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
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SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 21

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

A circular pizza having a radius of 21 cm. If 75% of it is eaten, what is the area of pizza remaining?

  1. AOption (a)

    173.25 cm²

  2. BOption (b)

    346.5 cm²

  3. COption (c)

    432.25 cm²

  4. DOption (d)

    115.5 cm²

View answer & explanation

Correct answer

Answer: (b)B.

346.5 cm²

Why this answer is correct

Shortcut Trick

Remaining pizza percentage = 100% − 75% = 25% = 1/4 of total pizza.

Area remaining = (1/4) × π × r2

⇒ Area = (1/4) × (22/7) × 21 × 21

⇒ Area = (1/4) × 1386 = 346.5 cm2

∴ The correct answer is 346.5 cm2.

Alternate Method

Given:

Radius of the circular pizza (r) = 21 cm

Percentage of pizza eaten = 75%

Formula Used:

Total Area of a circle = πr2 (where π ≈ 22/7)

Remaining Area = Total Area × (Remaining Percentage / 100)

Eaten (75%) 25% Left r = 21 cm Pizza Area Remaining = 25%

Calculation:

⇒ Total Area of the pizza = (22/7) × 21 × 21

⇒ Total Area = 22 × 3 × 21 = 1386 cm2

⇒ Remaining Percentage = 100% − 75% = 25%

⇒ Remaining Area = 25% of Total Area

⇒ Remaining Area = (25 / 100) × 1386

⇒ Remaining Area = 1386 ÷ 4 = 346.5 cm2

∴ The correct answer is 346.5 cm2.

Additional Information

Area of a Circle

The total area enclosed by a circle is given by A = πr2, where r is the radius of the circle.

Area of a Sector

A sector is a portion of a circle enclosed by two radii and an arc. Its area is calculated as (θ/360) × πr2, where θ is the central angle.

Percentage to Fraction Conversion

Converting standard percentages to fractions simplifies calculations. For example, 25% = 1/4, 50% = 1/2, and 75% = 3/4.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 22

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

A ring-shaped disc has outer radius 10 cm and inner radius 7 cm. What is the approximate ratio of the ring's area to the whole outer circle?

  1. AOption (a)

    1:2

  2. BOption (b)

    2:3

  3. COption (c)

    3:4

  4. DOption (d)

    4:5

View answer & explanation

Correct answer

Answer: (a)A.

1:2

Why this answer is correct

Shortcut Trick

Ratio of ring area to outer circle area = (R2 − r2) : R2

⇒ Ratio = (102 − 72) : 102 = (100 − 49) : 100

⇒ Exact Ratio = 51 : 100

⇒ Approximate Ratio ≈ 50 : 100 = 1 : 2

∴ The correct answer is 1:2.

Alternate Method

Given:

Outer radius (R) = 10 cm

Inner radius (r) = 7 cm

Formula Used:

Area of a circle = πR2

Area of a ring = π(R2 − r2)

R = 10 cm

r = 7 cm

Ring-Shaped Disc

Calculations:

⇒ Area of the outer circle = π × (10)2 = 100π cm2

⇒ Area of the inner circle = π × (7)2 = 49π cm2

⇒ Area of the ring = Area of outer circle − Area of inner circle

⇒ Area of the ring = 100π − 49π = 51π cm2

⇒ Ratio of ring's area to outer circle's area = 51π : 100π

⇒ Exact ratio = 51 : 100

⇒ Approximating 51 to 50 for simplification

⇒ Approximate ratio ≈ 50 : 100 = 1 : 2

∴ The correct answer is 1:2.

Additional Information

Area of a Ring

The region bounded by two concentric circles of outer radius R and inner radius r is called a ring. Its area is calculated using the formula π(R2 − r2).

Concentric Circles

Circles having the same center but different radii are called concentric circles. The perpendicular distance between the boundaries of these circles is uniform and equal to (R − r).

Area of a Sector

The area of a sector of a circle with radius r and central angle θ (in degrees) is calculated using the formula (θ ÷ 360) × πr2.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 23

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

A bicycle wheel having a radius of 35 cm. What percentage of its circumference is covered in a quarter turn? (Use π=22/7)

  1. AOption (a)

    15%

  2. BOption (b)

    25%

  3. COption (c)

    30%

  4. DOption (d)

    35%

View answer & explanation

Correct answer

Answer: (b)B.

25%

Why this answer is correct

Shortcut Trick

The percentage of circumference covered depends solely on the fraction of the turn, making the radius extraneous data.

A full turn (1 complete revolution) always covers exactly 100% of the circumference.

Therefore, a quarter turn covers exactly 1/4 of a full revolution.

Percentage covered = (1 ÷ 4) × 100% = 25%.

∴ The correct answer is 25%.

Alternate Method

Given:

Radius (r) = 35 cm

π = 22/7

Formula Used:

Circumference of a circle = 2 × π × r

Distance covered in a quarter turn = Circumference ÷ 4

90°

r = 35 cm Full Turn (360°) = 100% of Circumference Quarter Turn = 1/4 of Circle = 25% 1/4 Turn = 25% of Circumference

Calculations:

⇒ Total Circumference = 2 × (22/7) × 35

⇒ Total Circumference = 2 × 22 × 5 = 220 cm

⇒ Distance covered in a quarter turn = Total Circumference ÷ 4

⇒ Distance covered = 220 ÷ 4 = 55 cm

⇒ Percentage of circumference covered = (Distance ÷ Total Circumference) × 100

⇒ Percentage = (55 ÷ 220) × 100

⇒ Percentage = (1 ÷ 4) × 100 = 25%

∴ The correct answer is 25%.

Additional Information

Extraneous Data in Math Problems

When a question asks for a ratio, fraction, or percentage of a total, exact numerical measurements (like radius = 35 cm) are often not required to solve the problem and can be bypassed to save time.

