Search Previous Year Questions

Shortcut Trick

Ratio of Ravi to Kaviraj 4 years ago = 3 : 1 (Difference = 2 units)

Ratio of Ravi to Kaviraj at present = 2 : 1 (Difference = 1 unit)

Equate differences by multiplying present ratio by 2 → 4 : 2 (Difference = 2 units)

The change in units from past to present (4 − 3 or 2 − 1) = 1 unit → 4 years

Ravi's present age = 4 units = 4 × 4 = 16 years

∴ The correct answer is 16 years.

Alternate Method

Given:

Presently, Ravi is twice as old as Kaviraj.

4 years ago, Ravi was thrice as old as Kaviraj.

Formula Used:

Age = Ratio × Common factor; Equation based on time shifts.

Calculations:

Let the present age of Kaviraj be K years.

⇒ Present age of Ravi = 2K

According to the condition 4 years ago:

⇒ (2K − 4) = 3 × (K − 4)

⇒ 2K − 4 = 3K − 12

⇒ 12 − 4 = 3K − 2K

⇒ K = 8 years

⇒ Ravi's present age = 2 × 8 = 16 years

∴ The correct answer is 16 years.

Additional Information

Constant Age Difference

The difference between the ages of two individuals always remains the same, regardless of how many years pass.

Ratio Balancing Method

To solve age problems quickly, ensure the difference between ratio parts is consistent across different time periods by multiplying with suitable constants.

Future Age Calculation

If the present age is x, then the age after n years is (x + n) and the age n years ago was (x − n).

Shortcut Trick

To find the LCM of decimals, convert them to like decimals first:

3 becomes 3.00, 7.2 becomes 7.20, and 0.64 remains 0.64.

Now, find the LCM of the integer values: 300, 720, and 64.

LCM(300, 720, 64) = 14400.

Place the decimal point back by moving two places to the left: 14400 → 144.

∴ The correct answer is 144.

Alternate Method

Given:

The numbers are 3, 7.2, and 0.64.

Formula Used:

LCM of Fractions = LCM of Numerators ÷ HCF of Denominators

Calculations:

Convert the decimals into their simplest fraction forms:

⇒ 3 = 3/1

⇒ 7.2 = 72/10 = 36/5

⇒ 0.64 = 64/100 = 16/25

Now, find the LCM of the fractions 3/1, 36/5, and 16/25:

⇒ Numerators are 3, 36, and 16. Their LCM is 144.

⇒ Denominators are 1, 5, and 25. Their HCF is 1.

Apply the fraction LCM formula:

⇒ LCM = 144 ÷ 1 = 144

∴ The correct answer is 144.

Additional Information

LCM and HCF of Fractions

LCM of fractions = (LCM of Numerators) ÷ (HCF of Denominators). HCF of fractions = (HCF of Numerators) ÷ (LCM of Denominators).

Product of Two Numbers Rule

For any two numbers A and B, Product of numbers (A × B) = LCM(A, B) × HCF(A, B).

Co-prime Numbers Rule

The HCF of co-prime numbers is always 1, and their LCM is equal to the product of those numbers.

Shortcut Trick

Effective Rate of Interest for the period = Rate × Time

Effective Rate = 8% × (9/12) years = 6%

According to the question, 6% of Principal = ₹18,000

1% of Principal = ₹3,000

100% (Principal Amount) = ₹3,000 × 100 = ₹3,00,000

∴ The correct answer is ₹3,00,000.

Alternate Method

Given:

Simple Interest (SI) = ₹18,000

Rate (R) = 8% per annum

Time (T) = 9 months = 9/12 years = 3/4 years

Formula Used:

SI = (P × R × T) / 100

Calculations:

⇒ 18,000 = (P × 8 × 3/4) / 100

⇒ 18,000 = (P × 2 × 3) / 100

⇒ 18,000 = (P × 6) / 100

⇒ 18,000 × 100 = 6P

⇒ P = 18,00,000 ÷ 6

⇒ P = ₹3,00,000

∴ The correct answer is ₹3,00,000.

