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Given 

Pipe A usually fills a tank = 2 hours.

Due to leakage tank fill in = 2.5 hours

Formula used 

Work = Time × Efficiency 

Calculation 

LCM of (2, 2.5) = 10 

Let the total work be 10 

A's efficiency = 10/2 = 5/hrs

(A + leakage)'s efficiency = 10/2.5 = 4/hrs

Leakage's efficiency = 1/hrs 

Leakage empty the tank in = 10/1 = 10 hours 

∴ The required answer is 10 hours

Shortcut Trick

Abhinav time = 4 hours | Praveen time = 3.5 hours (7/2 hours).

Assume Distance = LCM(4, 7) = 28 units.

Abhinav speed = 28 ÷ 4 = 7 units/hr | Praveen speed = 28 ÷ (7/2) = 8 units/hr.

At 8 am, Abhinav has covered 7 × 2 = 14 units. Remaining distance = 14 units.

Meeting time after 8 am = Remaining Distance ÷ Relative Speed = 14 ÷ (7 + 8) = 14/15 hours.

14/15 hours = 14/15 × 60 = 56 minutes. Time = 8 : 56 am.

∴ The correct answer is 8 : 56 am.

Alternate Method

Given:

Time taken by Abhinav (T_A) = 10 am − 6 am = 4 hours.

Time taken by Praveen (T_P) = 11 : 30 am − 8 am = 3.5 hours = 7/2 hours.

Let the distance between Mumbai and Bangalore be D.

⇒ Speed of Abhinav (S_A) = D/4

⇒ Speed of Praveen (S_P) = D/(7/2) = 2D/7

⇒ Distance covered by Abhinav by 8 am (2 hours) = (D/4) × 2 = D/2

⇒ Remaining distance at 8 am = D − D/2 = D/2

⇒ Relative speed = S_A + S_P = D/4 + 2D/7 = (7D + 8D)/28 = 15D/28

⇒ Time to meet after 8 am = (D/2) ÷ (15D/28) = (D/2) × (28/15D) = 14/15 hours

⇒ 14/15 hours = (14/15) × 60 minutes = 56 minutes

⇒ Meeting time = 8 : 00 am + 56 minutes = 8 : 56 am

∴ The correct answer is 8 : 56 am.

Additional Information

Relative Speed

When two objects move in opposite directions, their relative speed is the sum of their individual speeds (S₁ + S₂).

Meeting Time Formula

Meeting Time = (Initial Distance between them at the same start time) ÷ (Relative Speed).

Time-Speed Relationship

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

Shortcut Trick

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

Efficiency: A = 60 ÷ 20 = 3 units/day, B = 60 ÷ 60 = 1 unit/day.

Work done by B alone in the last 12 days = 12 × 1 = 12 units.

Remaining work done by A and B together from the beginning = 60 − 12 = 48 units.

Days A and B worked together = 48 ÷ (3 + 1) = 12 days.

∴ The correct answer is 12 days.

Alternate Method

Given:

Time taken by A alone to complete the work = 20 days

Time taken by B alone to complete the work = 60 days

Time taken by B to complete the remaining work = 12 days

Formula Used:

Total Work = Efficiency × Time Taken

Calculations:

⇒ Work completed by B alone in the last 12 days = Efficiency of B × Days = 1 × 12 = 12 units

⇒ Remaining work to be completed by A and B together = Total Work − Work done by B

⇒ Remaining work = 60 − 12 = 48 units

⇒ Combined efficiency of A and B = 3 + 1 = 4 units/day

⇒ Number of days A and B worked together initially = Remaining work ÷ Combined efficiency

⇒ Days A and B worked together = 48 ÷ 4 = 12 days

This means A worked for 12 days from the beginning before leaving the work.

∴ The correct answer is 12 days.

Additional Information

Total Work & Efficiency Trick

Total work is universally taken as the Lowest Common Multiple (LCM) of individual completion times. Efficiency is inversely proportional to the time taken.

Alternate Days Working

When individuals work on alternate days, calculate the work done in a complete cycle (e.g., 2 days for 2 persons) and scale it up to reach the closest multiple near the total work.

Work & Wages Distribution

Wages are always distributed in proportion to the total units of work completed by each individual. If they work for the same duration, wages are directly distributed based on their efficiencies.

Shortcut Trick

When a value is first increased by x% and then decreased by x%, there is always a net decrease.

Net Decrease % = (x² ÷ 100)%

Given x = 13%, Net Decrease % = (13 × 13) ÷ 100 = 169 ÷ 100 = 1.69%

∴ The correct answer is Decrease of 1.69%.

Alternate Method

Given:

Increase in salary = 13%

Decrease in salary = 13%

Calculations:

⇒ Let the initial salary of Neeta be 100 units.

⇒ Salary after 13% increase = 100 + (13% of 100) = 113 units.

