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SSC CGL 16 September 2025 · Shift 2 · Quantitative Aptitude — questions with answers · ShikshaSphere

Chapter 55 of 192

16 September 2025 · Shift 2

Quantitative Aptitude

25 questions ~25 min readFree

25 questions

If a : b = 5 : 3, find (3a + 4b) : (3a − 4b).

A.9 : 1
B.1 : 9
C.7 : 1
D.1 : 7

Solution

Ratio substitution Take a = 5k, b = 3k. 3a + 4b = 15k + 12k = 27k. 3a − 4b = 15k − 12k = 3k. Ratio = 27 : 3 = 9 : 1 — option (a).

What is the value of: (

\sqrt{9}

\sqrt{4}

)² − (

\sqrt{8}

\sqrt{2}

)²

Solution

Surds — squares

\sqrt{9}

\sqrt{4}

= 3 + 2 = 5, so the first square is 25.

\sqrt{8}

\sqrt{2}

= 2

\sqrt{2}

\sqrt{2}

= 3

\sqrt{2}

, so the second square is 18. 25 − 18 = 7. Hence option (d).

Simplify: (

\frac{5}{6}

\frac{3}{4}

) ÷ (

\frac{7}{8}

−

\frac{2}{3}

)

A. $\frac{38}{5}$
B. $\frac{48}{5}$
C. $\frac{58}{5}$
D. $\frac{68}{5}$

Solution

Fraction simplification

\frac{5}{6}

\frac{3}{4}

\frac{10 + 9}{12}

\frac{19}{12}

\frac{7}{8}

−

\frac{2}{3}

\frac{21 - 16}{24}

\frac{5}{24}

\frac{19}{12}

\frac{5}{24}

\frac{19}{12}

\frac{24}{5}

\frac{38}{5}

. Hence option (a).

P, Q, and R invested ₹12,000, ₹16,000, and ₹20,000 respectively. P's investment was for 12 months, Q's for 8 months, and R's for 4 months. The total profit is ₹55,000. What is Q's share of the profit?

A.₹20,000
B.₹30,000
C.₹25,000
D.₹32,000

Solution

Partnership — 3 partners Capital × time: P = 12000×12 = 144000, Q = 16000×8 = 128000, R = 20000×4 = 80000. Ratio = 144 : 128 : 80 = 9 : 8 : 5 (total 22 parts). Q’s share =

\frac{8}{22}

× 55000. = ₹20,000 — option (a).

The average age of a committee of 8 members increases by 4 years when two women aged 34 and 44 years are replaced by two men. What is the average age of these two men?

A.50 years
B.46 years
C.55 years
D.65 years

Solution

Average — replacement Total increase in age = 8 × 4 = 32 years. Sum of the two women = 34 + 44 = 78. Sum of the two men = 78 + 32 = 110. Average =

\frac{110}{2}

= 55 years — option (c).

There are 5 consecutive integers and 4 consecutive integers. The average of the 5 integers is 2 less than the average of the 4 integers. The sum of the 4 integers is 2 more than the sum of the 5 integers. Find the average of the 5 integers.

Solution

Consecutive integers Let the 5 integers be a…a+4: sum 5a+10, average a+2. Let the 4 be b…b+3: sum 4b+6, average b+1.5. Averages: a + 2 = (b + 1.5) − 2 ⇒ b = a + 2.5. Sums: 4b + 6 = 5a + 12 ⇒ 4(a + 2.5) + 6 = 5a + 12 ⇒ 4a + 16 = 5a + 12 ⇒ a = 4. Average of the 5 = a + 2 = 6 — option (a).

Seventy-five percent of a number is 20 more than two-thirds of that number. Find the number.

A.140
B.150
C.240
D.210

Solution

Percent equation Let the number be x:

\frac{3 x}{4}

\frac{2 x}{3}

+ 20.

\frac{3 x}{4}

−

\frac{2 x}{3}

\frac{9 x - 8 x}{12}

\frac{x}{12}

\frac{x}{12}

= 20 ⇒ x = 240. Hence option (c).

A Mobile phone is available for ₹19,650 in cash or for ₹3,100 as a cash down payment and three equal annual installments. If the shopkeeper charges interest at the rate of 10% per annum compounded annually, calculate the amount of each instalment.

A.₹8,455
B.₹7,955
C.₹6,655
D.₹5,655

Solution

CI — equal instalments Amount financed = 19,650 − 3,100 = ₹16,550. Let each instalment be x. Present value of three instalments at 10%: x(

\frac{10}{11}

\frac{100}{121}

\frac{1000}{1331}

) = x ×

\frac{1210 + 1100 + 1000}{1331}

\frac{3310 x}{1331}

\frac{3310 x}{1331}

= 16550 ⇒ x =

\frac{16550 \times 1331}{3310}

= 5 × 1331. = ₹6,655 — option (c).

