SSC CGL 23 September 2025 · Shift 2 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 139 of 192
23 September 2025 · Shift 2
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
The number of students in two sections A and B having different heights is shown in the table below.
Height 1.56 m — A: 5, B: 4 | 1.60 m — A: 9, B: 0 | 1.68 m — A: 8, B: 9 | 1.64 m — A: 15, B: 10 | 1.69 m — A: 8, B: 7 | 1.72 m — A: 6, B: 5 | 1.78 m — A: 4, B: 3.
What is the ratio of the number of students with a height of 1.68 metres in Section A to the total number of students in Section A?
A.8 : 55
B.15 : 62
C.6 : 4
D.4 : 33
Solution
Ratio from a table
Students of height 1.68 m in Section A = 8.
Total in Section A = 5 + 9 + 8 + 15 + 8 + 6 + 4.
= 55.
Ratio = 8 : 55 — option (a).
52
Solve:
A.
B.
C.
D.
Solution
BODMAS under a root
Inside the root, do division first: 225 ÷ 15 = 15.
Then 4 × 15 = 60.
60 + 25 − 4 = 81.
So the value is — option (a).
53
A sum amounts to ₹5,800 in 2 years and ₹6,200 in 4 years at simple interest. What is the rate of interest per annum?
A.2%
B.4.1%
C.3.7%
D.1.5%
Solution
SI from two amounts
Interest for 2 years = 6,200 − 5,800 = ₹400, so one year’s interest = ₹200.
Principal = 5,800 − 2 × 200 = ₹5,400.
R = × 100.
≈ 3.7% — option (c).
54
A man reduces his speed from 16 km/h to 14 km/h, and so takes 6 minutes more than the normal time. What is the distance travelled by him?
A.10 km
B.9.2 km
C.18 km
D.11.2 km
Solution
Speed and time
Let the distance be x km. Then − = hour.
= ⇒ = .
20x = 224.
x = 11.2 km — option (d).
55
A kite is soaring at an altitude of 45 metres above the ground, with its string temporarily secured to a point on the ground. The angle of elevation of the string relative to the ground is 30°. Given that the string remains taut, what is the length of the string?
A.60 m
B.45 m
C.90 m
D.55 m
Solution
Angle of elevation
The string is the hypotenuse and the height is the side opposite the angle.
sin30° = = .
= .
L = 90 m — option (c).
56
A rectangular fish tank with a base of 80 cm × 60 cm contains water up to a height of 25 cm. If 4 identical spherical stones, each of radius 5 cm, are dropped into the tank and fully submerged, what is the rise in the water level?
A. cm
B. cm
C. cm
D. cm
Solution
Rise in water level
Volume of one sphere = πr³ = π(125) = cm³; four of them = cm³.
Base area of the tank = 80 × 60 = 4800 cm².
Rise = = .
= cm — option (b).
57
Partners X and Y invest ₹40,000 and ₹60,000 respectively. X works for 8 months and Y for 6 months. Profit is divided on the basis of capital × time. Find their profit ratio.
A.3 : 6
B.8 : 9
C.4 : 2
D.6 : 4
Solution
Capital × time
X: 40,000 × 8 = 3,20,000.
Y: 60,000 × 6 = 3,60,000.
Ratio = 320 : 360.
= 8 : 9 — option (b).
58
If the difference between the interior and exterior angles of a regular polygon is 60°, find the number of sides of the polygon.
A.6
B.9
C.13
D.11
Solution
Interior and exterior angles
Interior + exterior = 180° and interior − exterior = 60°.
Adding: 2 × interior = 240° ⇒ interior = 120°, so exterior = 60°.
Number of sides = = .
= 6 — option (a).
59
A value is increased by 10% and then decreased by an unknown percentage to return to the original value. What is the percentage decrease required?
A.9.09%
B.6.92%
C.3.26%
D.2.22%
Solution
Reverse percentage
Take the value as 100; after a 10% rise it becomes 110.
To come back it must fall by 10, but that fall is measured on 110.
× 100.
≈ 9.09% — option (a).
60
A square field has a diagonal of m. What is its area?
A.46 m²
B.24 m²
C.40 m²
D.12 m²
Solution
Square from diagonal
For a square, area = .
d² = ()² = 48.
Area = .
= 24 m² — option (b).
61
A 30-litre solution contains acid and water in the ratio 2 : 1. How much water must be added to make the ratio 1 : 1?
A.11 L
B.10 L
C.12 L
D.21 L
Solution
Change a ratio
Acid = × 30 = 20 L and water = 10 L.
