SSC Combined Graduate Level (CGL) · Topic Collection

Quantitative Aptitude PYQs

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

9600 Total Questions309 Official PapersYears: 2016–2025Search in Quantitative Aptitude
Showing questions 621–640 of 9600Page 32 of 100
SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 621

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

If x = 2 + √3, find x - \(\frac{1}{x}\).

  1. AOption (a)

    √3

  2. BOption (b)

    2√3

  3. COption (c)

    3√2

  4. DOption (d)

    5√3

View answer & explanation

Correct answer

Answer: (b)B.

2√3

Why this answer is correct

Shortcut Trick

If x = a + √b and a2 − b = 1, then 1/x = a − √b.

Here, a = 2 and √b = √3. We check: 22 − (√3)2 = 4 − 3 = 1.

Thus, 1/x = 2 − √3.

x − 1/x = (2 + √3) − (2 − √3) = 2√3.

∴ The correct answer is 2√3.

Alternate Method

Given:

x = 2 + √3

Formula Used:

(a + b)(a − b) = a2 − b2

Calculations:

⇒ 1/x = 1 / (2 + √3)

Rationalising the denominator by multiplying the numerator and denominator by the conjugate (2 − √3):

⇒ 1/x = (2 − √3) / [(2 + √3) × (2 − √3)]

⇒ 1/x = (2 − √3) / (22 − (√3)2)

⇒ 1/x = (2 − √3) / (4 − 3)

⇒ 1/x = 2 − √3

Now, substituting the values of x and 1/x into the required expression:

⇒ x − 1/x = (2 + √3) − (2 − √3)

⇒ x − 1/x = 2 + √3 − 2 + √3

⇒ x − 1/x = 2√3

∴ The correct answer is 2√3.

Additional Information

Conjugate Surds

The conjugate of a binomial surd (a + √b) is (a − √b). Multiplying a binomial surd by its conjugate yields a rational number: (a + √b)(a − √b) = a2 − b.

Rationalisation

Rationalisation is the process of removing a surd from the denominator of a fraction. This is done by multiplying both the numerator and the denominator by the conjugate of the denominator.

Direct Reciprocal Rule

If x = √a + √b and the difference between the bases is 1 (i.e., a − b = 1), then the reciprocal is directly 1/x = √a − √b. This is a crucial timesaver for competitive exams.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 622

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A circle that has a radius of 5 centimeters, and there is a chord measuring 8 centimeters in length. What is the distance from the center of the circle to the chord?

  1. AOption (a)

    3 centimeters

  2. BOption (b)

    4 centimeters

  3. COption (c)

    5 centimeters

  4. DOption (d)

    6 centimeters

View answer & explanation

Correct answer

Answer: (a)A.

3 centimeters

Why this answer is correct

Shortcut Trick

The perpendicular from the center of a circle bisects the chord into two equal halves of 4 cm each.

The radius (5 cm), half-chord (4 cm), and the perpendicular distance form a right-angled triangle.

Using the standard Pythagorean triplet (3, 4, 5), the missing perpendicular distance is directly 3 cm.

∴ The correct answer is 3 cm.

Alternate Method

Given:

Radius of the circle (r) = 5 cm

Length of the chord (AB) = 8 cm

Formula Used:

Perpendicular drawn from the center to a chord bisects the chord.

Pythagoras Theorem: (Hypotenuse)2 = (Perpendicular)2 + (Base)2

O

A B

d = ? M

r = 5 cm 4 cm 4 cm

Calculations:

⇒ Length of half-chord (MB) = 8 ÷ 2 = 4 cm

⇒ In right-angled triangle OMB, using Pythagoras theorem:

⇒ OB2 = OM2 + MB2

⇒ 52 = d2 + 42

⇒ 25 = d2 + 16

⇒ d2 = 25 − 16 = 9

⇒ d = √9 = 3 cm

∴ The correct answer is 3 cm.

Additional Information

Perpendicular to a Chord

A line drawn from the center of a circle perpendicular to a chord bisects the chord into two equal segments.

Equal Chords Property

Chords of equal length subtend equal angles at the center of the circle.

Pythagorean Triplets

Common standard triplets like (3, 4, 5), (5, 12, 13), and (8, 15, 17) are extremely useful for solving right-angled triangle problems directly without calculation.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 623

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A circle of radius of 12 cm. Two radii are drawn such that the length of the chord connecting their endpoints is 12 cm. What is the area of the minor segment formed?

  1. AOption (a)

    (12π–18√3) sq. cm

  2. BOption (b)

    (24π–36√3) sq. cm

  3. COption (c)

    (12π–36√3) sq. cm

  4. DOption (d)

    (24π–18√3) sq. cm

View answer & explanation

Correct answer

Answer: (b)B.

(24π–36√3) sq. cm

Why this answer is correct

Shortcut Trick

Since the radius and chord are equal (12 cm), they form an equilateral triangle with a central angle of 60°.

Area of minor segment = Area of 60° sector − Area of equilateral triangle.

Area = (60÷360) × π × r2 − (√3÷4) × side2

Area = (1÷6) × 144π − (√3÷4) × 144

Area = 24π − 36√3

∴ The correct answer is (24π−36√3) sq. cm.

Alternate Method

Given:

Radius of the circle (r) = 12 cm

Length of the chord = 12 cm

O A B

12 cm 12 cm 12 cm (Chord)

60°

Minor Segment

Formula Used:

Area of Sector = (θ÷360) × πr2

Area of Equilateral Triangle = (√3÷4) × a2

Area of Minor Segment = Area of Sector − Area of Triangle

Calculations:

⇒ In ΔOAB, OA = OB = AB = 12 cm. Thus, ΔOAB is an equilateral triangle and the central angle θ = 60°.

