SSC CGL 18 September 2025 · Shift 2 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 79 of 192
18 September 2025 · Shift 2
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
If 44% of P is equal to 24% of Q, what is the ratio P : Q?
A.6 : 11
B.6 : 13
C.13 : 7
D.11 : 6
Solution
Percentage equality
44% of P = 24% of Q ⇒ 44P = 24Q.
= .
Divide both by 4: .
P : Q = 6 : 11 — option (a).
52
Find the value of
A.44
B.37
C.27
D.33
Solution
Nested roots
Start from the innermost: = 21.
Next: = = 18.
Then: = .
= 27 — option (c).
53
Identify the incorrect relationship(s) from the list below:
(i) + = +
(ii) + < +
(iii) + > +
A.(ii)
B.(i)
C.(i) and (ii)
D.(ii) and (iii)
Solution
Comparing surds
Square both sides: ( + )² = 11 + 2.
( + )² = 11 + 2.
Since > , the first sum is GREATER, so (iii) is true.
Hence (i) and (ii) are incorrect — option (c).
54
A bookseller purchases 8 novels at the marked price of 6 novels from a distributor. If he then sells these novels to customers, offering a discount of 15% on the marked price, what is his profit percentage?
A.11%
B.21.75%
C.19.62%
D.13.34%
Solution
Novels — profit%
Let each novel be marked at 100 units. He pays for 6 but gets 8, so CP of 8 novels = 600 units.
SP per novel after 15% discount = 85, so SP of 8 novels = 8 × 85 = 680 units.
Profit = 680 − 600 = 80 units on a cost of 600.
Profit% = × 100 ≈ 13.34% — option (d).
55
A and B can finish a certain job in 24 and 30 days, respectively. They started working together, but A left after a while, and B managed to complete the remaining work in 3 days. How many days did A work before leaving?
A.18 days
B.15 days
C.12 days
D.17 days
Solution
Work — A leaves
Total work = LCM(24, 30) = 120 units ⇒ A = 5 units/day, B = 4 units/day.
Let them work together for y days: 9y + 4 × 3 = 120.
9y = 108.
y = 12 days — option (c).
56
Given y + = 3, find the value of: y⁵ + − 3(y³ + ) + 4(y + )
Reena borrowed ₹1,156 at simple interest for a number of years equal to the rate of interest. At the end of the loan term, she paid ₹289 in interest. What was the interest rate?
A.7%
B.5%
C.18%
D.10%
Solution
SI — rate = time
Let the rate be R% and the time also R years.
SI = ⇒ 289 = .
R² = = 25.
R = 5% — option (b).
58
A person invested ₹30,000 in two different schemes. He invested one part for 5 years at 6% simple interest and the other part for the same duration at 9% simple interest. If the interest earned from the second part was ₹1,950 more than the interest earned from the first part, how much money was invested in the first scheme?
A.₹9,000
B.₹10,500
C.₹13,000
D.₹15,400
Solution
SI — two parts
Let the first part be ₹y, so the second is ₹(30,000 − y).
− = 1950.
13,50,000 − 45y − 30y = 1,95,000 ⇒ 75y = 11,55,000.
y = ₹15,400 — option (d).
59
A rock climber uses a rope to ascend a cliff. If the vertical height of the cliff is 23 metres and the rope makes an angle of 45° with the level ground, what is the minimum length of the rope required to reach the top?
A.46 m
B.23 m
C.60 m
D.46 m
Solution
Rope at 45°
sin45° = .
= .
Rope = 23.
= 23 m — option (b).
60
A tile is spherical in shape with a radius of 6 cm. If the tile is placed inside a cube with side length 12 cm, what is the volume of the cube that is not occupied by the tile?
Two hemispheres of radii 9 cm and 10 cm respectively are melted and recast into another hemisphere. What is the approximate total surface area of the newly formed hemisphere?
A.1633 cm²
B.1425 cm²
C.1358 cm²
D.1385 cm²
Solution
Two hemispheres recast
Volumes add: R³ = 9³ + 10³ = 729 + 1000 = 1729 ⇒ R ≈ 12 cm.
