SSC CGL 19 September 2025 · Shift 3 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 95 of 192
19 September 2025 · Shift 3
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
If (A + B) : (A − B) = 7 : 3, find (A³ + B³) : (A³ − B³).
A.195 : 148
B.125 : 137
C.133 : 117
D.123 : 117
Solution
Ratio of cubes
Take A + B = 7 and A − B = 3; solving gives A = 5 and B = 2.
A³ + B³ = 125 + 8 = 133.
A³ − B³ = 125 − 8 = 117.
Ratio = 133 : 117 — option (c).
52
If you subtract the square of a number from 2796 and then multiply the result by 13, you get 35711. What is that number?
A.±8
B.±6
C.±9
D.±7
Solution
Equation with a square
13(2796 − x²) = 35711.
2796 − x² = = 2747.
x² = 2796 − 2747 = 49.
x = ±7 — option (d).
53
Choose the correct relation:
(i) < 3.1
(ii) <
(iii) <
A.(i) only
B.(i), (ii) and (iii)
C.(ii) and (iii)
D.(i) and (ii)
Solution
Comparing surds
= 3, and 3 < 3.1 ✓.
A bigger number under the root gives a bigger value, so < ✓.
Likewise < ✓.
All three hold — option (b).
54
A trader marked an article 30% above its cost price. He then allowed a discount of 10% on the marked price. If the selling price of the article was ₹702, what was its cost price?
A.₹600
B.₹880
C.₹950
D.₹550
Solution
Markup and discount
Let CP = 100 ⇒ MP = 130.
After 10% discount, SP = 130 × = 117 units.
117 units = ₹702 ⇒ 1 unit = ₹6.
CP = 100 units = ₹600 — option (a).
55
A and B together earn ₹960 for a job. A works the whole time; B joins after the job is 40% done. B is twice as efficient as A. Find A's share.
A.₹720
B.₹384
C.₹360
D.₹576
Solution
Work and wages
Take the work as 100 units; efficiencies A : B = 1 : 2.
A alone does the first 40 units. The remaining 60 units are shared in the ratio 1 : 2, so B does 40 and A does 20.
A’s total work = 40 + 20 = 60 units out of 100.
A’s share = 960 × = ₹576 — option (d).
56
Given m + = −3, determine the value of m⁵ + − 2(m³ + ) + 4(m + ).
A sum becomes 3.1 times itself in 7 years under simple interest. In how many years will it become 6 times itself at the same rate?
A. years
B. years
C. years
D. years
Solution
SI — multiple of itself
Interest in 7 years = 3.1P − P = 2.1P.
Rate = = 30% per annum.
To become 6 times, the interest needed is 5P: time = .
= years — option (b).
58
A person invested a total sum of ₹2,700 in three different simple-interest schemes at 3%, 4% and 6% per annum. At the end of one year he got the same interest from all three schemes. What was the amount invested at 6%?
A.₹1100
B.₹760
C.₹600
D.₹850
Solution
Equal interest schemes
For equal interest, principal ∝ .
Rates 3 : 4 : 6 give principals : : = 4 : 3 : 2 (9 parts).
9 parts = ₹2,700 ⇒ 1 part = ₹300.
6% scheme = 2 parts = ₹600 — option (c).
59
A ladder reaches a window 9 metres high on a vertical wall. If the ladder is positioned at a 30° angle to the ground, what is its length?
A.18 m
B.20 m
C.32 m
D.12 m
Solution
Ladder at 30°
sin30° = .
= .
Ladder = 9 × 2.
= 18 m — option (a).
60
A solid cylinder has a radius of 8 cm and a height of 15 cm. A smaller cylindrical hole of radius 4 cm is drilled coaxially through its entire length. What is the approximate volume of the remaining solid?
A tank is in the shape of a rectangular parallelepiped of size 30 m × 20 m × 5 m. Its capacity in kilolitres is:
A.7,000 kilolitres
B.3,000 kilolitres
C.5,500 kilolitres
D.3,500 kilolitres
Solution
Tank capacity
Volume = l × b × h = 30 × 20 × 5.
= 3000 m³.
1 m³ = 1000 litres = 1 kilolitre.
= 3,000 kilolitres — option (b).
62
A right circular cone of height 15 cm is cut by two parallel planes at heights 5 cm and 10 cm from the base. What is the ratio of the volumes of the three parts (from top to bottom)?