⇒ Area of sector OAB = (60÷360) × π × (12)2

⇒ Area of sector OAB = (1÷6) × 144π = 24π sq. cm

⇒ Area of equilateral ΔOAB = (√3÷4) × (12)2

⇒ Area of equilateral ΔOAB = (√3÷4) × 144 = 36√3 sq. cm

⇒ Area of minor segment = Area of sector − Area of triangle

⇒ Area of minor segment = (24π − 36√3) sq. cm

∴ The correct answer is (24π−36√3) sq. cm.

Additional Information

Area of Major Segment

It is the difference between the total area of the circle and the area of the minor segment. Formula: πr2 − Area of minor segment.

Length of an Arc

The length of an arc subtending angle θ at the center is given by (θ÷360) × 2πr.

Area of a Sector (using arc length)

If the length of the arc (l) is known, the area of the sector can be directly calculated as (l × r) ÷ 2.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 624

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

If sin x = 0.6 and x ∈ (0, π/2), find sin 2x + cos 2x.

  1. AOption (a)

    2,68

  2. BOption (b)

    0.52

  3. COption (c)

    0.68

  4. DOption (d)

    1.24

View answer & explanation

Correct answer

Answer: (d)D.

1.24

Why this answer is correct

Shortcut Trick

Given sin x = 0.6 = 3/5, this forms a 3-4-5 right-angled triangle where perpendicular = 3, hypotenuse = 5, and base = 4.

Thus, cos x = base ÷ hypotenuse = 4/5 = 0.8.

We know sin 2x = 2 × sin x × cos x = 2 × 0.6 × 0.8 = 0.96.

And cos 2x = cos2x − sin2x = 0.82 − 0.62 = 0.64 − 0.36 = 0.28.

Required sum = 0.96 + 0.28 = 1.24.

∴ The correct answer is 1.24.

Alternate Method

Given:

sin x = 0.6 and x ∈ (0, π/2) (First Quadrant, so all trigonometric ratios are positive)

Formula Used:

cos x = √(1 − sin2x)

sin 2x = 2 × sin x × cos x

cos 2x = 1 − 2sin2x

x Base = 4 Perpendicular = 3 Hypotenuse = 5

Calculations:

⇒ cos x = √(1 − 0.62)

⇒ cos x = √(1 − 0.36)

⇒ cos x = √0.64 = 0.8

⇒ sin 2x = 2 × 0.6 × 0.8 = 0.96

⇒ cos 2x = 1 − 2 × (0.6)2

⇒ cos 2x = 1 − 2 × 0.36 = 1 − 0.72 = 0.28

⇒ sin 2x + cos 2x = 0.96 + 0.28

⇒ sin 2x + cos 2x = 1.24

∴ The correct answer is 1.24.

Additional Information

Double Angle Identity for Sine

sin 2x = 2 × sin x × cos x

Double Angle Identity for Cosine

cos 2x = cos2x − sin2x = 2cos2x − 1 = 1 − 2sin2x

Pythagorean Trigonometric Identity

sin2x + cos2x = 1, which confirms cos x = √(1 − sin2x) for all angles in the first quadrant.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 625

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

Multiply 12.345 by 0.067 and round your answer to 4 decimal places.

  1. AOption (a)

    0.8271

  2. BOption (b)

    0.8764

  3. COption (c)

    0.5687

  4. DOption (d)

    0.8765

View answer & explanation

Correct answer

Answer: (a)A.

0.8271

Why this answer is correct

Shortcut Trick

Ignore the decimal points and multiply the numbers as integers: 12345 × 67 = 827115.

Count the total number of decimal places in both given numbers: 3 + 3 = 6 decimal places.

Place the decimal point 6 positions from the right in the product to get 0.827115.

To round to 4 decimal places, check the 5th digit (which is 1). Since 1 is less than 5, the 4th digit remains unchanged.

The rounded value is 0.8271.

∴ The correct answer is 0.8271.

Alternate Method

Given:

Expression: 12.345 × 0.067

Calculation:

⇒ Convert the given decimals into fractions:

⇒ 12.345 = 12345 ÷ 1000

⇒ 0.067 = 67 ÷ 1000

⇒ Multiply the fractions:

⇒ (12345 ÷ 1000) × (67 ÷ 1000)

⇒ (12345 × 67) ÷ (1000 × 1000)

⇒ 827115 ÷ 1000000

⇒ Convert back to a decimal by shifting the decimal point 6 places to the left:

⇒ 0.827115

⇒ Rounding off to 4 decimal places, we observe the 5th decimal digit is 1.

⇒ Since 1 is less than 5, the 4th decimal place remains exactly as it is.

⇒ Thus, the required value becomes 0.8271.

∴ The correct answer is 0.8271.

Additional Information

Rounding Rule for Decimals

Look at the digit immediately to the right of your target decimal place. If it is 5 or greater, round up by adding 1 to the target digit. If it is less than 5, round down by leaving the target digit unchanged.

Decimal Multiplication Rule

Multiply the numbers as standard integers, ignoring the decimals entirely. After finding the product, count the total number of decimal places in the original factors and insert the decimal point in the product so that it matches this total.

Significant Figures vs Decimal Places

Decimal places refer specifically to the count of digits after the decimal point, whereas significant figures include all digits starting from the first non-zero digit to denote overall precision.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 626

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

What is the value of √9801 - √1681

  1. AOption (a)

    58

  2. BOption (b)

    68

  3. COption (c)

    78

  4. DOption (d)

    88

View answer & explanation

Correct answer

Answer: (a)A.

58

Why this answer is correct

Shortcut Trick

For √9801: The nearest base square is 1002 = 10000. Since 9801 is slightly less and ends in 1, its square root is 99.

