Search Previous Year Questions

Shortcut Trick

Multiplying factor for 1 year = Amount₃ ÷ Amount₂ = 7920 ÷ 6600 = 6/5

Since the sum amounts to ₹6,600 in 2 years at CI:

Principal × (6/5)² = 6600

⇒ Principal = 6600 × (25/36) ≈ 4583.33

∴ The correct answer is ₹4,583.33.

Alternate Method

Given:

Amount after 2 years (A₂) = ₹6,600

Amount after 3 years (A₃) = ₹7,920

Year 0 P = ? Year 2 ₹6,600 Year 3 ₹7,920

× (1 + R/100)² × (1 + R/100) Compound Interest Timeline

Formula Used:

A = P × (1 + R/100)ⁿ

Calculations:

⇒ A₃ ÷ A₂ = [P × (1 + R/100)³] ÷ [P × (1 + R/100)²]

⇒ 7920 ÷ 6600 = 1 + R/100

⇒ 1.2 = 1 + R/100

⇒ R = 20%

Now, substitute R into the 2^nd year amount formula:

⇒ P × (1 + 20/100)² = 6600

⇒ P × (1.2)² = 6600

⇒ P × 1.44 = 6600

⇒ P = 6600 ÷ 1.44 = 4583.33

∴ The correct answer is ₹4,583.33.

Additional Information

Interest for 1 Year in CI

In compound interest, the amount accumulated at the end of a year acts as the principal for the next year. Thus, dividing the amount of any given year by the previous year's amount gives the direct multiplying factor for one year.

Simple Interest vs Compound Interest

For the first year (or the first compounding period), both Simple Interest and Compound Interest generate exactly the same interest. The difference arises from the second year onwards because Compound Interest charges interest on the previously accumulated interest.

Effective Rate for 2 Years

The effective compound interest rate for 2 consecutive years at R% per annum can be calculated directly using the successive percentage formula: R + R + (R × R) ÷ 100.

Shortcut Trick

Calculate the profit or loss per liter directly by comparing the cost prices with the final selling price of ₹48/liter.

Profit on the first quantity (25 liters) = 25 × (48 − 45) = 25 × 3 = +₹75

Loss on the second quantity (15 liters) = 15 × (48 − 50) = 15 × (−2) = −₹30

Net result = +75 − 30 = +₹45

Since the net result is positive, it represents an overall profit.

∴ The correct answer is Profit of ₹45.

Alternate Method

Given:

Quantity of first milk = 25 liters at ₹45/liter

Quantity of second milk = 15 liters at ₹50/liter

Selling price of the combined mixture = ₹48/liter

25 L @ ₹45/LCP = ₹112515 L @ ₹50/LCP = ₹750Total CP (40 L)₹1875Total SP @ ₹48₹1920Profit = ₹1920 − ₹1875 = ₹45

Formula Used:

Total CP = (Q₁ × CP₁) + (Q₂ × CP₂)

Total SP = Total Quantity × SP per liter

Profit = Total SP − Total CP

Calculations:

⇒ CP of 25 liters of milk = 25 × 45 = ₹1125

⇒ CP of 15 liters of milk = 15 × 50 = ₹750

⇒ Total CP of the mixture = 1125 + 750 = ₹1875

⇒ Total quantity of the mixture = 25 + 15 = 40 liters

⇒ Total SP of the mixture = 40 × 48 = ₹1920

⇒ Since Total SP > Total CP, Suman makes a profit.

⇒ Total Profit = 1920 − 1875 = ₹45

∴ The correct answer is Profit of ₹45.

Additional Information

Weighted Average (Mixtures)

The mean cost price of a mixture formed by mixing two distinct quantities can be calculated as: Mean CP = (Q₁ × P₁ + Q₂ × P₂) ÷ (Q₁ + Q₂).

