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Shortcut Trick

Effective Selling Price percentage = (100 − 18)% × (100 − 15)%

Effective Multiplier = 0.82 × 0.85 = 0.697

Final Selling Price = 1000 × 0.697 = 697

∴ The correct answer is ₹697.

Alternate Method

Given:

Marked Price (MP) = ₹1,000

Discount 1 (d₁) = 18%

Discount 2 (d₂) = 15%

Formula Used:

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

Calculations:

⇒ Price after 1^st discount = 1000 − (18% of 1000)

⇒ Price after 1^st discount = 1000 − 180 = ₹820

⇒ Final Selling Price = 820 − (15% of 820)

⇒ Final Selling Price = 820 − 123 = ₹697

∴ The correct answer is ₹697.

Additional Information

Successive Discount Formula

The single equivalent discount for two successive discounts x% and y% is given by (x + y − xy/100)%.

Marked Price and Cost Price Relation

The relationship is given by: MP/CP = (100 + Profit%)/(100 − Discount%).

Effective Percentage Change

For successive changes of a% and b%, the net percentage change is (a + b + ab/100)%, where negative values represent decreases/discounts.

Shortcut Trick

Number of sides (n) in a regular polygon − 2 = (Sum of interior angles) ÷ 180°

⇒ n − 2 = 1800 ÷ 180 = 10

⇒ n = 12

Number of diagonals = [n × (n − 3)] ÷ 2

⇒ [12 × (12 − 3)] ÷ 2 = 6 × 9 = 54

∴ The correct answer is 54 diagonals.

Alternate Method

Given: Sum of all interior angles of a regular polygon = 1800°

Formula Used:

1. Sum of interior angles of an n-sided polygon = (n − 2) × 180°
2. Number of diagonals = [n × (n − 3)] ÷ 2

Calculations:

⇒ (n − 2) × 180 = 1800

⇒ n − 2 = 1800 ÷ 180

⇒ n − 2 = 10

⇒ n = 10 + 2 = 12

Now, calculate the number of diagonals for a polygon with 12 sides:

⇒ Number of diagonals = [12 × (12 − 3)] ÷ 2

⇒ Number of diagonals = [12 × 9] ÷ 2

⇒ Number of diagonals = 108 ÷ 2

⇒ Number of diagonals = 54

∴ The correct answer is 54 diagonals.

Additional Information

Sum of Exterior Angles

The sum of the exterior angles of any convex polygon is always equal to 360°.

Individual Angle Formulas

For a regular polygon, each interior angle = [(n − 2) × 180°] ÷ n, and each exterior angle = 360° ÷ n.

Relationship between Interior and Exterior Angles

At each vertex of a polygon, the sum of the interior angle and the exterior angle is always 180°.

Shortcut Trick

In 5 hr, the company produces 1520 packets.

Find the packets produced in 1 hr: 1520 ÷ 5 = 304.

Find the production for 17 hr: 304 × 17.

Calculation: 304 × 17 = 5168.

∴ The correct answer is 5168 packets.

Alternate Method

Given:

Production in 5 hr (P₁) = 1520 packets

Time period 1 (T₁) = 5 hr

Time period 2 (T₂) = 17 hr

Since the rate of production is constant, this is a case of Direct Proportion.

Formula Used:

(Packets₁ / Time₁) = (Packets₂ / Time₂)

Calculations:

⇒ Rate = P₁ / T₁

⇒ Rate = 1520 ÷ 5

⇒ Rate = 304 packets/hr

⇒ P₂ = Rate × T₂

⇒ P₂ = 304 × 17

⇒ P₂ = 5168

∴ The correct answer is 5168 packets.

Additional Information

Direct Proportion

Two quantities are in direct proportion if an increase in one leads to a proportional increase in the other, expressed as x/y = k (constant).

Unitary Method

A technique where you find the value of a single unit first (e.g., 1 hour) and then multiply to find the value for the required number of units.

