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 1

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

Arrange the fractions 59,47,35\frac{5}9, \frac{4}7, \frac{3}5 and, 23\frac{2}3  in ascending order.

  1. AOption (a)

    35,47,23,59\frac{3}{5}, \frac{4}{7}, \frac{2}{3}, \frac{5}{9}

  2. BOption (b)

    59,47,35,23\frac{5}{9}, \frac{4}{7}, \frac{3}{5}, \frac{2}{3}

  3. COption (c)

    23,35,47,59 \frac{2}{3}, \frac{3}{5}, \frac{4}{7}, \frac{5}{9}

  4. DOption (d)

    47,59,23,35\frac{4}7, \frac{5}9, \frac{2}3, \frac{3}5

View answer & explanation

Correct answer

Answer: (b)B.

59,47,35,23\frac{5}{9}, \frac{4}{7}, \frac{3}{5}, \frac{2}{3}

Why this answer is correct

Shortcut Trick

Convert the fractions into percentages to easily compare their values.

5/9 ≈ 55.55%, 4/7 ≈ 57.14%, 3/5 = 60%, 2/3 ≈ 66.67%

Arranging the percentages in ascending order: 55.55% < 57.14% < 60% < 66.67%

This exactly corresponds to the fractions: 5/9 < 4/7 < 3/5 < 2/3

∴ The correct answer is 5/9, 4/7, 3/5, 2/3.

Alternate Method

Given:

Fractions to arrange: 5/9, 4/7, 3/5, 2/3

Method Used:

LCM Equalization Method

Comparing Fractions (Percentage Value) 5/9 55.55% 4/7 57.14% 3/5 60.00% 2/3 66.67%

Calculations:

⇒ LCM of denominators (9, 7, 5, 3) = 9 × 7 × 5 = 315

⇒ Convert each fraction so that the denominator becomes 315:

⇒ 5/9 = (5 × 35) ÷ (9 × 35) = 175/315

⇒ 4/7 = (4 × 45) ÷ (7 × 45) = 180/315

⇒ 3/5 = (3 × 63) ÷ (5 × 63) = 189/315

⇒ 2/3 = (2 × 105) ÷ (3 × 105) = 210/315

⇒ Comparing the numerators: 175 < 180 < 189 < 210

⇒ This implies: 175/315 < 180/315 < 189/315 < 210/315

⇒ Resulting ascending order: 5/9 < 4/7 < 3/5 < 2/3

∴ The correct answer is 5/9, 4/7, 3/5, 2/3.

Additional Information

Cross-Multiplication Method

To compare two fractions a/b and c/d, cross-multiply to get a×d and b×c. If a×d < b×c, then the first fraction is smaller (a/b < c/d).

Decimal Conversion Method

Convert each fraction to its decimal equivalent by simply dividing the numerator by the denominator. Compare the resulting decimal numbers up to 2 to 3 decimal places for accurate results.

Difference Method

When the difference between the numerator and denominator is the exact same for a set of proper fractions, the fraction with the largest numerator is strictly the greatest.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 2

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

Simplify: (2122\frac{1}2 + 3.6) – 1.9

  1. AOption (a)

    4.2

  2. BOption (b)

    5.2

  3. COption (c)

    6.2

  4. DOption (d)

    7.2

View answer & explanation

Correct answer

Answer: (a)A.

4.2

Why this answer is correct

Shortcut Trick

Apply the BODMAS rule and convert the fraction to a decimal directly.

Expression: (2½ + 3.6) − 1.9

⇒ (2.5 + 3.6) − 1.9

⇒ 6.1 − 1.9 = 4.2

∴ The correct answer is 4.2.

Alternate Method

Given:

Expression: (2½ + 3.6) − 1.9

Method Used:

BODMAS Rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction)

Calculations:

⇒ First, convert the mixed fraction into a decimal:

⇒ 2½ = 2.5

⇒ Substitute this value back into the original expression:

⇒ (2.5 + 3.6) − 1.9

⇒ According to the BODMAS rule, perform the addition inside the brackets first:

⇒ 2.5 + 3.6 = 6.1

⇒ Finally, subtract 1.9 from the sum obtained:

⇒ 6.1 − 1.9 = 4.2

∴ The correct answer is 4.2.

Additional Information

BODMAS Rule

It defines the correct order of operations to solve an expression: Brackets, Orders (powers/roots), Division, Multiplication, Addition, Subtraction.

Fraction to Decimal Conversion

Common fractions should be converted to decimals to make addition and subtraction easier (e.g., 1/2 = 0.5, 1/4 = 0.25).

Decimal Arithmetic

Always align the decimal points vertically when adding or subtracting numbers to ensure correct place value calculation.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 3

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

Evaluate: 7 \frac{1}{4} - \left[ \frac{5}{6} \div \left{ \frac{1}{3} - \left( \frac{1}{2} \times \left( \frac{3}{4} - \frac{1}{2} \right) \right) \right} \right]

  1. AOption (a)

    −314-3\frac{1}4

  2. BOption (b)

    3143\frac{1}4

  3. COption (c)

    −234-2\frac{3}4

  4. DOption (d)

    2342\frac{3}4

View answer & explanation

Correct answer

Answer: (b)B.

3143\frac{1}4

Why this answer is correct

Shortcut Trick

Using the BODMAS rule, evaluate from the innermost brackets outwards.

Inner terms: (3/4 − 1/2) = 1/4 ⇒ 1/2 × 1/4 = 1/8

Curly brackets: {1/3 − 1/8} = 8/24 − 3/24 = 5/24

Square brackets: [5/6 ÷ 5/24] = 5/6 × 24/5 = 4

Final step: 7 1/4 − 4 = 3 1/4

∴ The correct answer is 3 1/4.

