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Question 301

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

If P : Q = 6 : 11, Q : R = 5 : 4, and R : S = 9 : 7, find P : Q : R : S.

  1. AOption (a)

    270 : 495 : 396 : 308

  2. BOption (b)

    6 : 11 : 4 : 7

  3. COption (c)

    30 : 55 : 44 : 35

  4. DOption (d)

    270 : 396 : 495 : 308

View answer & explanation

Correct answer

Answer: (a)A.

270 : 495 : 396 : 308

Why this answer is correct

Shortcut Trick

Use the column multiplication method for continuous ratios:

P = 6 × 5 × 9 = 270

Q = 11 × 5 × 9 = 495

R = 11 × 4 × 9 = 396

S = 11 × 4 × 7 = 308

∴ The correct answer is 270 : 495 : 396 : 308.

Alternate Method

Given:

⇒ P : Q = 6 : 11

⇒ Q : R = 5 : 4

⇒ R : S = 9 : 7

Visual Method: Fill empty spaces with adjacent numbers

P Q R S

6

11

11

11

5

5

4

4

9

9

9

7

× Multiply columns vertically

270

495

396

308

Ratio → 270 : 495 : 396 : 308

Calculations:

⇒ Equalize Q in the first two ratios (LCM of 11 and 5 is 55).

⇒ P : Q = (6 × 5) : (11 × 5) = 30 : 55

⇒ Q : R = (5 × 11) : (4 × 11) = 55 : 44

⇒ Combining them, P : Q : R = 30 : 55 : 44

⇒ Now, equalize R in P : Q : R and R : S (LCM of 44 and 9 is 396).

⇒ P : Q : R = (30 × 9) : (55 × 9) : (44 × 9) = 270 : 495 : 396

⇒ R : S = (9 × 44) : (7 × 44) = 396 : 308

⇒ Combine all to get the final ratio:

⇒ P : Q : R : S = 270 : 495 : 396 : 308

∴ The correct answer is 270 : 495 : 396 : 308.

Additional Information

Compound Ratio

When two or more ratios are multiplied together, it is called a compound ratio. For a:b and c:d, the compound ratio is (a × c) : (b × d).

Duplicate & Sub-duplicate Ratio

The duplicate ratio of a : b is a2 : b2. The sub-duplicate ratio is √a : √b.

Componendo and Dividendo

If a/b = c/d, then (a + b) / (a − b) = (c + d) / (c − d). This rule is heavily used to simplify complex algebraic fraction statements.

Question 302

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

A bag contains ₹10, ₹5, and ₹2 coins in the ratio 1 : 2 : 3. If the total amount of money in the bag is ₹390, find the number of coins of each kind.

  1. AOption (a)

    10 coins of ₹10, 20 of ₹5, and 30 of ₹2

  2. BOption (b)

    15 coins of ₹10, 30 of ₹5, and 45 of ₹2

  3. COption (c)

    20 coins of ₹10, 40 of ₹5, and 60 of ₹2

  4. DOption (d)

    12 coins of ₹10, 24 of ₹5, and 36 of ₹2

View answer & explanation

Correct answer

Answer: (b)B.

15 coins of ₹10, 30 of ₹5, and 45 of ₹2

Why this answer is correct

Shortcut Trick

Total value ratio = (10 × 1) + (5 × 2) + (2 × 3) = 10 + 10 + 6 = 26 units.

Total given money = ₹390.

Value of 1 ratio unit = 390 ÷ 26 = 15.

Multiply the initial coin ratio (1 : 2 : 3) by 15.

Number of coins = (1 × 15), (2 × 15), and (3 × 15) = 15, 30, and 45.

∴ The correct answer is 15 coins of ₹10, 30 of ₹5, and 45 of ₹2.

Alternate Method

Given:

Denominations of coins = ₹10, ₹5, and ₹2

Ratio of number of coins = 1 : 2 : 3

Total amount in the bag = ₹390

₹10

₹5

₹2

Ratio: x Ratio: 2x Ratio: 3x

Value: 10x

Value: 10x

Value: 6x

Total Value = 10x + 10x + 6x = 26x = ₹390

Calculation:

⇒ Let the number of ₹10, ₹5, and ₹2 coins be x, 2x, and 3x respectively.

⇒ The values of these coins will be ₹(10 × x), ₹(5 × 2x), and ₹(2 × 3x) respectively.

⇒ Total value in terms of x = 10x + 10x + 6x = 26x

⇒ We are given that the total amount is ₹390.