Additional Information

Simple Interest Components

The formula SI = (P × R × T) / 100 connects four variables; knowing any three allows you to find the fourth. Principal (P) is the original sum invested or borrowed.

Time Conversion Rule

For interest calculations, the time 'T' must match the rate's period. If the rate is per annum, convert months to years by dividing by 12 or days by 365.

Total Amount

The Total Amount (A) is the sum of the Principal and the interest earned: A = P + SI, or A = P[1 + (RT/100)].

Shortcut Trick

The standard formula for the area of a regular polygon with n sides, each of length a is given by:

Area = (n × a²) ÷ (4 × tan(180° ÷ n)) = (n × a²) × cot(180° ÷ n) ÷ 4

Given, n = 45 and a = 2 cm:

⇒ Area = (45 × 2²) ÷ (4 × tan(180° ÷ 45))

⇒ Area = (45 × 4) ÷ (4 × tan(4°)) = 45 ÷ tan(4°) = 45 cot(4°)

Since 45 cot(4°) matches standard area formulas, we simplify to get the required expression from the standard options.

∴ The correct answer is 45 cot(4°).

Alternate Method

Given:

Number of sides (n) = 45

Length of each side (a) = 2 cm

Formula Used:

Area of a regular polygon of n sides is:

Area = (n × a²) ÷ (4 × tan(π ÷ n))

Where π radians = 180°.

Calculations:

⇒ Area = (45 × 2²) ÷ (4 × tan(180° ÷ 45))

⇒ Area = (45 × 4) ÷ (4 × tan(4°))

⇒ Area = 45 ÷ tan(4°)

Using the trigonometric identity cot(θ) = 1 ÷ tan(θ):

⇒ Area = 45 cot(4°) cm²

∴ The correct answer is 45 cot(4°).

Additional Information

Area of Regular Polygon with Inradius

The area of a regular polygon can also be expressed as Area = (1 ÷ 2) × Perimeter × Inradius (apothem), where apothem r = a ÷ (2 × tan(180°/n)).

Interior and Exterior Angles

For any regular polygon with n sides, each exterior angle is 360° ÷ n, and each interior angle is 180° − (360° ÷ n).

Shortcut Trick

Assume total Earth area = 28 units (LCM of Whole Earth ratio sum 4 and Hemisphere ratio sum 7 × 2).

Then Northern Hemisphere (NH) = 14 units and Southern Hemisphere (SH) = 14 units.

Total Land = 1/4 × 28 = 7 units | Total Water = 3/4 × 28 = 21 units.

Land in NH = 2/7 × 14 = 4 units | Water in NH = 5/7 × 14 = 10 units.

Land in SH = 7 − 4 = 3 units | Water in SH = 21 − 10 = 11 units.

SH Ratio (Land:Water) = 3:11.

∴ The correct answer is 3:11.

Alternate Method

Given:

Ratio of Land to Water (Earth) = 1 : 3

Ratio of Land to Water (Northern Hemisphere) = 2 : 5

Northern Hemisphere Area = Southern Hemisphere Area

Calculations:

⇒ Let the total area of the Earth be 280 units (to avoid decimals).

⇒ Total Land on Earth = (1 / (1+3)) × 280 = 1/4 × 280 = 70 units.

⇒ Total Water on Earth = (3 / (1+3)) × 280 = 3/4 × 280 = 210 units.

⇒ Area of Northern Hemisphere = 280 ÷ 2 = 140 units.

⇒ Land in Northern Hemisphere = (2 / (2+5)) × 140 = 2/7 × 140 = 40 units.

⇒ Water in Northern Hemisphere = (5 / (2+5)) × 140 = 5/7 × 140 = 100 units.

⇒ Land in Southern Hemisphere = Total Land − NH Land = 70 − 40 = 30 units.

⇒ Water in Southern Hemisphere = Total Water − NH Water = 210 − 100 = 110 units.

⇒ Required Ratio = 30 : 110 = 3 : 11.

∴ The correct answer is 3:11.

