SSC CGL 17 September 2025 · Shift 1 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 63 of 192
17 September 2025 · Shift 1
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
Simplify:
A.2 +
B.3 +
C. + 2
D.9 +
Solution
Nested surd
Write the inside as a perfect square: 14 + 6 = 9 + 5 + 2 × 3 × .
That is (3)² + ()² + 2(3)() = (3 + )².
Taking the square root gives 3 + .
Hence option (b).
52
Given that W : X = 3 : 5, X : Y = 5 : 7, and Y : Z = 7 : 9, find the compound ratio W : X : Y : Z.
A.20 : 24 : 56 : 63
B.3 : 5 : 7 : 9
C.5 : 6 : 14 : 9
D.20 : 24 : 42 : 63
Solution
Chained ratios
The linking terms already match: X is 5 in both, Y is 7 in both.
So the chain joins directly without scaling.
W : X : Y : Z = 3 : 5 : 7 : 9.
Hence option (b).
53
What is the value of
A.4
B.7
C.5
D.6
Solution
Infinite nested root
Let the value be x: x = ⇒ x² = x + 20.
x² − x − 20 = 0 ⇒ (x − 5)(x + 4) = 0.
Only the positive root is valid.
x = 5 — option (c).
54
A vendor purchases 5 shirts at the marked price of 4 shirts. If he sells each shirt after giving an 8% discount on its marked price, what is his profit percentage?
A.10%
B.11%
C.15%
D.13%
Solution
Shirts — profit%
Let the marked price be ₹100 each. CP of 5 shirts = price of 4 = ₹400.
SP of 5 shirts at 8% discount = 5 × 92 = ₹460.
Profit = 460 − 400 = ₹60 on ₹400.
Profit% = × 100 = 15% — option (c).
55
Pipe P is capable of filling a tank in 32 minutes while Pipe Q can fill the same tank in 48 minutes. Both pipes are opened together, but after 8 minutes, the rate of Pipe P doubles and the rate of Pipe Q becomes half of its original. How many more minutes will it take to fill the tank completely?
A.4 min
B.5 min
C.6 min
D.8 min
Solution
Pipes — changing rates
Take the tank as LCM(32, 48) = 96 units ⇒ P = 3 units/min, Q = 2 units/min.
In 8 minutes together: 8 × 5 = 40 units; remaining = 96 − 40 = 56 units.
New rates: P = 6 and Q = 1, i.e. 7 units/min.
56 ÷ 7 = 8 minutes — option (d).
56
If a + b = 7 and ab = 12, find: (a³ + b³)² − 8a²b²(a + b)².
A.−48167
B.68752
C.38761
D.38767
Solution
Algebra — substitution
a + b = 7 and ab = 12 give a = 4, b = 3.
a³ + b³ = 64 + 27 = 91, so the first term = 91² = 8281.
Second term = 8 × 16 × 9 × 49 = 56448.
8281 − 56448 = −48167 — option (a).
57
A sum of ₹15,000 is lent out in two parts, one at 14% simple interest and the other at 20% simple interest. If the annual interest is ₹2,400, the sum lent at 20% is:
A.₹5500
B.₹4400
C.₹4000
D.₹5000
Solution
SI — two parts
Overall rate = × 100 = 16%.
Alligation between 14% and 20% about 16%: (20 − 16) : (16 − 14) = 4 : 2 = 2 : 1.
The part at 20% is the smaller share = × 15,000.
= ₹5,000 — option (d).
58
₹1800 is lent at a certain rate of simple interest. After 9 months, another ₹1200 is lent at a rate that is 1.5 times the original rate. If the total simple interest after 1 year is ₹144, find the original rate.
A.6.4%
B.6.5%
C.4.2%
D.4.6%
Solution
SI — two loans
First loan: 1800 × r × 1. Second loan runs only 3 months = year at 1.5r: 1200 × 1.5r × = 450r.
Interest ratio = 1800r : 450r = 4 : 1, so 5 parts = ₹144.
Interest on the first sum = × 4 = ₹115.20.
r = × 100 = 6.4% — option (a).
59
A flagpole has a shadow measuring 20 metres. Given that the height of the flagpole is 20 metres, what is the angle at which the sun is elevated?
A.45°
B.30°
C.60°
D.90°
Solution
Angle of elevation
tan(angle) = .