⇒ Salary after subsequent 13% decrease = 113 − (13% of 113)

⇒ Decrease amount = (13 ÷ 100) × 113 = 14.69 units.

⇒ Final salary = 113 − 14.69 = 98.31 units.

⇒ Net change = Final salary − Initial salary = 98.31 − 100 = −1.69 units.

⇒ Net percentage change = (−1.69 ÷ 100) × 100% = −1.69%

⇒ The negative sign indicates a decrease in the final value.

∴ The correct answer is Decrease of 1.69%.

Additional Information

Successive Percentage Change

When a value is subjected to two successive percentage changes of a% and b%, the effective net percentage change is given by: a + b + (a × b) ÷ 100. Use a negative sign for any decrease.

Equal Increase and Decrease

If a number is first increased by x% and then decreased by the same x%, the net change is always a decrease, calculated as: Net Loss % = (x² ÷ 100)%.

Base Dependency

A fixed percentage increase followed by the same percentage decrease does not return the value to the original amount because the base for the decrease is larger than the original base.

Shortcut Trick

The LCM of 16, 21, 24, and 28 is 336.

Largest 4-digit number 9999 ÷ 336 leaves a remainder of 255.

Largest 4-digit multiple (x) = 9999 − 255 = 9744.

Sum of digits of 9744 = 9 + 7 + 4 + 4 = 24.

∴ The correct answer is 24.

Alternate Method

Given: Divisors are 16, 21, 24, and 28.

Formula Used: x = Largest 4-digit number − (Largest 4-digit number mod LCM of divisors).

⇒ Prime factorization of 16 = 2⁴

⇒ Prime factorization of 21 = 3 × 7

⇒ Prime factorization of 24 = 2³ × 3

⇒ Prime factorization of 28 = 2² × 7

⇒ LCM = 2⁴ × 3 × 7 = 16 × 21 = 336

⇒ Largest 4-digit number = 9999

⇒ 9999 ÷ 336 ⇒ Quotient = 29, Remainder = 255

⇒ Required number x = 9999 − 255 = 9744

⇒ Sum of digits = 9 + 7 + 4 + 4 = 24

∴ The correct answer is 24.

Additional Information

Least Common Multiple (LCM)

The LCM is the smallest positive integer that is exactly divisible by each of the given numbers. It is calculated by taking the highest power of each prime factor involved.

Divisibility Rule of 3

A number is divisible by 3 if the sum of its digits is a multiple of 3. Since 9744 is a multiple of 21 and 24, it must be divisible by 3.

Division Algorithm

For any dividend and divisor: Dividend = (Divisor × Quotient) + Remainder. To find the largest multiple below a limit, subtract the remainder from that limit.

Shortcut Trick

Difference in profit percentage = 20% − 16% = 4%

This 4% increase in profit corresponds to the increase in selling price of ₹ 100.

4% of CP = ₹ 100

1% of CP = 100 ÷ 4 = ₹ 25

100% of CP = 25 × 100 = ₹ 2500

∴ The correct answer is ₹ 2500.

Alternate Method

Given:

Initial Profit = 16%

Increased Selling Price = ₹ 100

New Profit = 20%

Formula Used:

Selling Price (SP) = CP × (100 + Profit%) ÷ 100

Calculations:

Let the cost price (CP) of the almirah be ₹ x.

⇒ Initial Selling Price (SP₁) = 116% of x = 1.16x

⇒ New Selling Price (SP₂) = 120% of x = 1.20x

⇒ SP₂ − SP₁ = 100

⇒ 1.20x − 1.16x = 100

⇒ 0.04x = 100

⇒ x = 100 ÷ 0.04

⇒ x = 10000 ÷ 4

⇒ x = 2500

∴ The correct answer is ₹ 2500.

Additional Information

Profit and Loss Basics

Profit occurs when SP > CP, and Profit = SP − CP. Profit percentage is always calculated on the Cost Price unless stated otherwise.

Successive Percentage Method

When two profits/losses are compared for the same base (CP), the difference in their percentages is directly proportional to the difference in their absolute values.

Marked Price and Discount

Discount is always calculated on the Marked Price (MP). SP = MP × (100 − Discount%) ÷ 100.

Shortcut Trick

Convert all bases to 3 and compare the powers directly.

LHS: (√3)⁵ × 9² = 3^2.5 × 3⁴ = 3^6.5

RHS: 3^x × 3√3 = 3^x × 3^1.5

Equate the powers: 6.5 = x + 1.5 ⇒ x = 5

∴ The correct answer is 5.