Arc Length Formula

The length of an arc is proportional to its central angle: Arc Length = (θ ÷ 360) × 2 × π × r. A quarter turn has an angle of 90°, so the arc length is exactly 1/4 of the full circumference.

Distance and Rotations

The total linear distance covered by a rolling wheel equals the number of rotations multiplied by its circumference (Distance = n × 2 × π × r).

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 24

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

The line y = mx + 5 passes through (1,8). Find m.

  1. AOption (a)

    5

  2. BOption (b)

    4

  3. COption (c)

    3

  4. DOption (d)

    2

View answer & explanation

Correct answer

Answer: (c)C.

3

Why this answer is correct

Shortcut Trick

If a line passes through a given point, the coordinates of the point directly satisfy the equation of the line.

Substitute the point (1, 8) into the line equation y = mx + 5.

⇒ 8 = m × 1 + 5

⇒ m = 8 − 5 = 3

∴ The correct answer is 3.

Alternate Method

Given:

Equation of the straight line: y = mx + 5

The line passes through the point (1, 8).

x y O

(0, 5)

(1, 8)

Δx = 1 Δy = 3

Line: y = mx + 5 y-intercept c = 5 → point (0, 5) Slope m = Δy ÷ Δx = 3 ÷ 1 = 3

Formula Used:

If a point (x, y) lies on a line, its values must satisfy the equation y = mx + c.

Calculations:

⇒ By placing x = 1 and y = 8 in the given equation y = mx + 5:

⇒ 8 = m(1) + 5

⇒ 8 = m + 5

⇒ m = 8 − 5

⇒ m = 3

∴ The correct answer is 3.

Additional Information

Slope-Intercept Form

The equation of a straight line is commonly written as y = mx + c, where m is the slope of the line and c is its y-intercept.

Calculating Slope

The slope m of a line passing through any two distinct points (x1, y1) and (x2, y2) is evaluated as m = (y2 − y1) ÷ (x2 − x1).

Point-Slope Form

If a line has a given slope m and passes through a specific point (x1, y1), its exact equation is found using y − y1 = m(x − x1).

X and Y Intercepts

The x-intercept is the point where the line touches the x-axis (y = 0). The y-intercept is the point where it crosses the y-axis (x = 0).

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 25

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

313+183−373+21031 ^ 3 + 18 ^ 3 - 37 ^ 3 + 210  is equal to:

  1. AOption (a)

    -36810

  2. BOption (b)

    -14820

  3. COption (c)

    -45670

  4. DOption (d)

    -23450

View answer & explanation

Correct answer

Answer: (b)B.

-14820

Why this answer is correct

Shortcut Trick

Apply the Digital Sum (DS) method to solve the equation quickly.

DS of 313 = 43 = 64 ⇒ 1 | DS of 183 = 93 = 729 ⇒ 0.

DS of 373 = 13 = 1 | DS of 210 = 3.

Overall Expression DS = 1 + 0 − 1 + 3 = 3.

Check options: DS of −14820 = −(1+4+8+2) = −15 ⇒ −6. Add 9 to get positive DS: −6 + 9 = 3.

∴ The correct answer is -14820.

Alternate Method

Given:

Expression: 313 + 183 − 373 + 210

Formula Used:

x3 = x × x × x

Calculations:

⇒ 313 = 31 × 31 × 31 = 29791

⇒ 183 = 18 × 18 × 18 = 5832

⇒ 373 = 37 × 37 × 37 = 50653

Substitute the obtained values back into the expression:

⇒ Expression = 29791 + 5832 − 50653 + 210

⇒ Expression = 35623 − 50653 + 210

⇒ Expression = −15030 + 210

⇒ Expression = −14820

∴ The correct answer is -14820.

Additional Information

Simplification Rules

Always follow the proper order of operations (BODMAS). First evaluate powers and exponents before proceeding with addition and subtraction.

Digital Sum Concept

The digital sum of a number is the single-digit sum of its digits. The digital sum of the problem expression must equal the digital sum of its exact solution, making it an excellent verification tool.

Unit Digit Concept

The unit digit of an arithmetic expression relies purely on the unit digits of the numbers involved. For example, 13 + 83 − 73 + 0 yields a unit digit calculation of 1 + 2 − 3 + 0 = 0.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 26

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

Let x = √3 + √5 and y = √8 + √2. Which is greater?

  1. AOption (a)

    x=y

  2. BOption (b)

    x>y

  3. COption (c)

    y>x

  4. DOption (d)

    Cannot be determined

View answer & explanation

Correct answer

Answer: (c)C.

y>x

Why this answer is correct

Shortcut Trick Square both values to compare them directly without estimating decimal roots. x2 = (√3 + √5)2 = 8 + 2√15 y2 = (√8 + √2)2 = 10 + 2√16 = 18 Since √15 < 4, the expression 8 + 2√15 is strictly less than 16. Because 18 > 16, it follows that y2 > x2, which means y > x. ∴ The correct answer is y > x. Alternate Method Given: x = √3 + √5 y = √8 + √2 Formula Used: (a + b)2 = a2 + b2 + 2ab Calculations: ⇒ Square the value of x: ⇒ x2 = (√3 + √5)2 ⇒ x2 = (√3)2 + (√5)2 + 2 × √3 × √5 ⇒ x2 = 3 + 5 + 2√15 ⇒ x2 = 8 + 2√15 ⇒ Simplify and square the value of y: ⇒ y = 2√2 + √2 = 3√2 ⇒ y2 = (3√2)2 = 9 × 2 = 18 ⇒ Compare the squared values: ⇒ We know that √15 is strictly less than √16 (which evaluates to 4). ⇒ So, x2 = 8 + 2√15 < 8 + 2 × 4 ⇒ x2 < 16 ⇒ Since x2 < 16 and y2 = 18, it is completely clear that y2 > x2. ⇒ y > x ∴ The correct answer is y > x. Additional Information Squaring Method to Compare Surds When comparing irrational expressions involving square roots, squaring both sides removes the root operations and provides whole numbers or simplified surds for easier comparison. Simplifying Like Surds Surds that can be reduced to have the exact same irrational part can be added or subtracted directly like standard algebraic terms, for example, √8 + √2 = 2√2 + √2 = 3√2. Comparing Roots of Different Orders For comparing surds of different roots (such as cube roots and square roots), find the LCM of the orders of the roots and raise all surds to that LCM power to compare exact integer values.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 27

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

The number of red, blue, and green marbles in a bag is in the ratio 3 : 4 : 6. If 20 red marbles, 15 blue marbles, and an unknown number of green marbles are added to the bag, the ratio of red, blue, and green marbles becomes 4 : 5 : 7. Determine the number of green marbles added.

