Search Previous Year Questions

Shortcut Trick

Rate = 16 2/3% = 1/6.

Let Principal = 6³ = 216 units.

Amount after 3 years = 7³ = 343 units ⇒ CI = 343 − 216 = 127 units.

Given CI = ₹1270 ⇒ 127 units = 1270 ⇒ 1 unit = ₹10.

SI for 3 years = 216 × (1/6) × 3 = 108 units = 108 × 10 = ₹1080.

∴ The correct answer is ₹1080.

Alternate Method

Given:

Rate (R) = 16 2/3% = 50/3% per annum

Time (T) = 3 years

Compound Interest (CI) = ₹1270

Formula Used:

Amount (A) = P × (1 + R/100)^T

CI = A − P

SI = (P × R × T) ÷ 100

Principal = ₹2160

Year 1 SI = ₹360

Year 2 SI = ₹360

Year 3 SI = ₹360

Total SI = ₹1080

Calculations:

⇒ Let the Principal be P.

⇒ Amount (A) = P × (1 + 1/6)³

⇒ A = P × (7/6)³ = P × (343/216)

⇒ CI = A − P = P × (343/216) − P = P × (127/216)

⇒ P × (127/216) = 1270

⇒ P = 1270 × (216/127) = 10 × 216 = 2160

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

⇒ SI = (2160 × 50/3 × 3) ÷ 100

⇒ SI = (2160 × 50) ÷ 100 = 1080

∴ The correct answer is ₹1080.

Additional Information

Difference between CI and SI for 2 years

The difference between compound interest and simple interest for 2 years on a certain sum is directly calculated by: D = P(R/100)².

Difference between CI and SI for 3 years

For a 3-year period, the difference formula extends to: D = P(R/100)² × (3 + R/100).

Multiplying Factor Method

When the rate of interest fraction is 1/x, the principal is usually assumed as x³ for 3 years to simplify compound interest calculations and avoid decimals.

Shortcut Trick Calculate the net profit or loss per unit directly using individual deviations from the final Selling Price: Loss on Darjeeling tea = (300 − 260) = ₹40/kg ⇒ Total = 15 × 40 = ₹600 (Loss) Profit on Assam tea = (260 − 240) = ₹20/kg ⇒ Total = 25 × 20 = ₹500 (Profit) Net overall effect = 500 (Profit) − 600 (Loss) = −100 (Net Loss) ∴ The correct answer is Loss of ₹100.

Alternate Method Given: Quantity of Darjeeling tea = 15 kg at ₹300/kg Quantity of Assam tea = 25 kg at ₹240/kg Selling price of the entire mixture = ₹260/kg

Formula Used: Total Cost Price (CP) = Sum of (Quantity × Rate per kg) Total Selling Price (SP) = Total Quantity × SP per kg Profit or Loss = Total SP − Total CP

Darjeeling Tea 15 kg × ₹300 = ₹4500

Assam Tea 25 kg × ₹240 = ₹6000

Total CP = ₹10500

Loss = ₹100

40 kg × ₹260 Total SP = ₹10400

Calculations: ⇒ Total CP of Darjeeling tea = 15 × 300 = ₹4500 ⇒ Total CP of Assam tea = 25 × 240 = ₹6000 ⇒ Total Cost Price (CP) of the mixture = 4500 + 6000 = ₹10500 ⇒ Total quantity of the mixture = 15 + 25 = 40 kg ⇒ Total Selling Price (SP) of the mixture = 40 × 260 = ₹10400 ⇒ Since Total CP > Total SP, the seller incurs a loss. ⇒ Loss = Total CP − Total SP = 10500 − 10400 = ₹100 ∴ The correct answer is Loss of ₹100.

Additional Information Alligation Method Used to find the ratio in which two ingredients are mixed to produce a mixture of a desired mean price. The ratio of quantities is inversely proportional to the differences in their prices from the mean price. Weighted Average To find the mean price directly without calculating overall totals, use the formula: Mean Price = (Q1 × P1 + Q2 × P2) ÷ (Q1 + Q2). Profit and Loss Percentage Profit % = (Profit ÷ Total CP) × 100. Loss % = (Loss ÷ Total CP) × 100. Both are always calculated on the total Cost Price unless specifically mentioned otherwise.

Shortcut Trick

To avoid decimal calculations, multiply the SP ratio 4 : 5 : 6 by 9 to get assumed SP values: 36, 45, and 54.

Total SP = 36 + 45 + 54 = 135 units.

CP of A = 36 ÷ 0.90 = 40, CP of B = 45 ÷ 1.50 = 30, and CP of C = 54 (no profit/loss).

Total CP = 40 + 30 + 54 = 124 units.

Profit = 135 − 124 = 11 units. ⇒ Profit % = (11 ÷ 124) × 100 ≈ 8.87% (approx. 8.88%).

∴ The correct answer is profit of 8.88%.