For √1681: The nearest base square is 402 = 1600. Since 1681 is slightly more and ends in 1, its square root is 41.

Expression value = 99 − 41 = 58.

∴ The correct answer is 58.

Alternate Method

Given:

The expression is √9801 − √1681

Calculation:

⇒ Let us find the square root of 9801 by prime factorization.

⇒ 9801 = 9 × 1089

⇒ 9801 = 9 × 9 × 121

⇒ 9801 = 92 × 112

⇒ √9801 = 9 × 11 = 99

⇒ Now let us find the square root of 1681.

⇒ Since 402 = 1600 and 1681 is slightly greater than 1600 with a unit digit of 1, the number must be 41.

⇒ 41 × 41 = 1681 ⇒ √1681 = 41

⇒ Now, substitute the extracted values back into the expression:

⇒ √9801 − √1681 = 99 − 41

⇒ 99 − 41 = 58

∴ The correct answer is 58.

Additional Information

Unit Digits of Perfect Squares

A perfect square can only end in 0, 1, 4, 5, 6, or 9. If a perfect square ends in 1, its square root will always have a unit digit of either 1 or 9.

Estimating Square Roots

Find the nearest multiples of 10 whose squares bracket the given number. For example, 1681 lies just above 1600 (which is 402), immediately indicating the root is close to 40.

Base 100 Square Trick

For a number written as (100 − a), its square is formed by writing (100 − 2a) followed by a2 (in two digits). For 99 (where a=1), (100 − 2) = 98 and 12 = 01, giving exactly 9801.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 627

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

The investments of three partners, P, Q, and R, are initially in the ratio 6 : 7 : 9. After a year, P adds ₹30,000, Q adds ₹25,000, and R adds some amount to their respective investments. The new ratio of their investments becomes 8 : 9 : 11. Find the amount (in ₹) added by partner R.

  1. AOption (a)

    ₹10,000

  2. BOption (b)

    ₹12,500

  3. COption (c)

    ₹15,000

  4. DOption (d)

    ₹18,000

View answer & explanation

Correct answer

Answer: (c)C.

₹15,000

Why this answer is correct

Shortcut Trick

Let the initial investments be 6x, 7x, and 9x.

Equating the scaling factors of the new ratio for P and Q: (6x + 30000) ÷ 8 = (7x + 25000) ÷ 9

⇒ 54x + 270000 = 56x + 200000 ⇒ 2x = 70000 ⇒ x = 35000

Value of 1 part in new ratio = (6 × 35000 + 30000) ÷ 8 = 30000

R's added amount = New Investment − Old Investment = (11 × 30000) − (9 × 35000) = 15000

∴ The correct answer is ₹15,000.

Alternate Method

Given:

Initial ratio of investments of P, Q, and R = 6 : 7 : 9

Amount added by P = ₹30,000

Amount added by Q = ₹25,000

New ratio of investments of P, Q, and R = 8 : 9 : 11

Formula Used:

Ratio = Value of Part 1 ÷ Value of Part 2

Partner P Partner Q Partner R

Initial: 6x = 210,000

Initial: 7x = 245,000

Initial: 9x = 315,000

  • 30,000

  • 25,000

  • 15,000

Final: 8 parts = 240,000

Final: 9 parts = 270,000

Final: 11 parts = 330,000

Amount added by R = 330,000 − 315,000 = 15,000

Calculations:

Let the initial investments of P, Q, and R be 6x, 7x, and 9x respectively.

After 1 year, new investments of P and Q are (6x + 30000) and (7x + 25000).

According to the given new ratio for P and Q:

⇒ (6x + 30000) ÷ (7x + 25000) = 8 ÷ 9

⇒ 9 × (6x + 30000) = 8 × (7x + 25000)

⇒ 54x + 270000 = 56x + 200000

⇒ 56x − 54x = 270000 − 200000

⇒ 2x = 70000

⇒ x = 35000

Initial investment of R:

⇒ 9x = 9 × 35000 = 315000

New investment of P:

⇒ 6x + 30000 = 6 × 35000 + 30000 = 240000

Since the new ratio for P is 8 parts, we can find the exact value of 1 ratio part:

⇒ 1 part = 240000 ÷ 8 = 30000

New investment of R:

⇒ 11 parts = 11 × 30000 = 330000

Amount added by R:

⇒ New investment − Initial investment

⇒ 330000 − 315000 = 15000

∴ The correct answer is ₹15,000.

Additional Information

Partnership Ratios

Profit in a partnership is distributed in proportion to the product of the capital invested and the time period of investment. (Profit Ratio = C1 × T1 : C2 × T2).

Changes in Capital

When a partner adds or withdraws capital midway through the term, the total investment weight is calculated by multiplying each active capital amount by the respective time duration it remained invested.

Ratio Equating Method

When fixed amounts are added to or subtracted from components of an original ratio to form a new ratio, setting up a cross-multiplication equation between any two quantities helps efficiently find the common scaling factor.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 628

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

An employee contributes 20% of their salary to charity. From the remaining amount, 40% is allocated to investments. The rest is divided for education and travel in the ratio 7: 5. If the expense on travel is ₹6000, what is the employee's total salary?

  1. AOption (a)

    ₹28,000

  2. BOption (b)

    ₹30,000

  3. COption (c)

    ₹32,000

  4. DOption (d)

    ₹35,000

View answer & explanation

Correct answer

Answer: (b)B.

₹30,000

Why this answer is correct

Shortcut Trick

Let the total salary of the employee be 100%.

Remaining after 20% charity = 100% − 20% = 80%.

Remaining after 40% investment = 80% × (100% − 40%) = 48%.

