Search Previous Year Questions

Shortcut Trick

Let the average of the 6 integers be A ⇒ Sum = 6A.

Average of 5 integers = A + 2 ⇒ Sum = 5(A + 2) = 5A + 10.

Difference in sums is given as 4: (5A + 10) − 6A = 4.

⇒ 10 − A = 4 ⇒ A = 6.

∴ The correct answer is 6.

Alternate Method

Given:

⇒ Number of integers in the first set = 6

⇒ Number of integers in the second set = 5

⇒ The average of the 6 integers is 2 less than the average of the 5 integers.

⇒ The sum of the 5 integers is 4 more than the sum of the 6 integers.

Formula Used:

⇒ Sum of observations = Average × Number of observations

Sum of 5 integers = 5 × (A + 2)

A+2

A+2

A+2

A+2

A+2

Sum of 6 integers + 4 = 6A + 4

A

A

A

A

A

A

+4

Equal Sum

5(A + 2) = 6A + 4

Calculations:

⇒ Let the average of the 6 integers be x.

⇒ Therefore, the average of the 5 integers = x + 2.

⇒ Sum of the 6 integers = 6 × x = 6x.

⇒ Sum of the 5 integers = 5 × (x + 2) = 5x + 10.

⇒ According to the problem statement: Sum of 5 integers = Sum of 6 integers + 4.

⇒ 5x + 10 = 6x + 4

⇒ 6x − 5x = 10 − 4

⇒ x = 6

∴ The correct answer is 6.

Additional Information

Average Formula

The average (or arithmetic mean) is defined as the sum of all observations divided by the total number of observations. Conversely, Sum = Average × Total Count.

Average of Consecutive Integers

For any sequence of consecutively spaced numbers, the average is always equal to the exact middle value of the data set. Additionally, the average equals (First Term + Last Term) ÷ 2.

Properties of Averages

If every observation in a dataset is increased, decreased, multiplied, or divided by a constant value 'k', the new resulting average also undergoes the exact same mathematical operation by 'k'.

Shortcut Trick

Let the total number be 12 units (the LCM of denominators 4 and 3).

75% (or 3/4) of 12 units = 9 units.

2/3 of 12 units = 8 units.

Difference = 9 units − 8 units = 1 unit.

Since 1 unit represents 15, the total number = 12 × 15 = 180.

∴ The correct answer is 180.

Alternate Method

Given:

75% of a number is 15 more than 2/3 of that number.

Method Used:

Convert percentage to fraction: 75% = 75/100 = 3/4

Total Number = 12 Units

12 Units (LCM of 4 and 3)

75% (or 3/4) = 9 Units

9 Units

2/3 = 8 Units

8 Units

Diff = 15

Calculations:

Let the required number be x.

⇒ 75% of x = 3x/4

⇒ Two-thirds of x = 2x/3

According to the problem statement:

⇒ 3x/4 − 2x/3 = 15

Taking the LCM of the denominators 4 and 3 (LCM = 12):

⇒ (9x − 8x) ÷ 12 = 15

⇒ x / 12 = 15

⇒ x = 15 × 12

⇒ x = 180

∴ The correct answer is 180.

Additional Information

Fractions and Percentages Equivalents

Percentages can be converted into fractions by dividing by 100. Important standard equivalents to remember: 75% = 3/4, 66.66% = 2/3, 25% = 1/4, and 50% = 1/2.

LCM Method for Fractions

When solving problems with multiple fractional parts, assuming the total initial value as the Least Common Multiple (LCM) of all denominators allows for faster, integer-based calculations.

Percentage Change

If a value changes from an initial base A to a new value B, the overall percentage change is given by the formula: ((B − A) ÷ A) × 100%.

Shortcut Trick

Interest rate = 10% = 1/10 ⇒ The amount to be paid back becomes 11/10 times the principal each year.

Since the rate multiplier has 11 in the numerator, the annual instalment value must be completely divisible by 11.

Checking the given options, only ₹6,655 is exactly divisible by 11 (6655 ÷ 11 = 605).

