SSC Combined Higher Secondary Level (CHSL) · Topic Collection

Quantitative Aptitude PYQs

Arithmetic, number systems, percentages, ratios, profit and loss, time and work, algebra, geometry, mensuration, and trigonometry.

13275 Total Questions531 Official PapersYears: 2018–2025Search in Quantitative Aptitude
Showing questions 21–40 of 13275Page 2 of 100
SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 21

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

A line passes through the point (5, 7) and has a slope of -1. What is the equation of the line?

  1. AOption (a)

    y = -x + 12

  2. BOption (b)

    y = \(\frac{x}{5}\) + 19

  3. COption (c)

    y = -2x - 4

  4. DOption (d)

    y = 4x - 6

View answer & explanation

Correct answer

Answer: (a)A.

y = -x + 12

Why this answer is correct

Given:

The line passes through the point (5, 7)

Slope of the line = −1

Calculation:

The general form of a straight line is:

y = mx + c

Substitute the given slope:

y = −x + c

Since the line passes through (5, 7), substitute x = 5 and y = 7:

7 = −5 + c

c = 12

Equation of the line:

y = −x + 12

Correct option: y = −x + 12

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 22

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

Find the value of k such that the quadratic equation x² + kx + 16 = 0 has real and distinct roots.

  1. AOption (a)

    k > 8

  2. BOption (b)

    k > 5

  3. COption (c)

    k < -7

  4. DOption (d)

    k < -9

View answer & explanation

Correct answer

Answer: (a)A.

k > 8

Why this answer is correct

Given:

The quadratic equation is x2 + kx + 16 = 0.

For the roots to be real and distinct:

Discriminant (D) > 0

Formula used:

Discriminant (D) = b2 - 4ac

Here, a = 1, b = k, and c = 16.

Calculation:

D = k2 - 4 × 1 × 16

⇒ D = k2 - 64

For real and distinct roots, D > 0:

⇒ k2 - 64 > 0

⇒ (k - 8)(k + 8) > 0

The critical points are k = 8 and k = -8.

Using the sign analysis of the quadratic inequality:

  • For (k - 8)(k + 8) > 0, k < -8 or k > 8.

∴ The correct answer is option (1): k > 8.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 23

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

Simplify:

\(\frac{1 + \tan^2(x)}{1 + \cot^2(x)} \times \frac{\sec(x) - \tan(x)}{\sec(x) + \tan(x)}\)

  1. AOption (a)

    \(\frac{1}{1 + \sin(x)}\)

  2. BOption (b)

    \(\frac{\tan^2(x)}{1 + \sin(x)}\)

  3. COption (c)

    \(\frac{1 - \sin(x)}{1 + \sin(x)}\)

  4. DOption (d)

    \(\frac{\tan^2(x)(1 - \sin(x))}{1 + \sin(x)}\)

View answer & explanation

Correct answer

Answer: (d)D.

\(\frac{\tan^2(x)(1 - \sin(x))}{1 + \sin(x)}\)

Why this answer is correct

Step 1: Simplify the first fraction

Given expression:

(1 + tan²x) / (1 + cot²x)

Using identities:

1 + tan²x = sec²x

1 + cot²x = csc²x

So, the expression becomes:

sec²x / csc²x

Now,

sec x = 1 / cos x

csc x = 1 / sin x

⇒ sec²x / csc²x = (1 / cos²x) ÷ (1 / sin²x)

⇒ (1 / cos²x) × (sin²x / 1)

⇒ sin²x / cos²x

⇒ tan²x

Step 2: Simplify the second fraction

(sec x − tan x) / (sec x + tan x)

Multiply numerator and denominator by (sec x − tan x)

Denominator:

(sec x + tan x)(sec x − tan x)

⇒ sec²x − tan²x

Using identity:

sec²x − tan²x = 1

So denominator = 1

Numerator:

(sec x − tan x)²

Substitute values:

sec x = 1 / cos x

tan x = sin x / cos x

⇒ ( (1 − sin x) / cos x )²

⇒ (1 − sin x)² / cos²x

Now,

cos²x = 1 − sin²x

⇒ cos²x = (1 − sin x)(1 + sin x)

So,

(1 − sin x)² / [(1 − sin x)(1 + sin x)]

Cancel (1 − sin x)

⇒ (1 − sin x) / (1 + sin x)

Step 3: Final Result

Multiply both simplified parts:

tan²x × (1 − sin x) / (1 + sin x)

∴ Final Answer = tan²x (1 − sin x) / (1 + sin x)

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 24

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

If sinθ= \(\frac{7}{25}\), find cos²θ.