Alternate Method

Given:

Expression: 7 1/4 − [ 5/6 ÷ { 1/3 − ( 1/2 × ( 3/4 − 1/2 ) ) } ]

Concept Used:

BODMAS Rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction)

Calculations:

⇒ 29/4 − [ 5/6 ÷ { 1/3 − ( 1/2 × ( 3/4 − 2/4 ) ) } ]

⇒ 29/4 − [ 5/6 ÷ { 1/3 − ( 1/2 × 1/4 ) } ]

⇒ 29/4 − [ 5/6 ÷ { 1/3 − 1/8 } ]

⇒ 29/4 − [ 5/6 ÷ { 8/24 − 3/24 } ]

⇒ 29/4 − [ 5/6 ÷ 5/24 ]

⇒ 29/4 − [ 5/6 × 24/5 ]

⇒ 29/4 − 24/6

⇒ 29/4 − 4

⇒ 29/4 − 16/4

⇒ 13/4 = 3 1/4

∴ The correct answer is 3 1/4.

Additional Information

BODMAS Rule

BODMAS stands for Brackets, Orders, Division, Multiplication, Addition, and Subtraction. It dictates the proper sequence to solve any mathematical expression systematically.

Fraction Division

To divide a number by a fraction, multiply the number by the reciprocal of the fraction: (a/b) ÷ (c/d) = (a/b) × (d/c).

Types of Brackets

Brackets are resolved in a specific order: first Line brackets (vinculum), then Round brackets ( ), followed by Curly brackets { }, and finally Square brackets [ ].

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 4

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

From a sample of 200 software engineers, determine the ratio of those proficient in Python to those proficient in Java using the given information:

CategoryNumber
Proficient in Python and Java50
Proficient in Python only70
Proficient in Java only60
Proficient in neither language20
  1. AOption (a)

    11:12

  2. BOption (b)

    12:11

  3. COption (c)

    7:6

  4. DOption (d)

    6:7

View answer & explanation

Correct answer

Answer: (b)B.

12:11

Why this answer is correct

Shortcut Trick

Total engineers proficient in Python = Python only + Both = 70 + 50 = 120

Total engineers proficient in Java = Java only + Both = 60 + 50 = 110

Required Ratio = 120 : 110 = 12 : 11

∴ The correct answer is 12:11.

Alternate Method

Given:

Engineers proficient in both Python and Java = 50

Engineers proficient in Python only = 70

Engineers proficient in Java only = 60

Formula Used:

Total elements in a set = Elements exclusively in that set + Elements in intersection

Calculations:

⇒ Total engineers proficient in Python = Proficient in Python only + Proficient in both

⇒ Total engineers proficient in Python = 70 + 50 = 120

⇒ Total engineers proficient in Java = Proficient in Java only + Proficient in both

⇒ Total engineers proficient in Java = 60 + 50 = 110

⇒ Required Ratio = Total Python : Total Java

⇒ Required Ratio = 120 : 110 = 12 : 11

∴ The correct answer is 12:11.

Additional Information

Set Theory (Venn Diagram) Formula

For any two overlapping sets A and B, the total number of unique elements is given by: n(A ∪ B) = n(A) + n(B) − n(A ∩ B).

Total Elements vs 'Only' Elements

The total number of elements in Set A is the sum of elements exclusively in Set A and the elements shared with Set B: n(A) = n(Only A) + n(A ∩ B).

Simplifying Ratios

A ratio indicates how many times one number contains another. It is simplified by dividing both quantities by their Highest Common Factor (HCF).

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 5

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

Arvind started a business by investing ₹80,000. After 4 months, Bhavin joined with ₹1,20,000. At the end of 8 months from the start, Chandan joined with ₹1,60,000. If the total profit is ₹1,05,000 at the end of the year, find the share of Chandan.

  1. AOption (a)

    ₹26,500

  2. BOption (b)

    ₹26,000

  3. COption (c)

    ₹26,200

  4. DOption (d)

    ₹26,250

View answer & explanation

Correct answer

Answer: (d)D.

₹26,250

Why this answer is correct

Shortcut Trick

Profit Ratio = (Capital × Time)

Arvind : Bhavin : Chandan = (80 × 12) : (120 × 8) : (160 × 4)

⇒ Profit Ratio = 960 : 960 : 640 = 3 : 3 : 2

Total parts = 3 + 3 + 2 = 8 parts ≡ ₹1,05,000

Chandan's share (2 parts) = ₹1,05,000 × (2 ÷ 8) = ₹26,250.

∴ The correct answer is ₹26,250.

Alternate Method

Given:

Arvind's investment = ₹80,000 for 12 months

Bhavin's investment = ₹1,20,000 for (12 − 4) = 8 months

Chandan's investment = ₹1,60,000 for (12 − 8) = 4 months

Total profit = ₹1,05,000

Arvind ₹80k × 12m = 960k (3 parts) Bhavin ₹120k × 8m = 960k (3 parts) Chandan ₹160k × 4m = 640k (2 parts) Total Profit = ₹1,05,000 (8 parts)

Formula Used:

Profit Ratio = (C1 × T1) : (C2 × T2) : (C3 × T3)

Calculations:

⇒ Profit Ratio of A : B : C = (80,000 × 12) : (1,20,000 × 8) : (1,60,000 × 4)

⇒ Ratio = 9,60,000 : 9,60,000 : 6,40,000

⇒ Dividing by 3,20,000, we get the simplified Ratio = 3 : 3 : 2

⇒ Sum of profit ratios = 3 + 3 + 2 = 8 parts

⇒ Chandan's share = (2 ÷ 8) × Total Profit

⇒ Chandan's share = (1 ÷ 4) × 1,05,000

⇒ Chandan's share = ₹26,250

∴ The correct answer is ₹26,250.

Additional Information

Profit Ratio in Partnership

The profit share of any partner is directly proportional to the product of their capital invested and the time duration of the investment.

Working vs Sleeping Partner

A working partner actively manages the business and usually receives a fixed salary or percentage of total profit before the remaining profit is distributed among all partners based on their investments.

Joining Mid-way

When a partner joins a business mid-way, their effective time period is calculated from the date of joining to the end of the business cycle (typically 12 months).

