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Putting all the values in a table we get

 

Average number of 'Q' type of 'quantities' displaced by warehouses 'B', 'C', and 'D' = (350 + 250 + 450)/3 = 350

Hence, the correct answer is 350.

Putting all the values in a table we get

 

The total cost beared by all four warehouses from selling all 'P' type of quantities = 275 × 1350 = 371250 Rs.

Hence, the correct answer is 371250 Rs.

Putting all the values in a table we get

 

The number of 'Q' type of displaced 'quantities' for warehouse 'E' = (13/8) × 400 = 650

The number of 'P' type of displaced 'quantities' for warehouse 'E' = 650 + 250 = 900

Total number of quantities displaced by warehouse 'E' = 650 + 900 = 1550

Hence, the correct answer is 1550.

Putting all the values in a table we get

 

The total number of 'P' type of quantities displaced by warehouses 'A', 'C', and 'D' = (350 + 400 + 350) = 1100

Required percentage = (1100/2800) × 100 = 39.28 ~ 39%

Hence, the correct answer is 39%.

Putting all the values in a table we get

 

The total number of quantities that are accepted and not returned back to all four warehouses = (4/7) × 2800 = 1600

Hence, the correct answer is 1600.

Shortcut Trick

Ratio of A : B is 7 : 8 (Present) and 8 : 9 (After 6 years).

Change in ratio units for both is 8 − 7 = 1 unit and 9 − 8 = 1 unit.

Since the time gap is 6 years, 1 unit → 6 years.

A's present age = 7 units × 6 = 42 years.

C's present age = 42 + 10 = 52 years.

∴ The correct answer is 52 years.

Alternate Method

Given: Present age ratio A : B = 7 : 8. After 6 years, ratio A : B = 8 : 9. C = A + 10.

Calculation:

⇒ Let present ages of A and B be 7x and 8x.

⇒ (7x + 6) ÷ (8x + 6) = 8 ÷ 9

⇒ 9 × (7x + 6) = 8 × (8x + 6)

⇒ 63x + 54 = 64x + 48

⇒ 64x − 63x = 54 − 48

⇒ x = 6

⇒ A's present age = 7 × 6 = 42 years

⇒ C's present age = 42 + 10 = 52 years

∴ The correct answer is 52 years.

Additional Information

Age Difference Property

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

Ratio Balancing Method

If the difference in ratio parts is not equal, multiply the ratios by the difference of the other ratio to balance them before solving.

Linear Equation in Ages

For any 'n' years after/before, add/subtract 'n' to the current age variables to form the equation: (Ax + n) / (Bx + n) = New Ratio.

Shortcut Trick

Let the Initial Amount be X.

Total percentage given to Son and Wife = 50% + 30% = 80%. Remaining = 20%.

Donation to trust = 62.5% of 20% = 5/8 × 20% = 12.5%. Final remaining = 20% − 12.5% = 7.5%.

7.5% of X = 15,000 ⇒ (3/40) × X = 15,000 ⇒ X = 2,00,000.

∴ The correct answer is ₹2,00,000.

Alternate Method

Given: Son = 50%, Wife = 30%, Trust = 62.5% of remaining, Left = ₹15,000.

Calculations:

⇒ Let total money = 100 units

⇒ Spent on Son and Wife = 50 + 30 = 80 units

⇒ Remaining money = 100 − 80 = 20 units

⇒ Donation to trust = 62.5% of 20 = (5/8) × 20 = 12.5 units

⇒ Final remaining amount = 20 − 12.5 = 7.5 units

⇒ 7.5 units = 15,000

⇒ 1 unit = 15,000 ÷ 7.5 = 2,000

⇒ Total money (100 units) = 100 × 2,000 = 2,00,000

∴ The correct answer is ₹2,00,000.

Additional Information

Percentage to Fraction Conversion

62.5% can be written as 5/8 and 12.5% as 1/8 to speed up calculations in competitive exams.

Successive Percentage Remainder

When spending from the 'remaining' amount, multiply the initial value by the remaining fractions: Final = Initial × (1 − r₁) × (1 − r₂).

Shortcut Trick

Profit ratio is the product of Capital and Time ratios.

Capital ratio (Akshay : Bikash) = 24,000 : 18,000 = 4 : 3.

Time ratio = Profit ratio ÷ Capital ratio.