  1. AOption (a)

    \(\frac{576}{625}\)

  2. BOption (b)

    \(\frac{718}{525}\)

  3. COption (c)

    \(\frac{736}{525}\)

  4. DOption (d)

    \(\frac{347}{125}\)

View answer & explanation

Correct answer

Answer: (a)A.

\(\frac{576}{625}\)

Why this answer is correct

Given:

sinθ = 7 / 25

Calculation:

Using the identity:

sin2θ + cos2θ = 1

So,

cos2θ = 1 − sin2θ

cos2θ = 1 − (7 / 25)2

cos2θ = 1 − 49 / 625

cos2θ = (625 − 49) / 625

cos2θ = 576 / 625

Correct Answer: 576 / 625

Correct Option: Option 1

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 25

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

In a right triangle ABC, right-angled at B, if tan A = 1, what is the value of 2 sin A cos A?

  1. AOption (a)

    2

  2. BOption (b)

    5

  3. COption (c)

    6

  4. DOption (d)

    1

View answer & explanation

Correct answer

Answer: (d)D.

1

Why this answer is correct

Shortcut Trick

Given tan A = 1, we know that A = 45° since tan 45° = 1.

The expression 2 sin A cos A is the trigonometric identity for sin 2A.

Substituting A = 45°: 2 sin 45° cos 45° = sin(2 × 45°) = sin 90°.

Since sin 90° = 1, the required value is 1.

∴ The correct answer is 1.

Alternate Method

Given: In ΔABC, ∠B = 90° and tan A = 1.

Formula Used: tan A = Perpendicular / Base; sin A = P / H; cos A = B / H

⇒ tan A = BC / AB = 1. Let BC = 1 and AB = 1.

⇒ By Pythagoras Theorem: AC = √(AB2 + BC2)

⇒ AC = √(12 + 12) = √2.

⇒ sin A = BC / AC = 1 / √2.

⇒ cos A = AB / AC = 1 / √2.

⇒ 2 sin A cos A = 2 × (1 / √2) × (1 / √2).

⇒ 2 sin A cos A = 2 × (1 / 2) = 1.

∴ The correct answer is 1.

Additional Information

Double Angle Identities

The expression 2 sin θ cos θ is identical to sin 2θ, which simplifies calculations for specific angles.

Standard Trigonometric Values

For θ = 45°, sin 45° = 1/√2, cos 45° = 1/√2, and tan 45° = 1.

Pythagorean Identity

For any angle A, the fundamental identity sin2 A + cos2 A = 1 always holds true.

Right Triangle Properties

In an isosceles right triangle (where tan A = 1), the two acute angles are both 45° and the legs are equal.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 26

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

Three partners share the profit in a business in the ratio 6:8:9. They invested for 14 months, 8 months, and 7 months, respectively. What is the ratio of their investments?

  1. AOption (a)

    3:7:9

  2. BOption (b)

    20:49:64

  3. COption (c)

    38:28:21

  4. DOption (d)

    6:8:9

View answer & explanation

Correct answer

Answer: (a)A.

3:7:9

Why this answer is correct

Shortcut Trick

Investment Ratio = (Profit Ratio) ÷ (Time Ratio)

Investment Ratio = (6/14) : (8/8) : (9/7)

Investment Ratio = (3/7) : 1 : (9/7)

Multiply the entire ratio by 7 to simplify: 3 : 7 : 9

∴ The correct answer is 3:7:9.

Alternate Method

Given:

Profit ratio of three partners = 6 : 8 : 9

Time of investment for each partner = 14 months, 8 months, and 7 months

Formula Used:

Profit = Investment × Time

⇒ Investment = Profit ÷ Time

Let the investments of the three partners be I1, I2, and I3 respectively.

⇒ I1 : I2 : I3 = (6/14) : (8/8) : (9/7)

⇒ I1 : I2 : I3 = (3/7) : 1 : (9/7)

To eliminate the fractions, multiply the entire ratio by the LCM of denominators (7, 1, 7), which is 7:

⇒ I1 : I2 : I3 = (3/7 × 7) : (1 × 7) : (9/7 × 7)

⇒ I1 : I2 : I3 = 3 : 7 : 9

∴ The correct answer is 3:7:9.