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 6

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

A and B start a business. A invests ₹80,000 for 9 months, B invests ₹1,20,000 for 6 months. What is B's share of a ₹45,000 profit?

  1. AOption (a)

    ₹26,500

  2. BOption (b)

    ₹28,000

  3. COption (c)

    ₹36,000

  4. DOption (d)

    ₹22,500

View answer & explanation

Correct answer

Answer: (d)D.

₹22,500

Why this answer is correct

Shortcut Trick

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

Profit Ratio = (80000 × 9) : (120000 × 6)

Profit Ratio = 720000 : 720000 = 1 : 1

B's share = (1 ÷ 2) × 45000 = 22500

∴ The correct answer is ₹22,500.

Alternate Method

Given:

Investment by A = ₹80,000

Time period of A = 9 months

Investment by B = ₹1,20,000

Time period of B = 6 months

Total Profit = ₹45,000

Formula Used:

Profit Ratio = (Investment × Time)

Partner A80k × 9mPartner B120k × 6mProfit: 720kProfit: 720kRatio A : B = 1 : 1

Calculations:

⇒ Equivalent Capital of A = 80000 × 9 = 7,20,000

⇒ Equivalent Capital of B = 120000 × 6 = 7,20,000

⇒ Profit Ratio (A : B) = 720000 : 720000 = 1 : 1

⇒ Total ratio parts = 1 + 1 = 2

⇒ B's share in profit = (1 ÷ 2) × Total Profit

⇒ B's share = (1 ÷ 2) × 45000 = 22500

∴ The correct answer is ₹22,500.

Additional Information

Simple Partnership

When partners invest their capital for the exact same duration, profits or losses are distributed directly in the ratio of their original investments.

Compound Partnership

When capital is invested for different time periods, the actual profit sharing ratio is derived by multiplying each partner's invested capital with the respective time period.

Working vs Sleeping Partner

A working partner actively manages the business and often draws a fixed salary from the profit before it is divided. A sleeping partner strictly provides capital and gets a pure proportional share based on investment and time.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 7

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

What is the average of all integers between 100 and 250 that are exactly divisible by 11?

  1. AOption (a)

    176

  2. BOption (b)

    186

  3. COption (c)

    196

  4. DOption (d)

    146

View answer & explanation

Correct answer

Answer: (a)A.

176

Why this answer is correct

Shortcut Trick

For an Arithmetic Progression, Average = (First term + Last term) ÷ 2

First multiple of 11 after 100 = 110

Last multiple of 11 before 250 = 242

Average = (110 + 242) ÷ 2 = 352 ÷ 2 = 176

∴ The correct answer is 176.

Alternate Method

100110121176242250Average = 176Arithmetic Progression (Common Difference = 11)

Given:

⇒ Range = Integers between 100 and 250.

⇒ The numbers must be exactly divisible by 11.

Calculations:

⇒ First integer divisible by 11 = 110 (since 11 × 10 = 110).

⇒ Last integer divisible by 11 = 242 (since 11 × 22 = 242).

⇒ The series is 110, 121, 132, ..., 242.

⇒ Total number of terms (n) = [(Last term − First term) ÷ Common Difference] + 1

⇒ n = [(242 − 110) ÷ 11] + 1

⇒ n = [132 ÷ 11] + 1 = 12 + 1 = 13 terms

⇒ Sum of an AP = (n ÷ 2) × (First term + Last term)

⇒ Sum = (13 ÷ 2) × (110 + 242)

⇒ Sum = (13 ÷ 2) × 352 = 13 × 176 = 2288

⇒ Average = Total Sum ÷ n

⇒ Average = 2288 ÷ 13 = 176

∴ The correct answer is 176.

Additional Information

Average of an Arithmetic Progression

If a series has a constant common difference (like multiples of a number), it forms an Arithmetic Progression (AP). Its average is always the exact mean of the first and last terms: Average = (First Term + Last Term) ÷ 2.

Sum of an Arithmetic Progression

The sum of n terms in an AP is calculated as: Sn = (n ÷ 2) × (First Term + Last Term).

Finding the Number of Terms

To find the total number of terms (n) in any given range for a specific sequence: n = [(Last Term − First Term) ÷ Common Difference] + 1.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 8

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

The average of 15 numbers is 80. The average of the first 6 numbers is 72. The average of the next 6 numbers is 25% more than the average of the first 6 numbers. The 13th number is 8 more than the 15th number, and the 14th number is 10 less than the 15th number. What is the average of the 13th and 14th numbers?

  1. AOption (a)

    70.89

  2. BOption (b)

    85

  3. COption (c)

    75.67

  4. DOption (d)

    80.65

View answer & explanation

Correct answer

Answer: (c)C.

75.67

Why this answer is correct

Shortcut Trick

Net deviation of first 12 numbers = 6 × (72 − 80) + 6 × (90 − 80) = −48 + 60 = +12

Sum of the 13th, 14th, and 15th numbers = (3 × 80) − 12 = 228

Assuming the 15th number is x: (x + 8) + (x − 10) + x = 228 ⇒ 3x = 230

Average of 13th and 14th numbers = [(x + 8) + (x − 10)] ÷ 2 = x − 1

Required Average = 230/3 − 1 = 227/3 ≈ 75.67

∴ The correct answer is 75.67.

Alternate Method

Given:

Total numbers = 15 (Average = 80)

Average of first 6 numbers = 72

Average of next 6 numbers = 72 + 25% of 72 = 90

13th number = 15th number + 8

14th number = 15th number − 10

Formula Used:

Total Sum = Average × Total Observations

Total 15 numbers (Sum = 1200)First 6 (Avg 72)Sum: 432Next 6 (Avg 90)Sum: 540Last 3Sum: 22813th + 14th + 15th = 1200 − (432 + 540) = 228

Calculations:

⇒ Total sum of 15 numbers = 15 × 80 = 1200

⇒ Sum of the first 6 numbers = 6 × 72 = 432

⇒ Sum of the next 6 numbers = 6 × 90 = 540

⇒ Sum of remaining 3 numbers (13th, 14th, 15th) = 1200 − (432 + 540) = 228

⇒ Let the 15th number be x.