  1. AOption (a)

    3

  2. BOption (b)

    5

  3. COption (c)

    7

  4. DOption (d)

    9

View answer & explanation

Correct answer

Answer: (b)B.

5

Why this answer is correct

Shortcut Trick

By applying the cross-multiplication method on Red and Blue marbles:

Initial Ratio → 3 : 4 | Final Ratio → 4 : 5 | Quantity Added → +20, +15

1 initial ratio unit = [(4 × 15) − (5 × 20)] ÷ [(3 × 5) − (4 × 4)] = (60 − 100) ÷ (15 − 16) = 40.

Initial Green marbles = 6 × 40 = 240.

New Red marbles = (3 × 40) + 20 = 140, which represents 4 parts in the final ratio ⇒ 1 final unit = 140 ÷ 4 = 35.

New Green marbles = 7 × 35 = 245 ⇒ Green marbles added = 245 − 240 = 5.

∴ The correct answer is 5.

Alternate Method

Given:

Initial ratio of red, blue, and green marbles = 3 : 4 : 6

Marbles added = 20 red, 15 blue, and an unknown number of green marbles (y).

Final ratio of red, blue, and green marbles = 4 : 5 : 7

Initial RatioRed = 3xBlue = 4xGreen = 6xFinal RatioRed = 4 partsBlue = 5 partsGreen = 7 parts+20+15+y

Calculations:

⇒ Let the initial number of red, blue, and green marbles be 3x, 4x, and 6x respectively.

⇒ After adding 20 red and 15 blue marbles, the new counts become (3x + 20) and (4x + 15).

⇒ Given the new ratio of red to blue is 4 : 5.

⇒ (3x + 20) ÷ (4x + 15) = 4 ÷ 5

⇒ 5 × (3x + 20) = 4 × (4x + 15)

⇒ 15x + 100 = 16x + 60

⇒ 16x − 15x = 100 − 60

⇒ x = 40

⇒ Initial red marbles = 3 × 40 = 120

⇒ Initial green marbles = 6 × 40 = 240

⇒ New red marbles = 120 + 20 = 140

⇒ Let the number of green marbles added be y. The new count of green marbles is (240 + y).

⇒ Given the new ratio of red to green is 4 : 7.

⇒ 140 ÷ (240 + y) = 4 ÷ 7

⇒ 240 + y = (140 × 7) ÷ 4

⇒ 240 + y = 35 × 7 = 245

⇒ y = 245 − 240 = 5

∴ The correct answer is 5.

Additional Information

Ratio and Proportion

A ratio compares two quantities, indicating how many times one number contains another. For quantities a and b, it is represented as a : b or the fraction a/b.

Cross-Multiplication Method in Ratios

When a given ratio a : b changes to c : d by adding quantities x and y respectively, the value of 1 initial ratio unit is directly obtained using the formula (cy − dx) ÷ (ad − bc).

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 28

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

Evaluate the continued fraction: x=2+13+14+12x = 2 + \frac{1}{3+\frac{1}{4+\frac{1}{2}}}

  1. AOption (a)

    6729 \frac{67}{29}

  2. BOption (b)

    4117 \frac{41}{17 }

  3. COption (c)

    4519 \frac{45}{19}

  4. DOption (d)

    4720 \frac{47}{20}

View answer & explanation

Correct answer

Answer: (a)A.

6729 \frac{67}{29}

Why this answer is correct

Shortcut Trick

For continued fractions, solve directly from the bottommost denominator to the top.

Bottom fraction: 4 + 1/2 = 9/2

Middle fraction: 3 + 2/9 = 29/9

Top fraction: 2 + 9/29 = 67/29

∴ The correct answer is 67/29.

Alternate Method

Given:

x = 2 + 1 / [3 + 1 / (4 + 1/2)]

Calculations:

Solve the innermost fraction first (bottom-up approach):

⇒ 4 + 1/2 = (8 + 1) / 2 = 9/2

Substitute this back into the expression:

⇒ x = 2 + 1 / [3 + 1 / (9/2)]

⇒ x = 2 + 1 / (3 + 2/9)

Now, simplify the next denominator:

⇒ 3 + 2/9 = (27 + 2) / 9 = 29/9

Substitute this back into the expression:

⇒ x = 2 + 1 / (29/9)

⇒ x = 2 + 9/29

⇒ x = (58 + 9) / 29

⇒ x = 67/29

∴ The correct answer is 67/29.

Additional Information

BODMAS Rule in Fractions

Always follow the order of Brackets, Orders, Division, Multiplication, Addition, and Subtraction. In continued fractions, treat the bottommost denominator as an implicit bracket.

Reciprocal of a Fraction

Dividing by a fraction is the same as multiplying by its reciprocal. For example, 1 ÷ (a/b) = b/a.

Continued Fractions

A continued fraction is evaluated through a step-by-step upward process of simplifying the lowest mixed fraction and reciprocating it.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 29

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

A and B invest ₹40,000 and ₹60,000 respectively in a business. What is A's share in a profit of ₹50,000?

  1. AOption (a)

    ₹15,000

  2. BOption (b)

    ₹20,000

  3. COption (c)

    ₹25,000

  4. DOption (d)

    ₹30,000

View answer & explanation

Correct answer

Answer: (b)B.