Additional Information

Partnership Basic Rule

The profit earned in a business is distributed among partners in the ratio of the product of their capital and time duration (Profit = C × T).

Ratio of Capital/Investment

When the profit ratio and time ratio are given, the investment ratio is calculated as (P1/T1) : (P2/T2) : (P3/T3).

Working vs Sleeping Partner

A working partner manages the business and may receive a salary or extra profit share, while a sleeping partner only invests money.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 27

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

Simplify the expression: sin² x + cos² x + tan²x.

  1. AOption (a)

    1

  2. BOption (b)

    cot² x

  3. COption (c)

    sec² x

  4. DOption (d)

    596

View answer & explanation

Correct answer

Answer: (c)C.

sec² x

Why this answer is correct

Shortcut Trick

Combine the Pythagorean identity terms: sin2 x + cos2 x = 1.

Substitute this into the expression to get 1 + tan2 x.

Recognize that 1 + tan2 x is the standard identity for sec2 x.

∴ The correct answer is sec2 x.

Alternate Method

Given: sin2 x + cos2 x + tan2 x

Formula Used:

  1. sin2 θ + cos2 θ = 1
  2. 1 + tan2 θ = sec2 θ

⇒ Expression = (sin2 x + cos2 x) + tan2 x

⇒ Expression = 1 + tan2 x

⇒ Expression = sec2 x

∴ The correct answer is sec2 x.

Additional Information

Primary Pythagorean Identity

The most fundamental identity in trigonometry is sin2 θ + cos2 θ = 1, which relates the sine and cosine of any angle.

Secant-Tangent Relationship

The identity sec2 θ − tan2 θ = 1 is derived by dividing the primary identity by cos2 θ.

Cosecant-Cotangent Relationship

Similarly, dividing the primary identity by sin2 θ yields the identity 1 + cot2 θ = cosec2 θ.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 28

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

The ratio of the profit to the cost price is 3: 8. What is the ratio of the selling price to the cost price?

  1. AOption (a)

    8:5

  2. BOption (b)

    11:8

  3. COption (c)

    8:11

  4. DOption (d)

    5:8

View answer & explanation

Correct answer

Answer: (b)B.

11:8

Why this answer is correct

Given:

Profit to cost price ratio = 3:8

Formula Used:

Selling Price (SP) = Cost Price (CP) + Profit

Profit = (Profit to CP ratio) × CP

SP to CP ratio = SP ÷ CP

Calculation:

⇒ Profit = (3/8) × CP

⇒ SP = CP + Profit

⇒ SP = CP + (3/8) × CP

⇒ SP = (1 + 3/8) × CP

⇒ SP = (8/8 + 3/8) × CP

⇒ SP = (11/8) × CP

⇒ SP to CP ratio = 11:8

∴ The correct answer is 11:8.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 29

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

If a ratio of 7:9 represents the number of apples and oranges in a basket, and there are 63 apples, how many oranges are there?

  1. AOption (a)

    72

  2. BOption (b)

    81

  3. COption (c)

    90

  4. DOption (d)

    84

View answer & explanation

Correct answer

Answer: (b)B.

81

Why this answer is correct

Shortcut Trick

Ratio of apples to oranges = 7 : 9

According to the question, 7 units = 63 apples.

1 unit = 63 ÷ 7 = 9

Number of oranges = 9 units = 9 × 9 = 81

∴ The correct answer is 81.

Alternate Method

Given: Ratio of apples to oranges in a basket = 7 : 9

Given: Total number of apples = 63

⇒ Let the number of apples be 7x and the number of oranges be 9x.

⇒ Since the number of apples is 63, we can write:

⇒ 7x = 63

⇒ x = 63 ÷ 7

⇒ x = 9

⇒ Now, number of oranges = 9x

⇒ Number of oranges = 9 × 9 = 81

∴ The correct answer is 81.

Additional Information

Direct Proportion

Two quantities are in direct proportion if the ratio of their values remains constant. If x ∝ y, then x1/y1 = x2/y2.

Simplification of Ratios

A ratio a : b can be simplified by dividing both terms by their Highest Common Factor (HCF). For example, 63 : 81 = (63/9) : (81/9) = 7 : 9.

Compound Ratio

The compound ratio of two ratios a : b and c : d is defined as (a × c) : (b × d).

