SSC CGL 21 September 2025 · Shift 3 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 119 of 192
21 September 2025 · Shift 3
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
What will come in place of ? to satisfy the equation:
( − 1)² = ? − 2
A.9
B.3
C.5
D.4
Solution
Expanding a square
Expand the left side: (a − b)² = a² − 2ab + b².
()² − 2()(1) + 1² = 8 + 1 − 2.
So 9 − 2 = ? − 2.
? = 9 — option (a).
52
A vendor lends ₹84,000 at a rate of 12% compound interest per annum, compounded annually. Find the interest for the 3rd year.
A.10,000.35
B.11,000
C.12,644.35
D.12,000
Solution
CI for the 3rd year
Interest of the 3rd year = interest earned on the amount standing at the end of 2 years.
Amount after 2 years = 84,000 × (1.12)² = 84,000 × 1.2544 = ₹1,05,369.6.
Third-year interest = 12% of that = 1,05,369.6 × .
= ₹12,644.35 — option (c).
53
If the simple interest on a sum of ₹A at 12% per annum for 4 years is equal to the simple interest on ₹B at 8% per annum for 6 years, then what is the ratio of A to B?
A.1 : 1
B.2 : 3
C.2 : 1
D.1 : 2
Solution
Equal simple interest
SI = , so = .
48A = 48B.
Hence = 1.
A : B = 1 : 1 — option (a).
54
Which of these is a number that is real but NOT rational?
A.
B.7
C.Square root of 5
D.−3
Solution
Real but not rational
A rational number can be written as with q ≠ 0; , 7 and −3 all qualify.
= 2.2360679… — its decimal never ends and never repeats, so it cannot be written as such a fraction.
Yet it does sit on the number line, so it IS a real number — an irrational one.
Hence Square root of 5 — option (c).
55
Manoj crosses an 800 m long street in 8 minutes. What is his speed in km per hour?
A.5.5 km/h
B.5 km/h
C.6 km/h
D.7 km/h
Solution
Speed in km/h
Speed = = = 100 m per minute.
In one hour he covers 100 × 60 = 6,000 m.
That is 6 km.
= 6 km/h — option (c).
56
A pyramid with an equilateral triangular base of side 15 cm and height 20 cm is placed on top of a cube with side length 15 cm. Find the total volume of the composite solid.
A and B invested ₹80,000 and ₹60,000 respectively for 12 months. A takes 10% of the profit for managing. The remaining profit of ₹1,40,000 is divided based on capital. What is B's share?
A.₹60,000
B.₹30,000
C.₹40,000
D.₹35,000
Solution
Partnership — B's share
Both invested for the same 12 months, so the capital ratio decides the split.
80,000 : 60,000 = 4 : 3, i.e. 7 parts in all.
B’s share = × 1,40,000.
= ₹60,000 — option (a).
58
A triangular field with base 80 m and height 60 m shares a side with a rectangle of length 80 m and breadth 50 m. Find the combined area and the percentage of the triangle's area in the total.
A.1,450 m² and 24.5%
B.2,450 m² and 25.5%
C.6,400 m² and 37.5%
D.2,500 m² and 25%
Solution
Combined area and %
Triangle = × 80 × 60 = 2400 m².
Rectangle = 80 × 50 = 4000 m².
Total = 2400 + 4000 = 6400 m²; share of the triangle = × 100.
= 6400 m² and 37.5% — option (c).
59
A regular hexagon is inscribed inside a circle of radius 18 cm. Find the area of the hexagon.
A.438.5 cm²
B.841.8 cm²
C.598.4 cm²
D.638.7 cm²
Solution
Hexagon in a circle
In a regular hexagon inscribed in a circle, each side equals the radius, so a = 18 cm.
Area = 6 × a² = 6 × × 324.
= 486 = 486 × 1.732.
≈ 841.8 cm² — option (b).
60
Three numbers are such that the second is 160% of the first, and the third is 90% of the second. What is the ratio of the first to the third?
A.25 : 36
B.22 : 23
C.25 : 14
D.15 : 8
Solution
Successive percentages
Take the first as 100 ⇒ second = 160.
Third = 90% of 160 = 144.
First : third = 100 : 144.
= 25 : 36 — option (a).
61
A steel plate is in the shape of a right triangle. The lengths of the legs are in the ratio 8 : 15. If the hypotenuse is 17 m, what is the area?
A.30 m²
B.60 m²
C.78 m²
D.50 m²
Solution
Right triangle area
(8, 15, 17) is a Pythagorean triplet, so the legs are exactly 8 m and 15 m.
Area = × base × height.
= × 8 × 15.
= 60 m² — option (b).
