
= × 4 × 8 × sin120°.
sin120° = , so area = 16 × .
= 8 cm² — option (b).
Take income = 100, so expenditure = 60 and savings = 40.
New income = 115 with the same expenditure 60, so new savings = 55.
Increase = 55 − 40 = 15.
× 100 = 37.5% — option (a).
Efficiency ratio Ajay : Sameer = 180 : 100 = 9 : 5.
Total work = 18 × 5 = 90 units (taking Sameer as 5 units/day).
Ajay does 9 units a day.
Time = = 10 days — option (b).
Base area = (8 + 10) × 6 = × 18 × 6 = 54 cm².
Volume of a prism = base area × prism height.
= 54 × 18.
= 972 cm³ — option (a).
Total of all nine = 9 × 51 = 459.
First four = 4 × 47 = 188 and last four = 4 × 55 = 220.
Those eight add up to 408.
Fifth number = 459 − 408 = 51 — option (b).
A = × 80 = 30 L and B = × 80 = 50 L.
Total cost = 30 × 55 + 50 × 40 = 1650 + 2000 = ₹3,650.
Average = = 45.625.
≈ ₹46 — option (a).
cosA = gives the triplet (3, 4, 5), so sinA = and tanA = .
tan(90° − A) = cotA.
cotA = = .
Hence option (a).
The exterior angles of any polygon add up to 360°.
Number of sides = .
= .
= 12 — option (b).
Centroid = (, ).
x: 4 = ⇒ 12 = x + 9 ⇒ x = 3.
y: 4 = ⇒ 12 = y + 9 ⇒ y = 3.
A = (3, 3) — option (a).
A median bisects the opposite side, so QS = SR.
Both smaller triangles have the same height from P.
Equal bases with the same height give equal areas.
Ratio = 1 : 1 — option (a).