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 30

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

The difference between the Cl and Sl on a sum for 2 years at 15% p.a. is ₹360. What is the principal?

  1. AOption (a)

    ₹15,000

  2. BOption (b)

    ₹16,000

  3. COption (c)

    ₹18,000

  4. DOption (d)

    ₹20,000

View answer & explanation

Correct answer

Answer: (b)B.

₹16,000

Why this answer is correct

Given:

Difference between Compound Interest (CI) and Simple Interest (SI) = ₹360

Time = 2 years

Rate of Interest = 15% p.a.

Formula Used:

CI − SI = P × (r/100)2

Where P = Principal, r = Rate of Interest

Calculation:

⇒ 360 = P × (15/100)2

⇒ 360 = P × (225/10000)

⇒ P = 360 ÷ (225/10000)

⇒ P = 360 × (10000/225)

⇒ P = ₹16,000

∴ The correct answer is ₹16,000.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 31

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

An amount of ₹2,00,000 is invested at 10% per annum compound interest. After 1 year, the investor withdraws 20,000 from the balance. What will be the total amount remaining in the account at the end of the 2nd year?

  1. AOption (a)

    ₹2,18,000

  2. BOption (b)

    ₹2,20,000

  3. COption (c)

    ₹2,24,200

  4. DOption (d)

    ₹2,25,800

View answer & explanation

Correct answer

Answer: (b)B.

₹2,20,000

Why this answer is correct

Given:

Principal = 200000

Rate of interest = 10%

Amount withdrawn after 1 year = 20000

Total time = 2 years

Formula Used:

Compound amount = Principal × (1 + Rate ÷ 100)

Calculation:

⇒ Amount after 1 year = 200000 × (1 + 10 ÷ 100)

⇒ Amount after 1 year = 200000 × 1.10

⇒ Amount after 1 year = 220000

⇒ Balance after withdrawal = 220000 − 20000

⇒ Balance = 200000

⇒ Amount after 2nd year = 200000 × (1 + 10 ÷ 100)

⇒ Amount after 2nd year = 200000 × 1.10

⇒ Amount after 2nd year = 220000

∴ The correct answer is 220000.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 32

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

The average of 50 numbers was calculated as 38. Later it was found that two numbers were wrongly taken as 33 and 27 instead of 23 and 17. What is the correct average?

  1. AOption (a)

    39

  2. BOption (b)

    38.2

  3. COption (c)

    37.6

  4. DOption (d)

    36

View answer & explanation

Correct answer

Answer: (c)C.

37.6

Why this answer is correct

Given:

Average of 50 numbers = 38

Wrongly taken numbers = 33 and 27

Correct numbers = 23 and 17

Total numbers = 50

Formula Used:

Total sum of numbers = Average × Total numbers

Corrected total sum = Original total sum - Wrong values + Correct values

Corrected average = Corrected total sum ÷ Total numbers

Calculation:

⇒ Total sum of numbers = 38 × 50 = 1900

⇒ Corrected total sum = 1900 - (33 + 27) + (23 + 17)

⇒ Corrected total sum = 1900 - 60 + 40

⇒ Corrected total sum = 1880

⇒ Corrected average = 1880 ÷ 50

⇒ Corrected average = 37.6

∴ The correct answer is 37.6.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 33

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

When 13511, 13903, and 14589 are divided by the greatest number 'n', the remainder is 'm'. What is the value of (n + m)? 

  1. AOption (a)

    131

  2. BOption (b)

    183

  3. COption (c)

    151

  4. DOption (d)

    161

View answer & explanation

Correct answer

Answer: (b)B.

183

Why this answer is correct

Given:

First number = 13511

Second number = 13903

Third number = 14589

Same divisor = n

Same remainder = m

Formula Used:

If three numbers leave the same remainder when divided by a number, then that divisor is the H.C.F. of the pairwise differences of the numbers.

Remainder = Number mod Divisor

Calculation:

⇒ 13903 − 13511 = 392

⇒ 14589 − 13903 = 686

⇒ 14589 − 13511 = 1078

⇒ H.C.F. of 392 and 686

⇒ 686 = 392 × 1 + 294

⇒ 392 = 294 × 1 + 98

⇒ 294 = 98 × 3 + 0

⇒ H.C.F. = 98

⇒ n = 98

⇒ m = 13511 mod 98

⇒ 13511 = 98 × 137 + 85

⇒ m = 85

⇒ n + m = 98 + 85

⇒ n + m = 183

∴ The correct answer is 183.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 34

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

The price of a bike increases by 20% in the first year and then decreases by 10% in the second year. If its final price becomes 59,400, find the original price of the bike.