62
A number was mistakenly increased by 20% instead of decreased by 25%. By what percent is the final result more than the CORRECT value?
A.56.25%
B.50.5%
C.45%
D.60%
Solution
Percentage error
Take the number as 100. Correct value = 100 × = 75.
Wrong value = 100 × = 120.
Excess = 120 − 75 = 45, measured against the correct 75.
× 100 = 60% — option (d).
63
A wholesaler sells a packet of tea to a retailer at a profit of 25%. The retailer then marks up the price by 40% and offers a discount of 10% to the customer. If the customer pays ₹630, find the cost price of the packet for the wholesaler.
A.₹600
B.₹400
C.₹560
D.₹650
Solution
Chain of profit and discount
Take the wholesaler’s CP = 100 ⇒ the retailer buys at 125.
Marked price = 125 × = 175; after 10% discount the customer pays 175 × = 157.5.
157.5 units = ₹630 ⇒ 1 unit = ₹4.
CP = 100 units = ₹400 — option (b).
64
In a right prism with a square base, volume = 1296 cm³ and height = 9 cm. What is the side of the square base?
A.5 cm
B.8 cm
C.10 cm
D.12 cm
Solution
Square-base prism
Volume = base area × height.
1296 = a² × 9.
a² = 144.
a = 12 cm — option (d).
65
A tap can fill an oil tank in 28 hours. After half the tank is filled, 14 more similar taps are opened. What is the total time taken to fill the oil tank completely?
A.16 hr 55 min
B.15 hr 52 min 20 sec
C.14 hr 56 min
D.17 hr 52 min 30 sec
Solution
Taps in two stages
One tap fills half the tank in = 14 hours.
For the second half there are 15 taps working together, so they take hour.
hour = × 60 = 56 minutes.
Total = 14 hours 56 minutes — option (c).
66
The average of x and y is 60. The average of x, y and z is 80. What is the average of x, y and 2z?
A.250
B.155
C.120
D.165
Solution
Averages
x + y = 2 × 60 = 120 and x + y + z = 3 × 80 = 240.
So z = 240 − 120 = 120.
x + y + 2z = 120 + 240 = 360.
Average = = 120 — option (c).
67
A seller mixes two varieties of rice costing ₹80/kg and ₹120/kg in a certain ratio and sells the mixture at ₹125/kg, gaining 25%. What is the ratio in which the two types are mixed?
A.1 : 1
B.2 : 3
C.3 : 2
D.4 : 1
Solution
Mixture ratio
A 25% gain means CP of the mixture = 125 × = ₹100 per kg.
Alligation: (120 − 100) : (100 − 80) = 20 : 20.
That is 1 : 1.
Hence option (a).
68
If sinA = , then find the value of tanA + secA.
A.
B.
C.
D.
Solution
tanA + secA
Take perpendicular = 1 and hypotenuse = .
Base = = = 4.
tanA = and secA = .
Sum = — option (a).
69
Find the value of cos²90° + sin²60°.
A.
B.
C.
D.
Solution
Standard values
cos90° = 0, so cos²90° = 0.
sin60° = , so sin²60° = .
0 + .
= — option (a).
70
Find the x-intercept of 5x + 2y = 20.
A.2
B.4
C.5
D.7
Solution
x-intercept
At the x-intercept the line meets the x-axis, so y = 0.
5x + 2(0) = 20.
5x = 20.
x = 4 — option (b).
71
If sin(P) = cos(Q), cos(P) = sin(R), and P + Q + R = 90°, then what is the value of angle P?
A.90°
B.60°
C.30°
D.15°
Solution
Complementary angles
sinP = cosQ gives P + Q = 90°, and cosP = sinR gives P + R = 90°.
Adding: 2P + Q + R = 180°.
Substituting Q + R = 90° − P: 2P + 90° − P = 180°.
P = 90° — option (a).
72
The sum of all interior angles of an 18-sided polygon is:
A.2880°
B.3600°
C.3060°
D.3420°
Solution
Interior angles
Sum of interior angles = (n − 2) × 180°.
For n = 18: (18 − 2) × 180°.
= 16 × 180°.
= 2880° — option (a).
73
Point M is at an equal distance from the three vertices of a triangle with coordinates P(0, 0), Q(4, 0) and R(0, 3). What are the coordinates of point M?
A.(2, 1.5)
B.(2, 2)
C.(1.5, 2)
D.(2.5, 2.5)
Solution
Circumcentre
A point equidistant from all three vertices is the circumcentre.
PQ lies along the x-axis and PR along the y-axis, so the angle at P is 90° — the triangle is right-angled.
In a right triangle the circumcentre is the MIDPOINT of the hypotenuse QR.
Midpoint of (4, 0) and (0, 3) = (2, 1.5) — option (a).