Total surface area of a hemisphere = 3πR².
= 3 × × 144 ≈ 1357.71.
≈ 1358 cm² — option (c).
62
If all three dimensions of a cuboid are increased by 25%, the volume increases by what percentage?
A.95.31%
B.89.21%
C.75.01%
D.82.15%
Solution
Cuboid — 25% rise
A 25% rise multiplies each dimension by .
Volume factor = ()³ = .
Increase = × 100.
≈ 95.31% — option (a).
63
A cone is sliced into two sections by a plane that is parallel to its base, with the upper section representing of the overall volume. What is the ratio of the heights of the smaller cone to the original cone?
A.1 : 4
B.1 : 6
C.5 : 1
D.1 : 3
Solution
Similar cones
For similar solids, volumes vary as the cube of corresponding heights.
()³ = .
Take the cube root: = .
= 1 : 3 — option (d).
64
If tanA = and tanB = , with A and B ∈ (0, ), find tan(A + B).
A.3
B.1
C.7
D.9
Solution
tan(A + B)
tan(A + B) = .
Numerator = + = .
Denominator = 1 − = .
Ratio = 1 — option (b).
65
A tower is 60 metres tall. When the sun's angle of elevation is 60 degrees, what is the length of the shadow of the tower?
A.45 m
B.38 m
C.20 m
D.19 m
Solution
Shadow at 60°
tan60° = .
= .
Shadow = = 20.
= 20 m — option (c).
66
A line y = mx + 3 passes through (2, 9). Find the value of m.
A.3
B.4
C.6
D.9
Solution
Line through a point
Substitute the point: 9 = m(2) + 3.
2m = 9 − 3 = 6.
m = 3.
Hence option (a).
67
A sector of a circle has a central angle of 150° and a radius of 6 cm. Another sector of the same circle has a central angle of 180°. What is the ratio of the area of the first sector to the area of the second sector?
A.6 : 5
B.3 : 4
C.2 : 3
D.5 : 6
Solution
Sector areas
With the same radius, sector area is proportional to the central angle.
First : second = 150° : 180°.
Divide both by 30.
= 5 : 6 — option (d).
68
What is the measure of each interior angle in a regular twelve-sided polygon?
A.120°
B.150°
C.180°
D.90°
Solution
Interior angle — 12-gon
Each interior angle = .
For n = 12: .
= .
= 150° — option (b).
69
In a trapezoid, the two parallel sides are in the ratio 2 : 3. If the height of the trapezoid is 12 cm and its total area is 360 cm², determine the lengths of the parallel sides.
A.26 cm and 38 cm
B.12 cm and 24 cm
C.24 cm and 36 cm
D.17 cm and 23 cm
Solution
Trapezium sides
Let the sides be 2x and 3x.
Area = (sum of parallel sides) × height ⇒ 360 = (5x)(12).
360 = 30x ⇒ x = 12.
Sides = 24 cm and 36 cm — option (c).
70
△PQR has sides PQ = 8 cm, QR = 9 cm, RP = 10 cm. △MNO is congruent to △PQR. What is the perimeter of △MNO?
A.27 cm
B.32 cm
C.25 cm
D.21 cm
Solution
Congruent triangles
Congruent triangles have equal corresponding sides.
So MN = 8, NO = 9 and OM = 10 cm.
Perimeter = 8 + 9 + 10.
= 27 cm — option (a).
71
Given + = 7, then what is m + ?
A.55
B.29
C.47
D.36
Solution
Square the surd
Square both sides: ( + )² = 49.
m + + 2 = 49.
m + = 49 − 2.
= 47 — option (c).
72
If secA = 2, then tanA − cosecA = ?
A.
B.
C.0
D.
Solution
secA = 2
sec60° = 2, so A = 60°.
tan60° = and cosec60° = .
− = .
= — option (b).
73
A circle has radius 9 cm. A tangent is drawn from an external point A. If the length of the tangent is 12 cm, what is the distance from A to the centre of the circle?
