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John salary = ₹12000. Rahul salary = 80% of 12000 = ₹9600.

Spent on Food (20%) + Furniture (25%) + Rent (30%) = 75%.

Remaining for Books = 100% − 75% = 25%.

Total for Books + Furniture = 25% + 25% = 50%.

Required Amount = 50% of ₹9600 = ₹4800.

∴ The correct answer is ₹4800.

Alternate Method

Given: John salary = ₹12000, Rahul salary = 20% less than John, Expenses: Food = 20%, Furniture = 25%, Rent = 30%.

Calculations:

⇒ Rahul salary = 12000 − (20% of 12000)

⇒ Rahul salary = 12000 − 2400 = ₹9600

⇒ Percentage spent on Books = 100 − (20 + 25 + 30) = 25%

⇒ Total % for Books and Furniture = 25% + 25% = 50%

⇒ Required Amount = 50% of 9600 = 9600 ÷ 2 = ₹4800

∴ The correct answer is ₹4800.

Additional Information

Percentage Change

To find a value x% less than a base B, use the formula: B × (100 − x)/100.

Successive Percentage Expenditure

If expenses are given as a % of the 'total' salary, they are additive. If given as % of 'remaining', they are multiplicative.

Base Value Concept

Always identify the correct base (Rahul's salary vs John's salary) before applying percentage calculations.

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Total Work (LCM of 15, 18, 10) = 90 units.

Efficiency of (A+B+C) = 90 ÷ 10 = 9 units/day.

Efficiency of (B+C) = 90 ÷ 18 = 5 units/day ⇒ Efficiency of A = 9 − 5 = 4 units.

Efficiency of (A+B) = 90 ÷ 15 = 6 units/day ⇒ Efficiency of B = 6 − 4 = 2 units.

Wage ratio = Efficiency ratio ⇒ A : B = 4 : 2 = 2 : 1.

If 2 units = ₹600, then 1 unit (B’s wage) = ₹300.

∴ The correct answer is ₹300.

Alternate Method

Given: Time taken by (A+B) = 15 days, (B+C) = 18 days, (A+B+C) = 10 days. A’s wage = ₹600.

Formula: Efficiency = Total Work ÷ Time; Wage ∝ Efficiency.

⇒ Efficiency of A = Efficiency(A+B+C) − Efficiency(B+C) = 9 − 5 = 4

⇒ Efficiency of B = Efficiency(A+B) − Efficiency(A) = 6 − 4 = 2

⇒ Wage is distributed in the ratio of efficiencies of A : B = 4 : 2 = 2 : 1

⇒ Since A’s wage (2 units) = ₹600

⇒ B’s wage (1 unit) = 600 ÷ 2 = ₹300

∴ The correct answer is ₹300.

Additional Information

Work and Wages Relationship

Wages are always distributed in the ratio of work done. If all persons work for the same duration, the wage ratio is identical to their efficiency ratio.

LCM Method for Total Work

Total work is assumed as the Least Common Multiple (LCM) of the individual days to ensure efficiencies are whole numbers, simplifying calculations.

Efficiency of Individual in a Group

Efficiency of one person can be found by subtracting the combined efficiency of others from the total combined efficiency: Eff(A) = Eff(A+B+C) − Eff(B+C).

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Initial average = 56.2 kg.

Calculate deviations of new weights from 56.2: (62.3 − 56.2) + (59.7 − 56.2) + (59.5 − 56.2) = 6.1 + 3.5 + 3.3 = 12.9 kg.

Net change in average = Total Deviation ÷ Total children = 12.9 ÷ 63 ≈ 0.20 kg.

New average = 56.2 + 0.20 = 56.40 kg.

∴ The correct answer is 56.40 kg.

Alternate Method

Given: Initial average weight of 60 children = 56.2 kg. Weights of 3 new children = 62.3 kg, 59.7 kg, and 59.5 kg.

Formula Used: Average = (Sum of all observations) ÷ (Total number of observations)

⇒ Total weight of 60 children = 60 × 56.2 = 3372 kg.