Profit and Loss Fundamentals

A business transaction yields a profit when the Selling Price (SP) is greater than the Cost Price (CP). Conversely, it results in a loss if the CP exceeds the SP. The absolute profit is computed as: Profit = SP − CP.

Rule of Alligation

This rule provides a direct method to find the ratio in which two quantities of varying prices must be combined to yield a required average price: Quantity of Cheaper ÷ Quantity of Dearer = (Price of Dearer − Mean Price) ÷ (Mean Price − Price of Cheaper).

Shortcut Trick

Remaining cars after charity = 1200 − 200 = 1000.

Effective paid cars = 1000 × [8 ÷ (8 + 2)] = 800.

Selling Price (SP) per car = ₹120 − 10% = ₹108.

Total Revenue (SP) = 800 × ₹108 = ₹86,400.

Loss % = [(90,000 − 86,400) ÷ 90,000] × 100 = 4%.

∴ The correct answer is 4% loss.

Alternate Method

Given:

Total cost price (CP) = ₹90,000

Total toy cars produced = 1200

Cars given to charity = 200

Market price (MP) per car = ₹120

Discount = 10%

Total Produced = 1200 cars (Total CP = ₹90,000)DonatedFreePaid (800 cars)200200800Selling Price per paid car = ₹120 − 10% = ₹108Total Revenue (SP) = 800 × ₹108 = ₹86,400Loss = ₹90,000 − ₹86,400 = ₹3,600

Calculations:

⇒ Remaining cars left for sale = 1200 − 200 = 1000

⇒ Selling Price per car after discount = 120 − (10% of 120) = 120 − 12 = ₹108

⇒ The offer states 2 free cars for every 8 cars purchased. This means out of every 10 cars handed out, 8 are paid for.

⇒ Number of billed/paid cars = 1000 × (8 ÷ 10) = 800 cars

⇒ Total Selling Price (SP) = 800 × 108 = ₹86,400

⇒ Overall Loss = Total CP − Total SP = 90000 − 86400 = ₹3600

⇒ Loss % = (Loss ÷ CP) × 100

⇒ Loss % = (3600 ÷ 90000) × 100 = 4%

∴ The correct answer is 4% loss.

Additional Information

Buy X Get Y Free Concept

This means for a total of (X + Y) articles, the customer only pays for X articles. The effective discount percentage for the customer is calculated as: [Y ÷ (X + Y)] × 100.

Marked Price (MP) & Discount

Discount is always calculated on the Marked Price. Selling Price (SP) = MP − Discount. The final Selling Price can also be calculated directly using SP = MP × (100 − D%) ÷ 100.

Profit & Loss Percentage

Profit or loss is always calculated over the Cost Price (CP) unless explicitly stated otherwise. Loss % = (Loss ÷ CP) × 100 and Profit % = (Profit ÷ CP) × 100.

Shortcut Trick

Let the Cost Price be CP.

Fraction values: +80% markup = 180/100 = 9/5.

Fraction values: 25% discount = 3/4, and 10% discount = 9/10.

Final Selling Price = CP × 9/5 × 3/4 × 9/10 = CP × 243/200

⇒ 15552 = CP × 243/200 ⇒ CP = 15552 × 200 ÷ 243 = 12800

∴ The correct answer is ₹12,800.

Alternate Method

Given:

Markup = 80%, 1^st Discount = 25%, 2^nd Discount = 10%

Final Selling Price = ₹15,552

CP (100x)

+80%

MP (180x)

-25%

SP1 (135x)

-10%

Final SP (121.5x)

Final Selling Price (121.5x) = ₹15,552

Calculations:

⇒ Let the Cost Price (CP) be 100x.

⇒ Marked Price (MP) = 100x + 80% of 100x = 180x

⇒ Selling Price after 1^st discount (25%) = 180x × (1 − 25/100) = 135x

⇒ Final Selling Price after 2^nd discount (10%) = 135x × (1 − 10/100) = 121.5x

⇒ Given Final SP = ₹15,552

⇒ 121.5x = 15552

⇒ x = 15552 ÷ 121.5 = 128

⇒ Cost Price = 100x = 100 × 128 = 12800

∴ The correct answer is ₹12,800.

