SSC CGL 22 September 2025 · Shift 2 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 127 of 192
22 September 2025 · Shift 2
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
What will come in place of ? to satisfy the equation:
( − 2)² = ? − 4
A.12
B.11
C.10
D.13
Solution
Expanding a square
Expand the left side: (a − b)² = a² − 2ab + b².
()² − 2()(2) + 2² = 7 + 4 − 4.
So 11 − 4 = ? − 4.
? = 11 — option (b).
52
What will be the compound interest on a sum of ₹21,000 after 3 years at the rate of 10% per annum?
A.₹6,951
B.₹6,421
C.₹6,541
D.₹6,351
Solution
CI for 3 years
Amount = 21,000(1 + )³.
= 21,000 × = ₹27,951.
CI = 27,951 − 21,000.
= ₹6,951 — option (a).
53
If the simple interest on a certain sum of money for 3 years at 8% per annum is ₹480, what would be the simple interest on the same sum for 5 years at 12% per annum?
A.₹960
B.₹1,200
C.₹1,000
D.₹1,440
Solution
Simple interest
SI = , so 480 = ⇒ P = ₹2,000.
New SI = .
= .
= ₹1,200 — option (b).
54
The decimal 0.785 is best classified as:
A.Whole Number
B.Irrational Number
C.Integer
D.Rational Number
Solution
Rational number
0.785 is a terminating decimal, so it can be written as .
Any number expressible as with q ≠ 0 is RATIONAL.
It is not a whole number or an integer since it has a fractional part.
Hence option (d).
55
A cyclist covers a distance of 450 metres in 25 seconds. What is his speed in km/h?
A.64.8 km/h
B.42 km/h
C.45 km/h
D.48 km/h
Solution
Speed conversion
Speed = = 18 m/s.
To convert m/s into km/h, multiply by .
18 × 3.6.
= 64.8 km/h — option (a).
56
A pyramid is inscribed inside a cube with edge 9 cm, sharing the base and with its apex at the centre of the top face. Find the volume of the pyramid.
A.180 cm³
B.220 cm³
C.243 cm³
D.270 cm³
Solution
Pyramid in a cube
Base area = 9² = 81 cm² and the height equals the cube edge, 9 cm.
Volume of a pyramid = × base area × height.
= × 81 × 9 = 81 × 3.
= 243 cm³ — option (c).
57
A, B and C invest in the ratio 5 : 6 : 7. The time ratio is 4 : 3 : 2. What is the ratio of their profits?
A rectangular park has an area of 720 square metres. If its length is increased by 8 metres and its width decreased by 3 metres, the area remains unchanged. Find the original dimensions of the park.
A.Length = 45 m, Width = 16 m
B.Length = 40 m, Width = 18 m
C.Length = 48 m, Width = 15 m
D.Length = 36 m, Width = 20 m
Solution
Unchanged area
We need l × b = 720 and (l + 8)(b − 3) = 720.
Test option (b): 40 × 18 = 720 ✓.
(40 + 8)(18 − 3) = 48 × 15 = 720 ✓.
Both conditions hold — option (b).
59
A garden is in the shape of a regular hexagon with each side 30 m. If 5% of it is occupied by a circular fountain, find the remaining area.
A.2,131.5 m²
B.2,351.9 m²
C.2,221.3 m²
D.2,421.7 m²
Solution
Hexagonal garden
Hexagon area = a² = × 900 = 1350 m².
Remaining = 95% of that = 1350 × = 1282.5.
1282.5 × 1.732.
≈ 2,221.3 m² — option (c).
60
The weights of three friends are in the ratio 4 : 5 : 6. If each gains 9 kg, the new ratio becomes 13 : 14 : 15. What is the original weight of the heaviest friend?
A.5 kg
B.7 kg
C.6 kg
D.9 kg
Solution
Ratio after adding
Let the weights be 4x, 5x and 6x, so the total is 15x.
After each gains 9 kg the total becomes 15x + 27, and the new parts add to 42.
Matching: 15x corresponds to 15 parts and 27 kg to the extra 27 parts ⇒ 1 part = 1 kg, so x = 1.
The heaviest was 6x = 6 kg — option (c).
61
A triangular field has sides in the ratio 3 : 5 : 7 and a perimeter of 90 m. What is the area of the field?
A.196 m²
B.234 m²
C.186 m²
D.182 m²
Solution
Heron's formula
3x + 5x + 7x = 90 ⇒ 15x = 90 ⇒ x = 6, so the sides are 18, 30 and 42 m.
s = = 45 m.
Area = = .
≈ 234 m² — option (b).
62
The value of a machine depreciates 10% in the first year, 20% in the second year and 33.33% in the third year. If the initial value was ₹20,000, what is its value at the end of 3 years?
