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Shortcut Trick

Average of ratio parts (7, 9, 11) = (7 + 9 + 11) ÷ 3 = 9 units.

Given average price = Rs. 14,265.

9 units = Rs. 14,265 ⇒ 1 unit = Rs. 1,585.

Price of the costliest item (11 units) = 11 × 1,585 = Rs. 17,435.

∴ The correct answer is Rs. 17,435.

Alternate Method

Given: Average price of 3 items = Rs. 14,265; Ratio of prices = 7 : 9 : 11.

Formula Used: Sum of values = Average × Number of values.

7x 9x 11x Total Ratio Parts = 27x Average (9x) = Rs. 14,265 Costliest Item

⇒ Total Sum of prices = 14,265 × 3 = 42,795

⇒ Let the prices be 7x, 9x, and 11x.

⇒ 7x + 9x + 11x = 42,795

⇒ 27x = 42,795

⇒ x = 42,795 ÷ 27 = 1,585

⇒ Price of costliest item = 11x = 11 × 1,585 = 17,435

∴ The correct answer is Rs. 17,435.

Additional Information

General Average Formula

Average = (Sum of all observations) ÷ (Total number of observations).

Ratio Distribution

To find a specific share in ratio a:b:c, use Share = [Total Sum × specific part] ÷ (a + b + c).

Arithmetic Progression Average

For terms in AP (like our ratio parts 7, 9, 11 with difference 2), the average is always the middle term.

Shortcut Trick

Step 1: Find the LCM of 12, 15, and 20, which is 60.

Step 2: Prime factorisation of 60 is 2² × 3 × 5.

Step 3: To make it a perfect square, multiply by the missing unpaired factors (3 × 5 = 15).

Step 4: The required smallest perfect square is 60 × 15 = 900.

∴ The correct answer is 900.

Alternate Method

Given:

The numbers are 12, 15, and 20.

Concept Used:

The smallest number exactly divisible by a set of numbers is their LCM. A perfect square must have exclusively even powers in its prime factorisation.

Calculations:

⇒ Prime factorisation of 12 = 2² × 3

⇒ Prime factorisation of 15 = 3 × 5

⇒ Prime factorisation of 20 = 2² × 5

⇒ LCM(12, 15, 20) = 2² × 3 × 5 = 60

⇒ In the LCM, the prime factors 3 and 5 have an odd power of 1.

⇒ To convert the LCM into a perfect square, we must multiply it by 3 and 5.

⇒ Required number = 60 × 3 × 5

⇒ Required number = 900

∴ The correct answer is 900.

Additional Information

LCM (Least Common Multiple)

The smallest non-zero common number which is a multiple of all the given numbers. Finding the LCM is the first step in solving simultaneous divisibility problems.

Perfect Squares

A number is considered a perfect square when its prime factorisation consists exclusively of even powers. Example: 36 = 2² × 3².

Divisibility Rules

A number is divisible by 12 if it is divisible by both 3 and 4. Similarly, a number is divisible by 15 if it is divisible by both 3 and 5.

Shortcut Trick

The extra time taken to cross the tunnel compared to the pole is used to cover the tunnel length.

Time difference = 20 s − 5 s = 15 s

Tunnel Length = Speed × Time difference = 90 × 15 = 1350 m.

∴ The correct answer is 1350 m.

Alternate Method

Given:

Speed of the train = 90 m/s

Time taken to cross a pole (t₁) = 5 s

Time taken to cross the tunnel (t₂) = 20 s

Formula Used:

Distance = Speed × Time

When crossing a pole, Distance = Length of Train (L_T)

When crossing a tunnel, Distance = Length of Train (L_T) + Length of Tunnel (L_Tunnel)

Train

Pole

Tunnel

t1 = 5s

t2 - t1 = 15s Speed = 90 m/s

Calculations:

⇒ L_T = 90 × 5 = 450 m

⇒ L_T + L_Tunnel = 90 × 20

⇒ 450 + L_Tunnel = 1800

⇒ L_Tunnel = 1800 − 450

⇒ L_Tunnel = 1350 m

∴ The correct answer is 1350 m.

Additional Information

Train Crossing Stationary Objects

When a train crosses a pole or a standing person, it covers a distance equal to its own length. When it crosses a platform or tunnel, it covers Distance = Length of Train + Length of Object.

Unit Conversion

To convert speed from km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Relative Speed

If two trains of lengths L₁ and L₂ cross each other, the total distance covered is always L₁ + L₂. Relative speed is S₁ + S₂ in opposite directions and |S₁ − S₂| in the same direction.

Shortcut Trick

Use the constant ratio formula: SP₁ / (Qty₁ × %Value₁) = SP₂ / (Qty₂ × %Value₂)

Here, %Value = (100 − Loss%) or (100 + Gain%)

60 / (50 × 90) = 20 / (n × 120)

1 / 75 = 1 / 6n

6n = 75 ⇒ n = 12.5

∴ The correct answer is 12.5.