Rate of Work

Rate = Total Work ÷ Total Time. In this problem, the rate of production is the number of packets produced per unit time.

Shortcut Trick

Find the LCM of divisors 15, 12, and 14, which is 420.

Divide the largest 4-digit number (9999) by 420 to find the remainder: 9999 ÷ 420 ⇒ Remainder = 339.

Largest 4-digit multiple of 420 = 9999 − 339 = 9660.

Required number = Multiple + Common Remainder = 9660 + 5 = 9665.

∴ The correct answer is 9665.

Alternate Method

Given:

Divisors: 15, 12, 14

Common Remainder (R) = 5

Target: Largest 4-digit number (N ≤ 9999)

Formula Used:

Required Number = [LCM(Divisors) × k] + Remainder

Step-by-step Calculations:

⇒ Prime factorization: 15 = 3 × 5, 12 = 2² × 3, 14 = 2 × 7

⇒ LCM = 2² × 3 × 5 × 7 = 4 × 3 × 35 = 420

⇒ Divide 9999 by 420 to find the nearest multiple:

⇒ 9999 = 420 × 23 + 339

⇒ Largest multiple = 420 × 23 = 9660

⇒ Required number = 9660 + 5

⇒ Required number = 9665

∴ The correct answer is 9665.

Additional Information

LCM and Common Remainder

When a number N is divided by a, b, and c and leaves the same remainder R in each case, the general form is N = LCM(a, b, c) × k + R.

LCM and Different Remainders

If the remainders are r₁, r₂, r₃ such that (a−r₁) = (b−r₂) = (c−r₃) = d (constant difference), then N = LCM(a, b, c) × k − d.

HCF and Remainder

To find the largest number that divides x, y, and z leaving the same remainder R, find the HCF of (x−R), (y−R), and (z−R).

Shortcut Trick

Profit ratio of A, B, and C = 60,000 : 80,000 : 1,20,000 = 3 : 4 : 6

Total units = 3 + 4 + 6 = 13 units

Given: 13 units = ₹78,000 ⇒ 1 unit = ₹6,000

Required difference (C − B) = (6 − 4) = 2 units

Value = 2 × 6,000 = ₹12,000

∴ The correct answer is ₹12,000.

Alternate Method

Given:

Capital of A = ₹60,000

Capital of B = ₹80,000

Capital of C = ₹1,20,000

Total Profit = ₹78,000

Formula Used:

In a partnership business, Profit Ratio = Capital × Time

Calculations:

Since the time period for all partners is 1 year, the profit will be divided in the ratio of their investments.

⇒ Profit Ratio (A : B : C) = 60,000 : 80,000 : 1,20,000

⇒ Ratio = 6 : 8 : 12 = 3 : 4 : 6

⇒ Total sum of ratios = 3 + 4 + 6 = 13 units

⇒ 13 units = ₹78,000

⇒ 1 unit = 78,000 ÷ 13 = ₹6,000

⇒ B’s share = 4 × 6,000 = ₹24,000

⇒ C’s share = 6 × 6,000 = ₹36,000

⇒ Difference = C’s share − B’s share = 36,000 − 24,000 = ₹12,000

∴ The correct answer is ₹12,000.

Additional Information

Partnership Profit Division

If partners invest C₁, C₂, C₃ capitals for times T₁, T₂, T₃, the profit is shared in the ratio (C₁ × T₁) : (C₂ × T₂) : (C₃ × T₃).

Working vs Sleeping Partner

A Working Partner receives a fixed salary or percentage for managing the business, while the remaining profit is shared according to investment ratios.

Compound Partnership

When partners invest capital for different time durations, it is termed a compound partnership, and time must be multiplied with capital for calculations.

Shortcut Trick

Relative speed (opposite direction) = 22 + 26 = 48 km/h.

Meeting time = Total distance ÷ Relative speed.

Meeting time = 264 ÷ 48 = 5.5 hours.

∴ The correct answer is 5.5 hr.

