Quantitative Aptitude PYQs

Shortcut Trick

By using the Alligation Method on average salaries:

Ratio of Officers to Non-officers = (Overall Avg − Non-officers Avg) : (Officers Avg − Overall Avg)

Ratio = (16000 − 10000) : (28000 − 16000)

Ratio = 6000 : 12000 = 1 : 2

Since 1 ratio part = 15 officers, 2 ratio parts = 15 × 2 = 30 non-officers.

∴ The correct answer is 30.

Alternate Method

Given:

Average salary of officers = ₹28,000

Average salary of non-officers = ₹10,000

Overall average salary = ₹16,000

Number of officers = 15

Officers Non-Officers

28,000 10,000

Overall Avg 16,000

6,000 12,000

Ratio = 1 : 2

Calculations:

Let the number of non-officers be x.

Total Salary = Overall Average × Total Employees

⇒ (15 × 28000) + (x × 10000) = 16000 × (15 + x)

⇒ 420000 + 10000x = 240000 + 16000x

⇒ 16000x − 10000x = 420000 − 240000

⇒ 6000x = 180000

⇒ x = 180000 ÷ 6000

⇒ x = 30

∴ The correct answer is 30.

Additional Information

Weighted Average

When two or more groups are combined, the overall average is calculated as (n₁A₁ + n₂A₂) ÷ (n₁ + n₂), where n is the number of elements and A is the respective average of each group.

Alligation Rule

A rapid visual method to find the ratio in which two quantities are mixed to produce a mixture of known mean. The ratio equals (Dearer − Mean) : (Mean − Cheaper).

Average Deviation Method

The sum of net deviations of all the observations from their exact arithmetic mean is always mathematically equal to zero.

Shortcut Trick

The sum of the three given results is 210 + 198 + 192 = 600.

Each number is added twice as a full unit and once as a half unit across the three equations, giving a combined sum of 2.5 times the total sum of the three numbers.

Sum of the three numbers = 600 ÷ 2.5 = 240.

Required Average = 240 ÷ 3 = 80.

∴ The correct answer is 80.

Alternate Method

Given:

Let the three numbers be a, b, and c.

Calculations:

a + b + c/2=210b + c + a/2=198c + a + b/2=1922.5 × (a + b + c)=600Average = (a + b + c) ÷ 3 = 240 ÷ 3 = 80

According to the given conditions:

⇒ a + b + c/2 = 210

⇒ b + c + a/2 = 198

⇒ c + a + b/2 = 192

Adding all three equations:

⇒ (a + b + c/2) + (b + c + a/2) + (c + a + b/2) = 210 + 198 + 192

⇒ 2(a + b + c) + 0.5(a + b + c) = 600

⇒ 2.5 × (a + b + c) = 600

⇒ a + b + c = 600 ÷ 2.5

⇒ a + b + c = 240

The average of the three numbers is their sum divided by 3:

⇒ Average = (a + b + c) ÷ 3

⇒ Average = 240 ÷ 3 = 80

∴ The correct answer is 80.

Additional Information

Average (Arithmetic Mean)

The average of a given set of numbers is equal to the sum of all the observations divided by the total number of observations. Average = Sum of observations ÷ Total number of observations.

Properties of Average

If each observation in a dataset is increased, decreased, multiplied, or divided by a constant value 'k', the new average will also be correspondingly increased, decreased, multiplied, or divided by 'k'.

Weighted Average

If values x₁, x₂, ..., x_n have associated weights w₁, w₂, ..., w_n, the weighted average is calculated as (w₁x₁ + w₂x₂ + ... + w_nx_n) ÷ (w₁ + w₂ + ... + w_n).

Shortcut Trick

Target questions to pass = 60% of 120 = 72

Current correct questions = (70% of 20) + (50% of 50) + (30% of 50)

⇒ 14 + 25 + 15 = 54

Questions needed to pass = 72 − 54 = 18

∴ The correct answer is 18 questions.

Alternate Method

Given:

Total questions = 120

General Knowledge questions = 20, correctly answered = 70%

Reasoning questions = 50, correctly answered = 50%

English questions = 50, correctly answered = 30%

Minimum passing score = 60% of total questions

Total: 120 questionsCorrect: 5418NeededPassing Score: 72Target = 60% of 120 = 72

Calculations:

⇒ Passing score = 60% of 120 = (60 ÷ 100) × 120 = 72

⇒ Correct GK questions = 70% of 20 = (70 ÷ 100) × 20 = 14

⇒ Correct Reasoning questions = 50% of 50 = (50 ÷ 100) × 50 = 25

⇒ Correct English questions = 30% of 50 = (30 ÷ 100) × 50 = 15

⇒ Total correctly answered questions = 14 + 25 + 15 = 54

⇒ Additional questions needed = Passing score − Total correctly answered

⇒ Additional questions needed = 72 − 54 = 18

∴ The correct answer is 18 questions.

