SSC CGL 23 September 2025 · Shift 1 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 135 of 192
23 September 2025 · Shift 1
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
An electronics store orders 180 portable speakers at a cost of ₹420 each. The owner plans to price them so that selling only 140 speakers would guarantee a 30% profit on the total investment. However, a local event leads to the sale of 155 speakers. If the remaining speakers have no resale value, calculate the store's actual profit percentage on the total investment.
A.15.52%
B.16.83%
C.20.23%
D.43.93%
Solution
Profit on total investment
Total cost = 180 × 420 = ₹75,600; a 30% profit needs a revenue of 75,600 × = ₹98,280.
That revenue is to come from 140 speakers, so SP per speaker = = ₹702.
Actual revenue on 155 speakers = 155 × 702 = ₹1,08,810.
Profit = 1,08,810 − 75,600 = ₹33,210 ⇒ × 100 = 43.93% — option (d).
52
The set of integers includes which of the following?
A.Only positive numbers
B.Both positive and negative numbers
C.Only whole numbers
D.Only natural numbers
Solution
Set of integers
Integers are …, −3, −2, −1, 0, 1, 2, 3, …
They therefore take in the negatives, zero and the positives.
Whole numbers and natural numbers are only parts of this larger set.
Hence option (b).
53
Riya crosses a road 150 metres wide in 75 seconds. Her speed in km/h is:
A.7.0 km/h
B.7.1 km/h
C.7.2 km/h
D.7.5 km/h
Solution
Speed conversion
Speed = = 2 m/s.
To convert m/s into km/h, multiply by .
2 × 3.6.
= 7.2 km/h — option (c).
54
A sphere is inscribed inside a cube. The volume of the cube is 3375 cm³. What is the total surface area of the sphere?
A.225π cm²
B.254π cm²
C.256π cm²
D.298π cm²
Solution
Sphere in a cube
Cube volume 3375 = a³, so the edge a = 15 cm.
An inscribed sphere has diameter equal to the edge, so r = 7.5 cm.
Surface area = 4πr² = 4π × 56.25.
= 225π cm² — option (a).
55
A large wooden box, open at the top, needs to be lined with fabric on its inner floor and inner four side walls. The inner dimensions of the box are 5 m length, 3.6 m width and 2.4 m height. If the fabric comes in square pieces of side 100 cm, and 30% extra fabric is needed for cuts and overlaps, approximately how many fabric pieces should be purchased?
A.77
B.88
C.62
D.98
Solution
Lining an open box
Floor = 5 × 3.6 = 18 m²; four walls = 2 × 2.4 × (5 + 3.6) = 41.28 m².
Total to be covered = 18 + 41.28 = 59.28 m².
With 30% extra: 59.28 × = 77.064 m².
Each piece is 1 m × 1 m = 1 m², so about 77 pieces — option (a).
56
P and Q start a business with ₹2,40,000 and ₹3,60,000 respectively. After 4 months P adds ₹60,000 more, and Q withdraws ₹60,000. At the end of 1 year they make a profit of ₹2,40,000. What is Q's share of the profit?
A.₹1,28,000
B.₹1,25,000
C.₹2,00,000
D.₹2,90,000
Solution
Partnership with change
P: 2,40,000 × 4 + 3,00,000 × 8 = 9,60,000 + 24,00,000 = 33,60,000.
Q: 3,60,000 × 4 + 3,00,000 × 8 = 14,40,000 + 24,00,000 = 38,40,000.
Ratio P : Q = 336 : 384 = 7 : 8, i.e. 15 parts in all.
Q’s share = × 2,40,000 = ₹1,28,000 — option (a).
57
A square of side 12 cm has a smaller square (of side 5 cm) removed from one side. What is the ratio of the remaining area to the original area?
A.129 : 136
B.119 : 144
C.114 : 119
D.117 : 125
Solution
Remaining area ratio
Large square = 12² = 144 cm².
Small square removed = 5² = 25 cm².
Remaining = 144 − 25 = 119 cm².
Ratio = 119 : 144 — option (b).
58
A rectangular park is 80 m by 40 m. A semicircular pond of diameter 20 m is inside it. What percent of the park's area does the pond occupy?
A.6.91%
B.4.91%
C.5.81%
D.3.00%
Solution
Semicircular pond
Park area = 80 × 40 = 3200 m².
Diameter 20 m gives r = 10 m, so pond = πr² = × × 100 = 157.14 m².
Percentage = × 100.
≈ 4.91% — option (b).
59
A regular hexagon has a perimeter of 84 cm. What is its area?
A.508.62 cm²
B.450.00 cm²
C.600.12 cm²
D.480.36 cm²
Solution
Hexagon area
Side = = 14 cm.
Area = a² = × 196 = 294.
Taking = 1.73: 294 × 1.73.
≈ 508.62 cm² — option (a).
60
Out of her total monthly income, a woman spends 50% on household expenses and 20% on savings. The remaining is spent on other items. What percentage of her income is spent on these other items?