Travel portion = 48% × [5 ÷ (7 + 5)] = 20%.

Since 20% ≡ ₹6000, 100% = ₹6000 × 5 = ₹30,000.

∴ The correct answer is ₹30,000.

Alternate Method

Given:

Contribution to charity = 20%

Investment allocation = 40% of the remaining amount

Education to travel ratio = 7 : 5

Expense on travel = ₹6000

Formula Used:

Remaining Value = Initial Value × (1 − Percentage ÷ 100)

Salary 100%

−20%

Rem. 80%

−40%

Rest 48%

Edu (28%)

Travel (20%)

Travel = 20% = ₹6000 → Total (100%) = ₹30,000

Calculations:

⇒ Let total salary be S.

⇒ Remaining amount after charity = S × (100% − 20%) = 0.8S

⇒ Amount left after investment = 0.8S × (100% − 40%) = 0.48S

⇒ Share of travel in the rest amount = 5 ÷ (7 + 5) = 5/12

⇒ Expense on travel = 0.48S × (5 ÷ 12) = 0.2S

⇒ 0.2S = 6000

⇒ S = 6000 ÷ 0.2 = 30000

∴ The correct answer is ₹30,000.

Additional Information

Successive Percentage Change

When multiple percentage reductions occur successively on a remaining value, apply the multiplying factor method: Final Value = Initial × (1 − R1/100) × (1 − R2/100).

Base Change in Percentages

Always note whether a percentage is applied to the original total or the remaining amount. A 40% reduction on an 80% remaining balance means a net reduction of 32% (40% of 80%) from the original base.

Ratio Distribution

If an amount is distributed in the ratio a : b, the individual parts are calculated as (a ÷ (a + b)) and (b ÷ (a + b)) respectively of the total distributed amount.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 629

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A certain amount increases to five times its initial value after 10 years of compounding interest. How many years will it take for it to grow to twenty-five times its original amount?

  1. AOption (a)

    22

  2. BOption (b)

    20

  3. COption (c)

    19

  4. DOption (d)

    18

View answer & explanation

Correct answer

Answer: (b)B.

20

Why this answer is correct

Shortcut Trick

In Compound Interest, if a sum becomes n times in T years, it becomes nm times in m × T years.

Here, the amount becomes 5 times in 10 years (n = 5, T = 10).

We need to find the time it takes to become 25 times, which is 52 times (m = 2).

Required time = m × T = 2 × 10 = 20 years.

∴ The correct answer is 20 years.

Alternate Method

P 5P 25P

0 Years 10 Years 20 Years

× 5 10 Years

× 5 10 Years

Compound Interest Timeline

Given:

Amount after 10 years = 5 × Initial Amount

Formula Used:

A = P(1 + R/100)T

Where A is amount, P is principal, R is rate of interest, and T is time in years.

Calculations:

⇒ Let the initial amount be P.

⇒ After 10 years, Amount A = 5P

⇒ 5P = P(1 + R/100)10

⇒ 5 = (1 + R/100)10     --- (Equation 1)

⇒ We need to find the time T when Amount A = 25P

⇒ 25P = P(1 + R/100)T

⇒ 25 = (1 + R/100)T

⇒ 52 = (1 + R/100)T

⇒ Substituting the value of 5 from Equation 1:

⇒ [(1 + R/100)10]2 = (1 + R/100)T

⇒ (1 + R/100)20 = (1 + R/100)T

⇒ By comparing the powers on both sides, T = 20 years.

∴ The correct answer is 20 years.

Additional Information

Compound Interest Multiplier Rule

If a given sum becomes n times in T years at a certain rate of compound interest, it will become nm times in m × T years.

Simple Interest Equivalent

In simple interest, becoming 5 times means earning an interest of 4P in 10 years. Becoming 25 times means earning an interest of 24P, which would take (24 ÷ 4) × 10 = 60 years.

Rule of 72

It is a quick approximation formula to estimate the doubling time of an investment. Doubling time ≈ 72 ÷ R years, where R is the annual compound interest rate.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 630

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A software company developed 800 licenses for a new application at a total cost of ₹4,000,000. They provided 50 licenses free to their top beta testers. For the remaining licenses, they offered a 15% discount on the market price of ₹6000 per license. Additionally, they gave 1 license free for every 9 licenses bought. If all 800 licenses were distributed, what is the company's overall gain or loss percentage?

  1. AOption (a)

    16.64% profit

  2. BOption (b)

    14.94% profit

  3. COption (c)

    13.94% loss

  4. DOption (d)

    15.64% profit

View answer & explanation

Correct answer

Answer: (c)C.

13.94% loss

Why this answer is correct

Shortcut Trick

Remaining licenses for sale = 800 − 50 = 750.

The offer 'Buy 9 get 1 free' implies 9 paid licenses out of every 10 distributed.

Total paid licenses = 750 × (9 ÷ 10) = 675.

Selling Price per license = ₹6000 × (1 − 0.15) = ₹5100.

Total SP = 675 × 5100 = ₹34,42,500. Total CP = ₹40,00,000.

Loss % = ((40,00,000 − 34,42,500) ÷ 40,00,000) × 100 = 13.9375% ≈ 13.94%.

∴ The correct answer is 13.94% loss.

Alternate Method

Given:

Total licenses = 800, Total CP = ₹4,000,000, Free to top beta testers = 50.

Remaining licenses = 750, Marked Price (MP) per license = ₹6000.

Discount = 15%, Promotional Offer = 1 free license for every 9 licenses bought.