Additional Information

Passing Percentage

The minimum benchmark score set by examination bodies to qualify. Formula: Passing Marks = (Passing % ÷ 100) × Total Marks.

Aggregate Marks

The sum total of marks obtained across all sections or subjects. Formula: Aggregate % = (Total Marks Obtained ÷ Total Maximum Marks) × 100.

Weighted Average

Used when combining groups with different base values or sections with different maximum marks. Formula: Weighted Average = (N₁x₁ + N₂x₂ + ... ) ÷ (N₁ + N₂ + ... ).

Shortcut Trick

Rate for the 1^st year = 9%

Rate for the next 8 months = 9% × (8/12) = 6%

Effective CI rate for 1 year 8 months = 9 + 6 + (9 × 6)/100 = 15.54%

Total Compound Interest = 15.54% of ₹7,000 = ₹1,087.80

∴ The correct answer is ₹1,087.80.

Alternate Method

Given:

Principal (P) = ₹7,000

Rate of Interest (R) = 9% p.a.

Time (T) = 1 year 8 months = 1 + (8/12) years = 1 + (2/3) years

Formula Used:

When time is in a fraction like n + a/b years:

Amount (A) = P × (1 + R/100)ⁿ × [1 + (R × a/b)/100]

Compound Interest (CI) = A − P

Start ₹7,000

End of 1 Year ₹7,630

End of 1 Yr 8 Mo ₹8,087.80

• 9% (₹630)

• 6% (₹457.80) Total CI = ₹630 + ₹457.80 = ₹1,087.80

Calculations:

⇒ Amount (A) = 7000 × (1 + 9/100)¹ × [1 + (9 × 2/3)/100]

⇒ A = 7000 × (109/100) × (1 + 6/100)

⇒ A = 7000 × 1.09 × 1.06

⇒ A = 7000 × 1.1554 = ₹8,087.80

⇒ CI = Amount − Principal

⇒ CI = 8087.80 − 7000 = ₹1,087.80

∴ The correct answer is ₹1,087.80.

Additional Information

Compound Interest Formula

When interest is compounded annually, the total amount is calculated as A = P × (1 + R/100)ⁿ, where P is the principal, R is the rate of interest, and n is the time in years.

Effective Rate of Interest

For two different continuous time periods with rates a% and b%, the combined effective rate is given by the successive percentage change formula: Effective Rate = a + b + (a × b)/100.

Compounding Half-Yearly

When the interest is compounded half-yearly, the rate of interest becomes half (R/2) and the number of time periods becomes double (2n).

Shortcut Trick

In compound interest, the amount from the previous year becomes the principal for the next year.

Interest for the 3^rd year = Amount of 3^rd year − Amount of 2^nd year

⇒ Interest = 13200 − 12000 = ₹1200

Rate of Interest (R) = Interest ÷ Principal of that year

⇒ R = 1200 ÷ 12000 = 0.1

∴ The correct answer is 0.1.

Alternate Method

Year 0 Year 2 Year 3

P = ? ₹12,000 ₹13,200

CI = ₹1,200

Rate = R

Given:

Amount at the end of the 2^nd year (A₂) = ₹12,000

Amount at the end of the 3^rd year (A₃) = ₹13,200

Formula Used:

Amount for n years, A = P(1 + R)ⁿ

Calculations:

⇒ For 2 years: P(1 + R)² = 12000   --- (Equation 1)

⇒ For 3 years: P(1 + R)³ = 13200   --- (Equation 2)

⇒ Dividing Equation 2 by Equation 1:

⇒ [P(1 + R)³] ÷ [P(1 + R)²] = 13200 ÷ 12000

⇒ 1 + R = 1.1

⇒ R = 1.1 − 1

⇒ R = 0.1

∴ The correct answer is 0.1.

Additional Information

Compound Interest Formula

The amount after n years is given by A = P(1 + R)ⁿ, where P is the principal and R is the decimal rate of interest.

Simple vs Compound Interest

For the first year, both Simple Interest (SI) and Compound Interest (CI) are equal if compounded annually on a given principal.

Successive Amounts in CI

The ratio of amounts for any two consecutive years is always constant and equal to the multiplying factor (1 + R).

Shortcut Trick Let the quantities bought be 3, 4, and 3 respectively, making a total of 10 notebooks. Cost of 3 notebooks of Type 1 = 3 × (20 ÷ 5) = ₹12 Cost of 4 notebooks of Type 2 = 4 × (9 ÷ 2) = ₹18 and 3 notebooks of Type 3 = 3 × (45 ÷ 10) = ₹13.5 Total CP of 10 notebooks = 12 + 18 + 13.5 = ₹43.5 Total SP of 10 notebooks = 10 × (19 ÷ 4) = ₹47.5 Profit % = [(47.5 − 43.5) ÷ 43.5] × 100 = (4 ÷ 43.5) × 100 ≈ 9.2% Profit. ∴ The correct answer is Profit of 9.2%.

