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Shortcut Trick

Effective percentage change = 20 − 24 − (20 × 24)/100 = −8.8%

Final cost as percentage of original = 100% − 8.8% = 91.2%

Given that 91.2% = ₹1,938

Original cost (100%) = (1,938 ÷ 91.2) × 100 = ₹2,125

∴ The correct answer is ₹2,125.

Alternate Method

Given:

First year increase = 20%

Second year decrease = 24%

Final cost of the article = ₹1,938

Calculations:

⇒ Let the original cost of the article be ₹x.

⇒ After 20% increase, cost = x × (1 + 20/100) = 1.2x

⇒ After 24% decrease on the increased cost, final cost = 1.2x × (1 − 24/100)

⇒ 1,938 = 1.2x × 0.76

⇒ 1,938 = 0.912x

⇒ x = 1,938 ÷ 0.912

⇒ x = 2,125

∴ The correct answer is ₹2,125.

Additional Information

Successive Percentage Change Formula

When a value is changed by a% and then by b%, the net change is given by (a + b + ab/100)%. Use positive for increase and negative for decrease.

Multiplying Factor Method

For an increase of r%, multiply by (1 + r/100). For a decrease of r%, multiply by (1 − r/100).

Reverse Percentage Calculation

To find the original value when the final value and percentage change are known: Original Value = Final Value ÷ (Net Multiplying Factor).

Shortcut Trick

Check the sum: 23 + (−11) + (−12) = 23 − 23 = 0.

Using the conditional identity: If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Result = 3 × 23 × (−11) × (−12) = 69 × 132 = 9108.

∴ The correct answer is 9108.

Alternate Method

Given: a = 23, b = −11, c = −12

Formula Used: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

⇒ First, evaluate the sum of the terms: a + b + c

⇒ 23 + (−11) + (−12) = 23 − 11 − 12 = 0

⇒ Since a + b + c = 0, the entire right-hand side of the formula becomes zero.

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

⇒ Therefore, a³ + b³ + c³ = 3abc

⇒ 23³ + (−11)³ + (−12)³ = 3 × 23 × (−11) × (−12)

⇒ 3 × 23 × 132 = 69 × 132 = 9108

∴ The correct answer is 9108.

Additional Information

Special Case of Cubic Identity

When the sum of three variables is zero, the identity a³ + b³ + c³ = 3abc simplifies complex cubic calculations into simple multiplication.

Alternative Form of Identity

The expression a² + b² + c² − ab − bc − ca can also be written as ½ [(a−b)² + (b−c)² + (c−a)²], which is always non-negative for real numbers.

BODMAS Rule in Simplification

Always resolve brackets and orders (powers) before addition or multiplication unless an identity is applicable.

Shortcut Trick

Difference in the number of notebooks = 7 − 4 = 3 notebooks.

Since the number of pens is the same (6 in both cases), the price difference is solely due to these 3 notebooks.

Price difference = ₹ 333 − ₹ 228 = ₹ 105.

Cost of 1 notebook = 105 ÷ 3 = 35.

∴ The correct answer is ₹ 35.

Alternate Method

Given:

Cost of 4 notebooks and 6 pens = ₹ 228

Cost of 7 notebooks and 6 pens = ₹ 333

Formula Used:

The Elimination Method for solving a System of Linear Equations.

Calculations:

⇒ Let the cost of one notebook be N and the cost of one pen be P.

⇒ 4N + 6P = 228 — (Equation 1)

⇒ 7N + 6P = 333 — (Equation 2)

⇒ Subtract Equation 1 from Equation 2:

⇒ (7N + 6P) − (4N + 6P) = 333 − 228

⇒ (7N − 4N) + (6P − 6P) = 105

⇒ 3N = 105

⇒ N = 105 ÷ 3

⇒ N = 35

∴ The correct answer is ₹ 35.

Additional Information

System of Linear Equations

A set of equations with the same variables. The solution is the set of values that satisfy all equations simultaneously.

Elimination Method

A process of solving a system by adding or subtracting equations to remove one variable, making it easier to find the value of the other.

Unitary Method

A technique where we first find the value of a single unit (like 1 notebook) and then multiply it to find the value of any required number of units.

Shortcut Trick

Direct application of Speed-Time-Distance relation.