Alternatively, the remaining principal = 19650 − 3100 = ₹16,550.

Each Instalment = 16550 × (1331 / 3310) = 5 × 1331 = ₹6,655.

∴ The correct answer is ₹6,655.

Alternate Method

PV = ₹16,550

Year 1: ₹X

Year 2: ₹X

Year 3: ₹X

PV&sub1; = X / 1.1 PV&sub2; = X / 1.21 PV&sub3; = X / 1.331

Given:

⇒ Total price of TV = ₹19,650

⇒ Down payment = ₹3,100

⇒ Rate of interest (R) = 10% per annum

⇒ Number of equal instalments (n) = 3

Formula Used:

⇒ Present Value (P) = X / (1 + R/100) + X / (1 + R/100)² + X / (1 + R/100)³

Calculations:

⇒ Principal to be paid in instalments (P) = 19,650 − 3,100 = ₹16,550

⇒ Let the equal annual instalment be ₹X.

⇒ 16550 = X / (1 + 10/100) + X / (1 + 10/100)² + X / (1 + 10/100)³

⇒ 16550 = X / 1.1 + X / 1.21 + X / 1.331

⇒ 16550 = X × [10/11 + 100/121 + 1000/1331]

⇒ 16550 = X × [(1210 + 1100 + 1000) / 1331]

⇒ 16550 = X × (3310 / 1331)

⇒ X = (16550 × 1331) / 3310

⇒ X = 5 × 1331 = 6655

∴ The correct answer is ₹6,655.

Additional Information

Compound Interest Instalments

When a loan P is paid back in n equal annual instalments X at a compound interest rate R%, the formula used is: P = X / (1 + R/100) + X / (1 + R/100)² + ... + X / (1 + R/100)ⁿ.

Simple Interest Instalments

If a debt A is discharged in n equal annual instalments X at a simple interest rate R%, the formula is: A = nX + [RX × n(n − 1)] / 200.

Present Value Concept

The sum of the present values of all future instalment payments is always exactly equal to the total principal loan amount taken initially.

Shortcut Trick

Let the Cost Price (CP) of the mobile phone be 100 units.

Since the initial profit is 20%, the original selling price P = 120 units.

The Marked Price (MP) for the festive offer is 1.5P = 1.5 × 120 = 180 units.

After a 10% discount on MP, the final Selling Price = 180 − (10% of 180) = 162 units.

Net Profit % = [(162 − 100) ÷ 100] × 100 = 62%.

∴ The correct answer is 62%.

Alternate Method

Given:

Initial Profit = 20%

Initial Selling Price = ₹P

Marked Price (MP) = 1.5P

Discount = 10%

Formula Used:

Profit = Final Selling Price − Cost Price

Profit % = (Profit ÷ Cost Price) × 100

  CP = 100 P = 120 New SP = 162 MP = 180   +20% Profit   Marked at 1.5P   -10% Discount   Final Profit = 62%

Calculations:

⇒ Let Cost Price (CP) = 100x

⇒ Original Selling Price (P) = CP + 20% of CP = 100x + 20x = 120x

⇒ Marked Price (MP) = 1.5 × P = 1.5 × 120x = 180x

⇒ Discount given = 10% of MP = (10 ÷ 100) × 180x = 18x

⇒ Final Selling Price = MP − Discount = 180x − 18x = 162x

⇒ Profit = Final Selling Price − CP = 162x − 100x = 62x

⇒ Profit % = (62x ÷ 100x) × 100 = 62%

∴ The correct answer is 62%.

Additional Information

Marked Price and Discount

Marked price is the price printed on an article. Discount is always provided on the Marked Price. Formula: SP = MP × (100 − Discount%) ÷ 100

Mark-up Percentage

It is the percentage by which the cost price is increased to set the marked price. Formula: MP = CP × (100 + Mark-up%) ÷ 100

Successive Discounts

When two discounts d₁% and d₂% are given consecutively, the equivalent single discount is [d₁ + d₂ − (d₁ × d₂) ÷ 100]%.