Additional Information

Marked Price (MP)

It is the price printed on the article. Markup percentage is always calculated on the base of Cost Price (CP).

Successive Discount

When multiple discounts are given sequentially, the final Selling Price is calculated by multiplying the successive discount factors: Final SP = MP × (1 − D₁/100) × (1 − D₂/100).

Equivalent Single Discount

If two successive discounts of x% and y% are given, the single equivalent discount is calculated using the formula: (x + y − xy/100)%.

Shortcut Trick

Since only water is added, the juice quantity is constant. Make juice units equal in both ratios:

Initial Juice : Water = 5 : 3 ⇒ (× 2) ⇒ 10 : 6

Final Juice : Water = 2 : 3 ⇒ (× 5) ⇒ 10 : 15

Initial quantity = 10 + 6 = 16 units = 40 L ⇒ 1 unit = 2.5 L.

Water added = 15 − 6 = 9 units ⇒ 9 × 2.5 = 22.5 L.

∴ The correct answer is 22.5 litres.

Alternate Method

Given:

Initial mixture = 40 liters | Initial ratio (Juice : Water) = 5 : 3

Final ratio (Juice : Water) = 2 : 3

Initial MixtureWater: 37.5%Added WaterWater: 100%Final60%100 − 60 = 4060 − 37.5 = 22.5Ratio of Quantities (Initial : Added) = 40 : 22.5Since Initial Mixture is 40 L   →   Added Water = 22.5 L

Calculations:

⇒ Quantity of Juice = (5 ÷ 8) × 40 = 25 liters

⇒ Quantity of Water = (3 ÷ 8) × 40 = 15 liters

Let the amount of water added be x liters, so the new amount of water = (15 + x) liters.

According to the new ratio of Juice to Water:

⇒ 25 ÷ (15 + x) = 2 ÷ 3

⇒ 75 = 30 + 2x

⇒ 2x = 45 ⇒ x = 22.5

∴ The correct answer is 22.5 litres.

Additional Information

Rule of Alligation

When two ingredients of different concentrations are mixed, the ratio of their quantities is: (Quantity of Cheaper) ÷ (Quantity of Dearer) = (Price of Dearer − Mean Price) ÷ (Mean Price − Price of Cheaper).

Constant Term Logic in Ratios

When only one component (like water) is added to a mixture, the quantity of the other component (like juice) remains unchanged. Equating the ratio units of the constant component naturally streamlines the calculation without needing algebra.

Replacement Formula

If V is the initial volume of pure liquid, and x volume is removed and replaced by water n times, the final volume of pure liquid left = V × (1 − x/V)ⁿ.

Shortcut Trick

Total work = LCM(15, 20) = 60 units.

Combined efficiency = (60 ÷ 15) + (60 ÷ 20) = 4 + 3 = 7 units/day.

Work done in 4 days = 7 × 4 = 28 units.

Remaining work = 60 − 28 = 32 units.

Fraction of work remaining = 32 ÷ 60 = 8/15.

∴ The correct answer is 8/15.

Alternate Method

Given:

Time taken by A to complete the task = 15 days

Time taken by B to complete the task = 20 days

Time they worked together = 4 days

⇒ Total efficiency of A and B = 4 + 3 = 7 units/day

⇒ Work done by A and B together in 4 days = 7 × 4 = 28 units

⇒ Remaining work = Total Work − Work Done

⇒ Remaining work = 60 − 28 = 32 units

⇒ Fraction of work remaining = Remaining Work ÷ Total Work

⇒ Fraction of work remaining = 32 ÷ 60 = 8/15

∴ The correct answer is 8/15.