= = .
tan60° = .
So the elevation is 60° — option (c).
60
A hollow metallic sphere has outer radius 14 cm and is melted to make 25 smaller solid spheres of radius 4 cm. What is the inner radius of the original sphere?
A.10.45 cm
B.17.48 cm
C.15.46 cm
D.16.44 cm
Solution
Hollow sphere → small spheres
Metal volume is conserved: π(R³ − x³) = 25 × πr³.
14³ − x³ = 25 × 4³ ⇒ 2744 − x³ = 1600.
x³ = 1144 ⇒ x = ∛1144.
≈ 10.45 cm — option (a).
61
The volume of a hemisphere is numerically equal to five times its curved surface area. Find its radius.
If the height of a cone is made four times and the radius is doubled, how does the volume change?
A.Doubles
B.Becomes sixteen times of the original
C.Becomes one-third of the original
D.Remains unchanged
Solution
Cone — volume change
Volume of a cone = πr²h.
New volume = π(2r)² × 4h = π × 4r² × 4h.
= 16 × (πr²h).
Sixteen times the original — option (b).
64
If sinθ = , and θ ∈ (0, ), then what is the value of tanθ?
A.
B.
C.
D.
Solution
Trigonometry — tanθ
(8, 15, 17) is a Pythagorean triplet.
sinθ = means opposite 15 and hypotenuse 17, so the adjacent side is 8.
tanθ = .
= — option (c).
65
What is the value of tanX + cotX if tanX = 3?
A.
B.
C.
D.
Solution
tanX + cotX
cotX is the reciprocal of tanX.
cotX = .
tanX + cotX = 3 + = .
= — option (a).
66
What is the slope of the line 4x + 6y = 12?
A.−
B.
C.−
D.
Solution
Slope of a line
Write it in the form y = mx + c.
6y = −4x + 12 ⇒ y = −x + 2.
Comparing, the slope m = −.
Hence option (a).
67
A sector of a circle has a central angle of 90° and a radius of 10 cm. Another sector of the same circle 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.2 : 3
B.1 : 2
C.1 : 1
D.1 : 3
Solution
Sectors — degrees vs radians
radians = 90° — the two angles are the same.
Both sectors lie in the same circle, so their radii are equal too.
Equal angle and equal radius mean equal area.
Ratio = 1 : 1 — option (c).
68
A triangle can have:
A.three right angles
B.One acute angle and one right angle
C.Only one right or one obtuse angle
D.Three obtuse angles each measuring less than 90°
Solution
Angles of a triangle
The three angles of a triangle always add up to 180°.
Two right angles alone would use up 180°, leaving nothing for the third.
So a triangle can contain at most ONE right angle or ONE obtuse angle.
Hence option (c).
69
A line L is the perpendicular bisector of the line segment connecting points A(4, 10) and B(16, −2). What is the y-intercept of line L?
A.−6
B.4
C.6
D.10
Solution
Perpendicular bisector
Midpoint of AB = (, ) = (10, 4).
Slope of AB = = −1, so the perpendicular slope = 1.
Line L: y − 4 = 1(x − 10) ⇒ y = x − 6.
At x = 0, y = −6 — option (a).
70
The area of a triangle PQR is 8 cm². If a similar triangle DEF has sides that are twice the length of △PQR's sides, what is the area of △DEF?
A.32 cm²
B.48 cm²
C.64 cm²
D.80 cm²
Solution
Similar triangles — area
In similar triangles the areas vary as the SQUARE of the sides.
Sides are doubled, so the area becomes 2² = 4 times.
Area = 4 × 8.
= 32 cm² — option (a).
71
42³ + 45³ − 63³ + 149 is equal to:
A.−84685
B.0
C.−78659
D.1
Solution
Cubes evaluation
42³ = 74,088 and 45³ = 91,125, so their sum is 1,65,213.
63³ = 2,50,047.
1,65,213 − 2,50,047 = −84,834; adding 149 gives −84,685.
Hence option (a).
From an outside point T, two tangents, TP and TQ, are drawn to a circle. Given that the length of TP is 15 cm, what is the length of TQ?
A.16 cm
B.10 cm
C.15 cm
D.24 cm
Solution
Equal tangents
Tangents drawn to a circle from the same external point are always equal.
So TP = TQ.
