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SSC CGL 19 September 2025 · Shift 1 · Quantitative Aptitude — questions with answers · ShikshaSphere

Chapter 87 of 192

19 September 2025 · Shift 1

Quantitative Aptitude

25 questions ~25 min readFree

25 questions

Simplify the expression:

\sqrt{10 + 4 \sqrt{6}}

A. $\sqrt{6}$ + 2
B.3 + $\sqrt{5}$
C.2 + $\sqrt{6}$
D.3 + $\sqrt{4}$

Solution

Nested surd

Write the inside as a perfect square: 10 + 4

\sqrt{6}

= 6 + 4 + 2

\sqrt{6 \times 4}

. That is (

\sqrt{6}

\sqrt{4}

)². Taking the square root gives

\sqrt{6}

\sqrt{4}

. =

\sqrt{6}

+ 2 — option (a).

If 75% of P = 60% of Q = 30% of R, find P : Q : R.

A.4 : 5 : 10
B.10 : 5 : 4
C.5 : 4 : 10
D.2 : 3 : 5

Solution

Ratio from percentages

75P = 60Q = 30R; divide throughout by 15: 5P = 4Q = 2R = k. So P =

\frac{k}{5}

, Q =

\frac{k}{4}

and R =

\frac{k}{2}

. Taking k = 20: P = 4, Q = 5, R = 10. P : Q : R = 4 : 5 : 10 — option (a).

Which of the following is correct? (i)

\sqrt{11}

\sqrt{3}

\sqrt{12}

\sqrt{2}

(ii)

\sqrt{11}

\sqrt{3}

\sqrt{12}

\sqrt{2}

(iii)

\sqrt{11}

\sqrt{3}

\sqrt{12}

\sqrt{2}

A.(i)
B.(ii)
C.(iii)
D.(ii) and (iii)

Solution

Comparing surds Square both sides: (

\sqrt{11}

\sqrt{3}

)² = 14 + 2

\sqrt{33}

. (

\sqrt{12}

\sqrt{2}

)² = 14 + 2

\sqrt{24}

. Since

\sqrt{33}

\sqrt{24}

, the first sum is greater. So (i) is correct — option (a).

A phone is priced at ₹44,000. A customer gets a discount of ₹3,000 on it. After the discount, the customer buys a charger for ₹1,600 at a 30% discount. What is the total amount the customer spends on both items?

A.₹42,120
B.₹41,820
C.₹42,180
D.₹42,108

Solution

Two purchases

Phone after discount = 44,000 − 3,000 = ₹41,000. Charger after 30% discount = 1,600 ×

\frac{70}{100}

= ₹1,120. Total = 41,000 + 1,120. = ₹42,120 — option (a).

Two pipes A and B can fill a tank in 36 and 48 minutes respectively. After both are opened together for 9 minutes, the rate of pipe A becomes double, and pipe B becomes one-third its original. In how many more minutes will the tank be full?

A.8 min
B.7 min
C.9 min
D.6 min

Solution

Pipes with changing rates

Take the tank as LCM(36, 48) = 144 units ⇒ A = 4 units/min, B = 3 units/min. In 9 minutes together: 9 × 7 = 63 units; remaining = 144 − 63 = 81 units. New rates: A = 8 and B = 1, i.e. 9 units per minute. 81 ÷ 9 = 9 minutes — option (c).

If x +

\frac{1}{x}

= −2, then compute: x⁴ +

\frac{1}{x ⁴}

+ 3x² +

\frac{3}{x ²}

A.−3
B.3
C.8
D.−4

Solution

Algebraic identity

Square: x² +

\frac{1}{x ²}

= (−2)² − 2 = 2. Square again: x⁴ +

\frac{1}{x ⁴}

= 2² − 2 = 2. Now the expression = 2 + 3(x² +

\frac{1}{x ²}

) = 2 + 3(2). = 8 — option (c).

Neha lent ₹14,000 to John on 10 July 2024 at a simple interest rate of 8% per annum. John plans to pay back the entire amount on 10 October 2024. What is the total amount John needs to return?

A.₹14,280
B.₹14,209
C.₹14,208
D.₹14,820

Solution

SI for three months

Time = 3 months =

\frac{3}{12}

\frac{1}{4}

year. SI =

\frac{14000 \times 8 \times 1}{100 \times 4}

= ₹280. Amount = 14,000 + 280. = ₹14,280 — option (a).

