SSC CGL 14 October 2025 · Shift 2 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 187 of 192
14 October 2025 · Shift 2
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
What is the least square number that is exactly divisible by 3, 5, 12 and 15?
A.3600
B.2304
C.3364
D.4489
Solution
Least square multiple
LCM(3, 5, 12, 15) = 60 = 2² × 3 × 5.
For a perfect square every prime must appear an even number of times.
Among the options 2304 = 48², 3364 = 58² and 4489 = 67², but none of these is divisible by 5.
Only 3600 = 60² is divisible by all four — option (a).
52
In how many years will a sum of ₹8,000 give a simple interest of ₹3,000 at 12.5% per annum?
A.3 years 3 months
B.2 years
C.3 years
D.2 years 6 months
Solution
Time from SI
SI = , so 3,000 = .
8,000 × 12.5 = 1,00,000, so 3,000 = 1,000 × T.
T = 3.
= 3 years — option (c).
Consider the following statements regarding simple interest:
1. The simple interest on ₹10,000 for 3 years at 15% per annum is ₹4,500.
2. If a sum of money doubles in 5 years at simple interest, then it will become 1.5 times in 3 years.
3. The simple interest rate required for an amount to become 4.5 times in 7 years is 50%.
Of these statements, which are correct?
A.Only 1 is correct
B.1 and 2 are correct
C.1 and 3 are correct
D.All 1, 2 and 3 are correct
Solution
Simple interest statements
Statement 1: SI = = ₹4,500 ✓.
Statement 2: doubling in 5 years means the interest equals the principal, so the rate is 20%. In 3 years the interest is 0.6 P, making the amount 1.6 times, not 1.5 ✗.
Statement 3: becoming 4.5 times means the interest is 3.5 P in 7 years, so the rate is × 100 = 50% ✓.
1 and 3 — option (c).
55
From a point on a bridge built over a river, the angles of depression to the two opposite banks are 30° and 45° respectively. If the width of the river is 50( + 1) metres, find the height of the bridge above the riverbanks.
A.50 m
B.45 m
C.30 m
D.55 m
Solution
Angles of depression
Let the height be h. From the 45° side the horizontal distance is h; from the 30° side it is h.
The two banks are on opposite sides, so the width = h + h = h( + 1).
h( + 1) = 50( + 1).
h = 50 m — option (a).
56
The areas of three adjacent faces of a cuboid are 48 cm², 54 cm² and 72 cm². What is the volume of the cuboid?
A.360 cm³
B.432 cm³
C.504 cm³
D.576 cm³
Solution
Cuboid from face areas
The three face areas are lb, bh and lh.
Their product = (lbh)², so the volume = .
48 × 54 × 72 = 1,86,624.
= 432 cm³ — option (b).
57
A solid hemisphere has a radius of 9 cm. It is melted into a cylinder of height 6 cm. Find the radius of the cylinder.
A.9 cm
B.6 cm
C.7 cm
D.5 cm
Solution
Hemisphere to cylinder
Melting keeps the volume the same.
Hemisphere = π(9)³ = π × 729 = 486π cm³.
Cylinder = πr²(6) = 6πr², so 6πr² = 486π ⇒ r² = 81.
r = 9 cm — option (a).
58
A hemisphere, a cone and a cylinder have the same base radius and equal volumes. If the height of the cylinder is 3h, find the height of the cone.
A.9h
B.10h
C.5h
D.6h
Solution
Equal volumes
Cylinder volume = πr²(3h) = 3πr²h.
Cone volume = πr²H, and this must equal 3πr²h.
H = 3h.
H = 9h — option (a).
59
The longest diagonal of a cuboid is equal to the diameter of the sphere in which it is inscribed. If the sphere's radius is 13 cm and two dimensions of the cuboid are 9 cm and 15 cm, what is the third dimension?
A.10.1 cm
B.12.3 cm
C.16.5 cm
D.19.2 cm
Solution
Cuboid in a sphere
Diagonal of a cuboid = , and it equals the diameter 2 × 13 = 26 cm.
26² = 9² + 15² + h².
676 = 81 + 225 + h² ⇒ h² = 370.
h = ≈ 19.2 cm — option (d).
60
A regular right pyramid has a square base with side length 10 cm and height 12 cm. Find the slant height of one triangular face.
A.12 cm
B.13 cm
C.14 cm
D.16 cm
Solution
Slant height of a pyramid
The slant height runs from the apex to the midpoint of a base edge.
That distance is half the base side, i.e. 5 cm, from the centre.
Slant height = = .
= = 13 cm — option (b).
61
A bag contains 25% red balls, 35% green balls and the rest blue balls. If there are 40 blue balls, what is the total number of balls?
A.100
B.160
C.200
D.240
Solution
Percentage of a total
Blue share = 100% − 25% − 35% = 40%.
40% of the total = 40 balls.
Total = .
= 100 — option (a).
62
A rhombus has diagonals in the ratio 5 : 6. If its area is 135 cm², find the diagonals.
A.18 cm and 24 cm
B.12 cm and 16 cm
C.15 cm and 18 cm
D.27 cm and 36 cm
Solution
Diagonals of a rhombus
Area of a rhombus = × d₁ × d₂.