Alternate Method

Given: SP₁ = ₹60, Qty₁ = 50 lemons, Loss = 10%, Target Profit = 20%, Target SP₂ = ₹20.

Formula: CP = SP × 100 / (100 − Loss%) | New SP = CP × (100 + Gain%) / 100

CP (100%) Loss 10% SP1 = 90% CP Gain 20% SP2 = 120% CP Quantity ∝ 1/Price (for same total SP)

⇒ SP of 1 lemon (Case 1) = 60 ÷ 50 = ₹1.2

⇒ CP of 1 lemon = 1.2 × 100 ÷ 90 = 120 ÷ 90 = ₹4/3

⇒ Required SP per lemon for 20% gain = (4/3) × 120 ÷ 100 = 4 × 1.2 ÷ 3 = ₹1.6

⇒ Total lemons to be sold for ₹20 = Total Amount ÷ SP per lemon

⇒ Number of lemons = 20 ÷ 1.6 = 200 ÷ 16 = 12.5

∴ The correct answer is 12.5.

Additional Information

Unitary Method for Quantity

When the total expenditure remains constant, the ratio of quantities is inversely proportional to the ratio of price per unit.

Profit/Loss on Selling Price

CP = (SP × 100) ÷ (100 ± P/L%). If SP and CP are known, Profit% = (SP − CP) ÷ CP × 100.

Successive Percentage for Profit

If an item is sold at a loss and then a profit is desired, the total change factor is (1 ± r₁/100) × (1 ± r₂/100).

Shortcut Trick

Time taken to cover Bridge length only = Total time − Time to cross board

Time for 204 m = 52 s − 18 s = 34 s

Speed of train = 204 ÷ 34 = 6 m/s

Length of train = Speed × Time taken to cross board

Length = 6 × 18 = 108 m

∴ The correct answer is 108 m.

Alternate Method

Given:

Length of bridge = 204 m

Time to cross bridge (t₁) = 52 s

Time to cross board (t₂) = 18 s

Formula Used:

Speed = Distance ÷ Time

Train (L)

Board (Point) Case 1: Crossing Board (18 s)

Train (L) Bridge (204 m) Case 2: Crossing Bridge (52 s)

Let the length of the train be L meters and speed be v m/s.

⇒ When crossing a board (point object): Speed (v) = L ÷ 18

⇒ When crossing a bridge: Speed (v) = (L + 204) ÷ 52

⇒ Since speed is constant: L ÷ 18 = (L + 204) ÷ 52

⇒ 52 × L = 18 × (L + 204)

⇒ 52L = 18L + 3672

⇒ 52L − 18L = 3672

⇒ 34L = 3672

⇒ L = 3672 ÷ 34

⇒ L = 108 m

∴ The correct answer is 108 m.

Additional Information

Crossing Point Objects

When a train crosses a pole, person, or board, the distance covered equals the length of the train (L).

Crossing Stationary Objects with Length

When a train crosses a bridge, platform, or tunnel, the total distance covered = length of train (L) + length of object (B).

Unit Conversion

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Shortcut Trick

Total ratio parts = 4 + 3 = 7 units.

Given, 7 units = 28 years ⇒ 1 unit = 4 years.

Since we need the ratio after 4 years, and 4 years = 1 unit, simply add 1 unit to each part.

New ratio = (4 + 1) : (3 + 1) = 5 : 4.

∴ The correct answer is 5 : 4.

Alternate Method

Given: Ratio Namrata (N) : Divya (D) = 4 : 3 and Sum of ages = 28.

Formula Used: Age = (Ratio part ÷ Total parts) × Total Sum

Namrata

Present: 16 yrs

+4 yrs Divya

Present: 12 yrs

+4 yrs

Ratio after 4 yrs: 20 : 16 = 5 : 4

⇒ Let current ages be 4x and 3x.

⇒ 4x + 3x = 28

⇒ 7x = 28 ⇒ x = 4

⇒ Namrata age = 4 × 4 = 16 years

⇒ Divya age = 3 × 4 = 12 years

⇒ Age after 4 years: Namrata = 16 + 4 = 20, Divya = 12 + 4 = 16

⇒ New ratio = 20 : 16 = 5 : 4

∴ The correct answer is 5 : 4.

Additional Information

Constant Age Difference

The difference between the ages of two people always remains the same, regardless of how many years pass.

Ratio Scaling

To find actual values from a ratio a : b when the sum S is known, multiply the total sum by a/(a+b) and b/(a+b).

Age Problems with Two Ratios

When two different ratios for two different time periods are given, use the cross-product method to find the multiplier 'x' quickly.