A Smart watch retailer sells a Smart watch for ₹P and earns a profit of 20%. For a special festive offer, he marks the same Smart watch at ₹1.5P. At the offer, he provides a discount of 10%. What will be the percentage profit that he will make during the festive offer?

A.36%
B.46%
C.62%
D.60%

Solution

Festive offer profit% CP =

\frac{P}{1.2}

\frac{5 P}{6}

. Festive SP = 1.5P × 0.9 = 1.35P. Profit% =

\frac{1.35 P - \frac{5 P}{6}}{\frac{5 P}{6}}

× 100 = (

\frac{1.35 \times 6}{5}

− 1) × 100 = (1.62 − 1) × 100. = 62% — option (c).

A gadget is sold at a 25% discount on its marked price. This sale leads to a profit of 10% on its cost price. If the profit earned is ₹300, what would have been the selling price if the gadget was sold at a 20% discount on its marked price?

A.₹3520
B.₹3720
C.₹2,879.45
D.₹3320

Solution

Discount → alternate SP 10% profit = ₹300 ⇒ CP = ₹3000, SP = ₹3300. SP = 75% of MP ⇒ MP =

\frac{3300 \times 100}{75}

= ₹4400. At 20% discount: SP = 4400 × 0.8. = ₹3520 — option (a).

P, Q, and R can complete a certain task in 20, 30, and 60 days, respectively. How many days will it take for P to finish the work if he receives help from Q and R every third day?

A.15 days
B.16 days
C.9 days
D.5 days

Solution

Work — help every 3rd day Total work = LCM(20, 30, 60) = 60 units ⇒ P = 3, Q = 2, R = 1 unit/day. In a 3-day cycle: 2 days of P alone = 6 units + 1 day of P+Q+R = 6 units ⇒ 12 units. Cycles needed =

\frac{60}{12}

= 5. 5 × 3 = 15 days — option (a).

Two distinct varieties of wheat, valued at ₹80 per kilogram and ₹120 per kilogram respectively, are combined in an unknown ratio. The resulting mixture is sold at ₹130 per kilogram, yielding a profit of 25% on the aggregate cost price. Determine the ratio in which the two wheat varieties were blended.

A.1 : 3
B.3 : 4
C.2 : 3
D.4 : 1

Solution

Alligation — wheat CP of mixture =

\frac{130}{1.25}

= ₹104 per kg. Alligation: (120 − 104) : (104 − 80). = 16 : 24 = 2 : 3. Hence option (c).

The ratio of the speeds of a car and a bus is 3 : 4. If the bus covers a distance of 480 km in 6 hours, what is the speed of the car in km/h?

A.60 km/h
B.65 km/h
C.70 km/h
D.80 km/h

Solution

Speed ratio Bus speed =

\frac{480}{6}

= 80 km/h. 4 units = 80 ⇒ 1 unit = 20. Car = 3 units = 60 km/h. Hence option (a).

An individual invested certain amounts in three different schemes A, B, and C at simple interest rates of 7% p.a., 9% p.a., and 11% p.a. respectively. If the total interest accrued in one year was ₹2520 and the amount invested in Scheme C was 250% of the amount invested in Scheme A and 300% of the amount invested in Scheme B, what was the amount invested in Scheme A?

A.₹3000
B.₹5500
C.₹6000
D.₹6500

Solution

SI — three schemes Let A = x. Then C = 2.5x, and B =

\frac{C}{3}

\frac{5 x}{6}

. One-year interest: 0.07x + 0.09 ×

\frac{5 x}{6}

+ 0.11 × 2.5x = 2520. 0.07x + 0.075x + 0.275x = 0.42x = 2520. x = ₹6000 — option (c).

A tower stands 60 meters tall. From its peak, the angles of depression to the top and bottom of a nearby building are measured at 45° and 60°, respectively. Determine the approximate height of the building as well as the horizontal distance separating the tower from the building.

A.height = 21 m, distance = 50 m
B.height = 25 m, distance = 35 m
C.height = 28 m, distance = 15 m
D.height = 18 m, distance = 15 m

Solution

Angles of depression

Let the horizontal distance be d and the building height h. To the base (60°): tan60° =

\frac{60}{d}

⇒ d =

\frac{60}{\sqrt{3}}

≈ 34.6 m. To the top (45°): tan45° =

\frac{60 - h}{d}

⇒ 60 − h = 34.6 ⇒ h ≈ 25.4 m. Height ≈ 25 m, distance ≈ 35 m — option (b).

A sphere is melted and recast into 8 identical cones, each with diameter 6 cm and height 4 cm. What was the radius of the original sphere?