Let the diagonals be 5x and 6x: × 5x × 6x = 135.
15x² = 135 ⇒ x² = 9 ⇒ x = 3.
Diagonals = 15 cm and 18 cm — option (c).
63
A circular plate has a radius of 12 cm and is surrounded by a decorative rim 3 cm wide. What is the area of the rim? (Use π = 3.14)
A.238.23 cm²
B.254.34 cm²
C.258.51 cm²
D.230.69 cm²
Solution
Area of a ring
Outer radius = 12 + 3 = 15 cm.
Area of the rim = π(R² − r²).
= 3.14 × (225 − 144) = 3.14 × 81.
= 254.34 cm² — option (b).
64
A number is increased by 100 and the result becomes a perfect square. The original number lies between the two consecutive square numbers 729 and 784. Find the original number.
A.700
B.732
C.741
D.745
Solution
Perfect square
729 = 27² and 784 = 28², so the number x satisfies 729 < x < 784.
Then x + 100 must be a perfect square lying between 829 and 884.
29² = 841 falls in that range, so x + 100 = 841.
x = 741 — option (c).
65
A square-shaped piece of metal has a circular hole cut out exactly from its centre, touching all four sides. What percentage of the metal sheet's area was removed by the hole? (Use π ≈ 3.14)
A.21.5%
B.25%
C.75%
D.78.5%
Solution
Circle in a square
Let the side of the square be 2r, so the circle has radius r.
Square area = 4r² and circle area = πr².
Percentage removed = × 100 = × 100.
= × 100 = 78.5% — option (d).
66
If two complementary angles are in the ratio 2 : 7, find the angles.
A.10°, 80°
B.20°, 70°
C.40°, 50°
D.35°, 55°
Solution
Complementary angles
Complementary angles add up to 90°.
2x + 7x = 90° ⇒ 9x = 90°.
x = 10°.
The angles are 20° and 70° — option (b).
67
From the top of a 60 m tower, the angles of depression of two objects on the same side of the tower are 30° and 45°. Find the distance between the objects.
A.21.13 m
B.28.7 m
C.43.92 m
D.58.87 m
Solution
Two angles of depression
The nearer object at 45° lies 60 m from the foot of the tower.
The farther object at 30° lies 60 m from the foot.
Distance between them = 60 − 60 = 60( − 1).
= 60 × 0.732 ≈ 43.92 m — option (c).
68
If tanθ = , what is secθ?
A.
B.
C.
D.
Solution
sec from tan
Perpendicular = 5 and base = 12.
Hypotenuse = = = 13.
secθ = .
= — option (d).
69
Find the y-intercept of 4x + 5y = 20.
A.3
B.4
C.5
D.−2
Solution
y-intercept
The y-intercept is the value of y where the line meets the y-axis, so put x = 0.
5y = 20.
y = 4.
Hence option (b).
70
Two lines intersect, forming four angles. One of the angles measures (4x − 5)° and its adjacent angle measures (6x + 25)°. What is the measure of the angle vertically opposite to the (4x − 5)° angle?
A.59°
B.55°
C.45°
D.60°
Solution
Adjacent and vertical angles
Adjacent angles on a straight line add up to 180°.
(4x − 5) + (6x + 25) = 180 ⇒ 10x + 20 = 180 ⇒ x = 16.
So the angle is 4(16) − 5 = 59°.
Vertically opposite angles are equal, so it is also 59° — option (a).
71
Which of the following statements is always true for a right-angled triangle?
A.The centroid is outside the triangle
B.The circumcentre is at the midpoint of the hypotenuse
C.The orthocentre, centroid, incentre and circumcentre all coincide
D.The incentre is one of the vertices of the triangle
Solution
Circumcentre of a right triangle
In a right triangle the hypotenuse is a diameter of the circumcircle.
So the circumcentre lies exactly at the midpoint of the hypotenuse.
The centroid is always inside, the incentre is always inside, and all four centres coincide only in an equilateral triangle.
Hence option (b).
72
In △KLM, NO ‖ LM and intersects KL and KM at N and O. If KN = 2, NL = 4 and KO = 3, find OM.
A.6
B.4.5
C.3
D.9
Solution
Basic proportionality
By the basic proportionality theorem, = .
= .
OM = 3 × 2.
= 6 — option (a).
73
The length of a direct common tangent to two circles is 24 cm. Their centres are 25 cm apart. If the larger circle has a radius of 10 cm, what is the radius of the smaller circle?
A.2 cm
B.3 cm
C.4 cm
D.5 cm
Solution
Direct common tangent
Direct common tangent = .
24 = ⇒ 576 = 625 − (10 − r)².
(10 − r)² = 49 ⇒ 10 − r = 7.
r = 3 cm — option (b).
74
If cos(2x + 15°) = sin(x − 5°), what is the value of x?
A.45.57°
B.60°
C.30°
D.26.67°
Solution
sin = cos
sinA = cosB holds when the two angles are complementary.
(2x + 15°) + (x − 5°) = 90°.
3x + 10° = 90° ⇒ 3x = 80°.
x = 26.67° — option (d).