Formula Used:

SP = MP × ((100 − Discount%) ÷ 100)

Loss = CP − SP

Loss % = (Loss ÷ CP) × 100

800 Licenses

-50 free

750 Remaining

9:1 ratio

675 Paid

CP: ₹4,000,000 SP: 675 × ₹5100 = ₹3,442,500 Loss = ₹557,500 (13.94%)

Calculations:

⇒ SP of each license after a 15% discount = 6000 × (85 ÷ 100) = ₹5100.

⇒ The offer '1 free for every 9 bought' means for every bundle of 10 licenses, only 9 are paid for.

⇒ Total bundles formed from 750 remaining licenses = 750 ÷ 10 = 75.

⇒ Number of paid licenses = 75 × 9 = 675.

⇒ Total Selling Price (SP) = 675 × 5100 = ₹3,442,500.

⇒ Total Loss = Total CP − Total SP = 4,000,000 − 3,442,500 = ₹557,500.

⇒ Loss % = (557,500 ÷ 4,000,000) × 100 = 13.9375%.

⇒ Rounding off to two decimal places, the loss is 13.94%.

∴ The correct answer is 13.94% loss.

Additional Information

Marked Price and Discount

Discount is always calculated on the Marked Price. Formula: SP = MP × (100 − D%) ÷ 100.

Buy X Get Y Free Concept

When a dealer offers Y items free on the purchase of X items, the total items given are (X + Y). The equivalent discount percentage is (Y ÷ (X + Y)) × 100.

Profit and Loss Percentage

Profit or Loss is always evaluated on the Cost Price (CP) unless specified otherwise. Loss % = (Loss ÷ CP) × 100.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 631

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

The speeds of two motorcyclists are in the ratio 5 : 3. If the slower motorcyclist has a 24 km head start, and his speed is 12 km/h, how far (in km) will the faster motorcyclist travel to overtake him?

  1. AOption (a)

    40 km

  2. BOption (b)

    48 km

  3. COption (c)

    60 km

  4. DOption (d)

    72 km

View answer & explanation

Correct answer

Answer: (c)C.

60 km

Why this answer is correct

Shortcut Trick

When the time taken to meet is constant, the ratio of distances covered is equal to the ratio of their speeds.

Distance ratio = Speed ratio = 5 : 3.

The difference in the distance ratio corresponds to the head start ⇒ 5 − 3 = 2 units.

Given head start = 24 km ⇒ 2 units = 24 km ⇒ 1 unit = 12 km.

Distance covered by the faster motorcyclist = 5 units = 5 × 12 = 60 km.

∴ The correct answer is 60 km.

Alternate Method

Start (Faster) Start (Slower) Overtake

Speed: 5x Speed: 3x

24 km Gap

Faster Motorcyclist Total Distance = ?

Given:

Ratio of speeds = 5 : 3

Speed of slower motorcyclist = 12 km/h

Head start distance = 24 km

Formula Used:

Relative Speed = Speed of faster − Speed of slower (moving in the same direction)

Time to Overtake = Head Start Distance ÷ Relative Speed

Distance = Speed × Time

Calculations:

⇒ Let the speeds be 5x and 3x respectively.

⇒ 3x = 12 km/h ⇒ x = 4 km/h.

⇒ Speed of faster motorcyclist = 5x = 5 × 4 = 20 km/h.

⇒ Relative speed = 20 − 12 = 8 km/h.

⇒ Time taken to overtake = 24 ÷ 8 = 3 hours.

⇒ Distance travelled by the faster motorcyclist = 20 × 3 = 60 km.

∴ The correct answer is 60 km.

Additional Information

Relative Speed

When two bodies travel in the same direction, their relative speed is the difference of their individual speeds (v1 − v2). When travelling in opposite directions, it is the sum of their speeds (v1 + v2).

Proportionality in Distance and Speed

When the time factor is constant for both objects, the distance travelled is directly proportional to their respective speeds (d1 ÷ d2 = s1 ÷ s2).

Overtaking Time

The exact time required to catch up to a moving target is calculated by dividing the initial gap distance by the relative speed of the chasing body.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 632

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A shopkeeper took a loan of ₹7500 from a bank on 1 April 2025 with simple interest at 9% per annum. He decided to clear the loan on 1 September 2025. How much amount did he pay back to the bank?

  1. AOption (a)

    ₹7775.00

  2. BOption (b)

    ₹7781.25

  3. COption (c)

    ₹7784.79

  4. DOption (d)

    ₹7790.00

View answer & explanation

Correct answer

Answer: (b)B.

₹7781.25

Why this answer is correct

Shortcut Trick

Time duration from 1st April to 1st September = 5 months = 5 ÷ 12 years.

Simple Interest = Principal × Rate × Time

SI = 7500 × 9% × (5 ÷ 12) = ₹281.25

Total Amount = Principal + SI = 7500 + 281.25 = ₹7781.25

∴ The correct answer is ₹7781.25.

Alternate Method

Given:

Principal (P) = ₹7500

Rate of Interest (R) = 9% per annum

Loan start date = 1 April 2025

Loan clear date = 1 September 2025

Formula Used:

Simple Interest (SI) = (P × R × T) ÷ 100

Amount (A) = P + SI

1 Apr 2025 P = ₹7500

1 Sep 2025 Amount = ?

Time = 5 Months (5/12 Year)

Rate = 9% p.a.

SI = ₹281.25 → Amount = ₹7781.25

Calculations:

⇒ Total duration from 1 April 2025 to 1 September 2025 is exactly 5 full months.

⇒ Time (T) in years = 5 ÷ 12 years.

⇒ SI = (7500 × 9 × 5) ÷ (100 × 12)

⇒ SI = (75 × 45) ÷ 12

⇒ SI = 3375 ÷ 12

⇒ SI = ₹281.25

⇒ Total Amount = Principal + SI

⇒ Total Amount = 7500 + 281.25

⇒ Total Amount = ₹7781.25

∴ The correct answer is ₹7781.25.