Shortcut Trick

Total capacity of bucket = LCM(4, 5) = 20 units.

Filling efficiency = 20 ÷ 4 = 5 units/min.

Net efficiency (Fill − Hole) = 20 ÷ 5 = 4 units/min.

Hole efficiency = 5 − 4 = 1 unit/min.

Time to empty = 20 ÷ 1 = 20 min.

∴ The correct answer is 20 min.

Alternate Method

Given: Time taken to fill = 4 min, Time taken with hole = 5 min.

Formula Used: Total Work = Efficiency × Time | Efficiency of Hole = Pipe efficiency − Net efficiency.

⇒ Let the total capacity of the bucket be the LCM of 4 and 5.

⇒ Total Capacity = 20 units.

⇒ Efficiency of the filling pipe = 20 ÷ 4 = 5 units/min.

⇒ Combined efficiency of pipe and hole = 20 ÷ 5 = 4 units/min.

⇒ Efficiency of the hole = 5 − 4 = 1 unit/min (Outflow).

⇒ Time required by the hole to empty the full bucket = Total Capacity ÷ Hole Efficiency

⇒ Time = 20 ÷ 1 = 20 min.

∴ The correct answer is 20 min.

Additional Information

Inlet and Outlet Pipes

Inlet pipes add water (positive efficiency) and outlet pipes (leaks/holes) remove water (negative efficiency).

Direct Formula for Leakage

If a pipe fills in X hours and with a leak it takes Y hours, the leak empties the tank in (X × Y) ÷ (Y − X) hours.

Work and Time Relation

Efficiency is inversely proportional to time taken when the total work (tank capacity) remains constant.

Shortcut Trick

Efficiency Ratio of A : B = 100 : 150 = 2 : 3

Since Time is inversely proportional to Efficiency, Time Ratio A : B = 3 : 2

A takes 15 days ⇒ 3 units = 15 days ⇒ 1 unit = 5 days

Time taken by B = 2 units = 2 × 5 = 10 days

∴ The correct answer is 10 days.

Alternate Method

Given:

Time taken by A = 15 days

B is 50% faster than A.

Formula Used:

Total Work = Days × Efficiency

Calculations:

⇒ Let efficiency of A = 2 units/day

⇒ Efficiency of B = 2 + (50% of 2) = 2 + 1 = 3 units/day

⇒ Total Work = Days of A × Efficiency of A

⇒ Total Work = 15 × 2 = 30 units

⇒ Days taken by B alone = Total Work ÷ Efficiency of B

⇒ Days = 30 ÷ 3 = 10 days

∴ The correct answer is 10 days.

Additional Information

Pipes & Cisterns

This follows the same logic as Time & Work, where inlet pipes have positive efficiency and outlet (leakage) pipes have negative efficiency.

Work & Wages

Total wages are distributed among workers in the ratio of the total work they have individually performed (Efficiency × Time).

Alternate Day Work

When workers work on alternate days, the work is calculated in cycles of 2 or more days based on the number of workers.

Given:

240 ÷ 40 ÷ 2 × 5 + 4 = ?

Concept used:

Follow BODMAS rule to solve this question, as per the order given below,

Calculations:

240 ÷ 40 ÷ 2 × 5 + 4 = ?

⇒ 6 ÷ 2 × 5 + 4 = ?

⇒ 3 × 5  + 4 = ?

⇒ 15 + 4 = ?

⇒ 19 = ?

∴ The value of ? is 19.

Shortcut Trick

Rate % = [(n − 1) × 100] ÷ Time

Given: n = 2 (sum doubles) and Time = 10 years

Rate % = [(2 − 1) × 100] ÷ 10 = 10%

In decimal form: 10 ÷ 100 = 0.1

∴ The correct answer is 0.1.

Alternate Method

Given: Time (T) = 10 years, Amount (A) = 2 × Principal (P)

Formula Used: SI = (P × R × T) ÷ 100; where R is rate per year

Principal (P) Amount (2P)

SI = 2P − P = P Time = 10 Years Simple Interest Growth

⇒ SI = A − P

⇒ SI = 2P − P = P

⇒ P = (P × R × 10) ÷ 100

⇒ 1 = (R × 10) ÷ 100

⇒ 1 = R ÷ 10

⇒ R = 10%

⇒ Decimal representation = 10 ÷ 100 = 0.1

∴ The correct answer is 0.1.

Additional Information

SI Multiplier Rule

If a sum becomes n times at simple interest in T years, the rate R is given by R = [(n − 1) ÷ T] × 100.

Finding Time (T)

Similarly, if the rate R is given, the time taken to become n times is T = [(n − 1) ÷ R] × 100.

SI vs CI Difference

For the first year, SI and CI on the same principal and rate are equal. The difference for 2 years is P(R ÷ 100)².