75
A contractor hires 12 workers who can complete a task in 20 days. After 5 days he lets 3 workers go. How many more days will the remaining workers take to complete the task?
A.20 days
B.14 days
C.15 days
D.16 days
Solution
Workers and days
Total work = 12 × 20 = 240 worker-days.
In 5 days: 12 × 5 = 60 worker-days done, leaving 180.
Only 9 workers remain.
= 20 days — option (a).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
LCM(3, 5, 12, 15) = 60 = 2² × 3 × 5.
For a perfect square every prime must appear an even number of times.
Among the options 2304 = 48², 3364 = 58² and 4489 = 67², but none of these is divisible by 5.
Only 3600 = 60² is divisible by all four — option (a).
SI = , so 3,000 = .
8,000 × 12.5 = 1,00,000, so 3,000 = 1,000 × T.
T = 3.
= 3 years — option (c).
a³ + b³ = (a + b)³ − 3ab(a + b).
407 = 11³ − 3ab(11) = 1331 − 33ab.
33ab = 1331 − 407 = 924.
ab = = 28 — option (c).
Let the height be h. From the 45° side the horizontal distance is h; from the 30° side it is h.
The two banks are on opposite sides, so the width = h + h = h( + 1).
h( + 1) = 50( + 1).
h = 50 m — option (a).
The three face areas are lb, bh and lh.
Their product = (lbh)², so the volume = .
48 × 54 × 72 = 1,86,624.
= 432 cm³ — option (b).
Melting keeps the volume the same.
Hemisphere = π(9)³ = π × 729 = 486π cm³.
Cylinder = πr²(6) = 6πr², so 6πr² = 486π ⇒ r² = 81.
r = 9 cm — option (a).
Cylinder volume = πr²(3h) = 3πr²h.
Cone volume = πr²H, and this must equal 3πr²h.
H = 3h.
H = 9h — option (a).
Diagonal of a cuboid = , and it equals the diameter 2 × 13 = 26 cm.
26² = 9² + 15² + h².
676 = 81 + 225 + h² ⇒ h² = 370.
h = ≈ 19.2 cm — option (d).
The slant height runs from the apex to the midpoint of a base edge.
That distance is half the base side, i.e. 5 cm, from the centre.
Slant height = = .
= = 13 cm — option (b).
Blue share = 100% − 25% − 35% = 40%.
40% of the total = 40 balls.
Total = .
= 100 — option (a).
Area of a rhombus = × d₁ × d₂.
Let the diagonals be 5x and 6x: × 5x × 6x = 135.
15x² = 135 ⇒ x² = 9 ⇒ x = 3.
Diagonals = 15 cm and 18 cm — option (c).
Outer radius = 12 + 3 = 15 cm.
Area of the rim = π(R² − r²).
= 3.14 × (225 − 144) = 3.14 × 81.
= 254.34 cm² — option (b).
729 = 27² and 784 = 28², so the number x satisfies 729 < x < 784.
Then x + 100 must be a perfect square lying between 829 and 884.
29² = 841 falls in that range, so x + 100 = 841.
x = 741 — option (c).
Let the side of the square be 2r, so the circle has radius r.
Square area = 4r² and circle area = πr².
Percentage removed = × 100 = × 100.
= × 100 = 78.5% — option (d).
Complementary angles add up to 90°.
2x + 7x = 90° ⇒ 9x = 90°.
x = 10°.
The angles are 20° and 70° — option (b).
The nearer object at 45° lies 60 m from the foot of the tower.
The farther object at 30° lies 60 m from the foot.
Distance between them = 60 − 60 = 60( − 1).
= 60 × 0.732 ≈ 43.92 m — option (c).
Perpendicular = 5 and base = 12.
Hypotenuse = = = 13.
secθ = .
= — option (d).
The y-intercept is the value of y where the line meets the y-axis, so put x = 0.
5y = 20.
y = 4.
Hence option (b).
Adjacent angles on a straight line add up to 180°.
(4x − 5) + (6x + 25) = 180 ⇒ 10x + 20 = 180 ⇒ x = 16.
So the angle is 4(16) − 5 = 59°.
Vertically opposite angles are equal, so it is also 59° — option (a).
In a right triangle the hypotenuse is a diameter of the circumcircle.
So the circumcentre lies exactly at the midpoint of the hypotenuse.
The centroid is always inside, the incentre is always inside, and all four centres coincide only in an equilateral triangle.
Hence option (b).
By the basic proportionality theorem, = .
= .
OM = 3 × 2.
= 6 — option (a).
Direct common tangent = .
24 = ⇒ 576 = 625 − (10 − r)².
(10 − r)² = 49 ⇒ 10 − r = 7.
r = 3 cm — option (b).
sinA = cosB holds when the two angles are complementary.
(2x + 15°) + (x − 5°) = 90°.
3x + 10° = 90° ⇒ 3x = 80°.
x = 26.67° — option (d).
Total work = 12 × 20 = 240 worker-days.
In 5 days: 12 × 5 = 60 worker-days done, leaving 180.
Only 9 workers remain.
= 20 days — option (a).