A.25%
B.30%
C.35%
D.40%
Solution
Remaining percentage
Household expenses and savings together = 50% + 20% = 70%.
The whole income is 100%.
Remaining = 100% − 70%.
= 30% — option (b).
61
X's salary is 50% more than Y's. If Y's salary increases by 30% and X's increases by p%, then X's new salary becomes 20% more than Y's new salary. What is the value of p?
A.2%
B.4%
C.5%
D.6%
Solution
Percentage of a percentage
Take Y = 100, so X = 150.
New Y = 100 × 1.30 = 130; the new X must be 20% more than this = 130 × 1.2 = 156.
So 150 × = 156 ⇒ 150 + 1.5p = 156.
1.5p = 6 ⇒ p = 4% — option (b).
62
The average of a certain number of quantities is 38. If 7 is added to each quantity, what will be the new average?
A.38
B.45
C.40
D.50
Solution
Average shift
Adding the same number to every quantity shifts the average by that number.
So the new average = old average + 7.
38 + 7.
= 45 — option (b).
63
A triangular prism has a base of area 56 cm² and height 15 cm. If 20% of the prism is hollowed out for wiring, what is the volume of the solid part?
A.672 cm³
B.680 cm³
C.690 cm³
D.700 cm³
Solution
Prism minus hollow
Volume of the prism = base area × height = 56 × 15 = 840 cm³.
20% is hollow, so the solid part is 80%.
840 × .
= 672 cm³ — option (a).
64
A group of 6 friends went to a cafe. If 5 of them paid ₹250 each and the sixth paid ₹x, the average bill per person came out to be ₹270. What is the value of x?
A.₹350
B.₹370
C.₹360
D.₹380
Solution
Average bill
Total bill = 6 × 270 = ₹1,620.
Five friends paid 5 × 250 = ₹1,250.
x = 1,620 − 1,250.
= ₹370 — option (b).
65
A seller offers a 30% discount on an item whose cost price is ₹800. Despite the discount, the seller earns a profit of 25%. What must be the marked price?
A.₹1,420
B.₹1,430
C.₹1,440
D.₹1,450
Solution
MP from CP and discount
A 25% profit on ₹800 gives SP = 800 × = ₹1,000.
This SP is what remains after a 30% discount, i.e. 70% of the marked price.
MP = = .
≈ ₹1,430 — option (b).
66
If sinA = 0.8, what is the value of tanA?
A.1.50
B.1.33
C.1.25
D.2.56
Solution
tanA from sinA
sin²A + cos²A = 1, so cos²A = 1 − 0.64 = 0.36.
cosA = 0.6.
tanA = = .
≈ 1.33 — option (b).
67
If cosecA + cotA = r, then what is cosecA in terms of r?
A.
B.
C.
D.
Solution
cosecA from r
The identity cosec²A − cot²A = 1 gives (cosecA + cotA)(cosecA − cotA) = 1.
So cosecA − cotA = .
Adding the two: 2 cosecA = r + = .
cosecA = — option (c).
68
A circular jogging track has a radius of 60 m. A runner covers a curved path that subtends an angle of 150° at the centre. What is the length of the arc she runs?
Which of the following is true for all acute angles A?
A.tan(90° − A) = sinA
B.cos(90° − A) = sinA
C.sin(90° − A) = tanA
D.cot(90° − A) = cosA
Solution
Complementary identity
For complementary angles, each ratio turns into its co-ratio.
So sin(90° − A) = cosA and cos(90° − A) = sinA.
Likewise tan(90° − A) = cotA and cot(90° − A) = tanA.
Only option (b) states a correct pair — option (b).
70
The compound interest on a certain sum for 2 years at 25% per annum is ₹625. Find the simple interest on the same sum at the same rate and period.
A.₹480.25
B.₹520.75
C.₹555.55
D.₹600.50
Solution
CI to SI
CI = P[(1.25)² − 1] = P(1.5625 − 1) = 0.5625P.
0.5625P = 625 ⇒ P = ₹1,111.11.
SI = = .
= ₹555.55 — option (c).
71
What is the distance between the centres of two circles having radii 7 cm and 4 cm such that exactly three common tangents exist?
A.11 cm
B.3 cm
C.7 cm
D.14 cm
Solution
Three common tangents
Exactly three common tangents occur when the two circles touch each other EXTERNALLY.
In that case the distance between the centres equals the sum of the radii.
d = 7 + 4.
= 11 cm — option (a).
72
In a circle, the angle subtended by chord AB at the centre is 100°. What is the measure of ∠ACB, where C lies on the circle?
A.40°
B.45°
C.50°
D.60°
Solution
Angle at circumference
The angle at the centre is twice the angle at any point on the remaining circumference.
∠ACB = × ∠AOB.
= .
= 50° — option (c).
73
In a circle with centre O, two chords AB and CD cross each other at right angles at the point P. If the distance from the centre O to chord AB is 6 cm and the distance from O to chord CD is 8 cm, what is the length of OP?