A.8 : 19 : 27
B.1 : 6 : 20
C.8 : 19 : 64
D.1 : 7 : 19
Solution
Cone cut by two planes
Measured from the apex the three cones have heights 5, 10 and 15 cm — a ratio of 1 : 2 : 3.
Their volumes go as the cubes: 1 : 8 : 27.
Top part = 1; middle = 8 − 1 = 7; bottom = 27 − 8 = 19.
Ratio = 1 : 7 : 19 — option (d).
63
A solid sphere of radius P is melted and recast into n smaller spheres, each of radius q. Find the value of n.
A.()³
B.()³
C.()³
D.()³
Solution
Sphere recast
Volume is conserved: πP³ = n × πq³.
Cancel π from both sides.
n = .
= ()³ — option (a).
64
If tanA = , what is the value of (1 + secA)(1 + cosecA)?
A.
B.
C.
D.
Solution
(1 + secA)(1 + cosecA)
tanA = means A = 30°.
sec30° = and cosec30° = 2.
(1 + )(1 + 2) = 3 × .
= — option (d).
65
If secx + tanx = , what is the value of secx − tanx?
Points on a line
Check (0, −5): 2(0) − 5 = −5 ✓.
Check (3, 1): 2(3) − 5 = 1 ✓.
Check (1, −3): 2(1) − 5 = −3 ✓.
All three lie on the line — option (d).
67
A sector has a central angle of 60° and a radius of 18 cm. Another sector has a central angle of radians. What is the ratio of the area of the first sector to the area of the second sector?
A.4 : 3
B.1 : 2
C.7 : 2
D.2 : 1
Solution
Sector areas
radians = 120°.
With the same radius, sector area is proportional to the angle.
60° : 120°.
= 1 : 2 — option (b).
68
A regular polygon has each interior angle measuring 120°. Find the number of its sides.
A.6
B.1
C.13
D.8
Solution
Interior angle 120°
Interior 120° means each exterior angle = 180° − 120° = 60°.
Number of sides = .
= .
= 6 — option (a).
69
A point P is 17 cm away from the centre of a circle. A tangent drawn from P to the circle has length 15 cm. What is the area of the circle?
A.121π cm²
B.512π cm²
C.64π cm²
D.16π cm²
Solution
Tangent and area
The radius meets the tangent at 90°: 17² = r² + 15².
r² = 289 − 225 = 64, so r = 8 cm.
Area = πr² = π × 64.
= 64π cm² — option (c).
70
In △PQR, an angle bisector from P meets QR at S. If PS bisects ∠QPR and PQ = PR, are △PQS and △PRS congruent? If so, by what rule?
A.Yes, by SSS
B.Yes, by SAS
C.Yes, by ASA
D.No, they are not congruent
Solution
SAS congruence
PQ = PR (given) and PS = PS (common side).
∠QPS = ∠SPR because PS bisects the angle at P.
Two sides and the INCLUDED angle are equal.
Congruent by SAS — option (b).
71
Simplify: +
A.
B.
C.
D.
Solution
Rationalising surds
Add the fractions: numerator = ( + ) + ( − ) = 2.
Denominator = ( − )( + ) = 5 − 2 = 3.
So the sum is .
Hence option (d).
72
If cosecA = y, then what is cot²A in terms of y?
A.y² − 3
B.y² − 2
C.y² − 1
D.2y² − 1
Solution
cot²A from cosecA
The identity is cosec²A − cot²A = 1.
So cot²A = cosec²A − 1.
With cosecA = y, cosec²A = y².
cot²A = y² − 1 — option (c).
73
The angle between two tangents drawn from an external point to a circle is 70°. What is the angle subtended by the chord connecting their points of contact at the centre?
A.80°
B.110°
C.115°
D.60°
Solution
Angle between tangents
The radii meet the tangents at 90° each, forming quadrilateral OMPN.
Its four angles add to 360°: 90° + 90° + 70° + ∠MON = 360°.
∠MON = 360° − 250°.
= 110° — option (b).
74
From a point L outside a circle, a tangent LK and a secant LMN are drawn. If LK = 10 cm and MN = 15 cm, what is the length of the segment LM?