  1. AOption (a)

    ₹50,000

  2. BOption (b)

    ₹55,000

  3. COption (c)

    ₹60,000

  4. DOption (d)

    ₹58,000

View answer & explanation

Correct answer

Answer: (b)B.

₹55,000

Why this answer is correct

Given:

Price increases by 20% in first year

Price decreases by 10% in second year

Final price = 59400

Formula Used:

New price = Original price × (1 + Increase%) × (1 − Decrease%)

Calculation:

⇒ Let original price = x

⇒ After 20% increase = x × 1.20

⇒ After 10% decrease = x × 1.20 × 0.90

⇒ Final price = x × 1.08

⇒ x × 1.08 = 59400

⇒ x = 59400 ÷ 1.08

⇒ x = 55000

∴ The correct answer is 55000.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 35

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

A truck travels to a town at 30 km/h and returns at 20 km/h. If the total time taken is 5 hours, what is the one-way distance to the town?

  1. AOption (a)

    30 km

  2. BOption (b)

    45 km

  3. COption (c)

    60 km

  4. DOption (d)

    75 km

View answer & explanation

Correct answer

Answer: (c)C.

60 km

Why this answer is correct

Given:

Speed while going = 30 km/h

Speed while returning = 20 km/h

Total time = 5 hours

Formula Used:

Time = Distance ÷ Speed

Total time = Time going + Time returning

Calculation:

⇒ Let one-way distance = x

⇒ Time going = x ÷ 30

⇒ Time returning = x ÷ 20

⇒ x ÷ 30 + x ÷ 20 = 5

⇒ (2x + 3x) ÷ 60 = 5

⇒ 5x ÷ 60 = 5

⇒ x ÷ 12 = 5

⇒ x = 60

∴ The correct answer is 60.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 36

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

A and B together can do a piece of work in 10 days. B alone can do it in 15 days. How long will A alone take to complete the work?

  1. AOption (a)

    20 days

  2. BOption (b)

    25 days

  3. COption (c)

    30 days

  4. DOption (d)

    35 days

View answer & explanation

Correct answer

Answer: (c)C.

30 days

Why this answer is correct

Shortcut Trick

If A and B together take x days and B alone takes y days, then A alone takes:

Time = (x × y) ÷ (y − x)

Time taken by A = (10 × 15) ÷ (15 − 10)

Time taken by A = 150 ÷ 5 = 30 days

∴ The correct answer is 30 days.

Alternate Method

Given:

Time taken by A and B together = 10 days

Time taken by B alone = 15 days

Formula Used:

Total Work = LCM of individual days

Efficiency = Total Work ÷ Time Taken

PersonDaysTotal WorkEfficiency
A + B10303
B152

Calculations:

⇒ Total work (LCM of 10 and 15) = 30 units

⇒ Efficiency of A + B = 3 units/day

⇒ Efficiency of B = 2 units/day

⇒ Efficiency of A = (Efficiency of A + B) − (Efficiency of B)

⇒ Efficiency of A = 3 − 2 = 1 unit/day

⇒ Time taken by A alone = Total Work ÷ Efficiency of A

⇒ Time taken by A = 30 ÷ 1 = 30 days

∴ The correct answer is 30 days.

Additional Information

Pipes and Cisterns

The concept is identical to time and work; filling pipes act like positive work, and drainage pipes (leaks) act like negative work.

Work and Wages

Wages are always distributed among workers in the direct ratio of their efficiencies or the share of work done by each.

Efficiency-Time Relation

Efficiency is inversely proportional to the time taken to complete the work (Efficiency ∝ 1 ÷ Time).

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 37

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

In a trapezium ABCD with AB || CD, the diagonals AC and BD intersect at O. The area of \(\Delta\)COD = 8cm2 . If AB = 2CD , what is the area of \(\Delta\)AΟΒ?

  1. AOption (a)

    32 cm²

  2. BOption (b)

    40 cm²

  3. COption (c)

    48 cm²

  4. DOption (d)

    56cm²

View answer & explanation

Correct answer

Answer: (a)A.

