SSC CGL 16 September 2025 · Shift 1 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 51 of 192
16 September 2025 · Shift 1
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
If A : B = 4 : 7, B : C = 3 : 5, and C : D = 8 : 9, find A : B : C : D.
A.96 : 168 : 280 : 315
B.168 : 96 : 280 : 315
C.96 : 168 : 315 : 280
D.96 : 280 : 168 : 315
Solution
Chained ratios
Link the ratios through the common terms.
A : B = 4 : 7 and B : C = 3 : 5 ⇒ A : B : C = 12 : 21 : 35.
With C : D = 8 : 9, multiply suitably: A : B : C : D = 96 : 168 : 280 : 315.
Hence option (a).
52
A box contains ₹10, ₹5, and ₹2 coins in the ratio 1 : 2 : 3. If the total amount of money in the box is ₹416, find the number of coins of each kind.
A.20 coins of ₹10, 10 of ₹5, and 30 of ₹2
B.16 coins of ₹10, 32 of ₹5, and 48 of ₹2
C.40 coins of ₹10, 10 of ₹5, and 60 of ₹2
D.10 coins of ₹10, 20 of ₹5, and 36 of ₹2
Solution
Coins in ratio
Let the counts be x, 2x and 3x.
Value: 10x + 5(2x) + 2(3x) = 10x + 10x + 6x = 26x.
26x = 416 ⇒ x = 16.
Coins = 16, 32 and 48 — option (b).
53
A tank holds 7 liters of water. If 3.375 liters are drained twice, how much water remains?
P and Q invested in a business in the ratio 4 : 5. The profit ratio was 8 : 9. If Q invested his money for 9 months, for how many months did P invest?
A.12 months
B.10 months
C.16 months
D.14 months
Solution
Partnership — time of P
Profit ∝ capital × time.
= .
36x = 360 ⇒ x = 10.
Hence 10 months — option (b).
57
The average of 8 numbers is 56. When one number is removed, the average becomes 54. What number was removed?
A.52
B.70
C.80
D.92
Solution
Average — number removed
Total of 8 numbers = 8 × 56 = 448.
Total of the remaining 7 = 7 × 54 = 378.
Removed number = 448 − 378.
= 70 — option (b).
58
What is the overall average daily diesel usage for both cars?
Vehicle
Mon
Tue
Wed
Thu
Fri
Car 1
19
22
24
16
21
Car 2
15
28
23
20
29
A.22.5 litres
B.21.7 litres
C.23.7 litres
D.25.5 litres
Solution
Table — daily average
Weekly Diesel Usage (litres). Add all ten readings.
Car 1: 19+22+24+16+21 = 102; Car 2: 15+28+23+20+29 = 115.
Total = 217 over 10 vehicle-days.
Average = = 21.7 litres — option (b).
59
Rohit invested a certain sum of money at 10% compound interest per annum. After 2 years, the amount became ₹24,200. Find the sum he had invested.
A.₹20000
B.₹21000
C.₹21500
D.₹10500
Solution
CI — find principal
At 10%, P : A over 2 years = 10² : 11² = 100 : 121.
121 units = ₹24,200 ⇒ 1 unit = ₹200.
P = 100 units = 100 × 200.
= ₹20,000 — option (a).
60
A grocer purchased 40 kg of rice at ₹40 per kg and 30 kg of another variety of rice at ₹55 per kg. He mixed the two varieties and sold the entire mixture at ₹48 per kg. Find his total profit or loss in this transaction.
The marked price of a TV is ₹30,000. Amit receives two successive discounts while buying it. If the first discount is 20% and the final price he pays is ₹16,800, what is the second discount rate?
A.30%
B.16%
C.14%
D.15%
Solution
Second discount
After the first discount: 30,000 × 0.80 = ₹24,000.
Let the second discount be x%: 24,000 × = 16,800.
100 − x = = 70.
x = 30% — option (a).
62
A manufacturer marked a product 20% above its cost price. He provided a discount to a customer, which resulted in him making a profit of 8%. What was the rate of discount given by the manufacturer?
A.22%
B.40%
C.30%
D.10%
Solution
Markup 20%, profit 8%
Take CP = ₹100 ⇒ MP = ₹120 and SP = ₹108.
Discount = 120 − 108 = ₹12 on the marked price.
Rate = × 100.
= 10% — option (d).
63
Two containers, A and B, contain mixtures of alcohol and water. Container A has them in the ratio 2 : 1, while Container B has them in the ratio 4 : 3. If 9 liters are drawn from Container A and 14 liters from Container B, and the contents are mixed in a third container, what is the ratio of alcohol to water in the new mixture?
A.3 : 11
B.14 : 9
C.9 : 14
D.11 : 3
Solution
Two containers mixed
From A (2 : 1) in 9 L: alcohol = × 9 = 6 L, water = 3 L.