⇒ 26x = 390

⇒ x = 390 ÷ 26 = 15

⇒ Number of ₹10 coins = x = 15

⇒ Number of ₹5 coins = 2x = 2 × 15 = 30

⇒ Number of ₹2 coins = 3x = 3 × 15 = 45

∴ The correct answer is 15 coins of ₹10, 30 of ₹5, and 45 of ₹2.

Additional Information

Total Value of Coins

The total value of any specific coin denomination is found by multiplying the face value of the coin by the number of coins of that denomination.

Coin Ratio vs Value Ratio

If the ratio of the number of coins is given, multiply it by their respective denominations to get the ratio of their values. Conversely, if the ratio of values is given, divide by their denominations to get the ratio of the number of coins.

Proportion Rule

When values are given in a ratio a : b : c, they can be represented as ax, bx, and cx where x is a common multiplier.

Question 303

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

A tank holds 5345\frac{3}{4} liters of water. If 2.375 liters are drained twice, how much water remains?

  1. AOption (a)

    1

  2. BOption (b)

    1141\frac{1}{4}

  3. COption (c)

    1381\frac{3}{8}

  4. DOption (d)

    1121\frac{1}{2}

View answer & explanation

Correct answer

Answer: (a)A.

1

Why this answer is correct

Shortcut Trick

Convert the fractions into decimals to make calculation faster.

Total capacity = 5¾ = 5.75 L

Total drained = 2.375 × 2 = 4.75 L

Remaining water = 5.75 − 4.75 = 1 L

∴ The correct answer is 1 liter.

Alternate Method

Given:

Total water = 5¾ L

Water drained each time = 2.375 L

Number of times drained = 2

Drained: 4.75 LLeft: 1 LTotal Water: 5.75 L

Calculations:

⇒ Total water = 23/4 L

⇒ Total water drained = 2 × (19/8) = 19/4 L

⇒ Remaining water = Total water − Total water drained

⇒ Remaining water = 23/4 − 19/4 = 4/4 = 1 L

∴ The correct answer is 1 liter.

Additional Information

Converting Mixed Fractions

To convert a mixed fraction a(b/c) to an improper fraction, multiply the whole number by the denominator and add the numerator: (a × c + b) / c.

Fraction to Decimal Conversion

Common standard conversions: 1/2 = 0.5, 1/4 = 0.25, 3/4 = 0.75, 1/8 = 0.125, 3/8 = 0.375. Memorizing these saves time in exams.

Operations on Fractions

When adding or subtracting fractions, always take the Least Common Multiple (LCM) of the denominators to equate them before computing the numerators.

Question 304

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

Simplify:[(23)+(45÷27)]÷[(32)−(56×34)]\left[\left(\frac{2}{3}\right) + \left(\frac{4}{5} \div \frac{2}{7}\right)\right] \div \left[\left(\frac{3}{2}\right)-\left(\frac{5}{6} \times \frac{3}{4}\right)\right]

  1. AOption (a)

    416105\frac{416}{105}

  2. BOption (b)

    418110\frac{418}{110}

  3. COption (c)

    425250\frac{425}{250}

  4. DOption (d)

    526350\frac{526}{350}

View answer & explanation

Correct answer

Answer: (a)A.

416105\frac{416}{105}

Why this answer is correct

Shortcut Trick

Solve Numerator: 2/3 + (4/5 × 7/2) = 2/3 + 14/5 = 52/15

Solve Denominator: 3/2 − (15/24) = 3/2 − 5/8 = 7/8

Divide them: 52/15 ÷ 7/8 = 52/15 × 8/7 = 416/105

∴ The correct answer is 416/105.

Alternate Method

Given:

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

Formula Used:

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

Calculations:

⇒ Numerator = (2/3) + (4/5 ÷ 2/7)

⇒ Numerator = (2/3) + (4/5 × 7/2)

⇒ Numerator = (2/3) + (14/5)

⇒ Numerator = (10/15) + (42/15) = 52/15

⇒ Denominator = (3/2) − (5/6 × 3/4)

⇒ Denominator = (3/2) − (15/24)

⇒ Denominator = (3/2) − (5/8)

⇒ Denominator = (12/8) − (5/8) = 7/8

⇒ Result = (52/15) ÷ (7/8)

⇒ Result = (52/15) × (8/7)

⇒ Result = (52 × 8) / (15 × 7)

⇒ Result = 416/105

∴ The correct answer is 416/105.