Shortcut Trick

If 20% profit corresponds to ₹450, the Selling Price (120% of CP) = ₹450 × 6 = ₹2700.

This SP is after a 15% discount, meaning 85% of Marked Price (MP) = ₹2700.

We need the new Selling Price at a 25% discount, which is 75% of MP.

New SP = ₹2700 × (75 ÷ 85) = ₹2700 × (15 ÷ 17)

New SP = 40500 ÷ 17 ≈ ₹2382.35

∴ The correct answer is ₹2382.35.

Alternate Method

Given:

Discount₁ = 15%

Profit = 20%

Profit amount = ₹450

Discount₂ = 25%

Formula Used:

Cost Price (CP) = Profit × (100 ÷ Profit%)

Selling Price (SP) = CP + Profit

Marked Price (MP) = SP × [100 ÷ (100 − Discount%)]

CP

SP₁ = 2700

MP

SP₂ = ?

+20% (₹450)

-15%

-25% Discount

Calculations:

⇒ Cost Price (CP) = 450 × (100 ÷ 20) = ₹2250

⇒ Selling Price (SP₁) = 2250 + 450 = ₹2700

⇒ SP₁ is after a 15% discount, meaning it is 85% of MP.

⇒ 85% of MP = ₹2700

⇒ MP = 2700 × (100 ÷ 85) = 54000 ÷ 17

⇒ New Selling Price (SP₂) at a 25% discount is 75% of MP.

⇒ SP₂ = (54000 ÷ 17) × (75 ÷ 100)

⇒ SP₂ = 40500 ÷ 17 ≈ ₹2382.35

∴ The correct answer is ₹2382.35.

Additional Information

Marked Price (MP)

The price printed on an article. Discount is always calculated on the MP unless explicitly stated otherwise.

Direct Relation (CP & MP)

The ratio between Marked Price and Cost Price can be directly found using: MP ÷ CP = (100 + Profit%) ÷ (100 − Discount%).

Successive Discounts

If two successive discounts of a% and b% are given, the equivalent single discount is (a + b − ab ÷ 100)%.

Shortcut Trick

Total work = LCM(20, 30, 60) = 60 units.

Efficiencies: A = 60 ÷ 20 = 3, B = 60 ÷ 30 = 2, C = 60 ÷ 60 = 1.

Pattern: A works for 2 days, and A + B + C work on the 3^rd day.

Work done in 3 days = (3) + (3) + (3 + 2 + 1) = 3 + 3 + 6 = 12 units.

Number of 3-day cycles needed = 60 ÷ 12 = 5 cycles.

Total time = 5 × 3 = 15 days.

∴ The correct answer is 15 days.

Alternate Method

Given:

Time taken by A = 20 days

Time taken by B = 30 days

Time taken by C = 60 days

Calculations:

Let the Total Work be the LCM of the given days.

⇒ Work done by A on the 1^st day = 3 units

⇒ Work done by A on the 2^nd day = 3 units

⇒ Work done by A, B, and C together on the 3^rd day = 3 + 2 + 1 = 6 units

⇒ Total work done in one 3-day cycle = 3 + 3 + 6 = 12 units

⇒ Number of cycles required to complete 60 units of work = 60 ÷ 12 = 5 cycles

⇒ Total days required = 5 cycles × 3 days/cycle = 15 days

∴ The correct answer is 15 days.

Additional Information

Total Work Rule

Total Work is often considered as the Least Common Multiple (LCM) of the days taken by different individuals to complete the given task.

Work Efficiency

Efficiency = Total Work ÷ Total Days. It represents the units of work done by a person in a single day.

Alternate Day Work

When people work in a specific periodic pattern, sum their combined work into a cycle and find the total time by finding the number of full cycles needed.

Combined Efficiency

When individuals work together on any given day, their combined efficiency is simply the sum of their individual daily efficiencies.