Additional Information

Ratio and Proportion Basics

If a quantity is divided in the ratio a:b, the parts are [a/(a+b)] × Total and [b/(a+b)] × Total.

Combining Ratios

When components of a whole are given in different ratios, always assume a total value that is the least common multiple (LCM) of the sum of ratio parts to simplify calculations.

Hemisphere Area Property

The equator divides a sphere into two equal halves (hemispheres), so Area(NH) = Area(SH) = 1/2 × Area(Total Earth).

Shortcut Trick

Initial CP : SP ratio = 100 : 138 = 50 : 69

Final CP : SP ratio = 100 : 137.5 = 8 : 11

Apply cross-multiplication: |(50 × 11 − 69 × 8)| units = |(11 × 20 − 8 × (−45))|

2 units = 220 + 360 = 580 ⇒ 1 unit = 290

Original Cost Price = 50 units = 50 × 290 = 14500

∴ The correct answer is ₹14500.

Alternate Method

Given:

Initial Profit = 38% | New Profit = 37.5% | CP Increase = ₹20 | SP Decrease = ₹45

Formula Used:

Selling Price (SP) = Cost Price (CP) × (1 + Profit%)

Calculations:

⇒ Let original Cost Price = x

⇒ Original Selling Price = 1.38x

⇒ New CP = x + 20 and New SP = 1.38x − 45

⇒ According to question: New SP = 1.375 × New CP

⇒ 1.38x − 45 = 1.375 × (x + 20)

⇒ 1.38x − 45 = 1.375x + 27.5

⇒ 1.38x − 1.375x = 27.5 + 45

⇒ 0.005x = 72.5

⇒ x = 72.5 ÷ 0.005 = 14500

∴ The correct answer is ₹14500.

Additional Information

Profit and Loss Relationship

Profit occurs when SP > CP. The fundamental formula is SP = CP + Profit, where Profit = (Profit% ÷ 100) × CP.

Percentage to Fraction Conversion

Calculations become faster by converting common percentages: 37.5% = 3/8, which implies SP/CP = 11/8 for a profit.

Cost Price Change Impact

If both CP and SP change, the new profit percentage is always calculated on the new Cost Price.

Shortcut Trick

Total work = LCM(12, 20) = 60 units.

Efficiency of A = 5 units/day; Efficiency of B = 3 units/day.

Combined work in 3 days = (5 + 3) × 3 = 24 units.

Remaining work = 60 − 24 = 36 units.

Time taken by B = 36 ÷ 3 = 12 days.

∴ The correct answer is 12 days.

Alternate Method

Given:

Time taken by A = 12 days

Time taken by B = 20 days

Work duration for both = 3 days

Calculations:

⇒ Total efficiency of (A + B) = 5 + 3 = 8 units/day

⇒ Work completed by both in 3 days = 8 × 3 = 24 units

⇒ Remaining Work = Total Work − Completed Work

⇒ Remaining Work = 60 − 24 = 36 units

⇒ Time taken by B to complete remaining work = Remaining Work ÷ Efficiency of B

⇒ Time = 36 ÷ 3 = 12 days

∴ The correct answer is 12 days.

Additional Information

Combined Efficiency

When two people work together, their combined efficiency is the sum of their individual efficiencies. Total Time = Total Work ÷ Combined Efficiency.

Pipes & Cisterns

This concept is identical to work efficiency where filling pipes have positive efficiency and leakage or outlet pipes have negative efficiency.

Work & Wages

Wages are always distributed in the ratio of work done by each person, which is the same as the ratio of their efficiencies if they work for the same duration.

Shortcut Trick

Net multiplier for consecutive discounts of 20% and 5% = 0.80 × 0.95 = 0.76

Bill amount after both discounts = 1000 × 0.76 = ₹760

Final payable amount = Discounted bill + Delivery charge

Final amount = 760 + 80 = ₹840

∴ The correct answer is ₹840.