Additional Information

Total Work & Efficiency

Efficiency is inversely proportional to the time taken. Total work is usually assumed as the LCM of the given individual days to avoid complex fractional calculations.

Work & Wages

Wages are completely distributed in proportion to the total actual work done by each person during the given period, not just by their efficiency ratios.

Pipes & Cisterns

This topic follows identical principles to Time & Work. The key difference is that filling pipes have a positive efficiency while emptying pipes (or leaks) have a negative efficiency.

Shortcut Trick

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

⇒ Mean CP = 144 × (100 ÷ 120) = ₹120/kg

Apply Alligation: Ratio = (Dearer Price − Mean Price) : (Mean Price − Cheaper Price)

⇒ Ratio = (150 − 120) : (120 − 90) = 30 : 30 = 1 : 1

∴ The correct answer is 1 : 1.

Alternate Method

Given:

Price of cheaper oil (C) = ₹90/kg

Price of dearer oil (D) = ₹150/kg

Selling price of mixture (SP) = ₹144/kg

Profit = 20%

Formula Used:

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

₹90 ₹150 ₹120 30 30

Cheaper Oil Dearer Oil Mean CP

Ratio = 30 : 30 = 1 : 1

Calculations:

⇒ Mean CP = 144 × [100 ÷ (100 + 20)]

⇒ Mean CP = 144 × (100 ÷ 120)

⇒ Mean CP = 144 × (5 ÷ 6) = ₹120/kg

⇒ By the rule of alligation:

⇒ Quantity of Cheaper : Quantity of Dearer = (150 − 120) : (120 − 90)

⇒ Quantity of Cheaper : Quantity of Dearer = 30 : 30

⇒ Ratio = 1 : 1

∴ The correct answer is 1 : 1.

Additional Information

Rule of Alligation

The ratio in which two ingredients are mixed to produce a mixture of a desired mean price is given by (Dearer Price − Mean Price) : (Mean Price − Cheaper Price).

Cost Price vs Selling Price in Mixtures

When applying alligation, all values must be of the same type. If the individual values are Cost Prices, the mean value must also be the Cost Price (CP), not the Selling Price (SP). CP = SP × 100 ÷ (100 + Profit%).

Weighted Average Method

The rule of alligation is a direct derivation of the weighted average formula: M = (Q₁ × C₁ + Q₂ × C₂) ÷ (Q₁ + Q₂), where Q represents quantity and C represents the respective cost.

Shortcut Trick

Total capacity = LCM(6, 8, 12) = 24 units.

Total efficiency of A, B and C = 4 + 3 + 2 = 9 units/hr.

Tank filled in 2 hours = 9 × 2 = 18 units.

Remaining capacity = 24 − 18 = 6 units.

Time taken by A and B to fill the rest = 6 ÷ (4 + 3) = 6/7 hours.

∴ The correct answer is 6/7 hours.

Alternate Method

Given:

Time taken by pipe A to fill the tank = 6 hours

Time taken by pipe B to fill the tank = 8 hours

Time taken by pipe C to fill the tank = 12 hours

Time all pipes operated together = 2 hours

Calculations:

⇒ Combined efficiency of A, B and C = 4 + 3 + 2 = 9 units/hr

⇒ Work done by A, B and C in 2 hours = 9 × 2 = 18 units

⇒ Remaining work = 24 − 18 = 6 units

⇒ Pipe C is closed, so the remaining work is done by pipes A and B.

⇒ Combined efficiency of pipes A and B = 4 + 3 = 7 units/hr

⇒ Additional time taken by A and B = Remaining work ÷ Combined efficiency

⇒ Additional time taken = 6/7 hours

∴ The correct answer is 6/7 hours.

Additional Information

Pipes and Cisterns Basics

If a pipe can fill a tank in x hours, the part filled in 1 hour is 1/x. This is analogous to the concept of Time and Work where efficiency is inversely proportional to time.