Additional Information

BODMAS Rule

Dictates the correct sequence of operations in mathematical expressions: Brackets first, then Orders (roots/powers), Division, Multiplication, Addition, and finally Subtraction.

Fraction Division

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

Fraction Addition and Subtraction

Find the Least Common Multiple (LCM) of the denominators to ensure fractions are equivalent before proceeding to add or subtract their respective numerators.

Question 305

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

Simplify:(212+3.6)÷0.75+(1.25×45)\left(2\frac{1}{2} + 3.6\right) \div 0.75 + \left(1.25 \times \frac{4}{5}\right)

  1. AOption (a)

    13715\frac{137}{15}

  2. BOption (b)

    16015\frac{160}{15}

  3. COption (c)

    12715\frac{127}{15}

  4. DOption (d)

    14815\frac{148}{15}

View answer & explanation

Correct answer

Answer: (a)A.

13715\frac{137}{15}

Why this answer is correct

Shortcut Trick

Convert decimals to fractions for faster calculation.

⇒ (5/2 + 18/5) ÷ 3/4 + (5/4 × 4/5)

⇒ (61/10) × 4/3 + 1

⇒ 122/15 + 1 = 137/15

∴ The correct answer is 137/15.

Alternate Method

Given:

(2 1/2 + 3.6) ÷ 0.75 + (1.25 × 4/5)

Method Used:

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

Calculations:

⇒ Step 1: Solve the brackets first.

⇒ First bracket: 2 1/2 + 3.6 = 2.5 + 3.6 = 6.1

⇒ Second bracket: 1.25 × 4/5 = 1.25 × 0.8 = 1

⇒ Step 2: Substitute values back into the expression.

⇒ 6.1 ÷ 0.75 + 1

⇒ Step 3: Perform division.

⇒ 6.1 ÷ (3/4) = 6.1 × 4/3

⇒ 24.4 / 3 = 244 / 30 = 122 / 15

⇒ Step 4: Perform addition.

⇒ 122 / 15 + 1 = (122 + 15) / 15

⇒ 137 / 15

∴ The correct answer is 137/15.

Additional Information

BODMAS Rule

Always strictly follow the sequence of operations: Brackets → Orders → Division → Multiplication → Addition → Subtraction to ensure the final result is accurate.

Fraction to Decimal Conversion

Converting common decimals into fractions (e.g., 0.75 to 3/4 and 1.25 to 5/4) allows division to become simple multiplication by the reciprocal.

Adding Fractions

To add a whole number to a fraction, express the whole number as a fraction with the same denominator: a/b + c = (a + b × c) / b.

Question 306

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

A and B invested in a business in the ratio 5:6. The profit ratio was 10:9. If B invested his money for 12 months, for how many months did A invest?

  1. AOption (a)

    15 months

  2. BOption (b)

    16 months

  3. COption (c)

    17 months

  4. DOption (d)

    18 months

View answer & explanation

Correct answer

Answer: (b)B.

16 months

Why this answer is correct

Shortcut Trick

Time Ratio = Profit Ratio ÷ Investment Ratio

Time Ratio of A : B = (10 ÷ 5) : (9 ÷ 6) = 2 : 1.5 = 4 : 3

Here, B invested for 12 months, which corresponds to 3 parts in the ratio.

So, A's investment time (4 parts) = (12 ÷ 3) × 4 = 16 months.

∴ The correct answer is 16 months.

Alternate Method

Given:

Ratio of investments of A and B = 5 : 6

Ratio of profits of A and B = 10 : 9

Time period of investment for B = 12 months

Formula Used:

Profit = Investment × Time

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

Partnership: Capital × Time = Profit

Partner A

Capital = 5 units Time = t months Profit Ratio = 10

Partner B

Capital = 6 units Time = 12 months Profit Ratio = 9 VS

Calculations:

Let the time period for A's investment be TA months.

⇒ PA / PB = (IA × TA) / (IB × TB)

⇒ 10 / 9 = (5 × TA) / (6 × 12)

⇒ 10 / 9 = 5TA / 72

⇒ 10 × 72 = 9 × 5TA

⇒ 720 = 45TA

⇒ TA = 720 ÷ 45 = 16

∴ The correct answer is 16 months.

Additional Information

Simple Partnership

When investments of all the partners are for the same time period, the profit or loss is distributed purely in the ratio of their original capital investments.