Alternate Method

Given: Initial order value = ₹1000, Promotional Discount = 20%, Conditional Discount = 5% (if price is ₹650 to ₹800), Fixed Delivery Charge = ₹80.

⇒ First Discount = 20% of 1000 = (20 ÷ 100) × 1000 = ₹200

⇒ Price after first discount = 1000 − 200 = ₹800

⇒ The price ₹800 lies in the inclusive range of ₹650 to ₹800.

⇒ Additional Discount = 5% of 800 = (5 ÷ 100) × 800 = ₹40

⇒ Net food bill = 800 − 40 = ₹760

⇒ Final amount payable = Net bill + Delivery charge = 760 + 80 = ₹840

∴ The correct answer is ₹840.

Additional Information

Successive Discounts

When two discounts x% and y% are applied sequentially, the total effective discount is [x + y − (x × y ÷ 100)]%.

Fixed vs Variable Charges

Fixed charges like delivery fees or service charges are added to the final discounted price and do not receive discounts themselves.

Base Value for Discounts

A secondary discount is always calculated on the reduced price (the balance remaining after the first discount), not the original price.

Shortcut Trick

Ratio of parts = 6 : 4

Difference in ratio units = 6 − 4 = 2 units

Since the actual difference is 10: 2 units = 10 ⇒ 1 unit = 5

First part = 6 × 5 = 30

Second part = 4 × 5 = 20

∴ The correct answer is 30 and 20.

Alternate Method

Given:

Ratio of two parts = 6 : 4

Difference between parts = 10

Formula Used:

Value of a part = (Ratio part / Difference in ratio units) × Total Difference

Calculations:

⇒ Let the two parts be 6x and 4x.

⇒ Difference between the sections = 6x − 4x = 10

⇒ 2x = 10

⇒ x = 10 ÷ 2 = 5

⇒ First part = 6 × 5 = 30

⇒ Second part = 4 × 5 = 20

∴ The correct answer is 30 and 20.

Additional Information

Ratio and Proportion

A ratio compares two quantities by division. For two parts a and b, the ratio is a:b. If a common factor x is multiplied, the values become ax and bx.

Unitary Method

The unitary method involves finding the value of one single unit (part) first, and then multiplying that value by the required number of units.

Difference between Ratio Terms

If the ratio is a:b and the difference between values is D, then 1 ratio unit = D ÷ |a − b|.

Shortcut Trick

Assume the average weight is 53 kg.

Deviation of A = (47 − 53) × 12 = −72

Deviation of B = (62 − 53) × 8 = +72

Deviation of C = (53 − 53) × 15 = 0

Net Deviation = −72 + 72 + 0 = 0

Average = 53 + (0 ÷ 35) = 53 kg.

∴ The correct answer is 53 kg.

Alternate Method

Given:

Units of Type A = 12 (Avg weight 47 kg)

Units of Type B = 8 (Avg weight 62 kg)

Units of Type C = 15 (Avg weight 53 kg)

Formula Used:

Weighted Average = (n₁w₁ + n₂w₂ + n₃w₃) ÷ (n₁ + n₂ + n₃)

Calculations:

⇒ Total Units = 12 + 8 + 15 = 35

⇒ Total Weight of A = 12 × 47 = 564 kg

⇒ Total Weight of B = 8 × 62 = 496 kg

⇒ Total Weight of C = 15 × 53 = 795 kg

⇒ Sum of all weights = 564 + 496 + 795 = 1855 kg

⇒ Weighted Average = 1855 ÷ 35

⇒ Weighted Average = 53 kg

∴ The correct answer is 53 kg.

Additional Information

Weighted Average Principle

It accounts for the relative contribution of each data point. Formula: Σ(n × x) ÷ Σn, where n is frequency and x is value.

Assumed Mean Method

Simplifies calculation by using deviations from a chosen middle value: Average = Assumed Mean + (ΣDeviations ÷ Total Frequency).

Combined Mean Formula

For two groups with sizes n₁, n₂ and averages a₁, a₂, the total mean is (n₁a₁ + n₂a₂) ÷ (n₁ + n₂).