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Shortcut Trick

Use the Ratio Method: 10% discount = 1/10 (MP = 10, SP = 9).

For three successive discounts: (10 × 10 × 10) : (9 × 9 × 9) ⇒ 1000 : 729.

Total discount = 1000 − 729 = 271 units.

Equivalent discount % = (271 ÷ 1000) × 100 = 27.1%.

∴ The correct answer is 27.1%.

Alternate Method

Given: Three successive discounts of 10% each.

Formula: Net Discount for two successive discounts a and b = (a + b − (a × b ÷ 100))%.

⇒ Let the Marked Price (MP) be ₹100.

⇒ First discount = 10% of 100 = ₹10. Price becomes = 100 − 10 = ₹90.

⇒ Second discount = 10% of 90 = ₹9. Price becomes = 90 − 9 = ₹81.

⇒ Third discount = 10% of 81 = ₹8.1. Price becomes = 81 − 8.1 = ₹72.9.

⇒ Total Single Discount = MP − Final Selling Price (SP)

⇒ Total Discount = 100 − 72.9 = 27.1.

⇒ Equivalent Discount Percentage = (27.1 ÷ 100) × 100 = 27.1%.

∴ The correct answer is 27.1%.

Additional Information

Successive Discount for n periods

The final multiplier for successive discounts d₁, d₂, ... d_n is (1 − d₁/100) × (1 − d₂/100) × ... × (1 − d_n/100).

Difference between Discount and Profit

Discount is always calculated on the Marked Price (MP), while Profit or Loss is calculated on the Cost Price (CP).

Mark-up Percentage

The percentage by which the cost price is increased to set the marked price: Mark-up % = [(MP − CP) ÷ CP] × 100.

Shortcut Trick

Let the total capacity of the tank be LCM(18, 9, 27) = 54 units.

Efficiency: M = +3, N = +6, and Leakage = −2.

Net efficiency when all are open = 3 + 6 − 2 = 7 units/hr.

Time required = Total capacity ÷ Net efficiency = 54 ÷ 7 = 7 5/7 hr.

∴ The correct answer is 7 5/7 hr.

Alternate Method

Given:

Time taken by Tap M = 18 hr

Time taken by Tap N = 9 hr

Time taken by Leakage to empty = 27 hr

Calculations:

⇒ Net efficiency of (M + N + Leakage) = 3 + 6 − 2

⇒ Net efficiency = 7 units/hr

⇒ Time taken to fill the tank = Total Work ÷ Net Efficiency

⇒ Time = 54 ÷ 7 hr

⇒ Time = 7 5/7 hr

∴ The correct answer is 7 5/7 hr.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to time. If a pipe fills a tank in 'x' hours, its 1-hour work is 1/x.

Negative Work (Leakage)

In pipes and cisterns, a filling tap does positive work, while a leakage or outlet pipe does negative work.

Combined Rate Formula

Net Rate = (1/M) + (1/N) − (1/L), where L is the time taken by the leakage to empty the tank.

Shortcut Trick

Check the divisibility rule of 4: the last two digits must be divisible by 4.

The last two digits of 2X2 are X2.

Test the options: If X = 1, the number is 12, which is divisible by 4 (12 ÷ 4 = 3).

Testing other digits: 02, 22, and 42 are not divisible by 4.

∴ The correct answer is 1.

Alternate Method

Given: A three-digit number 2X2.

Formula Used: A number is divisible by 4 if the number formed by its last two digits is a multiple of 4.

Calculations:

⇒ Let the digits be checked for X2:

⇒ If X = 1 → 12 ÷ 4 = 3 (Remainder = 0) ∴ Divisible

⇒ If X = 2 → 22 ÷ 4 = 5.5 (Remainder ≠ 0) ∴ Not Divisible

⇒ If X = 0 → 02 ÷ 4 = 0.5 (Remainder ≠ 0) ∴ Not Divisible

⇒ If X = 4 → 42 ÷ 4 = 10.5 (Remainder ≠ 0) ∴ Not Divisible

∴ The correct answer is 1.

Additional Information

Divisibility by 2

A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8).

Divisibility by 8

A number is divisible by 8 if the number formed by its last three digits is divisible by 8.

Divisibility by 3

A number is divisible by 3 if the sum of its digits is divisible by 3.

Divisibility by 5

A number is divisible by 5 if its last digit is either 0 or 5.

Shortcut Trick

Net efficiency of Tap B (filling in 11 hr) and Tap C (emptying in 11 hr) is zero.

Therefore, the effective rate of filling the cistern depends only on Tap A.

Rate of Tap A = 1 cistern per 4 hr ⇒ 1/4 cistern per hour.

To fill 1/4^th of the empty cistern, time = (1/4) ÷ (1/4) = 1 hr.

∴ The correct answer is 1 hr.

Alternate Method

Given: Tap A (filling) = 4 hr, Tap B (filling) = 11 hr, Tap C (emptying) = 11 hr.

Assume Total Capacity of the cistern = LCM of (4, 11, 11) = 44 units.