A.6 cm
B.8 cm
C.5 cm
D.11 cm
Solution
Tangent-secant
Tangent-secant relation: LK² = LM × LN.
Let LM = y, so LN = y + 15: 100 = y(y + 15).
y² + 15y − 100 = 0 ⇒ (y + 20)(y − 5) = 0.
y = 5 cm — option (c).
75
What is the value of (0.3³ + 0.03³) ÷ (0.6³ + 0.06³)?
A.0.065
B.0.005
C.0.225
D.0.125
Solution
Cubes — common factor
Numerator = 3³(0.1³ + 0.01³) and denominator = 6³(0.1³ + 0.01³).
The bracket cancels.
= .
= 0.125 — option (d).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Take A + B = 7 and A − B = 3; solving gives A = 5 and B = 2.
A³ + B³ = 125 + 8 = 133.
A³ − B³ = 125 − 8 = 117.
Ratio = 133 : 117 — option (c).
13(2796 − x²) = 35711.
2796 − x² = = 2747.
x² = 2796 − 2747 = 49.
x = ±7 — option (d).
Let CP = 100 ⇒ MP = 130.
After 10% discount, SP = 130 × = 117 units.
117 units = ₹702 ⇒ 1 unit = ₹6.
CP = 100 units = ₹600 — option (a).
Take the work as 100 units; efficiencies A : B = 1 : 2.
A alone does the first 40 units. The remaining 60 units are shared in the ratio 1 : 2, so B does 40 and A does 20.
A’s total work = 40 + 20 = 60 units out of 100.
A’s share = 960 × = ₹576 — option (d).
Square: m² + = 9 − 2 = 7. Cube: m³ + = −27 + 9 = −18.
Multiply: (m³ + )(m² + ) = m⁵ + + m + ⇒ m⁵ + = −126 + 3 = −123.
Now substitute: −123 − 2(−18) + 4(−3).
= −123 + 36 − 12 = −99 — option (a).
Interest in 7 years = 3.1P − P = 2.1P.
Rate = = 30% per annum.
To become 6 times, the interest needed is 5P: time = .
= years — option (b).
For equal interest, principal ∝ .
Rates 3 : 4 : 6 give principals : : = 4 : 3 : 2 (9 parts).
9 parts = ₹2,700 ⇒ 1 part = ₹300.
6% scheme = 2 parts = ₹600 — option (c).
sin30° = .
= .
Ladder = 9 × 2.
= 18 m — option (a).
Big cylinder = πr²h = π × 64 × 15 = 960π.
Hole = π × 16 × 15 = 240π.
Remaining = 960π − 240π = 720π = 720 × .
≈ 2263 cm³ — option (d).
Measured from the apex the three cones have heights 5, 10 and 15 cm — a ratio of 1 : 2 : 3.
Their volumes go as the cubes: 1 : 8 : 27.
Top part = 1; middle = 8 − 1 = 7; bottom = 27 − 8 = 19.
Ratio = 1 : 7 : 19 — option (d).
Volume is conserved: πP³ = n × πq³.
Cancel π from both sides.
n = .
= ()³ — option (a).
radians = 120°.
With the same radius, sector area is proportional to the angle.
60° : 120°.
= 1 : 2 — option (b).
Interior 120° means each exterior angle = 180° − 120° = 60°.
Number of sides = .
= .
= 6 — option (a).
The radius meets the tangent at 90°: 17² = r² + 15².
r² = 289 − 225 = 64, so r = 8 cm.
Area = πr² = π × 64.
= 64π cm² — option (c).
PQ = PR (given) and PS = PS (common side).
∠QPS = ∠SPR because PS bisects the angle at P.
Two sides and the INCLUDED angle are equal.
Congruent by SAS — option (b).
The radii meet the tangents at 90° each, forming quadrilateral OMPN.
Its four angles add to 360°: 90° + 90° + 70° + ∠MON = 360°.
∠MON = 360° − 250°.
= 110° — option (b).
Tangent-secant relation: LK² = LM × LN.
Let LM = y, so LN = y + 15: 100 = y(y + 15).
y² + 15y − 100 = 0 ⇒ (y + 20)(y − 5) = 0.
y = 5 cm — option (c).
Numerator = 3³(0.1³ + 0.01³) and denominator = 6³(0.1³ + 0.01³).
The bracket cancels.
= .
= 0.125 — option (d).