TP = 15 cm.
TQ = 15 cm — option (c).
74
A circle is inscribed within a right triangle. Considering that the lengths of the two legs measure 12 cm and 16 cm, what is the radius of the inscribed circle?
A.6 cm
B.9 cm
C.4 cm
D.5 cm
Solution
Inradius of right triangle
(12, 16, 20) is a Pythagorean triplet, so the hypotenuse is 20 cm.
For a right triangle, inradius r = .
= = .
= 4 cm — option (c).
75
If a = 0.04, b = 0.06, c = −0.10, and a + b + c = 0, find (a³ + b³ + c³) ÷ (3abc).
A.−1
B.0
C.1
D.−2
Solution
a³+b³+c³ when a+b+c=0
Check: 0.04 + 0.06 − 0.10 = 0 ✓.
When a + b + c = 0, the identity gives a³ + b³ + c³ = 3abc.
So the ratio = .
= 1 — option (c).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Let the marked price be ₹100 each. CP of 5 shirts = price of 4 = ₹400.
SP of 5 shirts at 8% discount = 5 × 92 = ₹460.
Profit = 460 − 400 = ₹60 on ₹400.
Profit% = × 100 = 15% — option (c).
Take the tank as LCM(32, 48) = 96 units ⇒ P = 3 units/min, Q = 2 units/min.
In 8 minutes together: 8 × 5 = 40 units; remaining = 96 − 40 = 56 units.
New rates: P = 6 and Q = 1, i.e. 7 units/min.
56 ÷ 7 = 8 minutes — option (d).
Overall rate = × 100 = 16%.
Alligation between 14% and 20% about 16%: (20 − 16) : (16 − 14) = 4 : 2 = 2 : 1.
The part at 20% is the smaller share = × 15,000.
= ₹5,000 — option (d).
First loan: 1800 × r × 1. Second loan runs only 3 months = year at 1.5r: 1200 × 1.5r × = 450r.
Interest ratio = 1800r : 450r = 4 : 1, so 5 parts = ₹144.
Interest on the first sum = × 4 = ₹115.20.
r = × 100 = 6.4% — option (a).
tan(angle) = .
= = .
tan60° = .
So the elevation is 60° — option (c).
Metal volume is conserved: π(R³ − x³) = 25 × πr³.
14³ − x³ = 25 × 4³ ⇒ 2744 − x³ = 1600.
x³ = 1144 ⇒ x = ∛1144.
≈ 10.45 cm — option (a).
Volume = πr³ and CSA = 2πr².
Given: πr³ = 5 × 2πr².
Cancel 2πr²: = 5.
r = 15 cm — option (b).
Minimum cuts = (nx − 1) + (ny − 1) + (nz − 1).
Split 16 as evenly as possible: 16 = 2 × 2 × 4.
Cuts = (2−1) + (2−1) + (4−1) = 1 + 1 + 3.
= 5 — option (c).
Volume of a cone = πr²h.
New volume = π(2r)² × 4h = π × 4r² × 4h.
= 16 × (πr²h).
Sixteen times the original — option (b).
(8, 15, 17) is a Pythagorean triplet.
sinθ = means opposite 15 and hypotenuse 17, so the adjacent side is 8.
tanθ = .
= — option (c).
Write it in the form y = mx + c.
6y = −4x + 12 ⇒ y = −x + 2.
Comparing, the slope m = −.
Hence option (a).
radians = 90° — the two angles are the same.
Both sectors lie in the same circle, so their radii are equal too.
Equal angle and equal radius mean equal area.
Ratio = 1 : 1 — option (c).
Midpoint of AB = (, ) = (10, 4).
Slope of AB = = −1, so the perpendicular slope = 1.
Line L: y − 4 = 1(x − 10) ⇒ y = x − 6.
At x = 0, y = −6 — option (a).
In similar triangles the areas vary as the SQUARE of the sides.
Sides are doubled, so the area becomes 2² = 4 times.
Area = 4 × 8.
= 32 cm² — option (a).
Tangents drawn to a circle from the same external point are always equal.
So TP = TQ.
TP = 15 cm.
TQ = 15 cm — option (c).
(12, 16, 20) is a Pythagorean triplet, so the hypotenuse is 20 cm.
For a right triangle, inradius r = .
= = .
= 4 cm — option (c).