The acid stays fixed, so the water must also become 20 L.
Water to be added = 20 − 10.
= 10 L — option (b).
62
If the selling price of an item is increased by 10% and the profit also increases from 15% to 20%, what is the percentage increase in the cost price?
A.3.5%
B.4.6%
C.1.28%
D.5.41%
Solution
SP up, profit up
Take the original CP as 100; at 15% profit the SP = 115.
New SP = 115 × 1.10 = 126.5, and this is a 20% profit.
New CP = = 105.41.
Increase = 5.41% — option (d).
63
A group of 5 students went on a trip and spent ₹1,000 in total. If one of them spent ₹400 while the rest spent equally, what was the average spending of the other 4 students?
A.₹150
B.₹180
C.₹210
D.₹230
Solution
Average of the rest
Amount left after one student’s ₹400 = 1,000 − 400 = ₹600.
This is shared equally by the remaining 4.
Average = .
= ₹150 — option (a).
64
A computer is sold at ₹6,800 after applying a 15% discount. What was its marked price?
A.₹7,500
B.₹6,800
C.₹8,000
D.₹6,500
Solution
MP from discount
After a 15% discount the buyer pays 85% of the marked price.
85% of MP = ₹6,800.
MP = .
= ₹8,000 — option (c).
65
If 4 men or 6 women can complete a work in 18 days, how long will 2 men and 3 women take to complete the same work?
A.12 days
B.10 days
C.18 days
D.13 days
Solution
Men and women work
4 men = 6 women, so 2 men = 3 women.
Hence 2 men + 3 women = 2 men + 2 men = 4 men.
But 4 men finish the work in 18 days.
So the time is again 18 days — option (c).
66
A square has a perimeter of 80 cm. What is the area of the square?
A.352 cm²
B.692 cm²
C.400 cm²
D.425 cm²
Solution
Square area
Perimeter = 4 × side, so side = = 20 cm.
Area = side².
= 20 × 20.
= 400 cm² — option (c).
67
A student took 6 tests and had an average of 52. If his average after 7 tests increased to 63, what was his score on the 7th test?
A.178
B.128
C.129
D.181
Solution
Seventh test score
Total of 6 tests = 6 × 52 = 312.
Total of 7 tests = 7 × 63 = 441.
7th score = 441 − 312.
= 129 — option (c).
68
A store announces a festive deal: "Get 20% off on purchases above ₹15,000, and an additional ₹2,000 off if the total after discount exceeds ₹20,000." A customer buys items worth ₹40,000. How much does the customer finally pay?
A.₹30,000
B.₹18,500
C.₹19,000
D.₹16,000
Solution
Two-stage discount
The bill of ₹40,000 is above ₹15,000, so a 20% discount applies: 40,000 × = ₹32,000.
₹32,000 exceeds ₹20,000, so the extra ₹2,000 comes off.
32,000 − 2,000.
= ₹30,000 — option (a).
69
A solution of 150 litres has a 40% salt concentration. How many litres of water must be added to reduce the concentration to 20%, and then how much pure salt must be added to bring the concentration back to 50%?
A.Add 150 L of water, then 180 L of salt
B.Add 90 L of water, then 170 L of salt
C.Add 170 L of water, then 30 L of salt
D.Add 150 L of water, then 10 L of salt
Solution
Dilute then strengthen
Salt = 40% of 150 = 60 L, so water = 90 L.
For 20% salt: = ⇒ 150 + x = 300 ⇒ x = 150 L of water.
Now the mixture is 300 L with 60 L of salt; adding y litres of salt: = .
120 + 2y = 300 + y ⇒ y = 180 L — option (a).
70
If cosθ = and θ lies in the 2nd quadrant, find the value of (1 + sinθ)(1 − sinθ).
If cos x = sin(3x − 10°), then what is the value of x?
A.25°
B.35°
C.28°
D.10°
Solution
cos = sin
sinA = cos(90° − A), so the two angles must be complementary.
x + (3x − 10°) = 90°.
4x = 100°.
x = 25° — option (a).
73
In an acute-angled triangle, the orthocentre lies at which position?
A.On one of the sides
B.Outside the triangle
C.Inside the triangle
D.At a vertex
Solution
Orthocentre
The orthocentre is where the three altitudes meet.
In an acute triangle all three altitudes fall inside, so their point of meeting is inside.
In a right triangle it sits at the right-angled vertex, and in an obtuse triangle it lies outside.
Hence option (c).