A.4.16 cm
B.3.26 cm
C.4.26 cm
D.5.56 cm

Solution

Sphere → 8 cones Cone: r = 3, h = 4 ⇒ volume =

\frac{1}{3}

π × 9 × 4 = 12π; eight cones = 96π. Sphere:

\frac{4}{3}

πR³ = 96π ⇒ R³ = 72. R = ∛72 ≈ 4.16. Hence 4.16 cm — option (a).

The area of a regular octagon is made of how many isosceles triangles?

Solution

Regular octagon Join the centre of a regular octagon to all 8 vertices. Each side then forms a triangle with two equal radii — an isosceles triangle. A regular n-gon splits into n such triangles. For an octagon: 8 — option (c).

A circular fountain with a diameter of 14 m is encircled by a grass bed that is 3.5 m wide. Determine the ratio of the area of the grass bed to the area of the fountain.

A.5 : 3
B.3 : 4
C.5 : 4
D.4 : 1

Solution

Ring : circle areas

Fountain radius r = 7 m; outer radius R = 7 + 3.5 = 10.5 m. Grass area = π(R² − r²) = π(110.25 − 49) = 61.25π. Fountain area = 49π. Ratio = 61.25 : 49 = 5 : 4 — option (c).

A tent designed as a triangular prism features a triangular base with an area of 30 m² and a height of 7 m. What is the volume of this tent?

A.210 m³
B.200 m³
C.220 m³
D.140 m³

Solution

Prism volume Volume of a prism = base area × height (length). = 30 × 7. = 210. Hence 210 m³ — option (a).

Which line has slope 2 and passes through (1, 4)?

A.y = 2x + 1
B.y = 2x + 2
C.y = 2x + 3
D.y = x + 3

Solution

Line — slope & point Slope 2 ⇒ y = 2x + c. Substitute (1, 4): 4 = 2 + c ⇒ c = 2. Line: y = 2x + 2. Hence option (b).

Simplify:

\sqrt{45}

\sqrt{20}

−

\sqrt{80}

A.4 $\sqrt{5}$
B.6 $\sqrt{5}$
C. $\sqrt{5}$
D.7 $\sqrt{5}$

Solution

Surds

\sqrt{45}

= 3

\sqrt{5}

\sqrt{20}

= 2

\sqrt{5}

\sqrt{80}

= 4

\sqrt{5}

. 3

\sqrt{5}

+ 2

\sqrt{5}

− 4

\sqrt{5}

. =

\sqrt{5}

. Hence option (c).

Four non-overlapping, non-touching circles are drawn in a plane. What is the maximum number of distinct common tangents that can be drawn to them in total?

A.20
B.24
C.28
D.32

Solution

Common tangents — 4 circles Two separate circles have 4 common tangents (2 direct + 2 transverse). Number of pairs from 4 circles = ⁴C₂ = 6. If no tangent is shared by three circles: 6 × 4 = 24. Hence option (b).

In a circle, chord AB and chord CD intersect at E such that AE : EB = 3 : 4 and CE : ED = 5 : 2. If AB = x and CD = y, which of the following is true?

A. $\frac{x ²}{y ²}$ = $\frac{5}{6}$
B. $\frac{x}{y}$ = $\frac{6}{15}$
C. $\frac{x}{y}$ = $\frac{10}{9}$
D. $\frac{x ²}{y ²}$ = $\frac{144}{100}$

Solution

Intersecting chords

Chord theorem: AE × EB = CE × ED. Let AE = 3k, EB = 4k, CE = 5m, ED = 2m. 12k² = 10m² ⇒

\frac{k ²}{m ²}

\frac{5}{6}

. AB = 7k = x and CD = 7m = y ⇒

\frac{x}{y}

\frac{k}{m}

. So

\frac{x ²}{y ²}

\frac{k ²}{m ²}

\frac{5}{6}

— option (a).

A chord in a circle with radius 20 cm subtends an angle of 90° at the center. What is the area of the minor segment?

A.(25π − 50) sq.cm
B.(100π − 150) sq.cm
C.(50π − 100) sq.cm
D.(100π − 200) sq.cm

Solution

Minor segment — 90° Sector (90°) =

\frac{90}{360}

× π × 20² = 100π. Triangle =

\frac{1}{2}

× 20 × 20 = 200. Minor segment = sector − triangle. = 100π − 200 — option (d).

Evaluate: (0.04³ + 0.01³) ÷ (0.16³ + 0.04³)

A.0.008
B.0.015
C.0.016
D.0.009

Solution

Cubes — common factor Notice 0.16 = 4 × 0.04 and 0.04 = 4 × 0.01. Denominator = 4³(0.04³ + 0.01³) = 64 × (numerator). Ratio =

\frac{1}{64}

= 0.0156… ≈ 0.015 — option (b).

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