⇒ 13th number = x + 8

⇒ 14th number = x − 10

⇒ (x + 8) + (x − 10) + x = 228

⇒ 3x − 2 = 228

⇒ 3x = 230 ⇒ x = 230/3

⇒ Average of 13th and 14th = [(x + 8) + (x − 10)] ÷ 2

⇒ Average = (2x − 2) ÷ 2 = x − 1

⇒ Average = 230/3 − 1 = 227/3 = 75.666... ≈ 75.67

∴ The correct answer is 75.67.

Additional Information

Net Deviation Method

The sum of deviations of all observations from their arithmetic mean is always zero. The deviation trick helps compute missing elements by finding how much the known parts differ from the combined average.

Weighted Average

When combining separate groups with varying numbers of elements, the new average is calculated as (N1A1 + N2A2 + ...) ÷ (N1 + N2 + ...).

Successive Substitution

When elements relate to one unknown variable directly, convert the sum into an algebraic expression and solve for the unknown to simplify calculation of required parts directly.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 9

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

A landlord bought a flat for ₹8,00,000. He wants to earn a 9% annual return on his investment after paying ₹2,000 per month for maintenance. What should be the monthly rent he charges?

  1. AOption (a)

    ₹7,000

  2. BOption (b)

    ₹7,500

  3. COption (c)

    ₹8,000

  4. DOption (d)

    ₹8,500

View answer & explanation

Correct answer

Answer: (c)C.

₹8,000

Why this answer is correct

Shortcut Trick

Monthly Return = (9% of ₹8,00,000) ÷ 12 = ₹72,000 ÷ 12 = ₹6,000

Monthly Rent = Monthly Return + Maintenance

Monthly Rent = ₹6,000 + ₹2,000 = ₹8,000

∴ The correct answer is ₹8,000.

Alternate Method

Given:

Investment (Cost of flat) = ₹8,00,000

Required annual return = 9%

Monthly maintenance = ₹2,000

Formula Used:

Monthly Rent = Monthly Net Return + Monthly Maintenance

Total Monthly Rent = ₹8,000

Maintenance ₹2,000

Net Return (9% p.a.) ₹6,000

9% of ₹8,00,000 ÷ 12 months

Calculations:

⇒ Annual Net Return = 9% of ₹8,00,000

⇒ Annual Net Return = (9 ÷ 100) × 8,00,000 = ₹72,000

⇒ Monthly Net Return = 72,000 ÷ 12 = ₹6,000

⇒ Total Monthly Rent = Monthly Net Return + Monthly Maintenance

⇒ Total Monthly Rent = 6,000 + 2,000 = ₹8,000

∴ The correct answer is ₹8,000.

Additional Information

Return on Investment (ROI)

ROI is a financial metric used to evaluate the profitability of an investment. It is calculated as the ratio of net profit to the initial investment cost, often expressed as an annual percentage.

Gross vs Net Income

Gross income refers to the total revenue generated (e.g., total rent collected), while net income is the profit remaining after deducting all operating expenses such as maintenance and taxes.

Annual to Monthly Equivalence

When cash flows or returns are uniform throughout the year, a monthly equivalent is obtained by simply dividing the annual expected value by 12. Monthly Return = Annual Return ÷ 12.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 10

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

An amount is said to double in 5 years with compound interest. How many years will it take for the amount to grow to 8 times its original value?

  1. AOption (a)

    15

  2. BOption (b)

    16

  3. COption (c)

    17

  4. DOption (d)

    18

View answer & explanation

Correct answer

Answer: (a)A.

15

Why this answer is correct

Shortcut Trick

If an amount becomes n times in t years at compound interest, it becomes nm times in m × t years.

Given: The amount becomes 2 times (n = 2) in 5 years (t = 5).

We need it to become 8 times, which is 23 times, so m = 3.

Required time = m × t = 3 × 5 = 15 years.

∴ The correct answer is 15 years.

Alternate Method

P

2P

4P

8P

5 years × 2

5 years × 2

5 years × 2

Total Time = 5 + 5 + 5 = 15 years

Given:

Amount doubles in 5 years at compound interest.

Let the initial principal be P.

Amount (A) = 2P, Time (t) = 5 years.

Formula Used:

A = P(1 + r/100)t

Calculations:

⇒ 2P = P(1 + r/100)5

⇒ 2 = (1 + r/100)5

We need to find the time T when the amount becomes 8 times (A = 8P).

⇒ 8P = P(1 + r/100)T

⇒ 8 = (1 + r/100)T

Since 8 can be written as 23:

⇒ 23 = (1 + r/100)T

Substituting the value of 2 from the first equation:

⇒ [(1 + r/100)5]3 = (1 + r/100)T

⇒ (1 + r/100)15 = (1 + r/100)T

Comparing the exponents on both sides:

⇒ T = 15

∴ The correct answer is 15 years.

Additional Information

Compound Interest Multiple Concept

If an amount becomes n times in t years at CI, it will become nm times in m × t years.

Simple Interest Multiple Concept

If an amount becomes n times in t years at SI, it will become m times in [(m−1)/(n−1)] × t years.

Rule of 72

A quick formula to approximate the number of years required to double the principal at a fixed annual CI rate (Time ≈ 72 ÷ r).

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 11

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

A sum becomes ₹6,600 in 2 years and ₹7,920 in 3 years at compound interest. What is the original principal?

  1. AOption (a)

    ₹4,000.33

  2. BOption (b)

    ₹5,583.33

  3. COption (c)

    ₹4,583.33

  4. DOption (d)

    ₹6,583.33

View answer & explanation

Correct answer

Answer: (c)C.

₹4,583.33

Why this answer is correct

Shortcut Trick

Multiplying factor for 1 year = Amount3 ÷ Amount2 = 7920 ÷ 6600 = 6/5

Since the sum amounts to ₹6,600 in 2 years at CI:

Principal × (6/5)2 = 6600

⇒ Principal = 6600 × (25/36) ≈ 4583.33

∴ The correct answer is ₹4,583.33.