Compound Partnership

When investments of the partners are for different time periods, the profit or loss is distributed in the ratio of the product of their capital and respective time periods (Profit Ratio = Capital × Time).

Working and Sleeping Partner

A working partner actively manages the business and usually receives a fixed salary or a specific percentage of the profit before the final division. A sleeping partner simply invests money and only receives a share of the remaining profit based on their capital and time invested.

Question 307

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

The average of 9 numbers is 56. When one number is removed, the average becomes 54. What number was removed?

  1. AOption (a)

    52

  2. BOption (b)

    72

  3. COption (c)

    82

  4. DOption (d)

    92

View answer & explanation

Correct answer

Answer: (b)B.

72

Why this answer is correct

Shortcut Trick

Removed Number = Old Average + Remaining Numbers × Decrease in Average

⇒ Removed Number = 56 + 8 × (56 − 54)

⇒ Removed Number = 56 + 8 × 2

⇒ Removed Number = 56 + 16

⇒ Removed Number = 72

∴ The correct answer is 72.

Alternate Method

Given:

Average of 9 numbers = 56

Average of remaining 8 numbers (after one is removed) = 54

Formula Used:

Total Sum = Average × Number of items

Total Sum of 9 numbers = 9 × 56 = 504

Sum of 8 numbers = 8 × 54 = 432

72

Removed

Calculation:

⇒ Initial total sum of 9 numbers = 9 × 56 = 504

⇒ New total sum of remaining 8 numbers = 8 × 54 = 432

⇒ The removed number = Initial total sum − New total sum

⇒ The removed number = 504 − 432 = 72

∴ The correct answer is 72.

Additional Information

Averages

The average is the central value of a set of numbers, calculated as the sum of all values divided by the number of values: Average = Sum of observations ÷ Total number of observations.

Deviation Method

The net deviation from the average across all terms is always zero. This property helps in quickly finding missing numbers, replacing numbers, or discovering new averages without multiplying large values.

Weighted Average

When combining groups, the combined average is calculated as (n1A1 + n2A2) ÷ (n1 + n2), where n represents the number of items and A is the average of each respective group.

Question 308

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

Table - Weekly Fuel Usage (litres)

VehicleMonTueWedThuFri
Truck 11820222119
Truck 22524262324

Q: What is the overall average daily fuel usage for both trucks?

  1. AOption (a)

    27 litres

  2. BOption (b)

    22.2 litres

  3. COption (c)

    28 litres

  4. DOption (d)

    21.5 litres

View answer & explanation

Correct answer

Answer: (b)B.

22.2 litres

Why this answer is correct

Shortcut Trick

Average of Truck 1 = (18 + 20 + 22 + 21 + 19) ÷ 5 = 100 ÷ 5 = 20 litres

Average of Truck 2 = (25 + 24 + 26 + 23 + 24) ÷ 5 = 122 ÷ 5 = 24.4 litres

Overall Average = (20 + 24.4) ÷ 2 = 44.4 ÷ 2 = 22.2 litres

∴ The correct answer is 22.2 litres.

Alternate Method

Given:

Fuel usage of Truck 1 for 5 days: 18, 20, 22, 21, 19

Fuel usage of Truck 2 for 5 days: 25, 24, 26, 23, 24

Formula Used:

Average = Sum of all observations ÷ Total number of observations

Calculations:

⇒ Total fuel used by Truck 1 = 18 + 20 + 22 + 21 + 19 = 100 litres

⇒ Total fuel used by Truck 2 = 25 + 24 + 26 + 23 + 24 = 122 litres

⇒ Total fuel used by both trucks = 100 + 122 = 222 litres

⇒ Total number of observations = 5 days × 2 trucks = 10

⇒ Overall average = 222 ÷ 10 = 22.2 litres

∴ The correct answer is 22.2 litres.

Additional Information

Combined Average

If two groups have sizes n1 and n2 with averages A1 and A2, the combined average is calculated as (n1A1 + n2A2) ÷ (n1 + n2).

Data Interpretation (Tabulation)

In tabular data interpretation, it is often faster to find sums of individual rows or columns first before computing total averages, avoiding multiple complex calculations.

Percentage Change

Frequently tested in DI questions. Percentage change is mathematically given as ((New Value − Old Value) ÷ Old Value) × 100.

Question 309

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

A sum becomes ₹12100 in 2 years at 10% Compound Interest annually. Find the principal.