⇒ Total weight of 63 children = 3372 + 62.3 + 59.7 + 59.5

⇒ Total weight of 63 children = 3372 + 181.5 = 3553.5 kg.

⇒ New Average = 3553.5 ÷ 63

⇒ New Average = 56.408... ≈ 56.40 kg.

∴ The correct answer is 56.40 kg.

Additional Information

Deviation Method for Average

New Average = Old Average + (Sum of deviations ÷ New total count). This avoids large multiplications.

Weighted Average

When two groups are combined, Combined Average = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂), where n is the count and x is the average.

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Total Interest = ₹640

Interest from First Sum = 600 × 12% × 6 = ₹432

Interest from Second Sum = 640 − 432 = ₹208

Second Sum = (208 × 100) ÷ (10 × 6) = 2080 ÷ 6 = ₹346.66

∴ The correct answer is ₹346.66.

Alternate Method

Given:

Principal 1 (P₁) = ₹600, Rate 1 (R₁) = 12%, Time (T) = 6 years

Rate 2 (R₂) = 10%, Total Simple Interest (SI_total) = ₹640

Formula Used:

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

⇒ SI₁ = (600 × 12 × 6) ÷ 100

⇒ SI₁ = 6 × 12 × 6 = ₹432

⇒ SI₂ = SI_total − SI₁

⇒ SI₂ = 640 − 432 = ₹208

⇒ SI₂ = (P₂ × 10 × 6) ÷ 100

⇒ 208 = (P₂ × 60) ÷ 100

⇒ P₂ = (208 × 100) ÷ 60

⇒ P₂ = 2080 ÷ 6 = ₹346.66

∴ The correct answer is ₹346.66.

Additional Information

Simple Interest Property

Simple interest remains constant every year because it is always calculated on the original Principal amount.

Amount Formula

Total Amount (A) = Principal (P) + Simple Interest (SI). In this question, the interest was additive.

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Ratio of Ankit : Son now is 3 : 1 (Difference = 2 units).

Ratio 4 years ago was 13 : 3 (Difference = 10 units).

Equating differences: Multiply present ratio by 5 → 15 : 5.

Change in units: 15 − 13 = 2 units → 4 years. So, 1 unit = 2 years.

Ankit's present age = 15 × 2 = 30 years.

∴ The correct answer is 30 years.

Alternate Method

Given: Present age ratio is 3 : 1. Ratio 4 years ago was 13 : 3.

Let the present age of the son be x and Ankit be 3x.

⇒ (3x − 4) / (x − 4) = 13 / 3

⇒ 3 × (3x − 4) = 13 × (x − 4)

⇒ 9x − 12 = 13x − 52

⇒ 13x − 9x = 52 − 12

⇒ 4x = 40

⇒ x = 10

⇒ Ankit's present age = 3x = 3 × 10 = 30 years.

∴ The correct answer is 30 years.

Additional Information

Concept of Age Difference

The difference between the ages of two people always remains constant regardless of the number of years passed.

Ratio Balancing Method

To solve age ratio problems quickly, multiply the ratios such that the difference between the parts (e.g., 15−5=10 and 13−3=10) becomes equal.

Calculations:

Total number of items brought by A in 2016 = 140 

Number of plastic items brought by A = 140 × 2 /7 = 40

Calculations:

Total number of items brought by A and B in 2015 = 120 + 130 = 250

Total number of items brought by A and B in 2016 = 140 + 160 = 300

Required ratio = 250 : 300 = 5:6

Calculations:

Total number of items brought by A and B in 2016 = 140 + 160 = 300

Average number of items brought by A and B in 2016 = 300/2 = 150

Number of items brought by C in the year 2016:

= 150 + 45 = 195

Calculations:

Number of items brought by A in 2017 = 180

Number of items brought by B in 2017 = 150

Total item brought by A and B together in the year 2018 is:

[180 × 125/100] + [150 ×120/100] = 225 + 180 = 405

Calculations:

Total number of items brought by A in the year 2015 and 2016 together = 120 + 140 = 260

Total number of items brought by B in the year 2015 and 2016 together = 130 + 160 = 290

So, required difference is 290 – 260 = 30