SSC CGL 25 September 2025 · Shift 1 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 159 of 192
25 September 2025 · Shift 1
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
Calculate the compound interest on ₹6,400 for 1 year at 25% per annum when compounded half-yearly.
A.₹2,300
B.₹1,700
C.₹2,250
D.₹1,300
Solution
Half-yearly CI
Half-yearly: rate = = 12.5% and periods = 2.
Amount = 6,400 × ()² = 6,400 × .
= ₹8,100.
CI = 8,100 − 6,400 = ₹1,700 — option (b).
52
Which set includes numbers like −7, 4 and 0?
A.Natural Numbers
B.Whole Numbers
C.Irrational Numbers
D.Integers
Solution
Integers
Natural numbers start from 1 and whole numbers from 0 — neither takes in −7.
Irrational numbers cannot be written as a fraction, so 4 and 0 are ruled out.
Integers are …, −2, −1, 0, 1, 2, … and cover all three.
Hence option (d).
53
A car travelling at a speed of 45 km/h can complete a journey in 9 hours. How long will it take to travel the same distance at 27 km/h?
A.19 h
B.16 h
C.15 h
D.25 h
Solution
Speed and time
Distance = 45 × 9 = 405 km.
Time = .
= .
= 15 hours — option (c).
54
The volume of a sphere is 36π cm³. If the radius is tripled, what is the new volume?
A.972π cm³
B.729π cm³
C.1296π cm³
D.1881π cm³
Solution
Volume and radius
Volume varies as the CUBE of the radius.
Tripling the radius multiplies the volume by 3³ = 27.
36π × 27.
= 972π cm³ — option (a).
55
A rectangular room measures 10 m long and 8 m wide. Its floor is to be covered with square tiles of side 50 cm. If 15% extra tiles are purchased for cutting and breakage, how many tiles are needed in total?
A.400
B.350
C.468
D.368
Solution
Tiling with extra
Floor area = 10 × 8 = 80 m².
Each tile is 0.5 m × 0.5 m = 0.25 m², so plain requirement = = 320 tiles.
Extra 15% = 320 × = 48.
Total = 320 + 48 = 368 — option (d).
56
A's salary is 25% more than B's. If B's salary increases by 30% and A's increases by y%, then A's new salary becomes 10% more than B's new salary. What is the value of y?
A.12.4%
B.8.5%
C.14.4%
D.16.6%
Solution
Percentage of a percentage
Take B = 100 units, so A = 125 units.
New B = 100 × = 130; new A must be 10% more = 130 × = 143.
Increase in A = 143 − 125 = 18 on an original 125.
y = × 100 = 14.4% — option (c).
57
Three friends A, B and C invested in a start-up in the ratio 3 : 4 : 1. C managed the business and was entitled to an additional 20% of the total annual profit as a management fee. If the profit at the end of the year was ₹5,00,000, how much did B receive?
A.₹2,00,000
B.₹2,12,000
C.₹1,10,000
D.₹1,20,000
Solution
Profit with a management fee
C’s management fee = 20% of 5,00,000 = ₹1,00,000, taken out first.
Remaining profit = 5,00,000 − 1,00,000 = ₹4,00,000, shared as 3 : 4 : 1 (8 parts).
1 part = = ₹50,000.
B has 4 parts = ₹2,00,000 — option (a).
58
A rectangular lawn measures 60 m by 40 m. A 5 m wide path is constructed all around the inside of the lawn. What percentage of the lawn's area is occupied by the path?
A.35%
B.37.5%
C.16.67%
D.47.5%
Solution
Path inside a lawn
Lawn area = 60 × 40 = 2400 m².
The path runs inside, so the inner rectangle is (60 − 10) × (40 − 10) = 50 × 30 = 1500 m².
Path area = 2400 − 1500 = 900 m².
× 100 = 37.5% — option (b).
59
An L-shaped floor is formed by adjoining an 8 m × 5 m and a 5 m × 3 m rectangle. Tiles cost ₹200 per m². What is the total cost to tile the floor?
A.₹19,600
B.₹21,700
C.₹22,000
D.₹11,000
Solution
L-shaped floor
First rectangle = 8 × 5 = 40 m².
Second rectangle = 5 × 3 = 15 m².
Total area = 40 + 15 = 55 m².
Cost = 55 × 200 = ₹11,000 — option (d).
60
An equilateral triangle and a square have equal perimeters. What is the ratio of their sides?
A.3 : 4
B.1 : 2
C.4 : 3
D.2 : 1
Solution
Equal perimeters
Let the sides be a and b, so 3a = 4b.
Therefore = .
The triangle’s side is the larger of the two.
Ratio = 4 : 3 — option (c).
61
A shopkeeper sells two shirts at ₹600 each. On one he gains 20% and on the other he loses 20%. What is his net gain or loss?