If P, Q, and R are three amounts of money such that Q is the simple interest on P, and R is the simple interest on Q, all for the same time and at the same rate of interest, then which of the following is true?

A.P² = QR
B.Q² = PR
C.R² = PQ
D.P + Q = R

Solution

SI relation Q =

\frac{P \times r \times t}{100}

and R =

\frac{Q \times r \times t}{100}

. Dividing the second by the first:

\frac{R}{Q}

\frac{Q}{P}

. Cross-multiplying gives Q² = PR. Hence option (b).

A tree is 40 metres tall. What is the length of the shadow cast by the tree when the sun's elevation is 30°?

A.20 m
B.30 m
C.40 $\sqrt{3}$ m
D.30 $\sqrt{3}$ m

Solution

Shadow at 30°

tan30° =

\frac{height}{shadow}

\frac{1}{\sqrt{3}}

\frac{40}{shadow}

. Shadow = 40

\sqrt{3}

. = 40

\sqrt{3}

m — option (c).

A hemisphere of radius 8 cm is carved from a sphere. What is the volume of the hemisphere?

A. $\frac{1024 π}{3}$ cm³
B. $\frac{1205 π}{6}$ cm³
C. $\frac{1250 π}{3}$ cm³
D. $\frac{275 π}{6}$ cm³

Solution

Hemisphere volume

Volume of a hemisphere =

\frac{2}{3}

πr³. =

\frac{2}{3}

× π × 8³. =

\frac{2 \times 512}{3}

π. =

\frac{1024 π}{3}

cm³ — option (a).

A solid ball is placed inside a cube such that it touches all six faces. What percentage of the cube's volume is occupied by the ball?

A.47.67%
B.48.56%
C.49.28%
D.52.33%

Solution

Sphere inside a cube

Let the cube’s side be s; the ball’s radius is

\frac{s}{2}

. Ball volume =

\frac{4}{3}

π(

\frac{s}{2}

)³ =

\frac{πs ³}{6}

; cube volume = s³. Percentage =

\frac{π}{6}

× 100. ≈ 52.33% — option (d).

The sides of a rectangular prism are in the ratio 2 : 5 : 3. If the volume of the prism is 240 cm³, what is the length of its longest side?

A.12 cm
B.10 cm
C.8 cm
D.16 cm

Solution

Cuboid from ratio

Let the sides be 2x, 5x and 3x. Volume = 2x × 5x × 3x = 30x³ = 240. x³ = 8 ⇒ x = 2. Longest side = 5 × 2 = 10 cm — option (b).

A conical bucket is cut at one-fourth its height from the top. What is the ratio of the volumes of the two parts (top part to bottom part)?

A.1 : 64
B.1 : 63
C.1 : 27
D.1 : 7

Solution

Cone cut near the top

The small top cone is similar to the whole cone with height ratio 1 : 4. Volumes vary as the cube of heights: 1³ : 4³ = 1 : 64. So the bottom frustum takes 64 − 1 = 63 parts. Top : bottom = 1 : 63 — option (b).

If sinA + cosA =

\frac{3}{2}

and A ∈ (0,

\frac{π}{2}

), then what is sinA cosA?

A.1
B.2
C. $\frac{5}{8}$
D.Cannot be determined

Solution

sinA cosA

Square both sides: sin²A + cos²A + 2 sinA cosA =

\frac{9}{4}

. Since sin²A + cos²A = 1, we get 2 sinA cosA =

\frac{9}{4}

− 1 =

\frac{5}{4}

. sinA cosA =

\frac{5}{8}

. Hence option (c).

What is the value of sin40° cos20° + cos40° sin20°?

A. $\frac{1}{2}$
B. $\frac{\sqrt{3}}{4}$
C. $\frac{\sqrt{3}}{2}$
D. $\frac{7}{8}$

Solution

sin(A + B) The identity is sin(A + B) = sinA cosB + cosA sinB. So the expression = sin(40° + 20°). = sin60°. =

\frac{\sqrt{3}}{2}

— option (c).

The graph of −x + y = 2 passes through which point?

A.(1, 3)
B.(2, 4)
C.(0, 2)
D.All the above

Solution

Points on a line

Check (1, 3): −1 + 3 = 2 ✓. Check (2, 4): −2 + 4 = 2 ✓. Check (0, 2): 0 + 2 = 2 ✓. All three lie on the line — option (d).