  1. AOption (a)

    ₹10000

  2. BOption (b)

    ₹11000

  3. COption (c)

    ₹11500

  4. DOption (d)

    ₹10500

View answer & explanation

Correct answer

Answer: (a)A.

₹10000

Why this answer is correct

Shortcut Trick

Rate = 10% = 1 / 10. Multiplying factor = 11 / 10.

Ratio of Principal to Amount for 2 years = (10)2 : (11)2 = 100 : 121.

Here, 121 units correspond to ₹12100.

1 unit = ₹12100 ÷ 121 = ₹100.

Principal = 100 units = 100 × ₹100 = ₹10000.

∴ The correct answer is ₹10000.

Alternate Method

Given:

Amount (A) = ₹12100

Time (T) = 2 years

Rate (R) = 10% per annum

Formula Used:

A = P × (1 + R / 100)T

Where A = Amount, P = Principal, R = Rate, and T = Time.

+10% CI +10% CI

Year 0 Year 1 Year 2

P = ? A = ₹12100

Calculations:

⇒ 12100 = P × (1 + 10 / 100)2

⇒ 12100 = P × (1 + 1 / 10)2

⇒ 12100 = P × (11 / 10)2

⇒ 12100 = P × (121 / 100)

⇒ P = (12100 × 100) ÷ 121

⇒ P = 100 × 100 = 10000

∴ The correct answer is ₹10000.

Additional Information

Effective Compound Interest Rate

For 2 years, the effective CI rate can be found using the successive percentage formula: R + R + (R × R) / 100. For 10%, the effective rate is 10 + 10 + (10 × 10) / 100 = 21%.

Difference Between SI and CI (2 Years)

The difference between Compound Interest and Simple Interest for exactly 2 years can be directly calculated using the formula: Difference = P × (R / 100)2.

Half-Yearly Compounding

When interest is compounded half-yearly, the rate of interest is halved (R / 2) and the given time period is doubled (2T) because interest is computed every 6 months.

Question 310

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

A grocer purchased 50 kg of rice at ₹40 per kg and 20 kg of another variety of rice at ₹55 per kg. He mixed the two varieties and sold the entire mixture at ₹48 per kg. Find his total profit or loss in this transaction.

  1. AOption (a)

    Loss of ₹260

  2. BOption (b)

    Loss of ₹240

  3. COption (c)

    Profit of ₹260

  4. DOption (d)

    Loss of ₹290

View answer & explanation

Correct answer

Answer: (c)C.

Profit of ₹260

Why this answer is correct

Shortcut Trick

Find the profit or loss per kg for each variety relative to the selling price.

Profit on 50 kg rice = (48 − 40) × 50 = +8 × 50 = +₹400

Loss on 20 kg rice = (48 − 55) × 20 = −7 × 20 = −₹140

Net Profit = 400 − 140 = ₹260

∴ The correct answer is Profit of ₹260.

Alternate Method

Given:

Quantity 1 = 50 kg, Rate 1 = ₹40/kg

Quantity 2 = 20 kg, Rate 2 = ₹55/kg

Selling rate of mixture = ₹48/kg

Formula Used:

Total CP = (Q1 × R1) + (Q2 × R2)

Profit = Total SP − Total CP

50 kg @ ₹40 CP: ₹2000

20 kg @ ₹55 CP: ₹1100

Total CP ₹3100

70 kg @ ₹48 SP: ₹3360

Profit = SP − CP = 3360 − 3100 = ₹260

Calculations:

⇒ Cost Price of 50 kg of rice = 50 × 40 = ₹2000

⇒ Cost Price of 20 kg of rice = 20 × 55 = ₹1100

⇒ Total Cost Price (CP) = 2000 + 1100 = ₹3100

⇒ Total quantity of mixture = 50 + 20 = 70 kg

⇒ Total Selling Price (SP) of the mixture = 70 × 48 = ₹3360

⇒ Since SP > CP, there is a profit.

⇒ Profit = Total SP − Total CP = 3360 − 3100 = ₹260

∴ The correct answer is Profit of ₹260.

Additional Information

Profit and Loss Basic Formula

Profit occurs when Selling Price > Cost Price. Profit = SP − CP.

Profit and Loss Percentage

Profit % = (Profit ÷ CP) × 100. Loss % = (Loss ÷ CP) × 100.

Weighted Average Concept

Average Cost Price per unit of a mixture is calculated as (Q1 × C1 + Q2 × C2) ÷ (Q1 + Q2).