Inlet vs Outlet Pipes

An inlet pipe fills the tank and its work is considered positive. An outlet pipe empties the tank and its work is considered negative. Net work = 1/x − 1/y.

Alternate Working of Pipes

When pipes are opened in alternate hours, work is calculated in cycles (e.g., 2 hours if 2 pipes are working alternately) rather than directly summing their efficiencies per hour.

Shortcut Trick

Since the travel time remains constant (30 minutes) in both cases, Speed is directly proportional to Distance.

Because the distance increases by 20%, the required average speed must also increase by exactly 20%.

New Speed = 240 + (20% of 240)

New Speed = 240 + 48 = 288 km/h

∴ The correct answer is 288 km/h.

Alternate Method

Given:

Initial Average Speed (S₁) = 240 km/h

Initial Time (T₁) = 30 minutes = 0.5 hours

New Time (T₂) = 30 minutes = 0.5 hours

Initial Distance: 120 km Speed: 240 km/h | Time: 30 min

New Distance: 144 km (+20%) New Speed: 288 km/h | Time: 30 min

Formula Used:

Distance = Speed × Time

Speed = Distance ÷ Time

Calculations:

⇒ Initial Distance (D₁) = 240 × 0.5 = 120 km

⇒ The distance is increased by 20% due to track maintenance.

⇒ New Distance (D₂) = 120 + (20 ÷ 100) × 120

⇒ D₂ = 120 + 24 = 144 km

⇒ The train needs to arrive in the same exact time window.

⇒ New Time (T₂) = 0.5 hours

⇒ New Speed (S₂) = D₂ ÷ T₂

⇒ S₂ = 144 ÷ 0.5 = 288 km/h

∴ The correct answer is 288 km/h.

Additional Information

Direct and Inverse Proportionality

When Time is constant, Distance and Speed are directly proportional (S₁ ÷ S₂ = D₁ ÷ D₂). When Distance is constant, Time and Speed are inversely proportional.

Unit Conversions for Speed

To convert an object's speed from km/h to m/s, multiply it by 5 ÷ 18. To change m/s back to km/h, multiply by 18 ÷ 5.

Average Speed

Average Speed over an entire journey equals Total Distance ÷ Total Time. If two equal distances are traveled at speeds a and b, the Average Speed is 2ab ÷ (a + b).

Shortcut Trick

Relative speed in opposite direction = 360 ÷ 4 = 90 km/h

Relative speed in same direction = 360 ÷ 12 = 30 km/h

Speed of the faster car (X) = (Sum of relative speeds + Difference of relative speeds) ÷ 2

Speed of X = (90 + 30) ÷ 2 = 120 ÷ 2 = 60 km/h

∴ The correct answer is 60 km/h.

Alternate Method

Given:

Distance between X and Y = 360 km

Time to meet moving towards each other = 4 hours

Time to meet moving in the same direction = 12 hours

Formula Used:

Speed = Distance ÷ Time

Opposite Direction: Relative Speed = X + Y

Same Direction: Relative Speed = X − Y

Calculations:

⇒ X + Y = 360 ÷ 4

⇒ X + Y = 90     ... (Equation 1)

⇒ X − Y = 360 ÷ 12

⇒ X − Y = 30     ... (Equation 2)

⇒ Adding Equation 1 and Equation 2:

⇒ (X + Y) + (X − Y) = 90 + 30

⇒ 2X = 120

⇒ X = 60 km/h

∴ The correct answer is 60 km/h.

Additional Information

Relative Speed in Opposite Direction

When two bodies move towards each other, their relative speed is calculated as the sum of their individual speeds (S₁ + S₂).

Relative Speed in Same Direction

When two bodies move in the same direction, their relative speed is calculated as the difference of their individual speeds (S₁ − S₂).

Distance Constant Rule

If the distance to be covered remains constant, the speed of an object is inversely proportional to the time taken: S₁ ÷ S₂ = T₂ ÷ T₁.