A sector of a circle has a central angle of 45° and a radius of 9 cm. Another sector of the same circle has a central angle of

\frac{π}{4}

radians. What is the ratio of the area of the first sector to the area of the second sector?

A.1 : 1
B.2 : 3
C.3 : 5
D.5 : 6

Solution

Sector areas

\frac{π}{4}

radians = 45° — the two angles are the same. Both sectors belong to the same circle, so their radii are equal. Equal angle and equal radius give equal area. Ratio = 1 : 1 — option (a).

In a parallelogram:

A.Only one diagonal is equal.
B.The diagonals are equal.
C.The diagonals bisect each other.
D.All angles are equal.

Solution

Parallelogram diagonals In a parallelogram each diagonal cuts the other into two equal parts. Equal diagonals occur only in a rectangle, and equal angles only in a rectangle or square. So the general property is that the diagonals bisect each other. Hence option (c).

A right-angled triangle has a hypotenuse of 17 cm and one leg of 8 cm. A second right-angled triangle is similar to the first, and its hypotenuse is 51 cm. What is the area of the second triangle?

A.390 cm²
B.460 cm²
C.260 cm²
D.540 cm²

Solution

Similar right triangles

(8, 15, 17) is a Pythagorean triplet, so the other leg is 15 cm. Area of the first =

\frac{1}{2}

× 8 × 15 = 60 cm². Scale factor =

\frac{51}{17}

= 3, so areas scale by 3² = 9. Area = 60 × 9 = 540 cm² — option (d).

In triangles ABC and PQR, if AB = PQ, AC = PR, and BC = QR, by which congruence rule are they congruent?

A.Yes, by SSS
B.Yes, by SAS
C.Yes, by ASA
D.No, they are not congruent

Solution

SSS congruence All three pairs of corresponding SIDES are given equal. That is exactly the side-side-side condition. No angle information is needed. Congruent by SSS — option (a).

17³ + 20³ − 19³ + 346 is equal to:

A.0
B.6400
C.−4860
D.5200

Solution

Cubes evaluation 17³ = 4913 and 20³ = 8000. 19³ = 6859. 4913 + 8000 + 346 = 13,259; 13,259 − 6,859 = 6,400. Hence option (b).

If sinx =

\frac{7}{25}

and x ∈ (0,

\frac{π}{2}

), then find

\frac{1 + tanx}{1 - tanx}

A.6
B. $\frac{31}{17}$
C. $\frac{24}{17}$
D.9

Solution

(1 + tanx)/(1 − tanx) (7, 24, 25) is a Pythagorean triplet, so tanx =

\frac{7}{24}

\frac{1 + tanx}{1 - tanx}

\frac{1 + 7 / 24}{1 - 7 / 24}

. =

\frac{31 / 24}{17 / 24}

. =

\frac{31}{17}

— option (b).

The length of each of two tangents drawn from an external point to a circle is 15 cm. What is the approximate distance from the external point to the centre of the circle, if the radius is 11 cm?

A.24 cm
B.15 cm
C.19 cm
D.16 cm

Solution

Tangent and radius

The radius meets the tangent at 90°, forming a right triangle. Distance =

\sqrt{15² + 11²}

\sqrt{225 + 121}

. =

\sqrt{346}

≈ 18.6. ≈ 19 cm — option (c).

A circle is inscribed in a right triangle with legs of length 9 and 12. What is the radius of the circle?

Solution

Inradius of right triangle

Hypotenuse =

\sqrt{9² + 12²}

\sqrt{225}

= 15. For a right triangle, inradius r =

\frac{a + b - c}{2}

. =

\frac{9 + 12 - 15}{2}

\frac{6}{2}

. = 3 — option (b).

What is the result of (0.4³ + 0.3³ − 0.7³) ÷ (3 × 0.4 × 0.3 × 0.7)?

A.−1
B. $\frac{7}{9}$
C. $\frac{8}{9}$
D. $\frac{6}{9}$

Solution

a³+b³+c³ when a+b+c=0

Take a = 0.4, b = 0.3, c = −0.7; then a + b + c = 0. When the sum is zero, a³ + b³ + c³ = 3abc = 3(0.4)(0.3)(−0.7) = −0.252. The denominator is 3 × 0.4 × 0.3 × 0.7 = +0.252.

\frac{- 0.252}{0.252}

= −1 — option (a).

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