Alternate Method

Given:

(√3)⁵ × 9² = 3^x × 3√3

Formula Used:

Law of Exponents: a^m × aⁿ = a^m+n and (a^m)ⁿ = a^m×n

Calculation:

Convert all terms to the common base of 3:

⇒ (√3)⁵ = (3^1/2)⁵ = 3^5/2

⇒ 9² = (3²)² = 3⁴

⇒ 3√3 = 3¹ × 3^1/2 = 3^3/2

Now substitute these values back into the equation:

⇒ 3^5/2 × 3⁴ = 3^x × 3^3/2

Apply the multiplication rule of exponents on both sides:

⇒ 3^{(5/2 + 4)} = 3^{(x + 3/2)}

⇒ 3^13/2 = 3^{(x + 3/2)}

Since the bases are the same, equate the exponents:

⇒ 13/2 = x + 3/2

⇒ x = 13/2 − 3/2

⇒ x = 10/2

⇒ x = 5

∴ The correct answer is 5.

Additional Information

Product Rule of Exponents

When multiplying terms with the same base, add their exponents: a^m × aⁿ = a^m+n.

Power of a Power Rule

When raising a power to another power, multiply the exponents: (a^m)ⁿ = a^m×n.

Fractional Exponents

Roots can be written as fractional powers: √a = a^1/2 and ⁿ√a = a^1/n.

Equality of Exponents

If two exponential expressions with the same base are equal, their exponents must be equal. If a^x = a^y, then x = y (provided a ≠ 0, 1, or −1).

Shortcut Trick

Total Interest for 3 years at 20% = 20% × 3 = 60%

Amount % = Principal % + Interest % = 100% + 60% = 160%

Given: 160% of Sum = ₹3120

Interest for 1 year = 20% of Sum

Required value = (3120 ÷ 160) × 20 = 3120 ÷ 8 = 390

∴ The correct answer is ₹390.

Alternate Method

Given:

Amount (A) = ₹3120, Rate (R) = 20%, Time (T) = 3 years

Formula used:

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

Amount (A) = P + SI = P × (1 + RT/100)

Year 0 (P) Year 1 Year 2 Year 3 (A=3120)

+20% SI

+20% SI

+20% SI Total Amount = 160% of P

Calculations:

⇒ 3120 = P × (1 + (20 × 3) / 100)

⇒ 3120 = P × (1 + 60/100)

⇒ 3120 = P × (160/100)

⇒ P = (3120 × 100) / 160

⇒ P = 3120 ÷ 1.6 = 1950

⇒ Simple Interest for 1 year = (1950 × 20 × 1) / 100

⇒ Simple Interest for 1 year = 1950 × 0.2 = 390

∴ The correct answer is ₹390.

Additional Information

Simple Interest Nature

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

Amount vs Interest

Total Amount = Principal + (Principal × Rate × Time) / 100. If R=20% and T=1, then SI = 0.2P.

Rate Calculation

The annual rate of interest can be found using R = (SI × 100) / (P × T).

Given:

\(\rm \left(3\frac{2}{3}\ of\ \frac{3}{4}-\frac{1}{4}\ of\ \frac{4}{3}\right)\div\left(\frac{1}{4}\div\frac{3}{2}\right)+1\frac{1}{2}\)

Concept used:

Calculation:

\(\rm \left(3\frac{2}{3}\ of\ \frac{3}{4}-\frac{1}{4}\ of\ \frac{4}{3}\right)\div\left(\frac{1}{4}\div\frac{3}{2}\right)+1\frac{1}{2}\)

⇒ \(\rm \left(\frac{11}{3}\ \times\ \frac{3}{4}-\frac{1}{4}\ \times\ \frac{4}{3}\right)\div\left(\frac{1}{4}\div\frac{3}{2}\right)+1\frac{1}{2}\)

⇒ \(\rm \left(\frac{11}{4}\ -\frac{1}{3}\right)\div\left(\frac{1}{4}\div\frac{3}{2}\right)+1\frac{1}{2}\)

⇒ \(\rm \left(\frac{33; -;4}{12}\right)\div\left(\frac{1}{4}\times\frac{2}{3}\right)+\frac{3}{2}\)

⇒ \(\rm \left(\frac{29}{12}\right)\div\left(\frac{1}{6}\right)+\frac{3}{2}\)

⇒ \(\rm \frac{29}{12}\times6+\frac{3}{2}\)

⇒ 29/2 + 3/2

⇒ 32/2

⇒ 16

∴ Required answer is 16

Given:

Cost price of articles = Rs.490

Loss % = 20%

Profit % = 16%

Formula:

Loss % = (CP - SP)/CP × 100

Profit % = (SP - CP)/CP × 100

Calculation:

Let cost price of one article be Rs.a and cost price of two article be Rs.(490 - a).

Selling price of one article = a × 80/100 = Rs.4a/5

Selling price of other article = (490 - a) × 116/100 = Rs.(568.4 - 29a/25)

Then,

⇒ 4a/5 = 568.4 - 29a/25

⇒ 49a/25 = 568.4

⇒ a = 290

∴ Cost price of article which sold at loss of 20% is Rs.290.