A.16 cm
B.35 cm
C.41 cm
D.15 cm
Solution
Tangent and radius
The radius meets the tangent at 90°, forming a right triangle.
Distance² = 9² + 12² = 81 + 144.
= 225.
Distance = 15 cm — option (d).
74
From a point outside a circle, two tangents are drawn to the circle. If one of the tangents measures 15 cm, what is the length of the other tangent?
A.15 cm
B.13 cm
C.12 cm
D.14 cm
Solution
Equal tangents
Tangents drawn from the same external point to a circle are always equal.
So the second tangent has the same length.
That is 15 cm.
Hence option (a).
75
Evaluate: (0.09³ + 0.03³) ÷ (0.3³ + 0.1³)
A.0.028
B.0.056
C.0.003
D.0.027
Solution
Cubes — common factor
Note that 0.09 = 0.3 × 0.3 and 0.03 = 0.3 × 0.1.
So the numerator = (0.3)³ × (0.3³ + 0.1³).
The bracket cancels with the denominator, leaving (0.3)³.
= 0.027 — option (d).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Start from the innermost: = 21.
Next: = = 18.
Then: = .
= 27 — option (c).
Let each novel be marked at 100 units. He pays for 6 but gets 8, so CP of 8 novels = 600 units.
SP per novel after 15% discount = 85, so SP of 8 novels = 8 × 85 = 680 units.
Profit = 680 − 600 = 80 units on a cost of 600.
Profit% = × 100 ≈ 13.34% — option (d).
Total work = LCM(24, 30) = 120 units ⇒ A = 5 units/day, B = 4 units/day.
Let them work together for y days: 9y + 4 × 3 = 120.
9y = 108.
y = 12 days — option (c).
Square: y² + = 9 − 2 = 7. Cube: y³ + = 27 − 9 = 18.
Multiply the two: (y³ + )(y² + ) = y⁵ + + y + ⇒ y⁵ + = 18 × 7 − 3 = 123.
Now put the values: 123 − 3(18) + 4(3).
= 123 − 54 + 12 = 81 — option (a).
Let the rate be R% and the time also R years.
SI = ⇒ 289 = .
R² = = 25.
R = 5% — option (b).
sin45° = .
= .
Rope = 23.
= 23 m — option (b).
Sphere volume = πr³ = × × 216 ≈ 905.14 cm³.
Cube volume = 12³ = 1728 cm³.
Empty space = 1728 − 905.14.
≈ 822.86 cm³ — option (d).
Volumes add: R³ = 9³ + 10³ = 729 + 1000 = 1729 ⇒ R ≈ 12 cm.
Total surface area of a hemisphere = 3πR².
= 3 × × 144 ≈ 1357.71.
≈ 1358 cm² — option (c).
A 25% rise multiplies each dimension by .
Volume factor = ()³ = .
Increase = × 100.
≈ 95.31% — option (a).
For similar solids, volumes vary as the cube of corresponding heights.
()³ = .
Take the cube root: = .
= 1 : 3 — option (d).
tan60° = .
= .
Shadow = = 20.
= 20 m — option (c).
With the same radius, sector area is proportional to the central angle.
First : second = 150° : 180°.
Divide both by 30.
= 5 : 6 — option (d).
Each interior angle = .
For n = 12: .
= .
= 150° — option (b).
Let the sides be 2x and 3x.
Area = (sum of parallel sides) × height ⇒ 360 = (5x)(12).
360 = 30x ⇒ x = 12.
Sides = 24 cm and 36 cm — option (c).
Square both sides: ( + )² = 49.
m + + 2 = 49.
m + = 49 − 2.
= 47 — option (c).
The radius meets the tangent at 90°, forming a right triangle.
Distance² = 9² + 12² = 81 + 144.
= 225.
Distance = 15 cm — option (d).
Note that 0.09 = 0.3 × 0.3 and 0.03 = 0.3 × 0.1.
So the numerator = (0.3)³ × (0.3³ + 0.1³).
The bracket cancels with the denominator, leaving (0.3)³.
= 0.027 — option (d).