Alternate Method

Given:

Selling price (SP) ratio of A, B, and C = 4 : 5 : 6

Loss on item A = 10%

Profit on item B = 50%

No profit or loss on item C

Formula Used:

Cost Price (CP) = (SP × 100) ÷ (100 ± Profit/Loss %)

Profit % = [(Total SP − Total CP) ÷ Total CP] × 100

Total Cost Price (CP) = 124 unitsTotal Selling Price (SP) = 135 unitsProfit = 11Overall Profit % = (11 ÷ 124) × 100 ≈ 8.87%

Calculations:

⇒ Let the SP of items A, B, and C be 36x, 45x, and 54x respectively. (Multiplying the ratio by 9x to avoid fractional CP)

⇒ Total SP = 36x + 45x + 54x = 135x

⇒ CP of A = (36x × 100) ÷ 90 = 40x

⇒ CP of B = (45x × 100) ÷ 150 = 30x

⇒ CP of C = 54x (Since he broke even on item C, CP = SP)

⇒ Total CP = 40x + 30x + 54x = 124x

⇒ Overall Profit = Total SP − Total CP

⇒ Overall Profit = 135x − 124x = 11x

⇒ Profit % = (11x ÷ 124x) × 100

⇒ Profit % = 1100 ÷ 124 ≈ 8.87%

⇒ The closest calculated value matches option 2.

∴ The correct answer is profit of 8.88%.

Additional Information

Marked Price & Discount

Discount is always calculated on the Marked Price (MP). SP = MP − Discount.

Dishonest Dealer

When a dealer sells goods at CP but uses false weights, the profit percentage is calculated as: Profit % = [(True Weight − False Weight) ÷ False Weight] × 100.

Successive Discounts

The equivalent single discount for two successive discounts of a% and b% is given by: Equivalent Discount % = (a + b − ab ÷ 100)%.

Shortcut Trick

Remaining mixture volume = 75 − 15 = 60 liters.

Since the ratio is 3:2, Milk = 60 × (3 / 5) = 36 liters and Water = 60 × (2 / 5) = 24 liters.

After adding 10 liters of pure milk, the new volume of Milk = 36 + 10 = 46 liters.

The volume of Water remains the same at 24 liters.

New ratio of Milk to Water = 46 : 24 = 23 : 12.

∴ The correct answer is 23:12.

Alternate Method

Given:

Total initial volume of mixture = 75 liters

Initial ratio of Milk to Water = 3 : 2

Volume of mixture removed = 15 liters

Volume of pure milk added = 10 liters

60 L Mix Milk: 36 L Water: 24 L

10 L Pure Milk

=

70 L Final Milk: 46 L Water: 24 L

Mixture Replacement Visualization Final Ratio = 46 : 24 = 23 : 12

Calculations:

⇒ Quantity of mixture remaining after removal = 75 − 15 = 60 liters

⇒ In the remaining 60 liters, the ratio of Milk to Water remains 3 : 2

⇒ Quantity of milk in the remaining mixture = 60 × (3 / 5) = 36 liters

⇒ Quantity of water in the remaining mixture = 60 × (2 / 5) = 24 liters

⇒ Now, 10 liters of pure milk is added to the mixture

⇒ Total milk in the final mixture = 36 + 10 = 46 liters

⇒ Total water in the final mixture remains strictly unchanged = 24 liters

⇒ Final ratio of Milk to Water = 46 : 24

⇒ Simplifying the ratio by dividing by 2 = 23 : 12

∴ The correct answer is 23:12.

Additional Information

Ratio Stability Principle

When any portion of a uniform mixture is drawn out, the ratio of its constituents remains exactly the same in both the drawn-off part and the remaining mixture.

Single Component Addition

If pure milk is explicitly added to a milk-water mixture, the absolute quantity of water naturally remains fully unchanged from its previous state, altering only the total total volume and the milk fraction.

Successive Replacement Formula

When a volume 'v' of a mixture of total volume 'V' is repeatedly drawn out and completely replaced by a pure component 'n' times, the remaining quantity of the other untouched component is given by Initial Quantity × (1 − v/V)ⁿ.

Shortcut Trick

Applying successive percentage multipliers on the base cost C yields the final selling price:

Final SP = C × 1.5 (markup) × 0.8 (discount) × 1.4 (markup) × 0.9 (discount)

⇒ 18144 = C × 1.2 × 1.26

⇒ 18144 = 1.512C

⇒ C = 18144 ÷ 1.512 = 12000

∴ The correct answer is ₹12,000.

Alternate Method

Given:

Factory base cost = ₹ C

Wholesaler markup = 50%, Trade discount = 20%

Retailer markup = 40%, Customer discount = 10%

Final amount paid by customer = ₹18,144

Base Cost₹ CRetailer CP1.2 CFinal SP1.512 CMarkup 50%Discount 20%Markup 40%Discount 10%1.512 C = ₹18,144

Calculations:

⇒ Wholesaler Marked Price (MP) = C × (1 + 50 ÷ 100) = 1.5C

⇒ Retailer Cost Price (CP) = 1.5C × (1 − 20 ÷ 100) = 1.5C × 0.8 = 1.2C

⇒ Retailer Marked Price (MP) = 1.2C × (1 + 40 ÷ 100) = 1.2C × 1.4 = 1.68C

⇒ Final Selling Price (SP) = 1.68C × (1 − 10 ÷ 100) = 1.68C × 0.9 = 1.512C

Equating this to the final amount paid:

⇒ 1.512C = 18144

⇒ C = 18144 ÷ 1.512

⇒ C = 12000

∴ The correct answer is ₹12,000.