Alternate Method

Given:

Amount after 2 years (A2) = ₹6,600

Amount after 3 years (A3) = ₹7,920

Year 0 P = ? Year 2 ₹6,600 Year 3 ₹7,920

× (1 + R/100)² × (1 + R/100) Compound Interest Timeline

Formula Used:

A = P × (1 + R/100)n

Calculations:

⇒ A3 ÷ A2 = [P × (1 + R/100)3] ÷ [P × (1 + R/100)2]

⇒ 7920 ÷ 6600 = 1 + R/100

⇒ 1.2 = 1 + R/100

⇒ R = 20%

Now, substitute R into the 2nd year amount formula:

⇒ P × (1 + 20/100)2 = 6600

⇒ P × (1.2)2 = 6600

⇒ P × 1.44 = 6600

⇒ P = 6600 ÷ 1.44 = 4583.33

∴ The correct answer is ₹4,583.33.

Additional Information

Interest for 1 Year in CI

In compound interest, the amount accumulated at the end of a year acts as the principal for the next year. Thus, dividing the amount of any given year by the previous year's amount gives the direct multiplying factor for one year.

Simple Interest vs Compound Interest

For the first year (or the first compounding period), both Simple Interest and Compound Interest generate exactly the same interest. The difference arises from the second year onwards because Compound Interest charges interest on the previously accumulated interest.

Effective Rate for 2 Years

The effective compound interest rate for 2 consecutive years at R% per annum can be calculated directly using the successive percentage formula: R + R + (R × R) ÷ 100.

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

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

Suman purchased 25 liters of milk at ₹45 per liter and another 15 liters of milk at ₹50 per liter. She combined both quantities and then sold the entire mixture at ₹48 per liter. What was her total profit or loss?

  1. AOption (a)

    Loss of ₹45

  2. BOption (b)

    Profit of ₹45

  3. COption (c)

    Profit of ₹90

  4. DOption (d)

    Loss of ₹90

View answer & explanation

Correct answer

Answer: (b)B.

Profit of ₹45

Why this answer is correct

Shortcut Trick

Calculate the profit or loss per liter directly by comparing the cost prices with the final selling price of ₹48/liter.

Profit on the first quantity (25 liters) = 25 × (48 − 45) = 25 × 3 = +₹75

Loss on the second quantity (15 liters) = 15 × (48 − 50) = 15 × (−2) = −₹30

Net result = +75 − 30 = +₹45

Since the net result is positive, it represents an overall profit.

∴ The correct answer is Profit of ₹45.

Alternate Method

Given:

Quantity of first milk = 25 liters at ₹45/liter

Quantity of second milk = 15 liters at ₹50/liter

Selling price of the combined mixture = ₹48/liter

25 L @ ₹45/LCP = ₹112515 L @ ₹50/LCP = ₹750Total CP (40 L)₹1875Total SP @ ₹48₹1920Profit = ₹1920 − ₹1875 = ₹45

Formula Used:

Total CP = (Q1 × CP1) + (Q2 × CP2)

Total SP = Total Quantity × SP per liter

Profit = Total SP − Total CP

Calculations:

⇒ CP of 25 liters of milk = 25 × 45 = ₹1125

⇒ CP of 15 liters of milk = 15 × 50 = ₹750

⇒ Total CP of the mixture = 1125 + 750 = ₹1875

⇒ Total quantity of the mixture = 25 + 15 = 40 liters

⇒ Total SP of the mixture = 40 × 48 = ₹1920

⇒ Since Total SP > Total CP, Suman makes a profit.

⇒ Total Profit = 1920 − 1875 = ₹45

∴ The correct answer is Profit of ₹45.

Additional Information

Weighted Average (Mixtures)

The mean cost price of a mixture formed by mixing two distinct quantities can be calculated as: Mean CP = (Q1 × P1 + Q2 × P2) ÷ (Q1 + Q2).

Profit and Loss Fundamentals

A business transaction yields a profit when the Selling Price (SP) is greater than the Cost Price (CP). Conversely, it results in a loss if the CP exceeds the SP. The absolute profit is computed as: Profit = SP − CP.

Rule of Alligation

This rule provides a direct method to find the ratio in which two quantities of varying prices must be combined to yield a required average price: Quantity of Cheaper ÷ Quantity of Dearer = (Price of Dearer − Mean Price) ÷ (Mean Price − Price of Cheaper).

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 13

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

A toy manufacturer produced 1200 toy cars at a total cost of ₹90,000. He donated 200 cars to a charity event. For the rest, he announced a 10% discount on the market price of ₹120 per car. He also offered 2 toy cars free for every 8 toy cars purchased. If all 1200 toy cars were distributed, what is his overall gain or loss percentage?

  1. AOption (a)

    4% loss

  2. BOption (b)

    4% profit

  3. COption (c)

    5% loss

  4. DOption (d)

    5% profit

View answer & explanation

Correct answer

Answer: (a)A.

4% loss

Why this answer is correct

Shortcut Trick

Remaining cars after charity = 1200 − 200 = 1000.

Effective paid cars = 1000 × [8 ÷ (8 + 2)] = 800.

Selling Price (SP) per car = ₹120 − 10% = ₹108.

Total Revenue (SP) = 800 × ₹108 = ₹86,400.

Loss % = [(90,000 − 86,400) ÷ 90,000] × 100 = 4%.

∴ The correct answer is 4% loss.

Alternate Method

Given:

Total cost price (CP) = ₹90,000

Total toy cars produced = 1200

Cars given to charity = 200

Market price (MP) per car = ₹120

Discount = 10%

Total Produced = 1200 cars (Total CP = ₹90,000)DonatedFreePaid (800 cars)200200800Selling Price per paid car = ₹120 − 10% = ₹108Total Revenue (SP) = 800 × ₹108 = ₹86,400Loss = ₹90,000 − ₹86,400 = ₹3,600

Calculations:

⇒ Remaining cars left for sale = 1200 − 200 = 1000

⇒ Selling Price per car after discount = 120 − (10% of 120) = 120 − 12 = ₹108

⇒ The offer states 2 free cars for every 8 cars purchased. This means out of every 10 cars handed out, 8 are paid for.