74
Simplify:
A.7 − 4
B.7 + 4
C.3 + 2
D.3 − 2
Solution
Rationalising a surd
Write it as and multiply inside by .
Denominator becomes 7² − (4)² = 49 − 48 = 1.
So the inside is (7 + 4)².
Taking the root gives 7 + 4 — option (b).
75
A right-angled triangle ABC has legs AB = 15 cm and BC = 20 cm. An altitude BD is drawn from the vertex B to the hypotenuse AC. What is the length of the altitude BD?
A.12 cm
B.16 cm
C.20 cm
D.15 cm
Solution
Altitude to hypotenuse
(15, 20, 25) is a Pythagorean triplet, so AC = 25 cm.
Area can be written two ways: × 15 × 20 = × 25 × BD.
BD = .
= 12 cm — option (a).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Expand the left side: (a − b)² = a² − 2ab + b².
()² − 2()(1) + 1² = 8 + 1 − 2.
So 9 − 2 = ? − 2.
? = 9 — option (a).
Interest of the 3rd year = interest earned on the amount standing at the end of 2 years.
Amount after 2 years = 84,000 × (1.12)² = 84,000 × 1.2544 = ₹1,05,369.6.
Third-year interest = 12% of that = 1,05,369.6 × .
= ₹12,644.35 — option (c).
SI = , so = .
48A = 48B.
Hence = 1.
A : B = 1 : 1 — option (a).
Speed = = = 100 m per minute.
In one hour he covers 100 × 60 = 6,000 m.
That is 6 km.
= 6 km/h — option (c).
Cube = a³ = 15³ = 3375 cm³.
Pyramid base = × 15² = cm².
Pyramid = × base × height = × × 20 = 375 ≈ 649.5 cm³.
Total = 3375 + 649.5 = 4024.5 cm³ — option (c).
Both invested for the same 12 months, so the capital ratio decides the split.
80,000 : 60,000 = 4 : 3, i.e. 7 parts in all.
B’s share = × 1,40,000.
= ₹60,000 — option (a).
Triangle = × 80 × 60 = 2400 m².
Rectangle = 80 × 50 = 4000 m².
Total = 2400 + 4000 = 6400 m²; share of the triangle = × 100.
= 6400 m² and 37.5% — option (c).
In a regular hexagon inscribed in a circle, each side equals the radius, so a = 18 cm.
Area = 6 × a² = 6 × × 324.
= 486 = 486 × 1.732.
≈ 841.8 cm² — option (b).
Take the first as 100 ⇒ second = 160.
Third = 90% of 160 = 144.
First : third = 100 : 144.
= 25 : 36 — option (a).
(8, 15, 17) is a Pythagorean triplet, so the legs are exactly 8 m and 15 m.
Area = × base × height.
= × 8 × 15.
= 60 m² — option (b).
Take the number as 100. Correct value = 100 × = 75.
Wrong value = 100 × = 120.
Excess = 120 − 75 = 45, measured against the correct 75.
× 100 = 60% — option (d).
Take the wholesaler’s CP = 100 ⇒ the retailer buys at 125.
Marked price = 125 × = 175; after 10% discount the customer pays 175 × = 157.5.
157.5 units = ₹630 ⇒ 1 unit = ₹4.
CP = 100 units = ₹400 — option (b).
One tap fills half the tank in = 14 hours.
For the second half there are 15 taps working together, so they take hour.
hour = × 60 = 56 minutes.
Total = 14 hours 56 minutes — option (c).
x + y = 2 × 60 = 120 and x + y + z = 3 × 80 = 240.
So z = 240 − 120 = 120.
x + y + 2z = 120 + 240 = 360.
Average = = 120 — option (c).
A 25% gain means CP of the mixture = 125 × = ₹100 per kg.
Alligation: (120 − 100) : (100 − 80) = 20 : 20.
That is 1 : 1.
Hence option (a).
Take perpendicular = 1 and hypotenuse = .
Base = = = 4.
tanA = and secA = .
Sum = — option (a).
At the x-intercept the line meets the x-axis, so y = 0.
5x + 2(0) = 20.
5x = 20.
x = 4 — option (b).
A point equidistant from all three vertices is the circumcentre.
PQ lies along the x-axis and PR along the y-axis, so the angle at P is 90° — the triangle is right-angled.
In a right triangle the circumcentre is the MIDPOINT of the hypotenuse QR.
Midpoint of (4, 0) and (0, 3) = (2, 1.5) — option (a).
(15, 20, 25) is a Pythagorean triplet, so AC = 25 cm.
Area can be written two ways: × 15 × 20 = × 25 × BD.
BD = .
= 12 cm — option (a).