A.11 cm
B.7 cm
C.9 cm
D.10 cm
Solution
Perpendicular chords — OP
Drop perpendiculars OM and ON from O onto AB and CD. Since AB ⊥ CD, the figure OMPN is a rectangle.
So OM = NP = 6 cm and ON = MP = 8 cm, and OP is the diagonal of that rectangle.
OP = = = .
= 10 cm — option (d).
74
If the angle subtended by a chord at the centre is 120°, what is the angle subtended at a point on the circle in the same segment?
A.60°
B.75°
C.90°
D.80°
Solution
Angle at circumference
Angle at the centre = 2 × angle at the circumference.
120° = 2 × angle at the circumference.
Angle = .
= 60° — option (a).
75
In a triangle PQR the sides are p, q and r, and the area of the triangle is T. What is the radius of the inscribed circle (inradius)?
A.
B.
C.
D.
Solution
Inradius formula
Area of a triangle in terms of its inradius: T = r × s, where s is the semi-perimeter.
s = .
So r = = T ÷ .
= — option (b).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Total cost = 180 × 420 = ₹75,600; a 30% profit needs a revenue of 75,600 × = ₹98,280.
That revenue is to come from 140 speakers, so SP per speaker = = ₹702.
Actual revenue on 155 speakers = 155 × 702 = ₹1,08,810.
Profit = 1,08,810 − 75,600 = ₹33,210 ⇒ × 100 = 43.93% — option (d).
Speed = = 2 m/s.
To convert m/s into km/h, multiply by .
2 × 3.6.
= 7.2 km/h — option (c).
Cube volume 3375 = a³, so the edge a = 15 cm.
An inscribed sphere has diameter equal to the edge, so r = 7.5 cm.
Surface area = 4πr² = 4π × 56.25.
= 225π cm² — option (a).
Floor = 5 × 3.6 = 18 m²; four walls = 2 × 2.4 × (5 + 3.6) = 41.28 m².
Total to be covered = 18 + 41.28 = 59.28 m².
With 30% extra: 59.28 × = 77.064 m².
Each piece is 1 m × 1 m = 1 m², so about 77 pieces — option (a).
P: 2,40,000 × 4 + 3,00,000 × 8 = 9,60,000 + 24,00,000 = 33,60,000.
Q: 3,60,000 × 4 + 3,00,000 × 8 = 14,40,000 + 24,00,000 = 38,40,000.
Ratio P : Q = 336 : 384 = 7 : 8, i.e. 15 parts in all.
Q’s share = × 2,40,000 = ₹1,28,000 — option (a).
Large square = 12² = 144 cm².
Small square removed = 5² = 25 cm².
Remaining = 144 − 25 = 119 cm².
Ratio = 119 : 144 — option (b).
Park area = 80 × 40 = 3200 m².
Diameter 20 m gives r = 10 m, so pond = πr² = × × 100 = 157.14 m².
Percentage = × 100.
≈ 4.91% — option (b).
Side = = 14 cm.
Area = a² = × 196 = 294.
Taking = 1.73: 294 × 1.73.
≈ 508.62 cm² — option (a).
Take Y = 100, so X = 150.
New Y = 100 × 1.30 = 130; the new X must be 20% more than this = 130 × 1.2 = 156.
So 150 × = 156 ⇒ 150 + 1.5p = 156.
1.5p = 6 ⇒ p = 4% — option (b).
Volume of the prism = base area × height = 56 × 15 = 840 cm³.
20% is hollow, so the solid part is 80%.
840 × .
= 672 cm³ — option (a).
A 25% profit on ₹800 gives SP = 800 × = ₹1,000.
This SP is what remains after a 30% discount, i.e. 70% of the marked price.
MP = = .
≈ ₹1,430 — option (b).
The identity cosec²A − cot²A = 1 gives (cosecA + cotA)(cosecA − cotA) = 1.
So cosecA − cotA = .
Adding the two: 2 cosecA = r + = .
cosecA = — option (c).
Arc length = 2πr × .
= 2π × 60 × .
= 120π × .
= 50π m — option (d).
CI = P[(1.25)² − 1] = P(1.5625 − 1) = 0.5625P.
0.5625P = 625 ⇒ P = ₹1,111.11.
SI = = .
= ₹555.55 — option (c).
Exactly three common tangents occur when the two circles touch each other EXTERNALLY.
In that case the distance between the centres equals the sum of the radii.
d = 7 + 4.
= 11 cm — option (a).
The angle at the centre is twice the angle at any point on the remaining circumference.
∠ACB = × ∠AOB.
= .
= 50° — option (c).
Drop perpendiculars OM and ON from O onto AB and CD. Since AB ⊥ CD, the figure OMPN is a rectangle.
So OM = NP = 6 cm and ON = MP = 8 cm, and OP is the diagonal of that rectangle.
OP = = = .
= 10 cm — option (d).
Area of a triangle in terms of its inradius: T = r × s, where s is the semi-perimeter.
s = .
So r = = T ÷ .
= — option (b).