⇒ Number of billed/paid cars = 1000 × (8 ÷ 10) = 800 cars

⇒ Total Selling Price (SP) = 800 × 108 = ₹86,400

⇒ Overall Loss = Total CP − Total SP = 90000 − 86400 = ₹3600

⇒ Loss % = (Loss ÷ CP) × 100

⇒ Loss % = (3600 ÷ 90000) × 100 = 4%

∴ The correct answer is 4% loss.

Additional Information

Buy X Get Y Free Concept

This means for a total of (X + Y) articles, the customer only pays for X articles. The effective discount percentage for the customer is calculated as: [Y ÷ (X + Y)] × 100.

Marked Price (MP) & Discount

Discount is always calculated on the Marked Price. Selling Price (SP) = MP − Discount. The final Selling Price can also be calculated directly using SP = MP × (100 − D%) ÷ 100.

Profit & Loss Percentage

Profit or loss is always calculated over the Cost Price (CP) unless explicitly stated otherwise. Loss % = (Loss ÷ CP) × 100 and Profit % = (Profit ÷ CP) × 100.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 14

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

A retailer marks an air conditioner 80% above its cost price. He offers a first discount of 25% on the marked price. During a festive offer, an additional discount of 10% is applied on the already discounted price. If the final selling price is ₹15,552, what is the approximate cost price of the air conditioner?"

  1. AOption (a)

    ₹9,000

  2. BOption (b)

    ₹9,200

  3. COption (c)

    ₹12,800

  4. DOption (d)

    ₹10,000

View answer & explanation

Correct answer

Answer: (c)C.

₹12,800

Why this answer is correct

Shortcut Trick

Let the Cost Price be CP.

Fraction values: +80% markup = 180/100 = 9/5.

Fraction values: 25% discount = 3/4, and 10% discount = 9/10.

Final Selling Price = CP × 9/5 × 3/4 × 9/10 = CP × 243/200

⇒ 15552 = CP × 243/200 ⇒ CP = 15552 × 200 ÷ 243 = 12800

∴ The correct answer is ₹12,800.

Alternate Method

Given:

Markup = 80%, 1st Discount = 25%, 2nd Discount = 10%

Final Selling Price = ₹15,552

CP (100x)

+80%

MP (180x)

-25%

SP1 (135x)

-10%

Final SP (121.5x)

Final Selling Price (121.5x) = ₹15,552

Calculations:

⇒ Let the Cost Price (CP) be 100x.

⇒ Marked Price (MP) = 100x + 80% of 100x = 180x

⇒ Selling Price after 1st discount (25%) = 180x × (1 − 25/100) = 135x

⇒ Final Selling Price after 2nd discount (10%) = 135x × (1 − 10/100) = 121.5x

⇒ Given Final SP = ₹15,552

⇒ 121.5x = 15552

⇒ x = 15552 ÷ 121.5 = 128

⇒ Cost Price = 100x = 100 × 128 = 12800

∴ The correct answer is ₹12,800.

Additional Information

Marked Price (MP)

It is the price printed on the article. Markup percentage is always calculated on the base of Cost Price (CP).

Successive Discount

When multiple discounts are given sequentially, the final Selling Price is calculated by multiplying the successive discount factors: Final SP = MP × (1 − D1/100) × (1 − D2/100).

Equivalent Single Discount

If two successive discounts of x% and y% are given, the single equivalent discount is calculated using the formula: (x + y − xy/100)%.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 15

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

A 40-liter mixture contains juice and water in the ratio 5:3. How much water (in liters) must be added to this mixture to change the ratio of juice to water to 2:3?

  1. AOption (a)

    15.5 litres

  2. BOption (b)

    22.5 litres

  3. COption (c)

    25 litres

  4. DOption (d)

    30 litres

View answer & explanation

Correct answer

Answer: (b)B.

22.5 litres

Why this answer is correct

Shortcut Trick

Since only water is added, the juice quantity is constant. Make juice units equal in both ratios:

Initial Juice : Water = 5 : 3 ⇒ (× 2) ⇒ 10 : 6

Final Juice : Water = 2 : 3 ⇒ (× 5) ⇒ 10 : 15

Initial quantity = 10 + 6 = 16 units = 40 L ⇒ 1 unit = 2.5 L.

Water added = 15 − 6 = 9 units ⇒ 9 × 2.5 = 22.5 L.

∴ The correct answer is 22.5 litres.

Alternate Method

Given:

Initial mixture = 40 liters | Initial ratio (Juice : Water) = 5 : 3

Final ratio (Juice : Water) = 2 : 3

Initial MixtureWater: 37.5%Added WaterWater: 100%Final60%100 − 60 = 4060 − 37.5 = 22.5Ratio of Quantities (Initial : Added) = 40 : 22.5Since Initial Mixture is 40 L   →   Added Water = 22.5 L

Calculations:

⇒ Quantity of Juice = (5 ÷ 8) × 40 = 25 liters

⇒ Quantity of Water = (3 ÷ 8) × 40 = 15 liters

Let the amount of water added be x liters, so the new amount of water = (15 + x) liters.

According to the new ratio of Juice to Water:

⇒ 25 ÷ (15 + x) = 2 ÷ 3

⇒ 75 = 30 + 2x

⇒ 2x = 45 ⇒ x = 22.5

∴ The correct answer is 22.5 litres.

Additional Information

Rule of Alligation

When two ingredients of different concentrations are mixed, the ratio of their quantities is: (Quantity of Cheaper) ÷ (Quantity of Dearer) = (Price of Dearer − Mean Price) ÷ (Mean Price − Price of Cheaper).

Constant Term Logic in Ratios

When only one component (like water) is added to a mixture, the quantity of the other component (like juice) remains unchanged. Equating the ratio units of the constant component naturally streamlines the calculation without needing algebra.