Time = Distance ÷ Speed

Time = 12 ÷ 6 = 2 hr.

∴ The correct answer is 2 hr.

Alternate Method

Given:

Walking speed of Rohit = 6 km/hr

Total Distance to be covered = 12 km

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Time = 12 km ÷ 6 km/hr

⇒ Time = 2 hr

∴ The correct answer is 2 hr.

Additional Information

Distance Formula

The total path length covered is given by Distance = Speed × Time.

Unit Conversion

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

Average Speed

If a person covers different distances at different speeds, Average Speed = Total Distance ÷ Total Time.

Shortcut Trick

Use the Ratio Method: Initial ratio of weights and the profit percentage.

Weight Ratio (CP : SP) = 950 : 1000

Profit Ratio (CP : SP) = 100 : 115

Effective CP : SP = (950 × 100) : (1000 × 115) = 95 : 115 = 19 : 23

Total Profit % = [(23 − 19) ÷ 19] × 100 = (4 ÷ 19) × 100 ≈ 21.05%

∴ The correct answer is 21.05%.

Alternate Method

Given: Profit on price = 15% | False weight = 950 g | True weight = 1000 g (1 kg).

Let the Cost Price (CP) of 1000 g be ₹1000.

⇒ CP of 1 g = ₹1.

⇒ The trader sells 950 g but charges for 1000 g plus a 15% profit.

⇒ Actual CP for the trader (for 950 g) = 950 × ₹1 = ₹950.

⇒ Selling Price (SP) = ₹1000 + 15% of ₹1000 = 1000 + 150 = ₹1150.

⇒ Total Profit = SP − CP = 1150 − 950 = ₹200.

⇒ Total Profit % = (Profit ÷ CP) × 100 = (200 ÷ 950) × 100.

⇒ Profit % = (4 ÷ 19) × 100 = 400 ÷ 19 ≈ 21.05%.

⇒ Therefore, the total profit percentage is 21.05%, which is Option 2.

∴ The correct answer is 21.05%.

Additional Information

Dishonest Dealer Formula

Profit % = [(True Value − False Value + % Profit on True Value) ÷ False Value] × 100. This accounts for both weight cheating and price markup.

Successive Percentage Method

Total Profit can also be viewed as successive gains. Net Profit % = [a + b + (a × b ÷ 100)] where 'a' is profit from price and 'b' is profit from weight manipulation.

Weight Gain Percentage

Profit due to weight cheating alone = [Error ÷ (True Value − Error)] × 100. In this case, (50 ÷ 950) × 100 ≈ 5.26%.

Given:

(26/7)⁵ × (7/26)¹¹ × (26/7)¹⁹ = (7/26)^{9m + 17}

Formula used:

Law of Exponents:

a^x × a^y = a^{x + y}

(a/b)ⁿ = (b/a)⁻ⁿ

Calculations:

(26/7)⁵ × (7/26)¹¹ × (26/7)¹⁹ = (7/26)^{9m + 17}

⇒ (26/7)^{(5 + 19)} × (7/26)¹¹ = (7/26)^{9m + 17}

⇒ (26/7)²⁴ × (7/26)¹¹ = (7/26)^{9m + 17}

Using the property (a/b)ⁿ = (b/a)⁻ⁿ to make the bases identical:

⇒ (7/26)⁻²⁴ × (7/26)¹¹ = (7/26)^{9m + 17}

⇒ (7/26)^{(−24 + 11)} = (7/26)^{9m + 17}

⇒ (7/26)⁻¹³ = (7/26)^{9m + 17}

On comparing the exponents on both sides of the equation:

⇒ −13 = 9m + 17

⇒ 9m = −13 − 17

⇒ 9m = −30

⇒ m = −30/9

⇒ m = −10/3

∴ The correct answer is option (3).

Shortcut Trick

Apply successive discounts as a single multiplying factor: 0.75 (for 25% off) and 0.90 (for 10% off).

Final Selling Price = 2,000 × 0.75 × 0.90

Final Selling Price = 2,000 × 0.675 = 1,350

∴ The correct answer is ₹1350.