Additional Information

Conversion of Time

When the time given is in months or days, it must be converted to years for simple interest formulas. Divide months by 12 and days by 365.

Counting Days

When calculating exact days for simple interest, it is standard practice to exclude either the day the money is borrowed or the day it is returned, but not both.

Simple Interest vs Compound Interest

For the first year (or unit time period), Simple Interest and Compound Interest are identical. From the second period onwards, Compound Interest is calculated on the accumulated interest as well as the principal.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 633

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

Which of these is both an Integer and a Whole number?

  1. AOption (a)

    -4

  2. BOption (b)

    0

  3. COption (c)

    3.5

  4. DOption (d)

    1/2

View answer & explanation

Correct answer

Answer: (b)B.

0

Why this answer is correct

Shortcut Trick

Whole numbers comprise 0 and all positive counting numbers (0, 1, 2, 3...).

Integers encompass all whole numbers along with their negative counterparts (...−2, −1, 0, 1, 2...).

Since the set of whole numbers is a subset of integers, any whole number is automatically an integer.

Scanning the options, 0 is the only number that fits into the whole number category.

∴ The correct answer is 0.

Alternate Method

Given:

Four numbers: −4, 0, 3.5, and 1/2

Concept Used:

Integers (Z) = {..., −3, −2, −1, 0, 1, 2, 3, ...}

Whole Numbers (W) = {0, 1, 2, 3, ...}

Calculation:

⇒ Evaluate Option 1 (−4): It is a negative number. It belongs to integers but not whole numbers.

⇒ Evaluate Option 2 (0): It is a non-negative number without a fractional or decimal part. It belongs to both integers and whole numbers.

⇒ Evaluate Option 3 (3.5): It is a decimal value. It is neither an integer nor a whole number.

⇒ Evaluate Option 4 (1/2): It is a fractional value. It is neither an integer nor a whole number.

∴ The correct answer is 0.

Additional Information

Natural Numbers

These are all positive counting numbers starting from 1. Set N = {1, 2, 3, 4, ...}. Remember that 0 is not a natural number.

Rational Numbers

Any number that can be expressed in the form of p/q where p and q are integers and q ≠ 0. For example, 1/2, −4/1, 0/1.

Irrational Numbers

Numbers that cannot be written as a simple fraction, such as √2, √3, or π. Their decimal expansions are non-terminating and non-repeating.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 634

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A, B, and C invested in a business in the ratio 2:3:5. After a year, the total profit is Rs. 1,00,000. What is B's share?

  1. AOption (a)

    Rs. 30,000

  2. BOption (b)

    Rs. 25,000

  3. COption (c)

    Rs. 20,000

  4. DOption (d)

    Rs. 40,000

View answer & explanation

Correct answer

Answer: (a)A.

Rs. 30,000

Why this answer is correct

Shortcut Trick

Since the investment duration is identical for all partners, the profit is distributed exactly in the ratio of their investments.

Profit Ratio of A : B : C = 2 : 3 : 5

Total ratio parts = 2 + 3 + 5 = 10 parts.

B's share = Total Profit × (B's parts ÷ Total parts) = 1,00,000 × (3 ÷ 10) = 30,000.

∴ The correct answer is Rs. 30,000.

Alternate Method

Given:

Ratio of investments of A, B, and C = 2 : 3 : 5

Total profit earned after a year = Rs. 1,00,000

Formula Used:

Profit Ratio = Capital × Time

When time periods are equal, Profit Ratio = Capital Ratio.

2 Parts A

3 Parts B

5 Parts C

Total Profit = Rs. 1,00,000 (10 Parts)

Calculations:

⇒ Let the individual profits of A, B, and C be 2x, 3x, and 5x respectively.

⇒ Sum of all parts (Total Profit) = 2x + 3x + 5x = 10x

⇒ We are given that the total profit is Rs. 1,00,000.

⇒ 10x = 1,00,000

⇒ x = 1,00,000 ÷ 10 = 10,000

⇒ B's share = 3x = 3 × 10,000 = Rs. 30,000

∴ The correct answer is Rs. 30,000.

Additional Information

Partnership Profit Ratio Formula

When investments and time periods differ, the profit is divided in the ratio of the product of their capitals and respective time periods: Profit Ratio = (C1 × T1) : (C2 × T2).

Active vs Sleeping Partner

An active partner manages the business and usually receives a fixed percentage of the total profit as a salary before the remaining profit is distributed among all partners according to their investment ratios. A sleeping partner only invests capital.

Joining Mid-Way

If a partner joins the business after a few months, their time period for investment is calculated only for the remaining months of the financial cycle (e.g., joining after 5 months means T = 7 months in a 1-year cycle).

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 635

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A solid pyramid is to be painted. If paint covers 2 m²/litre and total surface is 48 m², how much paint is needed, including 5% extra for retouch?

  1. AOption (a)

    25.2 litres

  2. BOption (b)

    35.8 litres

  3. COption (c)

    20.4 litres

  4. DOption (d)

    15.5 litres

View answer & explanation

Correct answer

Answer: (a)A.

25.2 litres

Why this answer is correct

Shortcut Trick

Total paint required = (Total Area ÷ Coverage per litre) × Increasing factor

Total paint = (48 ÷ 2) × 1.05

Total paint = 24 × 1.05 = 25.2 litres

∴ The correct answer is 25.2 litres.