Replacement Formula

If V is the initial volume of pure liquid, and x volume is removed and replaced by water n times, the final volume of pure liquid left = V × (1 − x/V)n.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 16

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

A is able to complete a task in 15 days, while B takes 20 days to finish the same task. If they collaborate and work together for 4 days, what fraction of the work will still remain?

  1. AOption (a)

    14\frac{1}4

  2. BOption (b)

    715\frac{7}{15}

  3. COption (c)

    815\frac{8}{15}

  4. DOption (d)

    1115\frac{11}{15}

View answer & explanation

Correct answer

Answer: (c)C.

815\frac{8}{15}

Why this answer is correct

Shortcut Trick

Total work = LCM(15, 20) = 60 units.

Combined efficiency = (60 ÷ 15) + (60 ÷ 20) = 4 + 3 = 7 units/day.

Work done in 4 days = 7 × 4 = 28 units.

Remaining work = 60 − 28 = 32 units.

Fraction of work remaining = 32 ÷ 60 = 8/15.

∴ The correct answer is 8/15.

Alternate Method

Given:

Time taken by A to complete the task = 15 days

Time taken by B to complete the task = 20 days

Time they worked together = 4 days

PersonDaysTotal WorkEfficiency
A15604
B203

⇒ Total efficiency of A and B = 4 + 3 = 7 units/day

⇒ Work done by A and B together in 4 days = 7 × 4 = 28 units

⇒ Remaining work = Total Work − Work Done

⇒ Remaining work = 60 − 28 = 32 units

⇒ Fraction of work remaining = Remaining Work ÷ Total Work

⇒ Fraction of work remaining = 32 ÷ 60 = 8/15

∴ The correct answer is 8/15.

Additional Information

Total Work & Efficiency

Efficiency is inversely proportional to the time taken. Total work is usually assumed as the LCM of the given individual days to avoid complex fractional calculations.

Work & Wages

Wages are completely distributed in proportion to the total actual work done by each person during the given period, not just by their efficiency ratios.

Pipes & Cisterns

This topic follows identical principles to Time & Work. The key difference is that filling pipes have a positive efficiency while emptying pipes (or leaks) have a negative efficiency.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 17

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

Two oils priced at ₹90 per kg and ₹150 per kg are blended and sold at ₹144 per kg, achieving a profit margin of 20%. What is the ratio of the two oils in the mixture?

  1. AOption (a)

    1 : 1

  2. BOption (b)

    2 : 1

  3. COption (c)

    3 : 2

  4. DOption (d)

    4 : 1

View answer & explanation

Correct answer

Answer: (a)A.

1 : 1

Why this answer is correct

Shortcut Trick

Mean Cost Price (CP) = SP × [100 ÷ (100 + Profit%)]

⇒ Mean CP = 144 × (100 ÷ 120) = ₹120/kg

Apply Alligation: Ratio = (Dearer Price − Mean Price) : (Mean Price − Cheaper Price)

⇒ Ratio = (150 − 120) : (120 − 90) = 30 : 30 = 1 : 1

∴ The correct answer is 1 : 1.

Alternate Method

Given:

Price of cheaper oil (C) = ₹90/kg

Price of dearer oil (D) = ₹150/kg

Selling price of mixture (SP) = ₹144/kg

Profit = 20%

Formula Used:

Cost Price (CP) = SP × [100 ÷ (100 + Profit%)]

₹90 ₹150 ₹120 30 30

Cheaper Oil Dearer Oil Mean CP

Ratio = 30 : 30 = 1 : 1

Calculations:

⇒ Mean CP = 144 × [100 ÷ (100 + 20)]

⇒ Mean CP = 144 × (100 ÷ 120)

⇒ Mean CP = 144 × (5 ÷ 6) = ₹120/kg

⇒ By the rule of alligation:

⇒ Quantity of Cheaper : Quantity of Dearer = (150 − 120) : (120 − 90)

⇒ Quantity of Cheaper : Quantity of Dearer = 30 : 30

⇒ Ratio = 1 : 1

∴ The correct answer is 1 : 1.

Additional Information

Rule of Alligation

The ratio in which two ingredients are mixed to produce a mixture of a desired mean price is given by (Dearer Price − Mean Price) : (Mean Price − Cheaper Price).

Cost Price vs Selling Price in Mixtures

When applying alligation, all values must be of the same type. If the individual values are Cost Prices, the mean value must also be the Cost Price (CP), not the Selling Price (SP). CP = SP × 100 ÷ (100 + Profit%).

Weighted Average Method

The rule of alligation is a direct derivation of the weighted average formula: M = (Q1 × C1 + Q2 × C2) ÷ (Q1 + Q2), where Q represents quantity and C represents the respective cost.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 18

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

Three pipes, A, B, and C are capable of filling a tank in 6, 8, and 12 hours, respectively. When all three pipes are opened together, they operate for 2 hours before pipe C is closed. How much additional time will it take to completely fill the tank after that?

  1. AOption (a)

    67\frac{6}7 hours

  2. BOption (b)

    26\frac{2}6 hours

  3. COption (c)

    35\frac{3}5 hours

  4. DOption (d)

    92\frac{9}2 hours

View answer & explanation

Correct answer

Answer: (a)A.

67\frac{6}7 hours

Why this answer is correct

Shortcut Trick

Total capacity = LCM(6, 8, 12) = 24 units.

Total efficiency of A, B and C = 4 + 3 + 2 = 9 units/hr.

Tank filled in 2 hours = 9 × 2 = 18 units.

Remaining capacity = 24 − 18 = 6 units.

Time taken by A and B to fill the rest = 6 ÷ (4 + 3) = 6/7 hours.

∴ The correct answer is 6/7 hours.