32 cm²

Why this answer is correct

Given:

AB ∥ CD

AB = 2CD

Area of \( \Delta COD \) = 8

Formula Used:

In a trapezium, diagonals divide each other in the ratio of the parallel sides

Area of triangles with same altitude is proportional to their bases

Calculation:

⇒ AB : CD = 2 : 1

⇒ AO : OC = 2 : 1

⇒ BO : OD = 2 : 1

⇒ Area(\( \Delta AOB \)) : Area(\( \Delta COD \)) = (AB / CD)2

⇒ Area ratio = 22 : 1

⇒ Area ratio = 4 : 1

⇒ Area(\( \Delta AOB \)) = 4 × 8

⇒ Area(\( \Delta AOB \)) = 32

∴ The correct answer is 32.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 38

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

In a scheme, a shopkeeper offers a 10% discount if you buy two or more books. Kiran buys two books for ₹540. If the shopkeeper still makes a 20% profit, find the profit he would make had he sold the books without the discount.

  1. AOption (a)

    ₹120

  2. BOption (b)

    ₹150

  3. COption (c)

    ₹180

  4. DOption (d)

    ₹200

View answer & explanation

Correct answer

Answer: (b)B.

₹150

Why this answer is correct

Given:

Discount offered = 10%

Selling price after discount for two books = 540

Profit after discount = 20%

Formula Used:

Selling Price = Cost Price × (1 + Profit%)

Discounted price = Marked Price × (1 − Discount%)

Calculation:

⇒ 540 = 120% of Cost Price

⇒ Cost Price = 540 × 100 ÷ 120

⇒ Cost Price = 450

⇒ 540 = 90% of Marked Price

⇒ Marked Price = 540 × 100 ÷ 90

⇒ Marked Price = 600

⇒ Profit without discount = 600 − 450

⇒ Profit = 150

∴ The correct answer is 150.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 39

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

The simple interest on a certain sum is one-fourth of the principal, and the number of years is equal to the rate of interest per annum. What is the rate of interest?

  1. AOption (a)

    10%

  2. BOption (b)

    5%

  3. COption (c)

    15%

  4. DOption (d)

    20%

View answer & explanation

Correct answer

Answer: (b)B.

5%

Why this answer is correct

Given:

Simple interest = Principal ÷ 4

Number of years = Rate of interest

Formula Used:

Simple Interest = (Principal × Rate × Time) ÷ 100

Calculation:

⇒ Let Principal = P

⇒ Rate = r

⇒ Time = r

⇒ Simple Interest = (P × r × r) ÷ 100

⇒ P ÷ 4 = (P × r²) ÷ 100

⇒ 1 ÷ 4 = r² ÷ 100

⇒ r² = 25

⇒ r = 5

∴ The correct answer is 5%.

SSC CHSL 2025 Tier-1 (Held On: 12 Nov, 2025 Shift 3) (Previous Year Paper)·Quantitative Aptitude
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Question 40

SSC Combined Higher Secondary Level (CHSL) 2025• Quantitative Aptitude• Quantitative Aptitude

A rectangular sheet of paper with dimensions 44 cm by 10 cm is rolled into a right circular cylinder with the 10 cm side becoming the height. What is the approximate volume of the cylinder?

  1. AOption (a)

    1890 cubic cm

  2. BOption (b)

    1978 cubic cm

  3. COption (c)

    1482 cubic cm

  4. DOption (d)

    1540 cubic cm

View answer & explanation

Correct answer

Answer: (d)D.

1540 cubic cm

Why this answer is correct

Given:

Length of sheet = 44 cm

Height of cylinder = 10 cm

Formula Used:

Circumference of base = 2 × π × r

Volume of cylinder = π × r2 × h

Calculation:

⇒ 2 × (22 ÷ 7) × r = 44

⇒ (44 ÷ 7) × r = 44

⇒ r = 7

⇒ Volume = (22 ÷ 7) × 72 × 10

⇒ Volume = (22 ÷ 7) × 49 × 10

⇒ Volume = 22 × 7 × 10

⇒ Volume = 1540

∴ The correct answer is 1540.

How should I prepare for Quantitative Aptitude in SSC Combined Higher Secondary Level (CHSL)?

Solving previous year questions by topic helps identify recurring concepts, question patterns, and frequently asked sub-topics. Attempt each question, review the step-by-step explanation, and note down key formulas or facts for quick revision before the exam.