Alternate Method Given: CP of 5 notebooks of Type 1 = ₹20 CP of 2 notebooks of Type 2 = ₹9 CP of 10 notebooks of Type 3 = ₹45 Mixing ratio of Type 1, Type 2, and Type 3 = 3 : 4 : 3 SP of 4 notebooks = ₹19 Formula Used: Profit = Selling Price (SP) − Cost Price (CP) Profit % = (Profit ÷ CP) × 100

Type 1: 3 CP = ₹12

Type 2: 4 CP = ₹18

Type 3: 3 CP = ₹13.5

Total CP = ₹43.5

Total 10 Notebooks Sold @ ₹19 for 4 SP = 10 × (19 ÷ 4) = ₹47.5

Total SP = ₹47.5

Profit = ₹4

Profit % = (4 ÷ 43.5) × 100 ≈ 9.2%

Calculations: ⇒ Let the number of notebooks of Type 1, Type 2, and Type 3 be 3x, 4x, and 3x respectively. ⇒ Total number of notebooks = 3x + 4x + 3x = 10x ⇒ CP of 1 notebook of Type 1 = 20 ÷ 5 = ₹4 ⇒ Total CP of 3x notebooks of Type 1 = 3x × 4 = ₹12x ⇒ CP of 1 notebook of Type 2 = 9 ÷ 2 = ₹4.5 ⇒ Total CP of 4x notebooks of Type 2 = 4x × 4.5 = ₹18x ⇒ CP of 1 notebook of Type 3 = 45 ÷ 10 = ₹4.5 ⇒ Total CP of 3x notebooks of Type 3 = 3x × 4.5 = ₹13.5x ⇒ Total CP of all 10x notebooks = 12x + 18x + 13.5x = ₹43.5x ⇒ SP of 1 notebook = 19 ÷ 4 = ₹4.75 ⇒ Total SP of all 10x notebooks = 10x × 4.75 = ₹47.5x ⇒ Profit = Total SP − Total CP = 47.5x − 43.5x = ₹4x ⇒ Profit % = (Profit ÷ Total CP) × 100 ⇒ Profit % = (4x ÷ 43.5x) × 100 = 400 ÷ 43.5 = 800 ÷ 87 ≈ 9.195% ⇒ Profit % ≈ 9.2% ∴ The correct answer is Profit of 9.2%.

Additional Information Basic Profit & Loss Rule Profit occurs when Selling Price (SP) > Cost Price (CP). The fundamental formula to calculate the percentage is: Profit % = (Profit ÷ CP) × 100. Mixture Replacement / Combined Cost Price When different quantities or qualities of items are mixed, calculate the per-unit Cost Price of each type first. Then multiply them by their respective assumed ratio amounts to find the combined Total CP, ensuring accurate overall profit/loss estimation.

Shortcut Trick

Let the Selling Prices be 300, 400, and 500 units.

Total SP = 300 + 400 + 500 = 1200 units.

Total CP = (300 ÷ 1.25) + (400 ÷ 1.1) + (500 ÷ 0.8) = 240 + 363.63 + 625 = 1228.63 units.

Net Loss = Total CP − Total SP = 1228.63 − 1200 = 28.63 units.

Loss % = (28.63 ÷ 1228.63) × 100 ≈ 2.33%.

∴ The correct answer is loss of 2.33%.

Alternate Method

Given:

Ratio of Selling Prices (SP) = 3 : 4 : 5

Profit on C1 = 25%, Profit on C2 = 10%, Loss on C3 = 20%

Calculations:

⇒ The fractional values are: 25% = 1/4, 10% = 1/10, 20% = 1/5.

⇒ For C1, CP : SP = 4 : 5

⇒ For C2, CP : SP = 10 : 11

⇒ For C3, CP : SP = 5 : 4

⇒ We need SPs to be in ratio 3 : 4 : 5. First, make SP parts equal by multiplying with LCM(5, 11, 4) = 220.

⇒ Now multiply each SP by its ratio share: SP₁ = 3 × 220 = 660, SP₂ = 4 × 220 = 880, SP₃ = 5 × 220 = 1100.

⇒ Calculate corresponding CPs: CP₁ = 660 × (4 ÷ 5) = 528

⇒ CP₂ = 880 × (10 ÷ 11) = 800

⇒ CP₃ = 1100 × (5 ÷ 4) = 1375

⇒ Total CP = 528 + 800 + 1375 = 2703 units

⇒ Total SP = 660 + 880 + 1100 = 2640 units

Total Cost Price = 2703 units

Total Selling Price = 2640 units

Loss = 63

Transaction Summary

⇒ Loss = Total CP − Total SP = 2703 − 2640 = 63 units

⇒ Loss % = (Loss ÷ Total CP) × 100 = (63 ÷ 2703) × 100 = (2100 ÷ 901)% ≈ 2.33%

∴ The correct answer is loss of 2.33%.