Shortcut Trick

Mean Cost Price = Selling Price × [100 ÷ (100 + Profit%)]

Mean Cost Price = 132 × (100 ÷ 120) = ₹110/kg

Using Alligation Cross-Multiplication directly:

Quantity Ratio = (120 − 110) : (110 − 80)

Quantity Ratio = 10 : 30 = 1 : 3

∴ The correct answer is 1 : 3.

Alternate Method

Given:

Cost price of Grade 1 rice (CP₁) = ₹80/kg

Cost price of Grade 2 rice (CP₂) = ₹120/kg

Selling price of the mixture (SP) = ₹132/kg

Overall profit margin = 20%

Formula Used:

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

Alligation Ratio = (CP₂ − Mean CP) : (Mean CP − CP₁)

Calculations:

⇒ Mean CP of mixture = (132 × 100) ÷ (100 + 20)

⇒ Mean CP = 13200 ÷ 120 = ₹110/kg

80 CP₁

120 CP₂

110 Mean CP

10

30

Ratio = 10 : 30

⇒ Quantity of Grade 1 : Quantity of Grade 2 = 10 : 30

⇒ Simplest Ratio = 1 : 3

∴ The correct answer is 1 : 3.

Additional Information

Rule of Alligation

It allows finding the specific ratio in which two or more ingredients at given prices must be mixed to produce a mixture of a desired average price. Ratio = (Dearer Price − Mean Price) : (Mean Price − Cheaper Price).

Mean Price Concept

The cost price of a unit quantity of the mixture is called its mean price. When using the alligation method, always ensure that all values (CP₁, CP₂, and Mean Price) are purely Cost Prices. Never use Selling Price directly in the alligation cross.

Repeated Replacement

If a vessel contains 'x' units of a pure liquid and 'y' units are taken out and replaced by water 'n' times, the quantity of pure liquid remaining = x × (1 − y ÷ x)ⁿ.

Shortcut Trick

Speed of the bus = 320 ÷ 4 = 80 km/h.

Given the speed ratio of Car : Bus = 3 : 4.

Car's speed = (3 ÷ 4) × 80 = 60 km/h.

∴ The correct answer is 60 km/h.

Alternate Method

Given:

Distance covered by the bus = 320 km

Time taken by the bus = 4 hours

Ratio of speeds (Car : Bus) = 3 : 4

Formula Used:

Speed = Distance ÷ Time

Speed Ratio → Car : Bus = 3 : 4

BUS 320 km ÷ 4 h = 80 km/h 4 units = 80

CAR Speed = ? 3 units = 60

Car Speed = 60 km/h

Calculations:

⇒ Speed of the bus = 320 ÷ 4 = 80 km/h

⇒ Let the speed of the car and the bus be 3x and 4x respectively.

⇒ Speed of the bus = 4x

⇒ 4x = 80

⇒ x = 20

⇒ Speed of the car = 3x = 3 × 20 = 60 km/h

∴ The correct answer is 60 km/h.

Additional Information

Relative Speed

When two bodies are moving in the same direction, relative speed = a − b. When moving in opposite directions, relative speed = a + b.

Average Speed Formula

When equal distances are covered at speeds a and b, Average Speed = (2 × a × b) ÷ (a + b). For varying distances, Average Speed = Total Distance ÷ Total Time.

Unit Conversion

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

Shortcut Trick

From the given conditions, R = 2.5P and R = 3Q. The ratio of investments P : Q : R is 6 : 5 : 15.

Let the investments in the three schemes be 600, 500, and 1500 units respectively.

Total interest in 1 year = 7% of 600 + 9% of 500 + 11% of 1500 = 42 + 45 + 165 = 252 units.

We are given total interest = ₹2520, so 252 units = 2520 ⇒ 1 unit = 10.

Amount invested in Scheme P = 600 units = 600 × 10 = ₹6000.

∴ The correct answer is ₹6000.

Alternate Method

Given:

Rate of interest for P = 7% p.a., Q = 9% p.a., R = 11% p.a.