₹20,000

Why this answer is correct

Shortcut Trick

Profit Ratio = Capital Ratio (since time is constant) = 40000 : 60000 = 2 : 3

Total profit ratio parts = 2 + 3 = 5 parts

5 parts = ₹50,000 ⇒ 1 part = ₹10,000

A's share (2 parts) = 2 × 10,000 = ₹20,000

∴ The correct answer is ₹20,000.

Alternate Method

Given:

A's investment = ₹40,000

B's investment = ₹60,000

Total Profit = ₹50,000

Formula Used:

Profit Ratio = (Capital of A × Time of A) : (Capital of B × Time of B)

Capital & Profit Distribution

Total Profit = ₹50,000 (5 units)

A ₹40,000

B ₹60,000

Ratio: 2 Units Ratio: 3 Units

A's Share = 2/5 × ₹50,000 = ₹20,000

Calculations:

⇒ Since the investment time is exactly the same for both A and B, their profit ratio will be directly equal to their investment ratio.

⇒ Profit Ratio = 40000 : 60000

⇒ Profit Ratio = 4 : 6 = 2 : 3

⇒ Total profit parts = 2 + 3 = 5

⇒ A's share = (2 ÷ 5) × Total Profit

⇒ A's share = (2 ÷ 5) × 50000

⇒ A's share = 2 × 10000 = 20000

∴ The correct answer is ₹20,000.

Additional Information

Simple Partnership

When partners invest their capital for the exact same time period, all final profits or losses are distributed in the direct ratio of their original investments.

Compound Partnership

When partners invest for varying time periods, their profit distribution ratio is calculated as (Capital1 × Time1) : (Capital2 × Time2).

Working vs Sleeping Partner

A working partner manages the business operations and generally receives a fixed salary or percentage of the profit before the remaining balance is divided among all partners based on their respective investments.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 30

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

A and B contribute to a business investment in the ratio 3:5. They earn a profit of ₹64,000. A is a working partner and receives 10% of the profit extra. What is A's total share?

  1. AOption (a)

    ₹26,500

  2. BOption (b)

    ₹28,000

  3. COption (c)

    ₹38,500

  4. DOption (d)

    ₹36,000

View answer & explanation

Correct answer

Answer: (b)B.

₹28,000

Why this answer is correct

Shortcut Trick

Total share of A = (Salary as working partner) + (Investment share of remaining profit)

⇒ A's total share = 10% of 64000 + (3 ÷ 8) × 90% of 64000

⇒ A's total share = 6400 + (3 ÷ 8) × 57600

⇒ A's total share = 6400 + 21600 = 28000

∴ The correct answer is ₹28,000.

Alternate Method

Given:

Total profit = ₹64,000

Investment ratio of A and B = 3 : 5

A's extra share (working partner) = 10% of total profit

Total Profit: ₹64,000A's 10% Extra₹6,400Remaining 90%₹57,600A (3 parts)₹21,600B (5 parts)₹36,000A's Total = ₹28,000

Calculations:

⇒ A's extra profit = 10% of ₹64,000

⇒ A's extra profit = (10 ÷ 100) × 64,000 = ₹6,400

⇒ Remaining profit = 64,000 − 6,400 = ₹57,600

⇒ Total ratio parts = 3 + 5 = 8 parts

⇒ A's share from remaining profit = (3 ÷ 8) × 57,600

⇒ A's share from remaining profit = 3 × 7,200 = ₹21,600

⇒ A's total share = Extra profit + Investment share

⇒ A's total share = 6,400 + 21,600 = ₹28,000

∴ The correct answer is ₹28,000.

Additional Information

Active vs Sleeping Partner

An active partner contributes both capital and time to manage the business, often receiving an extra salary or fixed percentage of profit before the remainder is distributed. A sleeping partner only contributes capital.

Profit Distribution Formula

Profit ratio is directly proportional to the product of invested capital and time period. Profit Ratio = (C1 × T1) : (C2 × T2).

Equivalent Capital

When investments are made for unequal times, we calculate the equivalent capital for 1 unit of time. Equivalent capital = Investment × Time invested.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 31

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

In a team of 75 employees, the average monthly sales per employee is ₹ 1.8 lakh. If 30 junior employees average ₹ 1.2 lakh each, what is the average monthly sales figure (in ₹ lakh) for the senior employees?

  1. AOption (a)

    ₹4.2 lakh

  2. BOption (b)

    ₹2.2 lakh

  3. COption (c)

    ₹2.8 lakh

  4. DOption (d)

    ₹2.7 lakh

View answer & explanation

Correct answer

Answer: (b)B.

₹2.2 lakh

Why this answer is correct

Shortcut Trick

Using the Net Deviation Method:

Ratio of Junior to Senior employees = 30 : 45 = 2 : 3

Junior group deviation from overall average = 1.2 − 1.8 = −0.6 lakh

Total deviation of Junior parts = 2 × (−0.6) = −1.2 lakh

To balance the mean, total deviation of 3 Senior parts must be +1.2 lakh ⇒ per part = +0.4 lakh

Average sales of Senior employees = Overall average + Senior deviation = 1.8 + 0.4 = 2.2 lakh

∴ The correct answer is ₹2.2 lakh.

Alternate Method

Given:

Total employees = 75

Overall average monthly sales = ₹1.8 lakh

Junior employees = 30

Average sales of junior employees = ₹1.2 lakh

Formula Used:

Total Value = Average × Number of Entities

1.2 Junior (₹ lakh)

x Senior (₹ lakh)

1.8 Overall Avg

x − 1.8 Emp: 30 (Ratio 2)

0.6 Emp: 45 (Ratio 3)

Calculations:

⇒ Number of senior employees = 75 − 30 = 45

⇒ Total sales of all employees = 75 × 1.8 = 135 lakh

⇒ Total sales of junior employees = 30 × 1.2 = 36 lakh

⇒ Total sales of senior employees = 135 − 36 = 99 lakh

⇒ Average sales of senior employees = 99 ÷ 45

⇒ Average sales of senior employees = 2.2 lakh

∴ The correct answer is ₹2.2 lakh.