A.₹9,200
B.₹9,700
C.₹9,500
D.₹9,600
Solution
Successive depreciation
Multiply the surviving fractions: × × .
= , i.e. 150 units become 72 units.
150 units = ₹20,000, so 72 units = .
= ₹9,600 — option (d).
63
An electronics dealer marks an item 50% above cost and offers a festival discount of 20%, plus a cashback of ₹200 on the final selling price. If he still makes a 15% profit, what is the cost price?
A.₹4,300
B.₹4,000
C.₹4,100
D.₹4,200
Solution
Markup, discount, cashback
Take CP = 100 units ⇒ MP = 150; after 20% discount the price is 120 units.
Cashback of ₹200 leaves 120 − 200, and this must equal a 15% profit, i.e. 115 units.
120 − 115 = 5 units = ₹200 ⇒ 1 unit = ₹40.
CP = 100 units = ₹4,000 — option (b).
64
Ravi can do a job in the same amount of time as Sonu and Monu together. If Ravi and Sonu together can do it in 12 days and Monu alone can do it in 48 days, in how many days can Sonu do it alone?
A.16 days
B.30 days
C.24 days
D.32 days
Solution
Work — three persons
Take the work as 48 units, so Monu does 1 unit a day and Ravi + Sonu do 4 units a day.
Ravi = Sonu + Monu, so Ravi − Sonu = Monu = 1 unit.
Adding to Ravi + Sonu = 4: Ravi = 2.5 and Sonu = 1.5 units a day.
Sonu alone: = 32 days — option (d).
65
A storage box is in the shape of a right rectangular prism with internal dimensions 60 cm × 50 cm × 40 cm. How many litres can it hold?
A trader bought 5 varieties of rice whose weights are in the ratio 3 : 4 : 5 : 6 : 7. Their respective prices per kg are ₹42, ₹45, ₹47, ₹48 and ₹50. What is the average cost per kg of the mixture?
An alloy contains gold and silver in the ratio 5 : 3. A second alloy contains gold and silver in the ratio 3 : 1. What quantity of the first alloy must be added to 40 kg of the second alloy to form a new alloy containing gold and silver in the ratio 4 : 2?
A.20 kg
B.60 kg
C.80 kg
D.40 kg
Solution
Mixing two alloys
Express the gold fractions: first alloy , second and the target = .
By alligation on gold: (first : second) = ( − ) : ( − ) = : = 2 : 1.
So the first alloy must be twice the second by weight.
2 × 40 = 80 kg — option (c).
68
If sinA = , find the value of (15sinA + 20cosA)² + (5sinA + 10cosA)².
sin³β + cos³β
sinβ = cosβ holds at β = 45°.
sin45° = cos45° = .
Each cube is , so the sum is = .
Rationalising gives — option (a).
70
Are these lines y = 3x + 2 and y = −x + 5 perpendicular?
A.Yes
B.No
C.Only if they intersect at the origin
D.Cannot be determined
Solution
Perpendicular lines
Two lines are perpendicular when the product of their slopes is −1.
Here m₁ = 3 and m₂ = −.
3 × (−) = −1 ✓.
So they are perpendicular — option (a).
71
If tan2θ = cot(3θ + 10°), then what is the value of θ?
Two angles are supplementary. The measure of one angle is 30° less than twice the measure of the other. What is the measure of the larger angle?
A.110°
B.125°
C.170°
D.105°
Solution
Supplementary angles
Let the angles be A and B with A + B = 180° and A = 2B − 30°.
(2B − 30°) + B = 180° ⇒ 3B = 210° ⇒ B = 70°.
A = 180° − 70° = 110°.
The larger angle is 110° — option (a).
73
In a triangle ABC, the exterior angle bisectors of ∠B and ∠C meet at point E (the excentre opposite A). If ∠BEC = 50°, what is the measure of ∠A?
A.110°
B.100°
C.80°
D.70°
Solution
Excentre angle
For the excentre opposite A, ∠BEC = 90° − ∠A.
50° = 90° − ∠A.
∠A = 40°.
∠A = 80° — option (c).
74
Simplify: ( + )²
A.8 + 2
B.8 − 2
C.3 + 2
D.3 − 2
Solution
Square of a surd sum
(a + b)² = a² + b² + 2ab.
()² + ()² + 2 × × .
= 5 + 3 + 2.
= 8 + 2 — option (a).
75
In triangle LMN, a line segment OP is drawn parallel to the side MN, with O on LM and P on LN. If LO = 4 cm, LM = 12 cm and LP = 5 cm, find the length of PN.