A.₹50 loss
B.₹75 gain
C.₹40 gain
D.₹80 loss
Solution
Equal SP, +20% and −20%
First shirt: CP = 600 × = ₹500.
Second shirt: CP = 600 × = ₹750.
Total CP = ₹1,250 while total SP = ₹1,200.
Loss = ₹50 — option (a).
62
If the selling price of an article is 85% of its marked price and the cost price is 75% of the marked price, what is the profit percentage?
A.14%
B.13%
C.16%
D.13%
Solution
Profit from MP
Take MP = 100 units, so SP = 85 and CP = 75.
Profit = 85 − 75 = 10 units.
Profit% = × 100 = .
= 13% — option (b).
63
If the average of eight numbers is 60 and the average of the first five numbers is 51, what is the average of the last three numbers?
A.80
B.65
C.95
D.75
Solution
Average of the rest
Total of eight numbers = 8 × 60 = 480.
Total of the first five = 5 × 51 = 255.
Last three together = 480 − 255 = 225.
Average = = 75 — option (d).
64
A prism has a base perimeter of 20 cm and height 8 cm. If its total surface area is 200 cm², what is the area of both bases combined?
A.52 cm²
B.46 cm²
C.40 cm²
D.62 cm²
Solution
Total surface area of a prism
TSA = lateral surface area + area of the two bases.
Lateral surface area = perimeter × height = 20 × 8 = 160 cm².
So the two bases together = 200 − 160.
= 40 cm² — option (c).
65
A bookshop advertises: "Buy 3 books and get the 4th at 40% discount." If the marked price of each book is ₹500 and all books are priced the same, what is the effective percentage discount on the total purchase?
A.10%
B.12%
C.18%
D.25%
Solution
Effective discount
Marked price of 4 books = 4 × 500 = ₹2,000.
The customer pays full price for three (₹1,500) and 60% for the fourth (₹300).
Total paid = ₹1,800, so the discount = ₹200.
× 100 = 10% — option (a).
66
An investor tracks 6 stocks. Five of them have an average return of 7%. If the overall average return is 10%, what was the return on the sixth stock?
A.18%
B.25%
C.20%
D.11%
Solution
Missing value from an average
Total of all six = 6 × 10 = 60%.
Total of the five = 5 × 7 = 35%.
Sixth stock = 60 − 35.
= 25% — option (b).
Perfect square identity
Notice that tan²A + cot²A + 2 tanA cotA = (tanA + cotA)².
So we simply square the given value.
()².
= — option (d).
68
If cos 2A = 3, then what is the value of cos A?
A.2
B.4
C.
D.
Solution
Double-angle formula
Use the identity cos 2A = 2cos²A − 1.
2cos²A − 1 = 3 ⇒ 2cos²A = 4.
cos²A = 2.
cos A = — option (c).
69
An angle of radians represents what percentage of a full circle?
A.33.33%
B.50%
C.43.33%
D.35%
Solution
Fraction of a circle
A full circle is 2π radians.
Fraction = = .
× 100.
≈ 33.33% — option (a).
70
If cosec x = sec(4x − 40°), find x.
A.45°
B.26°
C.30°
D.60°
Solution
cosec = sec
sec θ = cosec(90° − θ), so the two angles must be complementary.
x + (4x − 40°) = 90°.
5x = 130°.
x = 26° — option (b).
71
If two circles of radius 8 cm and 3 cm have their centres 13 cm apart, what is the length of the direct common tangent?
A. cm
B.114 cm
C.78 cm
D.12 cm
Solution
Direct common tangent
Direct common tangent = .
= .
= = .
= 12 cm — option (d).
72
A chord separates a circle into two segments. If the angle formed by the chord at the centre is 100°, what is the angle formed at a point located in the major segment?
A.90°
B.75°
C.50°
D.120°
Solution
Angle in the major segment
The angle at the centre is twice the angle at any point of the major segment.
So the angle at the circumference = × 100°.
= 50°.
Hence option (c).
73
In a circle, the chords PQ and RS are equal. The angle subtended by PQ at the circumference is 75°. What is the angle subtended by RS at the circumference?
A.75°
B.37.5°
C.120°
D.95°
Solution
Equal chords
Equal chords of a circle subtend equal angles at the centre.
Half of those equal central angles are the angles at the circumference.
So equal chords also subtend equal angles at the circumference.
= 75° — option (a).
74
A chord CD is drawn in a circle with centre O. The tangent at C and the tangent at D meet at a point Q. If ∠CDQ = 60°, what is the angle subtended by the chord CD at the circumference of the circle?
A.30°
B.60°
C.90°
D.120°
Solution
Tangents from a point
QC and QD are tangents from the same point, so QC = QD and ∠QCD = ∠CDQ = 60°.