Additional Information

Weighted Average

When combining two different groups, the overall average is calculated using the formula: A = (n1x1 + n2x2) ÷ (n1 + n2), where 'n' represents the number of elements and 'x' represents their respective averages.

Method of Deviation

This method assumes a central mean value and calculates the difference (deviation) of all group components from it. For any perfectly balanced average, the net sum of all weighted deviations must always be zero.

Alligation Cross Rule

A visual shortcut applied when two differently priced or valued sets are mixed to get a final average. The ratio of the quantities mixed is always given by (Higher Average − Target Average) : (Target Average − Lower Average).

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 32

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

The price of petrol shot up by 5%. Before the hike, the price was Rs. 82 per litre. A man travels 3045 km every month and his car gives a mileage of 15 km per litre. What is the increase in the monthly expenditure (to the nearest Rs.) on the man's travel due to the hike in the petrol prices?

  1. AOption (a)

    ₹820

  2. BOption (b)

    ₹832

  3. COption (c)

    ₹845

  4. DOption (d)

    ₹850

View answer & explanation

Correct answer

Answer: (b)B.

₹832

Why this answer is correct

Shortcut Trick

Total fuel consumed = Total distance ÷ Mileage = 3045 ÷ 15 = 203 litres

Increase in price per litre = 5% of ₹82 = ₹4.1

Total increase in expenditure = 203 × 4.1 = ₹832.3 ≈ ₹832

∴ The correct answer is ₹832.

Alternate Method

Given:

Initial price of petrol = ₹82 per litre

Price hike = 5%

Monthly distance travelled = 3045 km

Car mileage = 15 km/l

Petrol Consumed 3045 km ÷ 15 km/l = 203 litres

Price Increase 5% of ₹82 = ₹4.1 / litre

Total Increase 203 × 4.1 = ₹832.3 ≈ ₹832

Calculations:

⇒ Total petrol consumed per month = Total Distance ÷ Mileage

⇒ Total petrol consumed = 3045 ÷ 15 = 203 litres

⇒ Increase in petrol price = 5% of 82 = (5 ÷ 100) × 82 = ₹4.1 per litre

⇒ Total increase in monthly expenditure = Total petrol consumed × Increase per litre

⇒ Total increase = 203 × 4.1 = ₹832.3

⇒ Rounding to the nearest rupee, the increase is ₹832.

∴ The correct answer is ₹832.

Additional Information

Percentage Change

The formula to calculate the absolute change is: Change = Original Value × (Percentage Change ÷ 100).

Fuel and Distance Relationship

Total fuel required = Total Distance ÷ Fuel Efficiency (Mileage). Total cost is evaluated by multiplying the fuel required by the cost per unit of fuel.

Rounding Values

If the decimal part is less than 0.5, round down to the nearest integer. If it is 0.5 or greater, round up to the next integer.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 33

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

A company placed an order for 20 high-end laptops and some quantity of standard desktops. The price of a high-end laptop was 5 times that of a standard desktop. Due to a mistake, the number of laptops and desktops was interchanged in the delivery. This caused the total bill to increase by 80%. What was the ratio of the number of high-end laptops to standard desktops in the original order?

  1. AOption (a)

    1:2

  2. BOption (b)

    2:1

  3. COption (c)

    5:2

  4. DOption (d)

    2:5

View answer & explanation

Correct answer

Answer: (d)D.

2:5

Why this answer is correct

Shortcut Trick

Let desktop price = 1, laptop price = 5, and the number of desktops ordered = x.

Original bill = 20 × 5 + x × 1 = 100 + x

New interchanged bill = x × 5 + 20 × 1 = 5x + 20

Given the new bill is 80% more, so (5x + 20) = 1.8 × (100 + x)

⇒ 5x + 20 = 180 + 1.8x ⇒ 3.2x = 160 ⇒ x = 50

∴ The correct answer is 2:5.

Alternate Method

Original OrderLaptops = 20Desktops = xBill = 100 + x+80%New OrderLaptops = xDesktops = 20Bill = 5x + 20Required Ratio = 20 : x

Given:

⇒ Number of high-end laptops originally ordered = 20

⇒ Let the number of standard desktops originally ordered = x

⇒ Price of a high-end laptop = 5 × Price of a standard desktop

Formula Used:

⇒ Total Bill = (Quantity of Laptops × Price of Laptop) + (Quantity of Desktops × Price of Desktop)

⇒ New Bill = Original Bill × (100% + Increase%)

Calculations:

⇒ Let the price of a standard desktop be 1 unit. Then the price of a high-end laptop is 5 units.

⇒ Original Bill = (20 × 5) + (x × 1) = 100 + x

⇒ Due to the mistake, the new quantity of laptops = x, and desktops = 20.

⇒ New Bill = (x × 5) + (20 × 1) = 5x + 20

⇒ It is given that the new bill is 80% higher than the original bill. This means the new bill is 180% of the original bill.

⇒ 5x + 20 = (100 + x) × (180 ÷ 100)

⇒ 5x + 20 = (100 + x) × 1.8

⇒ 5x + 20 = 180 + 1.8x

⇒ 5x − 1.8x = 180 − 20

⇒ 3.2x = 160

⇒ x = 160 ÷ 3.2

⇒ x = 50

⇒ Number of standard desktops originally ordered = 50

⇒ Required Ratio = (Number of High-End Laptops) : (Number of Standard Desktops) = 20 : 50 = 2 : 5

∴ The correct answer is 2:5.

Additional Information

Interchange of Quantities Rule

When the quantities of two items are interchanged, the change in total value is equal to the product of the difference in their unit prices and the difference in their original quantities: Change in Value = (P1 − P2) × (Q2 − Q1).