Alternate Method

Given:

Total distance between two towns = 264 km

Speed of the first bus (S₁) = 22 km/h

Speed of the second bus (S₂) = 26 km/h

Formula Used:

Time = Distance ÷ Relative Speed

Calculations:

⇒ Since the buses are traveling towards each other, they are in opposite directions.

⇒ Relative Speed = S₁ + S₂

⇒ Relative Speed = 22 + 26 = 48 km/h

⇒ Time taken to meet = 264 ÷ 48

⇒ Time taken = 132 ÷ 24 = 66 ÷ 12 = 11 ÷ 2

⇒ Time taken = 5.5 hours

∴ The correct answer is 5.5 hr.

Additional Information

Relative Speed Rules

For two objects with speeds u and v: if moving in the opposite direction, Relative Speed = u + v; if moving in the same direction, Relative Speed = |u − v|.

Average Speed Formula

When a person travels equal distances at speeds x and y, the average speed for the whole journey is (2xy) ÷ (x + y).

Time and Distance Relation

Distance = Speed × Time. For a constant distance, speed is inversely proportional to time (S ∝ 1/T).

Shortcut Trick

Efficiency of Pipe P = 1 unit/hr and Pipe Q = 2 units/hr (Total 8 units).

Work done in 1 cycle (2 hours: Q then P) = 2 + 1 = 3 units.

Work done in 2 cycles (4 hours) = 3 × 2 = 6 units.

Remaining work = 8 − 6 = 2 units, which Pipe Q fills in exactly 1 hour.

Total time = 4 + 1 = 5 hr.

∴ The correct answer is 5 hr.

Alternate Method

Given: Filling time for Pipe P = 8 hr, Pipe Q = 4 hr.

Method: LCM method to find Total Work and Efficiency.

⇒ Total capacity of the tank = LCM(8, 4) = 8 units.

⇒ In the 1^st hour, Pipe Q is opened: Work done = 2 units.

⇒ In the 2^nd hour, Pipe P is opened: Work done = 1 unit.

⇒ Total work done in 2 hours = 2 + 1 = 3 units.

⇒ Total work done in 4 hours (2 such cycles) = 3 × 2 = 6 units.

⇒ Remaining work = 8 − 6 = 2 units.

⇒ In the 5^th hour, it is Pipe Q's turn again.

⇒ Time taken by Pipe Q to fill 2 units = 2 ÷ 2 = 1 hr.

⇒ Total time taken = 4 hr + 1 hr = 5 hr.

∴ The correct answer is 5 hr.

Additional Information

Pipes and Cisterns Basics

If a pipe fills a tank in x hours, its efficiency is 1/x of the tank per hour. For emptying pipes, efficiency is taken as negative.

Alternate Work Concept

When pipes work alternately, calculate the combined work in one full cycle (sum of individual hours) and then determine how many full cycles fit into the total work.

Total Time Calculation

Total Time = (Number of full cycles × hours per cycle) + (Remaining work ÷ Efficiency of the next pipe in sequence).

Shortcut Trick

Find the difference: (3 Adults + 4 Children) − (2 Adults + 3 Children) = ₹380 − ₹270

⇒ 1 Adult + 1 Child = ₹110

From the second group: 2(1 Adult + 1 Child) + 1 Child = ₹270

⇒ 2 × ₹110 + 1 Child = ₹270 ⇒ Child ticket = ₹270 − ₹220 = ₹50

⇒ Adult ticket = ₹110 − ₹50 = ₹60

∴ The correct answer is ₹60.

Alternate Method

Given:

Case 1: 3 adults and 4 children pay ₹380.

Case 2: 2 adults and 3 children pay ₹270.

Let the price of an adult ticket be ₹x and a child ticket be ₹y.