From B (4 : 3) in 14 L: alcohol = × 14 = 8 L, water = 6 L.
Total alcohol = 6 + 8 = 14 L; total water = 3 + 6 = 9 L.
Ratio = 14 : 9 — option (b).
64
If the radius of a hemisphere is increased to twice its original size, what is the ratio of the new surface area to the original surface area, as well as the proportion of the new volume to the original volume?
A.4 : 1 and 8 : 1
B.6 : 1 and 8 : 1
C.8 : 1 and 2 : 1
D.2 : 1 and 4 : 1
Solution
Hemisphere — radius doubled
Surface area of a hemisphere = 3πr² — it varies as r².
Doubling r multiplies the area by 2² = 4 ⇒ 4 : 1.
Volume = πr³ varies as r³, so it multiplies by 2³ = 8 ⇒ 8 : 1.
Hence 4 : 1 and 8 : 1 — option (a).
65
A hemisphere and a cone share the same base and have equal volumes. Given that their common radius is R, determine the height of the cone.
A.R
B.2R
C.5R
D.3R
Solution
Hemisphere = cone volume
Equate volumes: πR³ = πR²H.
Cancel πR²: 2R = H.
So the cone must be twice as tall as the radius.
H = 2R — option (b).
66
A solid sphere of radius 8 cm is melted into four cones of equal height. If each cone's radius is 4 cm, what is the height of each cone?
A.40 cm
B.32 cm
C.50 cm
D.52 cm
Solution
Sphere → 4 cones
Volume is conserved: π(8)³ = 4 × π(4)²h.
π × 512 = π × 16 × h.
512 = 16h ⇒ h = 32.
Hence 32 cm — option (b).
67
If base perimeter = 30 cm and slant height = 10 cm, what is the lateral surface area of a square pyramid?
A.160 cm²
B.200 cm²
C.150 cm²
D.152 cm²
Solution
Square pyramid LSA
LSA of a pyramid = × base perimeter × slant height.
= × 30 × 10.
= 150.
Hence 150 cm² — option (c).
68
Two circular tracks have radii in the ratio 2 : 5. If the smaller track has an area of 100π cm², what is the area of the larger track?
A.215π cm²
B.235π cm²
C.625π cm²
D.240π cm²
Solution
Areas ∝ r²
Area varies as the square of the radius.
Ratio of areas = 2² : 5² = 4 : 25.
= ⇒ x = 625π.
Hence 625π cm² — option (c).
69
In a triangle PQR, medians PS and QT intersect at O. If the length of median PS is 15 cm, what is the length of the segment PO?
A.10 cm
B.6 cm
C.8 cm
D.9 cm
Solution
Centroid divides 2 : 1
The centroid divides every median in the ratio 2 : 1 from the vertex.
So PS is made of 3 equal parts and PO takes 2 of them.
PO = × 15.
= 10 cm — option (a).
70
Two right-angled triangular blocks, PQR and XYZ, have ∠Q = ∠Y = 90°. If the lengths of the hypotenuses PR and XZ are equal, and the sides PQ and XY are equal, are the triangles congruent? If so, by what rule?
A.Yes, by RHS
B.Yes, by SAS
C.Yes, by SSS
D.Yes, by ASA
Solution
Congruence — RHS
Both triangles have a right angle, an equal hypotenuse and one equal side.
That is exactly the Right angle–Hypotenuse–Side condition.
By Pythagoras the third sides must then match too.
Congruent by RHS — option (a).
71
A right triangle ABC is inscribed in a circle with a diameter of 26 cm. An altitude BD is drawn from the vertex B to the hypotenuse AC. If the length of the side AB is 10 cm, what is the length of the segment AD?
A.3.8 cm
B.3 cm
C.5.2 cm
D.4 cm
Solution
Altitude on hypotenuse
The hypotenuse AC is the diameter = 26 cm (angle in a semicircle is 90°).
Geometric-mean relation: AB² = AD × AC.
100 = AD × 26 ⇒ AD = ≈ 3.85.
≈ 3.8 cm — option (a).
72
Two circles touch each other externally. Which of the following statements is true?
A.There are three common tangents to the circles
B.There are two common tangents to the circles
C.There are four common tangents to the circles
D.There is only one common tangent to the circles
Solution
Tangents — touching circles
Externally touching circles have two DIRECT common tangents.
At the point of contact there is one more common tangent.
Total = 2 + 1 = 3.
Hence option (a).
73
Two circles with radii R and r touch each other externally. If the length of their direct common tangent is D, which of the following is the correct relationship between D, R, and r?
A.D = R + r
B.D = 2(R + r)
C.D = 2
D.D = R² + r²
Solution
Tangent of touching circles
For a direct common tangent: D² = d² − (R − r)², where d is the distance between centres.
Touching externally means d = R + r.