⇒ Combined Efficiency of all taps = 11 + 4 − 4 = 11 units/hr

⇒ Target to fill one-fourth of the cistern = 1/4 × 44 = 11 units

⇒ Required Time = Target Work ÷ Combined Efficiency

⇒ Required Time = 11 ÷ 11 = 1 hr

∴ The correct answer is 1 hr.

Additional Information

Inlet and Outlet Pipes

Inlet pipes fill the tank (positive efficiency), while outlet pipes or leaks empty the tank (negative efficiency).

LCM Method for Capacity

Assuming the cistern capacity as the LCM of individual times makes calculation easier by providing integer efficiency values.

Total Time Formula

If multiple taps work together, the total time is calculated as: 1/T_total = 1/T₁ + 1/T₂ − 1/T₃.

Shortcut Trick

Direct Calculation: Weighted Average = Σ(Score × Weight) ÷ ΣWeights

Weighted Sum = (75 × 2) + (80 × 3) + (90 × 5)

Weighted Sum = 150 + 240 + 450 = 840

Total Weight = 2 + 3 + 5 = 10

Weighted Average = 840 ÷ 10 = 84

∴ The correct answer is 84.

Alternate Method

Given: Scores of the student are 75, 80, and 90 with respective weights 2, 3, and 5.

Formula Used: Weighted Average = [(w₁ × x₁) + (w₂ × x₂) + (w₃ × x₃)] ÷ (w₁ + w₂ + w₃)

⇒ Total weighted points = (75 × 2) + (80 × 3) + (90 × 5)

⇒ Total weighted points = 150 + 240 + 450

⇒ Total weighted points = 840

⇒ Total sum of weights = 2 + 3 + 5 = 10

⇒ Weighted Average = 840 ÷ 10 = 84

∴ The correct answer is 84.

Additional Information

Weighted Mean Formula

The weighted mean is calculated as x̄_w = (Σw_ix_i) ÷ Σw_i, where w is the weight and x is the value.

Simple Average vs Weighted Average

A simple average treats all data points equally (weights = 1), whereas a weighted average assigns relative importance to different values.

Deviation Method

Assume a mean 'A', then Average = A + [Σw_i(x_i − A) ÷ Σw_i]. This simplifies calculations with large numbers.

Shortcut Trick

Scheme A discount: 44 + 12 − (44 × 12) ÷ 100 = 50.72%

Scheme B discount: 1 − (0.82 × 0.57 × 0.77) ≈ 64% (highest)

Scheme C discount: (4 × 8 + 6 × 38) ÷ 10 = 26%

Scheme D discount: 3 ÷ 10 × 100 = 30%

Least beneficial scheme for customer is the one with the minimum discount.

∴ The correct answer is Scheme C.

Alternate Method

Given: Total toys to buy = 10. Let Marked Price (MP) of each toy = ₹100.

Formula Used: Total Discount % = (Total Discount ÷ Total MP) × 100

⇒ Total MP for 10 toys = 10 × 100 = ₹1000.

⇒ Scheme A: Successive 44% and 12% ⇒ Discount = 1000 − [1000 × 0.56 × 0.88] = ₹507.2 (50.72%)

⇒ Scheme B: Successive 18%, 43%, 23% ⇒ Final Price = 1000 × 0.82 × 0.57 × 0.77 ≈ ₹359.9 ⇒ Discount ≈ ₹640.1 (64.01%)

⇒ Scheme C: 8% on 1^st 4 toys + 38% on next 6 toys ⇒ Discount = (4 × 8) + (6 × 38) = 32 + 228 = ₹260 (26%)

⇒ Scheme D: Buy 10, pay for 7 ⇒ 3 items are free ⇒ Discount = (3 ÷ 10) × 100 = 30%

⇒ Comparing all discounts: 64.01% > 50.72% > 30% > 26%.

⇒ Since 26% is the lowest discount, Scheme C is the least beneficial to the customer.

∴ The correct answer is Scheme C.

Additional Information

Successive Discount Formula

For two successive discounts x% and y%, the effective discount is (x + y − xy/100)%.

Buy X Get Y Free

The discount percentage in "Buy X Get Y Free" schemes is calculated as: [Y ÷ (X + Y)] × 100.

Customer Benefit Rule

A scheme is most beneficial when the discount is maximum and least beneficial when the discount is minimum.

Shortcut Trick

Total reduction in the sum = 48 × 2 = 96.

Total reduction in the average = 12.

Number of terms (k) = (Total change in sum) ÷ (Change in average).

k = 96 ÷ 12 = 8.

∴ The correct answer is 8.

Alternate Method

Given:

Original average of k numbers = x

Reduction = 48 from any two numbers

New average = x − 12

Formula Used:

Average = Sum of observations ÷ Number of observations

Calculations:

⇒ Original Sum = k × x

⇒ Total reduction from the sum = 48 + 48 = 96

⇒ New Sum = (k × x) − 96

⇒ New Average = New Sum ÷ k

⇒ x − 12 = (kx − 96) ÷ k

⇒ k × (x − 12) = kx − 96

⇒ kx − 12k = kx − 96

⇒ 12k = 96

⇒ k = 96 ÷ 12

⇒ k = 8

∴ The correct answer is 8.