Additional Information

Successive Profit or Loss

When multiple profits/losses are given successively, the effective percentage change is calculated as: Net % = a + b + (a × b) ÷ 100. Use positive sign for profit and negative for loss.

Cost Price from Selling Price

When Profit % is given: CP = (SP × 100) ÷ (100 + P%). When Loss % is given: CP = (SP × 100) ÷ (100 − L%).

Dishonest Dealer Concept

If a dealer claims to sell goods at CP but uses false weight, Profit % = [Error ÷ (True Weight − Error)] × 100.

Shortcut Trick

Let CP = 100 units ⇒ MP = 150 units (50% markup).

Price after 20% discount = 150 − 30 = 120 units.

Overall profit is 8% ⇒ Final SP = 100 + 8 = 108 units.

Second discount (y) reduces price from 120 to 108 ⇒ y = (12 ÷ 120) × 100 = 10%.

108 units = ₹4320 ⇒ 1 unit = 40 ⇒ CP (100 units) = ₹4000.

∴ The correct answer is CP = ₹4000, y = 10%.

Alternate Method

CP = 100u

MP = 150u

SP1 = 120u

SP2 = 108u

+50%

−20%

−y%

Profit = +8%

108u = ₹4320 → 1u = ₹40 → CP (100u) = ₹4000

Given:

Markup = 50%, First discount = 20%, Overall profit = 8%, Final SP = ₹4320.

Calculations:

Let the Cost Price (CP) be x.

⇒ Marked Price (MP) = x + (50% of x) = 1.5x

⇒ Price after 1^st discount = 1.5x × (1 − 20÷100) = 1.5x × 0.8 = 1.2x

⇒ Price after 2^nd discount = 1.2x × (1 − y÷100) = Final SP

⇒ Overall Profit = 8%, so Final SP = x + (8% of x) = 1.08x

Equating both expressions for Final SP:

⇒ 1.2x × (1 − y÷100) = 1.08x

⇒ 1 − y÷100 = 1.08 ÷ 1.2 = 0.9

⇒ y÷100 = 1 − 0.9 = 0.1

⇒ y = 10

Now, calculating CP using the Final SP:

⇒ 1.08x = 4320

⇒ x = 4320 ÷ 1.08 = 4000

⇒ CP = ₹4000

∴ The correct answer is CP = ₹4000, y = 10%.

Additional Information

Successive Discounts

When two discounts d₁% and d₂% are given one after another, the equivalent single discount is given by (d₁ + d₂ − d₁×d₂÷100)%.

Relation between CP and MP

When an item is sold at a profit of P% after offering a discount of D%, the ratio of Cost Price to Marked Price is given directly by: CP ÷ MP = (100 − D) ÷ (100 + P).

Markup Percentage

Markup percentage represents how much higher the marked price is compared to the cost price. It is calculated as: Markup% = (MP − CP) ÷ CP × 100.

Shortcut Trick

Let total work = LCM of 12 and 18 = 36 units.

Efficiencies: A = 36 ÷ 12 = 3 units/day, B = 36 ÷ 18 = 2 units/day.

Work done by A and B together in 4 days = (3 + 2) × 4 = 20 units.

Remaining work = 36 − 20 = 16 units.

Time taken by A to finish the remaining work = 16 ÷ 3 = 5¹/₃ days.

∴ The correct answer is 5¹/₃ days.

Alternate Method

Given:

Time taken by A to complete the task = 12 days

Time taken by B to complete the task = 18 days

Time A and B worked together = 4 days

Formula Used:

Total Work = LCM of individual days

Efficiency = Total Work ÷ Days

Time = Work ÷ Efficiency

Calculations:

⇒ Combined efficiency of A and B = 3 + 2 = 5 units/day.

⇒ Work completed by both in 4 days = 5 × 4 = 20 units.

⇒ Remaining work = 36 − 20 = 16 units.

⇒ Time taken by A for remaining work = Remaining work ÷ Efficiency of A.

⇒ Time taken = 16 ÷ 3 = 5¹/₃ days.

∴ The correct answer is 5¹/₃ days.

Additional Information

Pipes & Cisterns

Functions similarly to Time & Work. Inlets (filling pipes) have positive efficiency, while outlets (emptying pipes) have negative efficiency.

Work & Wages

Wages are distributed in direct proportion to the actual amount of work done by each individual.

Alternate Day Work

Work is done in turns. Calculate the total work done in a complete cycle (e.g., 2 days) and scale it up to reach close to the total work, then find the remainder for the next turn.