Total Simple Interest (SI) earned in 1 year = ₹2520

Investment in R = 250% of P ⇒ R = (5 ÷ 2) × P ⇒ P ÷ R = 2 ÷ 5 = 6 ÷ 15

Investment in R = 300% of Q ⇒ R = 3 × Q ⇒ Q ÷ R = 1 ÷ 3 = 5 ÷ 15

Formula Used:

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

Scheme P P = 6x R = 7% SI = 0.42x

Scheme Q P = 5x R = 9% SI = 0.45x

Scheme R P = 15x R = 11% SI = 1.65x

Total Interest = 2.52x = ₹2520

Calculations:

⇒ Let the principal amounts be P = 6x, Q = 5x, and R = 15x.

⇒ SI from Scheme P = (6x × 7 × 1) ÷ 100 = 0.42x

⇒ SI from Scheme Q = (5x × 9 × 1) ÷ 100 = 0.45x

⇒ SI from Scheme R = (15x × 11 × 1) ÷ 100 = 1.65x

⇒ Total Simple Interest = 0.42x + 0.45x + 1.65x = 2.52x

⇒ We are given Total SI = ₹2520.

⇒ 2.52x = 2520

⇒ x = 2520 ÷ 2.52 = 1000

⇒ Amount invested in Scheme P = 6x = 6 × 1000 = ₹6000

∴ The correct answer is ₹6000.

Additional Information

Simple Interest (SI)

Simple interest is calculated only on the principal amount for the entire duration. Formula: SI = (P × R × T) ÷ 100.

Compound Interest (CI)

Compound interest is calculated on the principal plus any accumulated interest. Amount = P × (1 + R ÷ 100)^T.

Equal Simple Interest Rule

If a sum is divided into parts P₁, P₂, P₃ such that their simple interests are equal, their ratio is 1 ÷ (R₁T₁) : 1 ÷ (R₂T₂) : 1 ÷ (R₃T₃).

Shortcut Trick

Horizontal Distance = Height of tower × cot(bottom depression angle)

⇒ Distance = 50 × cot(45°) = 50 × 1 = 50 m

Height of building = Height of tower × [1 − cot(bottom angle) × tan(top angle)]

⇒ Height = 50 × [1 − cot(45°) × tan(30°)] = 50 × [1 − 1 × (1 ÷ √3)]

⇒ Height ≈ 50 × [1 − 0.577] = 50 × 0.423 = 21.15 m ≈ 21 m

∴ The correct answer is Height = 21 m, Distance = 50 m.

Alternate Method

30° 45° 30° 45°

50 m x = 50 m h ≈ 21 m

A B C D E F

Given:

Height of tower AB = 50 m

Angle of depression to the top of the building ∠FAC = 30°

Angle of depression to the bottom of the building ∠FAD = 45°

Calculations:

From the property of alternate interior angles, ∠ACE = ∠FAC = 30° and ∠ADB = ∠FAD = 45°.

In right-angled ΔABD:

⇒ tan(45°) = AB ÷ BD

⇒ 1 = 50 ÷ BD

⇒ BD = 50 m

So, the horizontal distance separating the tower from the building is 50 m.

Since CE is parallel to the ground BD, CE = BD = 50 m.

In right-angled ΔAEC:

⇒ tan(30°) = AE ÷ CE

⇒ 1 ÷ √3 = AE ÷ 50

⇒ AE = 50 ÷ √3 ≈ 50 ÷ 1.732 ≈ 28.87 m

The height of the building (CD) = EB = AB − AE

⇒ CD = 50 − 28.87 = 21.13 m

⇒ Approximated height of the building ≈ 21 m

∴ The correct answer is Height = 21 m, Distance = 50 m.

Additional Information

Angles of Elevation & Depression

An angle of elevation is measured upwards from the horizontal observer's line, whereas an angle of depression is measured downwards. They are numerically equivalent via the alternate interior angles property.

Common Trigonometric Values

tan(30°) = 1 ÷ √3 | tan(45°) = 1 | tan(60°) = √3. Standard approximate value used for √3 is 1.732.

Multi-Angle Height Formula

If top and bottom of an object are observed at respective depression angles α and β from height H, object height h = H × [1 − cot(β) × tan(α)].