Additional Information

Arithmetic Change in Average

If a value 'n' is subtracted from every observation in a set, the average also decreases by 'n'.

Sum of Observations

Total Sum = Average × Number of items. Any change in the sum affects the average proportionally to the number of items.

Net Deviation Method

The sum of deviations of all items from their average is always zero. New average can be found by adding (Total Change ÷ Count) to the old average.

Shortcut Trick

Average Speed = (Total Distance) ÷ (Total Time)

Total Distance = (72 × 3) + (60 × 1.5) = 216 + 90 = 306 km

Total Time = 3 + 1.5 = 4.5 hr

Average Speed = 306 ÷ 4.5 = 68 km/h

∴ The correct answer is 68 km/h.

Alternate Method

Given:

Speed₁ = 72 km/h, Time₁ = 3 hr

Speed₂ = 60 km/h, Time₂ = 1.5 hr

Formula Used:

Distance = Speed × Time

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Distance in first part (D₁) = 72 × 3 = 216 km

⇒ Distance in second part (D₂) = 60 × 1.5 = 90 km

⇒ Total Distance = 216 + 90 = 306 km

⇒ Total Time = 3 + 1.5 = 4.5 hr

⇒ Average Speed = 306 ÷ 4.5

⇒ Average Speed = 3060 ÷ 45

⇒ Average Speed = 68 km/h

∴ The correct answer is 68 km/h.

Additional Information

Average Speed for Equal Distance

If a journey is divided into two equal distances with speeds x and y, the average speed is given by (2xy) ÷ (x + y).

Unit Conversion

To convert speed from km/h to m/s, multiply the value by 5 ÷ 18. To convert from m/s to km/h, multiply by 18 ÷ 5.

Relative Speed

When two objects move in the opposite direction, relative speed is (u + v). In the same direction, relative speed is (u − v).

Shortcut Trick

Identify the values as a = 24, b = −7, and c = −17.

Check the sum: 24 + (−7) + (−17) = 24 − 24 = 0.

Apply the Algebraic Identity: If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Calculation: 3 × 24 × (−7) × (−17) = 72 × 119 = 8568.

∴ The correct answer is 8568.

Alternate Method

Given:

Expression: 24³ + (−7)³ + (−17)³

Formula Used:

The standard algebraic identity is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

Calculations:

⇒ Let a = 24, b = −7, and c = −17.

⇒ First, evaluate the sum (a + b + c).

⇒ a + b + c = 24 + (−7) + (−17) = 24 − 24 = 0.

⇒ Substituting (a + b + c) = 0 into the identity:

⇒ a³ + b³ + c³ − 3abc = 0 × (a² + b² + c² − ab − bc − ca) = 0.

⇒ a³ + b³ + c³ = 3abc.

⇒ Now, calculate the value of 3abc:

⇒ Value = 3 × 24 × (−7) × (−17).

⇒ Value = 72 × 119.

⇒ Value = 8568.

∴ The correct answer is 8568.

Additional Information

Special Case of Cubic Identity

If a + b + c = 0, then a³ + b³ + c³ = 3abc. This is widely used in simplifying sum of cubes when the bases sum to zero.

Alternate Condition for 3abc

The expression a³ + b³ + c³ also equals 3abc if a = b = c, as the second bracket (a² + b² + c² − ab − bc − ca) becomes zero.

Square of Trinomial

Related identity: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

Shortcut Trick

To make the number 92419x divisible by 13, first find the remainder of 924190 ÷ 13.

924190 ÷ 13 → Remainder is 7.

Required digit = Divisor − Remainder

Required digit = 13 − 7 = 6.

∴ The correct answer is 6.

Alternate Method

Given: Original transaction code = 92419. Divisor = 13.

Let the least digit to be appended at the end be x.

The new number becomes 924190 + x.

⇒ Perform division of 924190 by 13 to find the remainder:

⇒ 92 ÷ 13 = 7, Remainder = 1

⇒ 14 ÷ 13 = 1, Remainder = 1

⇒ 11 ÷ 13 = 0, Remainder = 11

⇒ 119 ÷ 13 = 9, Remainder = 2

⇒ 20 ÷ 13 = 1, Remainder = 7

⇒ So, 924190 ≡ 7 (mod 13).

⇒ For the number to be divisible by 13, (7 + x) must be a multiple of 13.

⇒ 7 + x = 13

⇒ x = 13 − 7 = 6.

∴ The correct answer is 6.

Additional Information

Divisibility Rule of 13

Multiply the last digit of the number by 4 and add it to the rest of the number. If the resulting sum is divisible by 13, the original number is also divisible.

Remainder Addition Rule

To make a number N divisible by D, the smallest number to add is (D − R), where R is the remainder obtained when N is divided by D.

Unit Digit Concept

Appending a digit to the end of a number N is equivalent to shifting the existing number one place to the left: N_new = 10 × N + x.