Shortcut Trick

By applying the Rule of Alligation directly:

Quantity Ratio = (Mean Price − Price of 2^nd quality) / (Price of 1^st quality − Mean Price)

⇒ 4 / 5 = (130 − x) / (150 − 130)

⇒ 4 / 5 = (130 − x) / 20

⇒ 130 − x = 16 ⇒ x = 114

∴ The correct answer is ₹114.

Alternate Method

Given:

Cost of the 1^st quality tea = ₹150 per kg

Cost of the mixture (at no profit, no loss) = ₹130 per kg

Ratio of 1^st to 2^nd quality mixed = 4 : 5

Formula Used:

Total Cost = (Quantity₁ × Price₁) + (Quantity₂ × Price₂)

₹150 1^st Quality

₹x 2^nd Quality

130 Mixture Price

4 parts 130 − x

5 parts 150 − 130 = 20

Calculations:

Let the quantities of the 1^st and 2^nd quality tea be 4k and 5k respectively.

Let the cost of the 2^nd quality tea be ₹x per kg.

⇒ Total quantity of the mixture = 4k + 5k = 9k

⇒ Total cost of the mixture = 9k × 130 = 1170k

⇒ Total cost of the 1^st quality = 4k × 150 = 600k

⇒ Total cost of the 2^nd quality = 5k × x = 5kx

Since the total cost of individual parts must equal the total mixture cost:

⇒ 600k + 5kx = 1170k

Dividing the entire equation by k:

⇒ 600 + 5x = 1170

⇒ 5x = 1170 − 600

⇒ 5x = 570

⇒ x = 570 ÷ 5 = 114

∴ The correct answer is ��114.

Additional Information

Rule of Alligation

It allows us to directly find the ratio in which two components at given prices must be mixed to obtain a specific mean price. Ratio = (Cheaper Quantity) : (Dearer Quantity) = (D − M) : (M − C).

Mean Price

The effective cost price of a unit quantity of the combined mixture is known as the mean price. It always lies strictly between the cost prices of the cheaper and dearer components being mixed.

Weighted Average Logic

Alligation is an alternative derivation of the weighted average formula: A_w = (w₁x₁ + w₂x₂) ÷ (w₁ + w₂), where w represents weights (quantities) and x represents values (prices).

Shortcut Trick

A working for 16 days and B working for 44 days is equivalent to A and B working together for 16 days, and B working alone for 28 days (44 − 16).

Work done by A and B together in 16 days = 16 ÷ 30 = 8/15 of total work.

Remaining work = 1 − 8/15 = 7/15, which B completes alone in 28 days.

Total time taken by B to complete the entire task = 28 × (15 ÷ 7) = 60 days.

∴ The correct answer is 60 days.

Alternate Method

Given:

Time taken by A and B together = 30 days

A works for 16 days and B finishes the remaining work in 44 days.

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ Let Total Work = 30 units (LCM of 30).

⇒ Efficiency of (A + B) = 30 ÷ 30 = 1 unit/day.

⇒ Total work done = (A's work for 16 days) + (B's work for 44 days) = 30 units.

⇒ Grouping the work: (A + B) working for 16 days + B working alone for (44 − 16) days = 30 units.

⇒ 16 × (Efficiency of A + B) + 28 × (Efficiency of B) = 30

⇒ 16 × 1 + 28 × Efficiency of B = 30

⇒ 28 × Efficiency of B = 30 − 16

⇒ 28 × Efficiency of B = 14

⇒ Efficiency of B = 14 ÷ 28 = 0.5 units/day.

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

⇒ Time = 30 ÷ 0.5 = 60 days.

∴ The correct answer is 60 days.

Additional Information

Pipes & Cisterns

Follows the exact same principles as Time & Work. Inlet pipes have positive efficiency as they fill the tank, while outlet pipes have negative efficiency as they empty the tank.

Work & Wages

Wages are always distributed in direct proportion to the actual amount of work done by each person. If people work for the exact same number of days, wages can be shared directly in the ratio of their efficiencies.

Alternate Day Work

When individuals work on alternate days, calculate the total work done in one complete cycle (e.g., 2 days) and multiply it to get close to the total work, then find the remaining work and assign it to the next person in sequence.

Shortcut Trick

Let the initial mixture be 16x. The initial parts are Orange = 5x, Apple = 7x.

Removing 24 litres takes out (5 / 16) × 24 = 7.5L Orange and (7 / 16) × 24 = 10.5L Apple.

Resultant Apple − Resultant Orange = (7x − 10.5) − (5x − 7.5 + 10) = 14.

⇒ 2x − 13 = 14 ⇒ 2x = 27.

Initial quantity = 16x = 27 × 8 = 216.

∴ The correct answer is 216 litres.