Percentage Increase Formula

The percentage increase between two values is calculated as: % Increase = [(New Value − Original Value) ÷ Original Value] × 100.

Base Value in Percentages

Whenever a value increases by R%, the new value becomes Original Value × (1 + R/100). Identifying the correct base value (the original order bill) is crucial to avoid calculation errors.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 34

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

Find the compound interest on ₹8,000 at 12% per annum for 3 years 4 months, compounded annually.

  1. AOption (a)

    ₹3,000

  2. BOption (b)

    ₹4,000

  3. COption (c)

    ₹3,689

  4. DOption (d)

    ₹3,600

View answer & explanation

Correct answer

Answer: (c)C.

₹3,689

Why this answer is correct

Shortcut Trick

Using the multiplying factor method for Compound Interest:

Rate for 3 full years = 12% (Multiplier = 1.12)

Rate for next 4 months = 12% × (4/12) = 4% (Multiplier = 1.04)

Amount = P × (1.12)3 × (1.04)

Amount = 8000 × 1.404928 × 1.04 ≈ 11689

Compound Interest = 11689 − 8000 = 3689

∴ The correct answer is ₹3,689.

Alternate Method

Given:

Principal (P) = ₹8,000

Rate (R) = 12% per annum

Time (T) = 3 years 4 months = 3 + 4/12 years = 10/3 years

Formula Used:

A = P × (1 + R/100)n × (1 + r/100)

Where n is the full years and r is the rate for the fractional year.

Start ₹8000

+12% (1Yr)

Year 1 ₹8960

+12% (1Yr)

Year 2 ₹10035.2

+12% (1Yr)

Year 3 ₹11239.4

+4% (4Mo)

3 Yr 4 Mo ₹11689

Total CI = ₹11689 − ₹8000 = ₹3689

Calculations:

⇒ Rate for the fractional 4 months (1/3 year) = 12% × 1/3 = 4%

⇒ Amount (A) = 8000 × (1 + 12/100)3 × (1 + 4/100)

⇒ A = 8000 × (112/100)3 × (104/100)

⇒ A = 8000 × 1.12 × 1.12 × 1.12 × 1.04

⇒ A = 8000 × 1.404928 × 1.04

⇒ A = 11239.424 × 1.04 = 11689.00

⇒ Compound Interest (CI) = A − P

⇒ CI = 11689 − 8000 = 3689

∴ The correct answer is ₹3,689.

Additional Information

Fractional Year Compounding

When time is in fractions (e.g., n years and m months), apply the full compounding rate for complete years and a proportionally reduced rate for the remaining fractional year.

Effective Compound Interest Rate

For two consecutive periods at rates x% and y%, the overall effective compound interest percentage can be calculated using the successive percentage formula: x + y + (x × y)/100.

Difference Between CI and SI

For 2 years, the difference between compound and simple interest is CI − SI = P(R/100)2. For 3 years, the difference is CI − SI = P(R/100)2(3 + R/100).

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 35

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

A candy shop owner buys three kinds of candies: red, blue, and green. Red candies are purchased at 3 for ₹15, blue candies at 4 for ₹18, and green candies at 5 for ₹22. He mixes them in the ratio 1 : 1 : 2. He sells all the mixed candies at 2 for ₹10. What is his approximate gain or loss percentage?

  1. AOption (a)

    Loss of 9.2%

  2. BOption (b)

    Profit of 9.29%

  3. COption (c)

    Profit of 10%

  4. DOption (d)

    Loss of 10%

View answer & explanation

Correct answer

Answer: (b)B.

Profit of 9.29%

Why this answer is correct

Shortcut Trick

CP per candy: Red = 15 ÷ 3 = ₹5, Blue = 18 ÷ 4 = ₹4.5, Green = 22 ÷ 5 = ₹4.4

Total CP for 4 mixed candies (in ratio 1:1:2) = 5 + 4.5 + (2 × 4.4) = ₹18.3

SP for 4 mixed candies = 4 × (10 ÷ 2) = ₹20

Profit = 20 − 18.3 = ₹1.7

Profit % = (1.7 ÷ 18.3) × 100 ≈ 9.29%

∴ The correct answer is Profit of 9.29%.

Alternate Method

Candy Mixture (1 : 1 : 2 Ratio)

1 Red @ ₹5

1 Blue @ ₹4.5

2 Green @ ₹4.4

Total CP = ₹18.3

  • Profit ₹1.7

Total SP = ₹20

Profit % = (1.7 ÷ 18.3) × 100 ≈ 9.29%

Given:

Red candies bought at 3 for ₹15

Blue candies bought at 4 for ₹18

Green candies bought at 5 for ₹22

Mixture ratio = 1 : 1 : 2

Selling price = 2 for ₹10

Calculations:

⇒ Let the number of Red, Blue, and Green candies mixed be 1, 1, and 2 respectively.

⇒ Total candies mixed = 1 + 1 + 2 = 4 candies

⇒ Cost Price (CP) of 1 Red candy = 15 ÷ 3 = ₹5

⇒ Cost Price (CP) of 1 Blue candy = 18 ÷ 4 = ₹4.5

⇒ Cost Price (CP) of 1 Green candy = 22 ÷ 5 = ₹4.4

⇒ Total Cost Price of 4 mixed candies = (1 × 5) + (1 × 4.5) + (2 × 4.4)

⇒ Total CP = 5 + 4.5 + 8.8 = ₹18.3

⇒ Selling Price (SP) of 1 candy = 10 ÷ 2 = ₹5

⇒ Total SP of 4 mixed candies = 4 × 5 = ₹20

⇒ Since SP is greater than CP, there is a profit.