Time ratio = (16 ÷ 4) : (7 ÷ 3) = 4 : (7/3) = 12 : 7.

Since Akshay invested for 12 months, Bikash invested for 7 months.

Bikash joined after = 12 − 7 = 5 months.

∴ The correct answer is 5 months.

Alternate Method

Given → Akshay Capital = Rs. 24,000, Bikash Capital = Rs. 18,000, Akshay Time = 12 months, Profit Ratio = 16 : 7. Formulas: Profit Ratio = (C₁ × T₁) : (C₂ × T₂).

⇒ Let Bikash invested for t months. ⇒ (24,000 × 12) ÷ (18,000 × t) = 16 ÷ 7. ⇒ (4 × 12) ÷ (3 × t) = 16 ÷ 7. ⇒ 48 ÷ 3t = 16 ÷ 7. ⇒ 16 ÷ t = 16 ÷ 7. ⇒ t = 7 months. ⇒ Bikash joined after = 12 − 7 = 5 months.

∴ The correct answer is 5 months.

Additional Information

Partnership Basics

In a partnership, the profit is distributed among partners in the ratio of their investment multiplied by the duration of investment.

Joining vs Duration

If a person joins after 'x' months in a one-year business, the investment duration is (12 − x) months.

Shortcut Trick

Efficiency of Pipe A = 15, Pipe B = 12, Pipe C = −10 (Total Work 180).

Water filled in 5 hours by (A + B) = (15 + 12) × 5 = 135 units.

Time taken by Pipe C to empty this water = 135 ÷ 10 = 13.5 hours.

∴ The correct answer is 13.5 hr.

Alternate Method

Given: Filling time of Pipe 1 = 12 hr, Pipe 2 = 15 hr. Emptying time of Pipe 3 = 18 hr.

Formula Used: Efficiency = Total Work ÷ Time

⇒ Combined efficiency of first two pipes = 15 + 12 = 27 units/hr.

⇒ Work done in 5 hours = 27 × 5 = 135 units.

⇒ Time taken by third pipe to empty = 135 ÷ 10 = 13.5 hours.

∴ The correct answer is 13.5 hr.

Additional Information

Negative Efficiency

In Pipes and Cisterns, an outlet or leak is represented by negative efficiency as it reduces the total volume of water.

LCM Method for Total Work

Taking the Least Common Multiple of individual times as the total capacity of the tank simplifies calculations to integers.

Emptying the Filled Part

If a tank is partially filled and then only the outlet pipe is opened, the time taken is calculated based on the current volume of water, not the full capacity.

Shortcut Trick

Let the usual speed be S. Distance (D) = 400 km. Speed increase (x) = 10 km/h. Time difference (t) = 2 hours.

Using the formula: Distance = [S × (S + x) / x] × t

⇒ 400 = [S × (S + 10) / 10] × 2

⇒ 2000 = S(S + 10). By observation, S = 40 km/h (as 40 × 50 = 2000).

⇒ Usual Time = Distance ÷ Usual Speed = 400 ÷ 40 = 10 hours.

∴ The correct answer is 10 hours.

Alternate Method

Given: Total Distance = 400 km, Speed difference = 10 km/h, Time difference = 2 hours.

Formula Used: Time = Distance / Speed

⇒ Let usual speed be x km/h.

⇒ 400/x − 400/(x + 10) = 2

⇒ 400 [(x + 10 − x) / (x(x + 10))] = 2

⇒ 400 [10 / (x² + 10x)] = 2

⇒ 4000 = 2(x² + 10x)

⇒ x² + 10x − 2000 = 0

⇒ x² + 50x − 40x − 2000 = 0 ⇒ (x + 50)(x − 40) = 0

⇒ x = 40 km/h (Speed cannot be negative).

⇒ Usual Time = 400 / 40 = 10 hours.

∴ The correct answer is 10 hours.

Additional Information

Speed-Time Relation

For a constant distance, speed is inversely proportional to time (S ∝ 1/T). If speed increases, time decreases.

General Distance Formula

Distance = [Product of Speeds / Difference of Speeds] × Time Difference. This is useful for all 'late/early' or speed-change problems.

Average Speed

When a journey is split into two equal distances with speeds V₁ and V₂, Average Speed = 2V₁V₂ / (V₁ + V₂).