74
A chord is drawn in a circle with a radius of 17 cm. If the distance of the chord from the centre is 15 cm, what is the length of the chord?
A.13 cm
B.14 cm
C.16 cm
D.31 cm
Solution
Chord length
The perpendicular from the centre bisects the chord, forming a right triangle.
Half-chord = = .
= = 8 cm.
Chord = 2 × 8 = 16 cm — option (c).
75
The tangent to a circle from a point P is 9 cm and the distance from P to the centre is 15 cm. Find the radius.
A.12 cm
B.11 cm
C. cm
D. cm
Solution
Tangent and radius
The radius meets the tangent at right angles, so we get a right triangle.
r = .
= = .
= 12 cm — option (a).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Students of height 1.68 m in Section A = 8.
Total in Section A = 5 + 9 + 8 + 15 + 8 + 6 + 4.
= 55.
Ratio = 8 : 55 — option (a).
Inside the root, do division first: 225 ÷ 15 = 15.
Then 4 × 15 = 60.
60 + 25 − 4 = 81.
So the value is — option (a).
Interest for 2 years = 6,200 − 5,800 = ₹400, so one year’s interest = ₹200.
Principal = 5,800 − 2 × 200 = ₹5,400.
R = × 100.
≈ 3.7% — option (c).
Let the distance be x km. Then − = hour.
= ⇒ = .
20x = 224.
x = 11.2 km — option (d).
The string is the hypotenuse and the height is the side opposite the angle.
sin30° = = .
= .
L = 90 m — option (c).
Volume of one sphere = πr³ = π(125) = cm³; four of them = cm³.
Base area of the tank = 80 × 60 = 4800 cm².
Rise = = .
= cm — option (b).
X: 40,000 × 8 = 3,20,000.
Y: 60,000 × 6 = 3,60,000.
Ratio = 320 : 360.
= 8 : 9 — option (b).
Interior + exterior = 180° and interior − exterior = 60°.
Adding: 2 × interior = 240° ⇒ interior = 120°, so exterior = 60°.
Number of sides = = .
= 6 — option (a).
Take the value as 100; after a 10% rise it becomes 110.
To come back it must fall by 10, but that fall is measured on 110.
× 100.
≈ 9.09% — option (a).
For a square, area = .
d² = ()² = 48.
Area = .
= 24 m² — option (b).
Acid = × 30 = 20 L and water = 10 L.
The acid stays fixed, so the water must also become 20 L.
Water to be added = 20 − 10.
= 10 L — option (b).
Take the original CP as 100; at 15% profit the SP = 115.
New SP = 115 × 1.10 = 126.5, and this is a 20% profit.
New CP = = 105.41.
Increase = 5.41% — option (d).
After a 15% discount the buyer pays 85% of the marked price.
85% of MP = ₹6,800.
MP = .
= ₹8,000 — option (c).
4 men = 6 women, so 2 men = 3 women.
Hence 2 men + 3 women = 2 men + 2 men = 4 men.
But 4 men finish the work in 18 days.
So the time is again 18 days — option (c).
The bill of ₹40,000 is above ₹15,000, so a 20% discount applies: 40,000 × = ₹32,000.
₹32,000 exceeds ₹20,000, so the extra ₹2,000 comes off.
32,000 − 2,000.
= ₹30,000 — option (a).
Salt = 40% of 150 = 60 L, so water = 90 L.
For 20% salt: = ⇒ 150 + x = 300 ⇒ x = 150 L of water.
Now the mixture is 300 L with 60 L of salt; adding y litres of salt: = .
120 + 2y = 300 + y ⇒ y = 180 L — option (a).
(1 + sinθ)(1 − sinθ) = 1 − sin²θ.
But 1 − sin²θ = cos²θ.
cos²θ = ()².
= — option (d).
Degrees = radians × .
= 0.6 × .
= .
≈ 34.39° — option (b).
sinA = cos(90° − A), so the two angles must be complementary.
x + (3x − 10°) = 90°.
4x = 100°.
x = 25° — option (a).
The orthocentre is where the three altitudes meet.
In an acute triangle all three altitudes fall inside, so their point of meeting is inside.
In a right triangle it sits at the right-angled vertex, and in an obtuse triangle it lies outside.
Hence option (c).
The perpendicular from the centre bisects the chord, forming a right triangle.
Half-chord = = .
= = 8 cm.
Chord = 2 × 8 = 16 cm — option (c).
The radius meets the tangent at right angles, so we get a right triangle.
r = .
= = .
= 12 cm — option (a).