⇒ 3x + 4y = 380 --- (i)

⇒ 2x + 3y = 270 --- (ii)

To find x, eliminate y by multiplying (i) by 3 and (ii) by 4:

⇒ 3 × (3x + 4y) = 3 × 380 ⇒ 9x + 12y = 1140 --- (iii)

⇒ 4 × (2x + 3y) = 4 × 270 ⇒ 8x + 12y = 1080 --- (iv)

Subtract equation (iv) from (iii):

⇒ (9x − 8x) + (12y − 12y) = 1140 − 1080

⇒ x = 60

∴ The correct answer is ₹60.

Additional Information

Linear Equations in Two Variables

Equations of the form ax + by + c = 0 where a and b are coefficients and x and y are variables.

Elimination Method

A process of solving a system of equations by adding or subtracting them to remove one variable.

Substitution Method

Solving one equation for one variable and replacing that expression in the other equation.

Shortcut Trick

Total profit is calculated using the weighted average of the profit percentages.

Overall Profit % = [ (3 × 16) + (4 × 24) ] ÷ 7 = (48 + 96) ÷ 7 = 144/7 %

Total Profit = (Total CP) × (Overall Profit %)

Total Profit = (7 × 375) × (144 ÷ 7 ÷ 100) = 375 × 1.44 = ₹540

∴ The correct answer is ₹540.

Alternate Method

Given:

Cost Price (CP) of 1 set = ₹375

Number of sets sold at 16% profit = 3

Remaining sets sold at 24% profit = 7 − 3 = 4

Formula Used:

Profit = CP × Profit%

Total Profit = (Profit per set₁ × n₁) + (Profit per set₂ × n₂)

Calculations:

⇒ Profit from 3 sets = 3 × (16 ÷ 100 × 375) = 3 × 60 = ₹180

⇒ Profit from 4 sets = 4 × (24 ÷ 100 × 375) = 4 × 90 = ₹360

⇒ Total Profit = ₹180 + ₹360 = ₹540

∴ The correct answer is ₹540.

Additional Information

Net Profit Percentage

It is the total profit earned on all items divided by the total cost price of all items, multiplied by 100: (Total Profit ÷ Total CP) × 100.

Weighted Average Profit

When quantities are different, the overall profit % = (n₁p₁ + n₂p₂ + ...) ÷ (n₁ + n₂ + ...), where n is the quantity and p is the profit %.

Relation between CP and SP

Selling Price (SP) = CP + Profit. If profit is given in percentage, SP = CP × (100 + Profit%) ÷ 100.

Shortcut Trick

Net percentage change formula = a + b + (a × b)/100.

Here, a = +15 (increase) and b = −12 (decrease).

Net change = 15 − 12 + (15 × −12)/100 = 3 − 1.8 = 1.2%.

Since the result is positive, it is a 1.2% increase.

∴ The correct answer is 1.2% increase.

Alternate Method

Given:

Initial increase in salary = 15%

Subsequent decrease in salary = 12%

Formula Used:

Final Value = Initial Value × (1 + r₁/100) × (1 + r₂/100)

Calculations:

⇒ Let the initial salary of the person be ₹100.

⇒ Salary after 15% increase = 100 + (15% of 100) = 100 + 15 = ₹115.

⇒ Salary after 12% decrease = 115 − (12% of 115)

⇒ New Salary = 115 − (12/100 × 115) = 115 − 13.8 = ₹101.2.

⇒ Net Change in salary = 101.2 − 100 = 1.2.

⇒ Net percentage change = (1.2 / 100) × 100 = 1.2%.

⇒ Since the final value is more than the initial value, it is an increase.

∴ The correct answer is 1.2% increase.

Additional Information

Successive Percentage Change

When a value is changed by a% and then by b%, the net change is given by [a + b + (ab/100)]%.

Multiplying Factor Method

For an increase of r%, the factor is (1 + r/100). For a decrease, it is (1 − r/100). Net change = (Factor₁ × Factor₂ − 1) × 100.

Applications in Revenue

This logic is widely used in questions involving Price × Consumption = Expenditure or Length × Breadth = Area.