Alternate Method

Given:

Time taken by pipe A to fill the tank = 6 hours

Time taken by pipe B to fill the tank = 8 hours

Time taken by pipe C to fill the tank = 12 hours

Time all pipes operated together = 2 hours

PipeHoursTotal WorkEfficiency
A62424 ÷ 6 = 4
B824 ÷ 8 = 3
C1224 ÷ 12 = 2

Calculations:

⇒ Combined efficiency of A, B and C = 4 + 3 + 2 = 9 units/hr

⇒ Work done by A, B and C in 2 hours = 9 × 2 = 18 units

⇒ Remaining work = 24 − 18 = 6 units

⇒ Pipe C is closed, so the remaining work is done by pipes A and B.

⇒ Combined efficiency of pipes A and B = 4 + 3 = 7 units/hr

⇒ Additional time taken by A and B = Remaining work ÷ Combined efficiency

⇒ Additional time taken = 6/7 hours

∴ The correct answer is 6/7 hours.

Additional Information

Pipes and Cisterns Basics

If a pipe can fill a tank in x hours, the part filled in 1 hour is 1/x. This is analogous to the concept of Time and Work where efficiency is inversely proportional to time.

Inlet vs Outlet Pipes

An inlet pipe fills the tank and its work is considered positive. An outlet pipe empties the tank and its work is considered negative. Net work = 1/x − 1/y.

Alternate Working of Pipes

When pipes are opened in alternate hours, work is calculated in cycles (e.g., 2 hours if 2 pipes are working alternately) rather than directly summing their efficiencies per hour.

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 19

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

A bullet train covers a fixed distance in 30 minutes at an average speed of 240 km/h. Due to track maintenance, it needs to be diverted, increasing the distance by 20%. If the train needs to arrive at its destination on time (i.e., in 30 minutes), what should its new average speed be in km/h?

  1. AOption (a)

    280 km/h

  2. BOption (b)

    300 km/h

  3. COption (c)

    288 km/h

  4. DOption (d)

    320 km/h

View answer & explanation

Correct answer

Answer: (c)C.

288 km/h

Why this answer is correct

Shortcut Trick

Since the travel time remains constant (30 minutes) in both cases, Speed is directly proportional to Distance.

Because the distance increases by 20%, the required average speed must also increase by exactly 20%.

New Speed = 240 + (20% of 240)

New Speed = 240 + 48 = 288 km/h

∴ The correct answer is 288 km/h.

Alternate Method

Given:

Initial Average Speed (S1) = 240 km/h

Initial Time (T1) = 30 minutes = 0.5 hours

New Time (T2) = 30 minutes = 0.5 hours

Initial Distance: 120 km Speed: 240 km/h | Time: 30 min

New Distance: 144 km (+20%) New Speed: 288 km/h | Time: 30 min

Formula Used:

Distance = Speed × Time

Speed = Distance ÷ Time

Calculations:

⇒ Initial Distance (D1) = 240 × 0.5 = 120 km

⇒ The distance is increased by 20% due to track maintenance.

⇒ New Distance (D2) = 120 + (20 ÷ 100) × 120

⇒ D2 = 120 + 24 = 144 km

⇒ The train needs to arrive in the same exact time window.

⇒ New Time (T2) = 0.5 hours

⇒ New Speed (S2) = D2 ÷ T2

⇒ S2 = 144 ÷ 0.5 = 288 km/h

∴ The correct answer is 288 km/h.

Additional Information

Direct and Inverse Proportionality

When Time is constant, Distance and Speed are directly proportional (S1 ÷ S2 = D1 ÷ D2). When Distance is constant, Time and Speed are inversely proportional.

Unit Conversions for Speed

To convert an object's speed from km/h to m/s, multiply it by 5 ÷ 18. To change m/s back to km/h, multiply by 18 ÷ 5.

Average Speed

Average Speed over an entire journey equals Total Distance ÷ Total Time. If two equal distances are traveled at speeds a and b, the Average Speed is 2ab ÷ (a + b).

SSC CGL 2025 (Held On: 12 Sept, 2025 Shift 1) (Previous Year Paper)·Quantitative Aptitude
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Question 20

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

Two cars, X and Y, start from points 360 km apart and travel at constant speeds. If they move towards each other, they meet in 4 hours. If they move in the same direction, they meet in 12 hours. What is the speed of car X? (Assume X is the faster car).

  1. AOption (a)

    60 km/h

  2. BOption (b)

    75 km/h

  3. COption (c)

    45 km/h

  4. DOption (d)

    90 km/h

View answer & explanation

Correct answer

Answer: (a)A.

60 km/h

Why this answer is correct

Shortcut Trick

Relative speed in opposite direction = 360 ÷ 4 = 90 km/h

Relative speed in same direction = 360 ÷ 12 = 30 km/h

Speed of the faster car (X) = (Sum of relative speeds + Difference of relative speeds) ÷ 2

Speed of X = (90 + 30) ÷ 2 = 120 ÷ 2 = 60 km/h

∴ The correct answer is 60 km/h.

Alternate Method

Given:

Distance between X and Y = 360 km

Time to meet moving towards each other = 4 hours

Time to meet moving in the same direction = 12 hours

Formula Used:

Speed = Distance ÷ Time

Opposite Direction: Relative Speed = X + Y

Same Direction: Relative Speed = X − Y

Calculations:

⇒ X + Y = 360 ÷ 4

⇒ X + Y = 90     ... (Equation 1)

⇒ X − Y = 360 ÷ 12

⇒ X − Y = 30     ... (Equation 2)

⇒ Adding Equation 1 and Equation 2:

⇒ (X + Y) + (X − Y) = 90 + 30

⇒ 2X = 120

⇒ X = 60 km/h

∴ The correct answer is 60 km/h.

Additional Information

Relative Speed in Opposite Direction

When two bodies move towards each other, their relative speed is calculated as the sum of their individual speeds (S1 + S2).

Relative Speed in Same Direction

When two bodies move in the same direction, their relative speed is calculated as the difference of their individual speeds (S1 − S2).

Distance Constant Rule

If the distance to be covered remains constant, the speed of an object is inversely proportional to the time taken: S1 ÷ S2 = T2 ÷ T1.

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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.