Alternate Method

Given:

Ratio of Orange : Apple : Guava = 5 : 7 : 4

Quantity of mixture taken out = 24 litres

Quantity of added juices = 10 litres Orange, 6 litres Guava

Resultant Apple juice − Resultant Orange juice = 14 litres

Initial Mix 16x Litres O : A : G 5 : 7 : 4

− 24L

Remaining 16x − 24 Ratio intact 5 : 7 : 4

• 10L Org
• 6L Guv

Final Mix Apple − Org = 14 Litres

Calculations:

⇒ Total ratio parts initially = 5 + 7 + 4 = 16 parts

⇒ Let the initial quantity of mixture be 16x litres.

⇒ The initial quantities are Orange = 5x, Apple = 7x, and Guava = 4x.

⇒ Quantity of Orange juice removed = (5 / 16) × 24 = 7.5 litres

⇒ Quantity of Apple juice removed = (7 / 16) × 24 = 10.5 litres

After taking out 24 litres and adding the respective juices, the new volumes are:

⇒ Resultant Orange juice = 5x − 7.5 + 10 = 5x + 2.5

⇒ Resultant Apple juice = 7x − 10.5

According to the given condition, the difference is 14 litres:

⇒ (7x − 10.5) − (5x + 2.5) = 14

⇒ 7x − 10.5 − 5x − 2.5 = 14

⇒ 2x − 13 = 14

⇒ 2x = 27

⇒ x = 13.5

⇒ Initial quantity of mixture = 16x = 16 × 13.5 = 216 litres.

∴ The correct answer is 216 litres.

Additional Information

Constant Ratio Property

When any volume of a mixture is drawn out, the remaining mixture still contains the original ingredients in the exact same ratio. Only adding or removing an individual component selectively will alter this final ratio.

Replacement Method

When a certain volume of mixture is replaced completely by one of its pure ingredients, the remaining quantity of the pure liquid left can be found using the formula: Final = Initial × (1 − Removed/Total)ⁿ.

Rule of Alligation

Used to find the required ratio when two ingredients at different prices or concentrations are mixed. Ratio = (Dearer − Mean) : (Mean − Cheaper).

Shortcut Trick

For a constant distance, Speed is inversely proportional to Time.

Initial time = 1 hr 45 min = 105 min.

New time = 105 + 15 = 120 min.

Time ratio = T₁ : T₂ = 105 : 120 = 7 : 8

∴ Speed ratio = S₁ : S₂ = 8 : 7

Decrease in speed = (8 − 7) ÷ 8 = 1/8 = 0.125 (or 12.5%)

∴ The correct answer is 12.5% (0.125).

Alternate Method

Given:

⇒ Initial speed (S₁) = 60 km/h

⇒ Initial time (T₁) = 1 hour 45 minutes = 105 minutes = 1.75 hours

⇒ New time (T₂) = 105 minutes + 15 minutes = 120 minutes = 2 hours

Distance = 105 km (Constant) Case 1: S = 60 km/h, T = 105 min

Case 2: S = 52.5 km/h, T = 120 min

Speed Decrease = 7.5 km/h Percentage Decrease = 12.5% (0.125)

Calculations:

⇒ Distance covered = S₁ × T₁ = 60 × 1.75 = 105 km

⇒ New speed (S₂) = Distance ÷ T₂ = 105 ÷ 2 = 52.5 km/h

⇒ Decrease in speed = S₁ − S₂ = 60 − 52.5 = 7.5 km/h

⇒ Fraction decrease = Decrease ÷ Original Speed

⇒ Fraction decrease = 7.5 ÷ 60 = 1/8 = 0.125

⇒ Percentage decrease = 0.125 × 100% = 12.5%

∴ The correct answer is 12.5% (0.125).

Additional Information

Inverse Proportionality of Speed and Time

When distance is kept constant, speed is inversely proportional to the time taken (S ∝ 1/T). If the time increases by a ratio of a : b, the required speed decreases by the ratio of b : a.

Average Speed

Average Speed = Total Distance ÷ Total Time. It must be noted that it is strictly calculated using this formula and is not merely the simple arithmetic average of individual speeds unless the travel times are identical.

Percentage Change

Percentage Change = (New Value − Initial Value) ÷ (Initial Value) × 100. A negative end result simply indicates a percentage decrease.

Shortcut Trick

Speed ratio = 4 : 6 = 2 : 3, so Time ratio = 3 : 2

Difference in time parts = 3 − 2 = 1 part

1 part = 15 min (late) + 5 min (early) = 20 mins = 1/3 hours

Time taken at 4 km/h (3 parts) = 3 × 1/3 = 1 hour

Distance = Speed × Time = 4 km/h × 1 hour = 4 km

∴ The correct answer is 4 km.

Alternate Method

Given:

Speed 1 (S₁) = 4 km/h (reaches 15 minutes late)

Speed 2 (S₂) = 6 km/h (reaches 5 minutes early)

Home

School

Distance = D km

Speed = 4 km/h | Time = t + 15 min

Speed = 6 km/h | Time = t - 5 min

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Let the distance from home to school be D km.