⇒ Profit = SP − CP = 20 − 18.3 = ₹1.7

⇒ Profit % = (Profit ÷ Total CP) × 100

⇒ Profit % = (1.7 ÷ 18.3) × 100

⇒ Profit % = 170 ÷ 18.3 ≈ 9.289% ≈ 9.29%

∴ The correct answer is Profit of 9.29%.

Additional Information

Profit & Loss Formulas

Profit occurs when Selling Price (SP) > Cost Price (CP). Profit = SP − CP. Profit Percentage is calculated on the total Cost Price: (Profit ÷ CP) × 100.

Mixture Average Cost

When different quantities of items are mixed, the Average Cost Price is simply the total cost of all components combined, divided by the total physical quantity.

Ratio Multipliers Trick

Instead of finding a large common multiple (LCM) to equate quantities, directly taking sample quantities equal to the given ratio (such as 1, 1, and 2 units) makes percentage-based calculations perfectly proportional and much faster.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 36

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

The ratio of boys to girls in a school is 7:5. If 20 more girls join the school, the new ratio of boys to girls becomes 7:6. What is the total number of students in the school initially?

  1. AOption (a)

    200

  2. BOption (b)

    240

  3. COption (c)

    300

  4. DOption (d)

    360

View answer & explanation

Correct answer

Answer: (b)B.

240

Why this answer is correct

Shortcut Trick

Initial Ratio of Boys : Girls = 7 : 5 and Final Ratio = 7 : 6

Since the number of boys remains the same, the ratio part for boys (7) is unchanged.

Increase in girls' ratio = 6 − 5 = 1 unit.

This 1 unit increase corresponds to the 20 girls who joined.

Total initial students = 7 + 5 = 12 units = 12 × 20 = 240.

∴ The correct answer is 240 students.

Alternate Method

Given:

Initial ratio of Boys : Girls = 7 : 5

Number of girls added = 20

Final ratio of Boys : Girls = 7 : 6

Formula Used:

If a ratio is given as a:b, the actual values can be taken as ax and bx.

Initial (7:5) Final (7:6)

Boys (7 units)

Girls (5 units)

Boys (7 units)

Girls (5 units)

+1

  • 20 Girls

Calculations:

⇒ Let the initial number of boys be 7x and girls be 5x.

⇒ Total initial students = 7x + 5x = 12x.

⇒ After 20 girls join, the new number of girls = 5x + 20.

⇒ According to the given condition, 7x ÷ (5x + 20) = 7 ÷ 6

⇒ Cross-multiplying: 6 × 7x = 7 × (5x + 20)

⇒ 42x = 35x + 140

⇒ 42x − 35x = 140

⇒ 7x = 140

⇒ x = 20

⇒ Total initial students = 12x = 12 × 20 = 240

∴ The correct answer is 240 students.

Additional Information

Ratio Balancing Method

When one quantity changes while the other remains constant, ensure the ratio parts of the constant quantity are made equal before directly equating the change in units.

Compound Ratio

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

Basic Proportion Property

If four quantities are in proportion such that a:b = c:d, then their cross products are equal (a × d = b × c).

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 37

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

A, B, and C are capable of completing a job in 30, 45, and 90 days, respectively. A works every day, while B and C join A every fifth day. How long will it take to finish the task?

  1. AOption (a)

    25 days

  2. BOption (b)

    18 days

  3. COption (c)

    16 days

  4. DOption (d)

    12 days

View answer & explanation

Correct answer

Answer: (a)A.

25 days

Why this answer is correct

Shortcut Trick

Total work = LCM(30, 45, 90) = 90 units.

Efficiencies: A = 3, B = 2, C = 1.

Work done in a 5-day cycle = (3 × 5) + 2 + 1 = 18 units.

Number of cycles = Total work ÷ Work per cycle = 90 ÷ 18 = 5 cycles.

Total days = 5 cycles × 5 days/cycle = 25 days.

∴ The correct answer is 25 days.

Alternate Method

Given:

Time taken by A = 30 days

Time taken by B = 45 days

Time taken by C = 90 days

PersonDaysTotal WorkEfficiency
A30903
B452
C901

Calculations:

⇒ In 1 cycle of 5 days, A works for all 5 days, while B and C work only on the 5th day.

⇒ Work done by A in 5 days = 5 × 3 = 15 units

⇒ Work done by B and C on the 5th day = 2 + 1 = 3 units

⇒ Total work done in 5 days = 15 + 3 = 18 units

⇒ Number of 5-day cycles required to complete 90 units = 90 ÷ 18 = 5 cycles

⇒ Total time taken = 5 cycles × 5 days/cycle = 25 days

∴ The correct answer is 25 days.

Additional Information

Total Work Concept

Total work is considered as the Least Common Multiple (LCM) of the days taken by individuals. Efficiency is the work done per unit time (Total Work ÷ Days).

Alternate or Cyclic Work

Find the work done in one complete cycle. Divide the total work by the work per cycle to find the number of cycles, then multiply by days per cycle to find the total time required.

Pipes and Cisterns

Works exactly on the same principle as Time and Work, but pipes that empty the tank are assigned a negative efficiency value.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 38

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

A 100-litre solution contains acid and water in the ratio 3:2. Some quantity of this solution is removed and replaced with pure acid. If the final ratio of acid to water becomes 7:3, how many litres of solution were replaced?

  1. AOption (a)

    35 litres

  2. BOption (b)

    25 litres

  3. COption (c)

    50 litres

  4. DOption (d)

    20 litres

View answer & explanation

Correct answer

Answer: (b)B.

25 litres

Why this answer is correct

Shortcut Trick

Apply the alligation method on the water concentration of the mixtures.

Initial mixture water = 40%, Added pure acid water = 0%, Final mixture water = 30%.

Ratio of remaining mixture to added pure acid = (30 − 0) : (40 − 30) = 30 : 10 = 3 : 1.

This means 1 part out of a total of (3 + 1) = 4 parts was replaced.

Replaced quantity = 1 / 4 of total volume = (1 / 4) × 100 = 25 litres.