Hence ∠CQD = 180° − 120° = 60°, and ∠COD = 180° − ∠CQD = 120°.
The angle at the circumference is half the angle at the centre.
= 60° — option (b).
75
A line is tangent to a circle at point P. If a chord PQ has a length of 10 cm and makes an angle of 45° with the tangent, what is the radius of the circle?
A.4 cm
B.3 cm
C.5 cm
D.5 cm
Solution
Tangent-chord angle
By the alternate segment theorem, the chord subtends 45° at the circumference.
So the angle at the centre is 90°, and the chord relation gives PQ = 2R sin45°.
10 = 2R × ⇒ R = .
= 5 cm — option (d).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Half-yearly: rate = = 12.5% and periods = 2.
Amount = 6,400 × ()² = 6,400 × .
= ₹8,100.
CI = 8,100 − 6,400 = ₹1,700 — option (b).
Distance = 45 × 9 = 405 km.
Time = .
= .
= 15 hours — option (c).
Volume varies as the CUBE of the radius.
Tripling the radius multiplies the volume by 3³ = 27.
36π × 27.
= 972π cm³ — option (a).
Floor area = 10 × 8 = 80 m².
Each tile is 0.5 m × 0.5 m = 0.25 m², so plain requirement = = 320 tiles.
Extra 15% = 320 × = 48.
Total = 320 + 48 = 368 — option (d).
Take B = 100 units, so A = 125 units.
New B = 100 × = 130; new A must be 10% more = 130 × = 143.
Increase in A = 143 − 125 = 18 on an original 125.
y = × 100 = 14.4% — option (c).
C’s management fee = 20% of 5,00,000 = ₹1,00,000, taken out first.
Remaining profit = 5,00,000 − 1,00,000 = ₹4,00,000, shared as 3 : 4 : 1 (8 parts).
1 part = = ₹50,000.
B has 4 parts = ₹2,00,000 — option (a).
Lawn area = 60 × 40 = 2400 m².
The path runs inside, so the inner rectangle is (60 − 10) × (40 − 10) = 50 × 30 = 1500 m².
Path area = 2400 − 1500 = 900 m².
× 100 = 37.5% — option (b).
First rectangle = 8 × 5 = 40 m².
Second rectangle = 5 × 3 = 15 m².
Total area = 40 + 15 = 55 m².
Cost = 55 × 200 = ₹11,000 — option (d).
Let the sides be a and b, so 3a = 4b.
Therefore = .
The triangle’s side is the larger of the two.
Ratio = 4 : 3 — option (c).
First shirt: CP = 600 × = ₹500.
Second shirt: CP = 600 × = ₹750.
Total CP = ₹1,250 while total SP = ₹1,200.
Loss = ₹50 — option (a).
Take MP = 100 units, so SP = 85 and CP = 75.
Profit = 85 − 75 = 10 units.
Profit% = × 100 = .
= 13% — option (b).
Total of eight numbers = 8 × 60 = 480.
Total of the first five = 5 × 51 = 255.
Last three together = 480 − 255 = 225.
Average = = 75 — option (d).
TSA = lateral surface area + area of the two bases.
Lateral surface area = perimeter × height = 20 × 8 = 160 cm².
So the two bases together = 200 − 160.
= 40 cm² — option (c).
Marked price of 4 books = 4 × 500 = ₹2,000.
The customer pays full price for three (₹1,500) and 60% for the fourth (₹300).
Total paid = ₹1,800, so the discount = ₹200.
× 100 = 10% — option (a).
Total of all six = 6 × 10 = 60%.
Total of the five = 5 × 7 = 35%.
Sixth stock = 60 − 35.
= 25% — option (b).
sec θ = cosec(90° − θ), so the two angles must be complementary.
x + (4x − 40°) = 90°.
5x = 130°.
x = 26° — option (b).
Direct common tangent = .
= .
= = .
= 12 cm — option (d).
The angle at the centre is twice the angle at any point of the major segment.
So the angle at the circumference = × 100°.
= 50°.
Hence option (c).
Equal chords of a circle subtend equal angles at the centre.
Half of those equal central angles are the angles at the circumference.
So equal chords also subtend equal angles at the circumference.
= 75° — option (a).
QC and QD are tangents from the same point, so QC = QD and ∠QCD = ∠CDQ = 60°.
Hence ∠CQD = 180° − 120° = 60°, and ∠COD = 180° − ∠CQD = 120°.
The angle at the circumference is half the angle at the centre.
= 60° — option (b).
By the alternate segment theorem, the chord subtends 45° at the circumference.
So the angle at the centre is 90°, and the chord relation gives PQ = 2R sin45°.
10 = 2R × ⇒ R = .
= 5 cm — option (d).