⇒ Time taken to cover D km at 4 km/h = D / 4 hours

⇒ Time taken to cover D km at 6 km/h = D / 6 hours

⇒ Time difference = 15 minutes (late) − (−5 minutes early) = 20 minutes

⇒ Convert time difference into hours = 20 / 60 = 1 / 3 hours

⇒ D / 4 − D / 6 = 1 / 3

⇒ (3D − 2D) / 12 = 1 / 3

⇒ D / 12 = 1 / 3

⇒ D = 12 / 3

⇒ D = 4 km

∴ The correct answer is 4 km.

Additional Information

Direct Formula for Early/Late Problems

Distance = [ (S₁ × S₂) ÷ (S₁ − S₂) ] × Total Time Difference in hours.

Average Speed

When a certain distance is covered at speed x and the same distance is covered at speed y, Average Speed = (2xy) ÷ (x + y).

Relative Speed

If two bodies move in opposite directions at speeds S₁ and S₂, Relative Speed = S₁ + S₂. In the same direction, Relative Speed = |S₁ − S₂|.

Shortcut Trick

Circumference of the ground = Total Cost ÷ Rate per meter

Circumference = 3768 ÷ 120 = 31.4 m

Since Circumference = 2 × π × r

2 × 3.14 × r = 31.4 ⇒ 6.28 × r = 31.4 ⇒ r = 5 m

∴ The correct answer is 5 m.

Alternate Method

Given:

Total cost of fencing = ₹3,768

Rate of fencing = ₹120/m

Formula Used:

Total Cost = Circumference × Rate

Circumference of a circle = 2πr (where r is the radius)

O

r = ? Circumference = Total Cost ÷ Rate Perimeter = 31.4 m

Total Cost ₹3768

Rate ₹120/m

Calculation:

⇒ Circumference = Total Cost ÷ Rate

⇒ Circumference = 3768 ÷ 120 = 31.4 m

⇒ 2 × π × r = 31.4

Taking π ≈ 3.14:

⇒ 2 × 3.14 × r = 31.4

⇒ 6.28 × r = 31.4

⇒ r = 31.4 ÷ 6.28

⇒ r = 5 m

∴ The correct answer is 5 m.

Additional Information

Area of a Circle

The area enclosed by a circle is given by the formula πr². If a question asks for the cost of levelling or plowing a circular field, the area must be calculated instead of the perimeter.

Perimeter of a Semi-Circle

The perimeter of a semi-circle includes both the curved arc and the straight diameter. The direct formula is πr + 2r.

Difference Between Fencing and Paving

Fencing always relates to the outer boundary (circumference of 2πr), whereas paving, painting, or laying grass relates to the inner flat surface (area of πr²).

Shortcut Trick

The unused area when a circle is perfectly inscribed in a square is directly given by the formula: r²(4 − π)

Substitute the given radius r = 14 cm, and use a precise value for π ≈ 3.1416:

Unused Area = 14² × (4 − 3.1416)

Unused Area = 196 × 0.8584 ≈ 168.25 cm²

∴ The correct answer is 168.25 cm².

Alternate Method

Given:

Radius of the circular logo (r) = 14 cm

Formula Used:

Side of the circumscribing square (a) = 2 × r

Area of square = a²

Area of circle = πr²

Unused square space = Area of square − Area of circle

r = 14 cm

Side = 28 cm Unused space (shaded blue)

Calculations:

⇒ Side of the square (a) = 2 × 14 = 28 cm

⇒ Area of the square = 28 × 28 = 784 cm²

⇒ Area of the circular logo = 3.1416 × 14 × 14 ≈ 615.75 cm²

⇒ Unused square space = 784 − 615.75 = 168.25 cm²

∴ The correct answer is 168.25 cm².

Additional Information

Area of Circumscribed Square

When a circle of radius r perfectly fits (is inscribed) inside a square, the side of the square is exactly the diameter of the circle (2r). The area of the square evaluates to 4r².

Area of Inscribed Square

Conversely, if a square is inscribed inside a circle of radius r, its diagonal matches the circle's diameter (2r), and its area calculates to 2r².

Inscribed Area Ratio

The ratio of the area of an inscribed circle to the area of its circumscribing square is π : 4. This implies the circle covers approximately 78.5% of the total square space.

Shortcut Trick

The minute hand covers a complete circle (360°) in 60 minutes.

In 30 minutes, it covers exactly half of the circular path.

Distance covered in 30 minutes = Half of circumference = π × r

Distance = (22/7) × 14 = 44 cm

∴ The correct answer is 44 cm.