Alternate Method

Given:

Total surface area of the solid pyramid = 48 m2

Paint coverage = 2 m2/litre

Extra paint required for retouch = 5%

Total Area 48 m²

÷ 2

Base Paint 24 L

  • 5% Extra

Total Paint 25.2 L

Formula Used:

Base paint = Total surface area ÷ Coverage

Total paint = Base paint + (5% of Base paint)

Calculations:

⇒ Base paint required = 48 ÷ 2 = 24 litres

⇒ Extra paint = 5% of 24 = (5 ÷ 100) × 24 = 1.2 litres

⇒ Total paint needed = 24 + 1.2 = 25.2 litres

∴ The correct answer is 25.2 litres.

Additional Information

Total Surface Area of a Pyramid

The total surface area is the sum of the base area and the lateral surface area (sum of the areas of all the triangular side faces).

Volume of a Pyramid

The space enclosed by a solid pyramid is given by the formula: Volume = (1 ÷ 3) × Base Area × Vertical Height.

Percentage Increase

When a base value needs an extra R% addition, the new value can be quickly computed by multiplying the base value by the multiplying factor (1 + R ÷ 100).

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 636

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

An L-shaped floor is made of two rectangles (6 m × 4 m and 8 m × 6 m). Tiles cost ₹450/m², but a 10% discount is given if more than 70 m² is tiled. What is the total cost after discount?

  1. AOption (a)

    ₹30, 600

  2. BOption (b)

    ₹29, 160

  3. COption (c)

    ₹29, 240

  4. DOption (d)

    ₹29, 540

View answer & explanation

Correct answer

Answer: (b)B.

₹29, 160

Why this answer is correct

Shortcut TrickTotal Area = (6 × 4) + (8 × 6) = 24 + 48 = 72 m2.Since 72 m2 > 70 m2, effective cost per m2 = 450 × 0.9 = ₹405.Total Cost = 72 × 405 = ₹29,160.∴ The correct answer is ₹29,160.Alternate MethodGiven:⇒ Dimensions of first rectangle = 6 m × 4 m⇒ Dimensions of second rectangle = 8 m × 6 m⇒ Rate of tiles = ₹450/m2⇒ Discount = 10% (applicable if total area > 70 m2)6 m4 m24 m²+8 m6 m48 m²=72 m²Total Area = 72 m² (Discount Applies)Calculations:⇒ Total Area = (6 × 4) + (8 × 6)⇒ Total Area = 24 + 48 = 72 m2⇒ Total cost without discount = 72 × 450 = ₹32,400⇒ Since 72 m2 > 70 m2, a 10% discount is applied to the total cost.⇒ Discount amount = 10% of 32,400 = (10 ÷ 100) × 32400 = ₹3,240⇒ Final Cost = 32400 − 3240 = ₹29,160∴ The correct answer is ₹29,160.Additional InformationArea of a RectangleThe total space enclosed by a rectangular shape is calculated as Length × Breadth.Calculating Total CostWhen a specific rate per unit area is given, Total Cost = Total Area × Rate per unit area.Discount PercentageWhen a discount of D% is applied on a value, Final Value = Initial Value × (100 − D) ÷ 100.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 637

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A large regular polygon is made by combining 6 regular hexagons around a common center. What is the ratio of area of the combined shape to one hexagon?

  1. AOption (a)

    5:1

  2. BOption (b)

    6:1

  3. COption (c)

    7:1

  4. DOption (d)

    8:1

View answer & explanation

Correct answer

Answer: (b)B.

6:1

Why this answer is correct

Shortcut Trick

Let the area of a single regular hexagon be 1 unit.

Since the combined shape is formed by exactly 6 such regular hexagons without overlapping, its total area is simply the sum of all their individual areas.

Total area of the combined shape = 6 × 1 = 6 units.

Ratio of the combined area to the area of one hexagon = 6 : 1.

∴ The correct answer is 6:1.

Alternate Method

Given:

Number of regular hexagons combined together = 6

Concept Used:

When multiple non-overlapping two-dimensional figures are joined together, the total area of the resulting figure is equal to the sum of the individual areas (Conservation of Area principle).

Center

Area = A

Combined Shape (Area = 6A) One Hexagon

Calculation:

⇒ Let the area of one regular hexagon = A

⇒ The large polygon is constructed by placing 6 regular hexagons around a common center in a tiled format.

⇒ Total Area of the new shape = Area of Hexagon 1 + Area of Hexagon 2 + ... + Area of Hexagon 6

⇒ Total Area = 6 × A = 6A

⇒ Required Ratio = (Total Area of combined shape) ÷ (Area of one hexagon)

⇒ Required Ratio = 6A / A = 6:1

∴ The correct answer is 6:1.

Additional Information

Area of a Regular Hexagon

The area of a regular hexagon with side length a is given by the formula: Area = (3√3 ÷ 2) × a2.

Conservation of Area

In two-dimensional geometry, if a planar figure is divided into several non-overlapping pieces, the total area of the whole figure exactly equals the sum of the areas of those distinct pieces.

Tessellation of Hexagons

Regular hexagons are one of the only three regular polygons (along with equilateral triangles and squares) that can perfectly tile a flat plane without leaving gaps or creating overlaps.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 638

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A triangular traffic sign has sides measuring 13 m, 14 m, and 15 m. If the cost to paint the sign is ₹5 per square meter, what will be the total expense to paint the entire sign?

  1. AOption (a)

    ₹420

  2. BOption (b)

    ₹380

  3. COption (c)

    ₹360

  4. DOption (d)

    ₹480

View answer & explanation

Correct answer

Answer: (a)A.

₹420

Why this answer is correct

Shortcut Trick

Semi-perimeter (s) = (13 + 14 + 15) ÷ 2 = 21 m.

Area using Heron's formula = √(21 × 8 × 7 × 6) = √7056 = 84 m2.

Total Cost = Area × Rate = 84 × 5 = ₹420.