A.11 cm
B.12 cm
C.15 cm
D.10 cm
Solution
Thales' theorem
OM = LM − LO = 12 − 4 = 8 cm.
Since OP ∥ MN, Thales’ theorem gives = .
= .
PN = 10 cm — option (d).
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()(2) + 2² = 7 + 4 − 4.
So 11 − 4 = ? − 4.
? = 11 — option (b).
Amount = 21,000(1 + )³.
= 21,000 × = ₹27,951.
CI = 27,951 − 21,000.
= ₹6,951 — option (a).
SI = , so 480 = ⇒ P = ₹2,000.
New SI = .
= .
= ₹1,200 — option (b).
Speed = = 18 m/s.
To convert m/s into km/h, multiply by .
18 × 3.6.
= 64.8 km/h — option (a).
Base area = 9² = 81 cm² and the height equals the cube edge, 9 cm.
Volume of a pyramid = × base area × height.
= × 81 × 9 = 81 × 3.
= 243 cm³ — option (c).
Profit share ∝ capital × time.
A: 5 × 4 = 20, B: 6 × 3 = 18, C: 7 × 2 = 14.
20 : 18 : 14, divided by 2.
= 10 : 9 : 7 — option (d).
We need l × b = 720 and (l + 8)(b − 3) = 720.
Test option (b): 40 × 18 = 720 ✓.
(40 + 8)(18 − 3) = 48 × 15 = 720 ✓.
Both conditions hold — option (b).
Hexagon area = a² = × 900 = 1350 m².
Remaining = 95% of that = 1350 × = 1282.5.
1282.5 × 1.732.
≈ 2,221.3 m² — option (c).
Let the weights be 4x, 5x and 6x, so the total is 15x.
After each gains 9 kg the total becomes 15x + 27, and the new parts add to 42.
Matching: 15x corresponds to 15 parts and 27 kg to the extra 27 parts ⇒ 1 part = 1 kg, so x = 1.
The heaviest was 6x = 6 kg — option (c).
3x + 5x + 7x = 90 ⇒ 15x = 90 ⇒ x = 6, so the sides are 18, 30 and 42 m.
s = = 45 m.
Area = = .
≈ 234 m² — option (b).
Multiply the surviving fractions: × × .
= , i.e. 150 units become 72 units.
150 units = ₹20,000, so 72 units = .
= ₹9,600 — option (d).
Take CP = 100 units ⇒ MP = 150; after 20% discount the price is 120 units.
Cashback of ₹200 leaves 120 − 200, and this must equal a 15% profit, i.e. 115 units.
120 − 115 = 5 units = ₹200 ⇒ 1 unit = ₹40.
CP = 100 units = ₹4,000 — option (b).
Take the work as 48 units, so Monu does 1 unit a day and Ravi + Sonu do 4 units a day.
Ravi = Sonu + Monu, so Ravi − Sonu = Monu = 1 unit.
Adding to Ravi + Sonu = 4: Ravi = 2.5 and Sonu = 1.5 units a day.
Sonu alone: = 32 days — option (d).
Volume = 60 × 50 × 40 = 1,20,000 cm³.
1 litre = 1000 cm³.
.
= 120 litres — option (b).
Express the gold fractions: first alloy , second and the target = .
By alligation on gold: (first : second) = ( − ) : ( − ) = : = 2 : 1.
So the first alloy must be twice the second by weight.
2 × 40 = 80 kg — option (c).
(3, 4, 5) is a Pythagorean triplet, so cosA = .
15 × + 20 × = 9 + 16 = 25.
5 × + 10 × = 3 + 8 = 11.
25² + 11² = 625 + 121 = 746 — option (a).
sinβ = cosβ holds at β = 45°.
sin45° = cos45° = .
Each cube is , so the sum is = .
Rationalising gives — option (a).
Two lines are perpendicular when the product of their slopes is −1.
Here m₁ = 3 and m₂ = −.
3 × (−) = −1 ✓.
So they are perpendicular — option (a).
tanA = cot(90° − A), so tan2θ = cot(90° − 2θ).
Hence 90° − 2θ = 3θ + 10°.
5θ = 80°.
θ = 16° — option (c).
Let the angles be A and B with A + B = 180° and A = 2B − 30°.
(2B − 30°) + B = 180° ⇒ 3B = 210° ⇒ B = 70°.
A = 180° − 70° = 110°.
The larger angle is 110° — option (a).
For the excentre opposite A, ∠BEC = 90° − ∠A.
50° = 90° − ∠A.
∠A = 40°.
∠A = 80° — option (c).
OM = LM − LO = 12 − 4 = 8 cm.
Since OP ∥ MN, Thales’ theorem gives = .
= .
PN = 10 cm — option (d).