Alternate Method

Given:

Marked Price (MP) = ₹2,000

First Discount = 25%

Second Discount = 10%

Formula Used:

Final Selling Price = MP × (100 − d₁)/100 × (100 − d₂)/100

⇒ Price after 25% discount = 2,000 − (25% of 2,000)

⇒ 2,000 − 500 = ₹1,500

⇒ Final Selling Price after 10% discount = 1,500 − (10% of 1,500)

⇒ 1,500 − 150 = ₹1,350

∴ The correct answer is ₹1350.

Additional Information

Successive Discount Formula

For two successive discounts of a% and b%, the net equivalent discount is (a + b − ab/100)%.

Single Equivalent Discount

In this case, 25 + 10 − (25 × 10)/100 = 35 − 2.5 = 32.5% single discount on ₹2,000.

Selling Price Relation

The Selling Price (SP) is always calculated as Marked Price (MP) − Total Discount.

Given:

Expression: 156 ÷ [12 + {7 × (15 − 9) − 28}]

Formula used:

BODMAS Rule (Brackets, Of, Division, Multiplication, Addition, Subtraction)

Calculations:

156 ÷ [12 + {7 × (15 − 9) − 28}]

⇒ 156 ÷ [12 + {7 × 6 − 28}]

⇒ 156 ÷ [12 + {42 − 28}]

⇒ 156 ÷ [12 + 14]

⇒ 156 ÷ 26

⇒ 6

∴ The correct answer is option (3).

Shortcut Trick

By inspection, the value 5/2 can be written as the sum of 2 + 1/2.

Since the equation is in the form y + 1/y = 5/2 (where y = x / (x + 1)), we can equate y = 2 or y = 1/2.

If x / (x + 1) = 2 ⇒ x = 2x + 2 ⇒ x = −2.

If x / (x + 1) = 1/2 ⇒ 2x = x + 1 ⇒ x = 1.

∴ The correct answer is 1 or −2.

Alternate Method

Given: x / (x + 1) + (x + 1) / x = 5/2

Method: Simplify the algebraic expression into a standard Quadratic Equation.

⇒ [x × x + (x + 1) × (x + 1)] / [x × (x + 1)] = 5/2

⇒ [x² + (x² + 2x + 1)] / [x² + x] = 5/2

⇒ [2x² + 2x + 1] / [x² + x] = 5/2

⇒ 2 × (2x² + 2x + 1) = 5 × (x² + x)

⇒ 4x² + 4x + 2 = 5x² + 5x

⇒ 5x² − 4x² + 5x − 4x − 2 = 0

⇒ x² + x − 2 = 0

⇒ x² + 2x − x − 2 = 0

⇒ x(x + 2) − 1(x + 2) = 0

⇒ (x − 1)(x + 2) = 0

⇒ x = 1 or x = −2

∴ The correct answer is 1 or −2.

Additional Information

Quadratic Roots Formula

For any equation of the form ax² + bx + c = 0, the roots are calculated using x = [−b ± √(b² − 4ac)] / 2a.

Factorization Method

This involves splitting the middle term bx into two terms such that their product equals ac and their sum equals b.

Reciprocal Sum Property

If y + 1/y = k, and k is in the form of a fraction a/b + b/a, then the solutions for y are simply a/b and b/a.

Shortcut Trick

The Mode is simply the value that appears most often in the Data Set.

Count the occurrences: 2 (twice), 3 (thrice), 4 (once), 5 (four times).

Since the number 5 has the highest Frequency, it is the mode.

∴ The correct answer is 5.

Alternate Method

Given: The number of books read by students: 3, 5, 2, 3, 4, 5, 3, 2, 5, 5.

Formula Used: Mode = Value with the maximum Frequency in the distribution.

⇒ Arrange observations: 2, 2, 3, 3, 3, 4, 5, 5, 5, 5

⇒ Frequency of 2 = 2

⇒ Frequency of 3 = 3

⇒ Frequency of 4 = 1

⇒ Frequency of 5 = 4

⇒ The value 5 occurs 4 times, which is the Highest Frequency.

∴ The correct answer is 5.

Additional Information

Empirical Formula

For a moderately skewed distribution, the relationship is: Mode = 3 × Median − 2 × Mean.

Bimodal Data

A data set that contains two values with the highest frequency is called Bimodal. If there are more than two, it is Multimodal.

Measures of Central Tendency

Mean is the average, Median is the middle value, and Mode is the most frequent value in a Data Set.