Alternate Method

r = 14 cm 30 min (180°) 12 6

Given:

Radius of the clock face (r) = 14 cm

Time duration = 30 minutes

Formula Used:

Arc length = (θ ÷ 360°) × 2πr

Angle covered by minute hand in 1 minute = 360° ÷ 60 = 6°

Calculations:

⇒ Angle covered in 30 minutes (θ) = 30 × 6° = 180°

⇒ Arc length = (180° ÷ 360°) × 2 × (22/7) × 14

⇒ Arc length = (1/2) × 2 × 22 × 2

⇒ Arc length = 1 × 44 = 44 cm

∴ The correct answer is 44 cm.

Additional Information

Area of a Sector

The region bounded by two radii and an arc is called a sector. Formula: Area = (θ ÷ 360°) × πr².

Speed of Clock Hands

The minute hand moves at a speed of 6° per minute, while the hour hand moves at 0.5° per minute.

Circumference of a Circle

The total boundary length of a circular path is its circumference. Formula: C = 2πr or C = πd.

Shortcut Trick

The given line is in the form y = mx + c, giving slope m = -3.

For perpendicular lines, the product of their slopes is always -1.

So, perpendicular slope = -1 ÷ (-3) = 1/3.

∴ The correct answer is 1/3.

Alternate Method

Given:

Equation of the line: y = -3x + 7

XYy = -3x + 7Slope (m1) = -3Perpendicular LineSlope (m2) = ?90°

Formula Used:

Slope-intercept form of a line: y = mx + c, where m is the slope.

Condition for two perpendicular lines: m₁ × m₂ = -1

Calculations:

⇒ The given line is y = -3x + 7

⇒ Comparing with y = mx + c, we get the slope of the line, m₁ = -3

⇒ Let the slope of the perpendicular line be m₂

⇒ According to the condition for perpendicular lines, m₁ × m₂ = -1

⇒ -3 × m₂ = -1

⇒ m₂ = 1/3

∴ The correct answer is 1/3.

Additional Information

Parallel Lines

Lines which are parallel to each other have equal slopes (m₁ = m₂).

General Equation of a Line

For a line given by Ax + By + C = 0, the slope is m = -A/B.

Angle Between Two Intersecting Lines

If m₁ and m₂ are the slopes, the acute angle θ between them is given by tan(θ) = |(m₁ − m₂) ÷ (1 + m₁ × m₂)|.

Shortcut Trick

If x − 1/x = a, then x² + 1/x² = a² + 2

If x² + 1/x² = b, then x⁴ + 1/x⁴ = b² − 2

Here, a = 3 ⇒ b = 3² + 2 = 9 + 2 = 11

Now, x⁴ + 1/x⁴ = 11² − 2 = 121 − 2 = 119

∴ The correct answer is 119.

Alternate Method

Given:

x − 1/x = 3

Formula Used:

(a − b)² = a² − 2ab + b²

(a + b)² = a² + 2ab + b²

Calculations:

Squaring both sides of the given equation:

⇒ (x − 1/x)² = 3²

⇒ x² − 2 × x × 1/x + 1/x² = 9

⇒ x² − 2 + 1/x² = 9

⇒ x² + 1/x² = 9 + 2 = 11

Squaring both sides again:

⇒ (x² + 1/x²)² = 11²

⇒ x⁴ + 2 × x² × 1/x² + 1/x⁴ = 121

⇒ x⁴ + 2 + 1/x⁴ = 121

⇒ x⁴ + 1/x⁴ = 121 − 2 = 119

∴ The correct answer is 119.

Additional Information

Square Identity 1

If x + 1/x = a, then x² + 1/x² = a² − 2

Cube Identity 1

If x + 1/x = a, then x³ + 1/x³ = a³ − 3a

Cube Identity 2

If x − 1/x = a, then x³ − 1/x³ = a³ + 3a

Power 5 Identity

x⁵ + 1/x⁵ = (x² + 1/x²) × (x³ + 1/x³) − (x + 1/x)

Shortcut Trick

Convert the percentage to a fraction to simplify the calculation directly.

2.5% = 2.5 ÷ 100 = 1 ÷ 40

Multiply the fraction by the given number:

Value = (1 ÷ 40) × 200 = 5

∴ The correct answer is 5.

Alternate Method

Given:

Base number = 200

Percentage = 2.5%

Formula Used:

Percentage Value = (Percentage ÷ 100) × Base Number

5 195

Total = 200 (100%)

2.5%

Calculations:

⇒ Value = (2.5 ÷ 100) × 200

⇒ Value = 2.5 × 2

⇒ Value = 5

∴ The correct answer is 5.

Additional Information

Percentage to Fraction Conversion

To convert any percentage to a fraction, divide it by 100 and simplify. Example: 2.5% = 2.5 ÷ 100 = 25 ÷ 1000 = 1 ÷ 40.

Successive Percentage

When a value changes by x% and then by y%, the net percentage change is given by the formula: Net Change = x + y + (x × y) ÷ 100.

Percentage Change

To find the percentage change between two values, use the formula: % Change = (New Value − Old Value) ÷ Old Value × 100.