∴ The correct answer is 25 litres.

Alternate Method

Given:

Initial volume of the solution = 100 litres

Initial ratio of acid to water = 3 : 2

Final ratio of acid to water = 7 : 3

Calculations:

⇒ Quantity of water in the initial solution = (2 / 5) × 100 = 40 litres

⇒ Let x litres of the solution be removed and replaced with pure acid.

⇒ Quantity of water removed = (2 / 5) × x = 0.4x litres

⇒ Remaining water in the solution = 40 − 0.4x litres

⇒ Since only pure acid is added back, the overall quantity of water in the final mixture does not increase.

⇒ Final volume of the solution is maintained at 100 litres.

⇒ Quantity of water in the final solution = (3 / 10) × 100 = 30 litres

⇒ Equating the remaining water to final water:

⇒ 40 − 0.4x = 30

⇒ 0.4x = 10

⇒ x = 10 ÷ 0.4 = 25 litres

∴ The correct answer is 25 litres.

Additional Information

Alligation Rule

It is a quick method used to find the ratio in which two ingredients at their given prices or concentrations must be mixed to produce a mixture of a desired average concentration.

Tracking the Untouched Component

When solving replacement questions, always frame your equation using the component that is NOT being added from the outside. In this case, since pure acid is added back, it is much faster to track the changing quantity of water.

Replacement Formula

If a container initially holds V units of pure liquid and x units are taken out and replaced by water n times sequentially, the final quantity of pure liquid remaining is given by V × (1 − x/V)n.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 39

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

The ratio of the efficiencies of two workers, P and Q, is 5 : 2. If P can complete a project in 12 days, how many days will Q take to complete the same project alone?

  1. AOption (a)

    24 days

  2. BOption (b)

    30 days

  3. COption (c)

    35 days

  4. DOption (d)

    40 days

View answer & explanation

Correct answer

Answer: (b)B.

30 days

Why this answer is correct

Shortcut Trick

Time and efficiency are inversely proportional to each other.

Ratio of Time taken by P and Q = 2 : 5

Here, 2 parts correspond to 12 days.

So, 1 part = 12 ÷ 2 = 6 days.

Time taken by Q (5 parts) = 5 × 6 = 30 days.

∴ The correct answer is 30 days.

Alternate Method

Given:

Ratio of the efficiencies of P and Q = 5 : 2

Time taken by P to complete the project = 12 days

Formula Used:

Total Work = Efficiency × Time

Time = Total Work ÷ Efficiency

PersonDaysTotal WorkEfficiency
P12605
Q60 ÷ 2 = 302

Calculations:

⇒ Let the efficiency of P be 5 units/day and Q be 2 units/day.

⇒ Total Work = Efficiency of P × Time taken by P

⇒ Total Work = 5 × 12 = 60 units

⇒ Time taken by Q = Total Work ÷ Efficiency of Q

⇒ Time taken by Q = 60 ÷ 2 = 30 days

∴ The correct answer is 30 days.

Additional Information

Time and Efficiency Relationship

Efficiency is inversely proportional to the time taken to complete a work. If the ratio of efficiencies of A and B is x : y, then the ratio of time taken by them is y : x.

Combined Work Rule

If A can do a piece of work in x days and B can do it in y days, their combined efficiency is (1/x + 1/y). Together they complete the work in (x × y) ÷ (x + y) days.

Work and Wages

Wages are distributed in proportion to the total work done by each person. If workers work for the exact same number of days, wages are distributed in the direct ratio of their efficiencies.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 2) (Previous Year Paper)·Quantitative Aptitude
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Question 40

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

A car starts from point P on a circular track and an SUV starts from point Q, which is 600 meters ahead of P in the direction of motion. The car's speed is 15 m/s, and the SUV's speed is 10 m/s. The circumference of the track is 1.5 km. How much distance will the car have traveled when it first overtakes the SUV?

  1. AOption (a)

    900 m

  2. BOption (b)

    1200 m

  3. COption (c)

    1500 m

  4. DOption (d)

    1800 m

View answer & explanation

Correct answer

Answer: (d)D.

1800 m

Why this answer is correct

Shortcut Trick

Relative speed of Car with respect to SUV = 15 − 10 = 5 m/s.

Time required to close the initial gap of 600 m = 600 ÷ 5 = 120 seconds.

Distance traveled by car in this time = Speed × Time = 15 × 120 = 1800 m.

∴ The correct answer is 1800 m.

Alternate Method

Point P Car (15 m/s)

Point Q SUV (10 m/s)

Meeting

Gap = 600 m

SUV dist = 1200 m

Car distance = 1800 m

Given:

⇒ Speed of car (SC) = 15 m/s

⇒ Speed of SUV (SS) = 10 m/s

⇒ Initial distance between them in direction of motion (d) = 600 m

⇒ Track circumference = 1.5 km = 1500 m (Extra information)

Formula Used:

⇒ Time taken to overtake = Distance ÷ Relative Speed

⇒ Distance = Speed × Time

Calculation:

⇒ Since both vehicles are moving in the same direction, Relative Speed = SC − SS

⇒ Relative Speed = 15 − 10 = 5 m/s

⇒ Time taken to overtake = 600 ÷ 5 = 120 seconds

⇒ Distance traveled by car = SC × Time taken

⇒ Distance = 15 × 120 = 1800 m

∴ The correct answer is 1800 m.

Additional Information

Relative Speed (Same Direction)

When two bodies move in the same direction, their relative speed is the difference of their individual speeds: SR = S1 − S2. The distance required to overtake is determined by the initial gap between them.

Relative Speed (Opposite Direction)

When two bodies move in opposite directions (e.g., towards each other), their relative speed is the sum of their individual speeds: SR = S1 + S2.

Circular Track Properties

The circumference of the track is only needed when finding the time taken for the vehicles to meet again at the starting point, or for a subsequent overtake cycle (which requires covering an extra 1 full lap).

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.