D² = (R + r)² − (R − r)² = 4Rr.
D = 2 — option (c).
74
A chord of a circle has a length of 10 cm. The angle subtended by the chord at a point on the circumference is 30°. What is the distance from the center of the circle to the chord?
A.5 cm
B.3 cm
C.5 cm
D.3 cm
Solution
Chord — distance from centre
Angle at the centre = 2 × angle at the circumference = 60°.
The perpendicular from O bisects both the chord and the central angle, giving a right triangle with half-chord 5 and angle 30° at O.
tan60° at the other vertex: = ⇒ OP = 5.
Distance = 5 cm — option (a).
75
What is the area of the segment formed by a chord in a circle of radius 4 cm, if the angle subtended at the center is 60°?
A.() − 4
B.() − 4
C.() − 8
D.() − 8
Solution
Segment — 60°
Sector = × π × 4² = .
With a 60° central angle the triangle is equilateral of side 4: area = × 16 = 4.
Segment = sector − triangle.
= − 4 — option (a).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Link the ratios through the common terms.
A : B = 4 : 7 and B : C = 3 : 5 ⇒ A : B : C = 12 : 21 : 35.
With C : D = 8 : 9, multiply suitably: A : B : C : D = 96 : 168 : 280 : 315.
Hence option (a).
Let the counts be x, 2x and 3x.
Value: 10x + 5(2x) + 2(3x) = 10x + 10x + 6x = 26x.
26x = 416 ⇒ x = 16.
Coins = 16, 32 and 48 — option (b).
Profit ∝ capital × time.
= .
36x = 360 ⇒ x = 10.
Hence 10 months — option (b).
Total of 8 numbers = 8 × 56 = 448.
Total of the remaining 7 = 7 × 54 = 378.
Removed number = 448 − 378.
= 70 — option (b).
At 10%, P : A over 2 years = 10² : 11² = 100 : 121.
121 units = ₹24,200 ⇒ 1 unit = ₹200.
P = 100 units = 100 × 200.
= ₹20,000 — option (a).
Cost = 40 × 40 + 30 × 55 = 1600 + 1650 = ₹3,250.
Total quantity = 70 kg, sold at ₹48 ⇒ SP = 70 × 48 = ₹3,360.
Profit = 3,360 − 3,250.
= ₹110 profit — option (c).
After the first discount: 30,000 × 0.80 = ₹24,000.
Let the second discount be x%: 24,000 × = 16,800.
100 − x = = 70.
x = 30% — option (a).
Take CP = ₹100 ⇒ MP = ₹120 and SP = ₹108.
Discount = 120 − 108 = ₹12 on the marked price.
Rate = × 100.
= 10% — option (d).
From A (2 : 1) in 9 L: alcohol = × 9 = 6 L, water = 3 L.
From B (4 : 3) in 14 L: alcohol = × 14 = 8 L, water = 6 L.
Total alcohol = 6 + 8 = 14 L; total water = 3 + 6 = 9 L.
Ratio = 14 : 9 — option (b).
Surface area of a hemisphere = 3πr² — it varies as r².
Doubling r multiplies the area by 2² = 4 ⇒ 4 : 1.
Volume = πr³ varies as r³, so it multiplies by 2³ = 8 ⇒ 8 : 1.
Hence 4 : 1 and 8 : 1 — option (a).
Equate volumes: πR³ = πR²H.
Cancel πR²: 2R = H.
So the cone must be twice as tall as the radius.
H = 2R — option (b).
Volume is conserved: π(8)³ = 4 × π(4)²h.
π × 512 = π × 16 × h.
512 = 16h ⇒ h = 32.
Hence 32 cm — option (b).
Area varies as the square of the radius.
Ratio of areas = 2² : 5² = 4 : 25.
= ⇒ x = 625π.
Hence 625π cm² — option (c).
The centroid divides every median in the ratio 2 : 1 from the vertex.
So PS is made of 3 equal parts and PO takes 2 of them.
PO = × 15.
= 10 cm — option (a).
The hypotenuse AC is the diameter = 26 cm (angle in a semicircle is 90°).
Geometric-mean relation: AB² = AD × AC.
100 = AD × 26 ⇒ AD = ≈ 3.85.
≈ 3.8 cm — option (a).
Externally touching circles have two DIRECT common tangents.
At the point of contact there is one more common tangent.
Total = 2 + 1 = 3.
Hence option (a).
Angle at the centre = 2 × angle at the circumference = 60°.
The perpendicular from O bisects both the chord and the central angle, giving a right triangle with half-chord 5 and angle 30° at O.
tan60° at the other vertex: = ⇒ OP = 5.
Distance = 5 cm — option (a).
Sector = × π × 4² = .
With a 60° central angle the triangle is equilateral of side 4: area = × 16 = 4.
Segment = sector − triangle.
= − 4 — option (a).