∴ The correct answer is ₹420.

Alternate Method

Given:

Sides of the triangular sign: a = 13 m, b = 14 m, c = 15 m.

Painting cost rate = ₹5 per m2.

Triangular Traffic Sign

14 m 13 m 15 m

Formula Used:

Heron's Formula for Area of a Triangle = √[s(s − a)(s − b)(s − c)]

Where semi-perimeter, s = (a + b + c) ÷ 2

Calculation:

⇒ s = (13 + 14 + 15) ÷ 2

⇒ s = 42 ÷ 2 = 21 m

⇒ Area = √[21 × (21 − 13) × (21 − 14) × (21 − 15)]

⇒ Area = √(21 × 8 × 7 × 6)

⇒ Area = √(7 × 3 × 4 × 2 × 7 × 3 × 2)

⇒ Area = √(72 × 32 × 42)

⇒ Area = 7 × 3 × 4 = 84 m2

⇒ Total Cost = Area × Rate per m2

⇒ Total Cost = 84 × 5 = 420

∴ The correct answer is ₹420.

Additional Information

Area of an Equilateral Triangle

For a triangle with all equal sides (a), Area = (√3 ÷ 4) × a2.

Area of a Right-Angled Triangle

When base and perpendicular height are given, Area = (1 ÷ 2) × Base × Height.

Inradius of a Triangle

The radius of the largest circle that can fit inside a triangle is calculated as r = Area ÷ s.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 639

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

If sin θ = cos θ, what is the value of θ (in degrees)?

  1. AOption (a)

    45°

  2. BOption (b)

    35°

  3. COption (c)

    30°

  4. DOption (d)

    40°

View answer & explanation

Correct answer

Answer: (a)A.

45°

Why this answer is correct

Shortcut Trick

Given, sin θ = cos θ

Divide both sides of the equation by cos θ:

⇒ sin θ ÷ cos θ = 1

⇒ tan θ = 1

We know from standard trigonometric values that tan 45° = 1.

∴ The correct answer is 45°.

Alternate Method

θ Perpendicular Base Hypotenuse Isosceles Right Triangle (Base = Perpendicular)

Given:

sin θ = cos θ

Formulas Used:

sin θ = Perpendicular ÷ Hypotenuse

cos θ = Base ÷ Hypotenuse

Calculation:

⇒ Perpendicular ÷ Hypotenuse = Base ÷ Hypotenuse

⇒ Perpendicular = Base

In a right-angled triangle, if the perpendicular equals the base, the triangle is an isosceles right-angled triangle.

⇒ The sum of angles in a triangle is 180°, and the two non-right angles are equal.

⇒ θ + θ + 90° = 180°

⇒ 2θ = 90°

⇒ θ = 45°

∴ The correct answer is 45°.

Additional Information

Complementary Angles Property

We know that sin θ = cos (90° − θ). If sin θ = cos θ, then equating the angles gives θ = 90° − θ ⇒ 2θ = 90° ⇒ θ = 45°.

Standard Trigonometric Values

For standard angles, sin 30° = 1/2 and cos 30° = √3/2. However, for 45°, sin 45° = 1/√2 and cos 45° = 1/√2, hence their values are exactly equal.

Pythagorean Identity Method

Using the identity sin2 θ + cos2 θ = 1, substituting sin θ for cos θ gives 2 × sin2 θ = 1 ⇒ sin θ = 1/√2, which yields θ = 45° in the first quadrant.

SSC CGL 2025 (Held On: 20 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 640

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

The base area of a prism is increased by 25%, while the height remains the same. By what percentage does the volume increase?

  1. AOption (a)

    10%

  2. BOption (b)

    20%

  3. COption (c)

    25%

  4. DOption (d)

    30%

View answer & explanation

Correct answer

Answer: (c)C.

25%

Why this answer is correct

Shortcut Trick

The Volume of a prism is the product of its Base Area and Height (V = A × h).

Since the height of the prism remains constant, the Volume is directly proportional to the Base Area.

Therefore, a 25% increase in the Base Area directly results in a 25% increase in the Volume.

∴ The correct answer is 25%.

Alternate Method

Given:

Increase in Base Area = 25%

Height remains unchanged.

Formula Used:

Volume of Prism (V) = Base Area (A) × Height (h)

Percentage Increase = [(New Value − Original Value) ÷ Original Value] × 100

Original Prism Area = A Height = h V = A × h

New Prism Area = 1.25A Height = h V = 1.25Ah

Area +25% Volume +25%

Calculations:

⇒ Let Original Base Area = A and Original Height = h.

⇒ Original Volume (V1) = A × h

⇒ New Base Area = A + (25 ÷ 100) × A = 1.25A

⇒ New Height = h

⇒ New Volume (V2) = 1.25A × h = 1.25(A × h)

⇒ Increase in Volume = V2 − V1 = 1.25(A × h) − 1(A × h) = 0.25(A × h)

⇒ Percentage Increase = [0.25(A × h) ÷ (A × h)] × 100 = 0.25 × 100 = 25%

∴ The correct answer is 25%.

Additional Information

Proportionality Rule

When a formula consists of a product of variables (like V = A × h), and one variable is kept constant, the percentage change in the result strictly equals the percentage change in the varying quantity.

Volume of a Prism

A prism is a 3D solid with identical parallel ends (bases) and flat side faces. The volume of any uniform solid is consistently calculated as: Volume = Base Area × Height.

Successive Percentage Change

If both the base area and the height were changed by x% and y% respectively, the overall percentage change in volume would be calculated using the successive change formula: x + y + (x × y) ÷ 100.

How should I prepare for Quantitative Aptitude in SSC Combined Graduate Level (CGL)?

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.