Additional Information

Successive Percentage Change

When a value undergoes multiple sequential percentage changes (like markup followed by discount), the final value is found by multiplying the initial base value by the combined multiplying factors (1 + R ÷ 100) or (1 − R ÷ 100).

Marked Price & Discount

Discount is always calculated on the Marked Price (also known as Printed Price or Label Price) and never on the Cost Price. Formula: Selling Price = Marked Price × (1 − Discount%).

Effective Discount Formula

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

Shortcut Trick

Total Work = LCM(16, 12, 4) = 48 units

Efficiencies: A = 48 ÷ 16 = 3, B = 48 ÷ 12 = 4, A + B + C = 48 ÷ 4 = 12

Efficiency of C = 12 − (3 + 4) = 5 units/day

Time taken by C alone = 48 ÷ 5 = 9 ³/₅ days

∴ The correct answer is 9 ³/₅ days.

Alternate Method

Given:

⇒ Time taken by A = 16 days

⇒ Time taken by B = 12 days

⇒ Time taken by A, B and C together = 4 days

Formula Used:

⇒ Total Work = Efficiency × Time

Calculations:

⇒ Total efficiency of A, B and C = 12 units/day

⇒ Efficiency of A = 3 units/day

⇒ Efficiency of B = 4 units/day

⇒ Efficiency of C = Total efficiency − (Efficiency of A + Efficiency of B)

⇒ Efficiency of C = 12 − (3 + 4) = 5 units/day

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

⇒ Time taken by C alone = 48 ÷ 5 = 9 ³/₅ days

∴ The correct answer is 9 ³/₅ days.

Additional Information

Work & Wages

Wages are distributed in direct proportion to the work done. If individuals work for the same number of days, their wages are divided exactly in the ratio of their daily efficiencies.

Pipes & Cisterns

This concept operates similarly to time and work, but inlet pipes have a positive efficiency (filling) while outlet pipes or leaks have a negative efficiency (emptying).

Alternate Days Work

When workers operate on alternate days, find the combined work done in a complete cycle (e.g., 2 days for 2 workers) and multiply to approach the total work limit before calculating the remaining days.

Shortcut Trick

Cost Price (CP) of mixture = 55 × (100 ÷ 110) = ₹50/kg.

Using the net deviation method from the average CP (₹50):

Net deviation = 2 × (40 − 50) + 3 × (50 − 50) + 5 × (x − 50) = 0

⇒ −20 + 0 + 5(x − 50) = 0

⇒ 5(x − 50) = 20 ⇒ x − 50 = 4 ⇒ x = 54

∴ The correct answer is ₹54.00.

Alternate Method

Given:

Prices of the first two varieties = ₹40/kg and ₹50/kg

Ratio of the three varieties = 2 : 3 : 5

Selling Price (SP) of the mixture = ₹55/kg at 10% profit

Mixture CP = ₹55 × (100 ÷ 110) = ₹50/kg2 parts₹40/kg3 parts₹50/kg5 parts₹x/kgTotal = 10 partsAvg CP = ₹50/kg

Formula Used:

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

Average Cost = (Total Cost) ÷ (Total parts)

Calculations:

⇒ CP of the mixture = (55 × 100) ÷ (100 + 10) = ₹50/kg

⇒ Let the price of the third variety be ₹x/kg.

⇒ Total parts of the mixture = 2 + 3 + 5 = 10 parts

⇒ Total CP for 10 parts = 10 × 50 = ₹500

⇒ Sum of individual costs = (2 × 40) + (3 × 50) + (5 × x)

⇒ 80 + 150 + 5x = 500

⇒ 230 + 5x = 500

⇒ 5x = 500 − 230 = 270

⇒ x = 270 ÷ 5 = 54

∴ The correct answer is ₹54.00.

Additional Information

Rule of Alligation

When two ingredients are mixed, the ratio of their quantities is inversely proportional to the difference in their prices from the average price. Quantity of cheaper / Quantity of dearer = (CP of dearer − Mean price) / (Mean price − CP of cheaper).

Weighted Average Method

For multiple mixtures, the weighted average cost is calculated as A_w = (w₁x₁ + w₂x₂ + ... + w_nx_n) ÷ (w₁ + w₂ + ... + w_n), where w represents the weight or ratio and x represents the value.

Cost Price from Selling Price

When profit percentage (P%) is given, Cost Price (CP) = SP × 100 ÷ (100 + P%). All mixture calculations must always be done on Cost Price, never on Selling Price directly.

Shortcut Trick

Let total work be LCM(8, 3) = 24 units.

Efficiency of A = 24 ÷ 8 = +3 units/day, Efficiency of B = 24 ÷ 3 = −8 units/day (negative as B breaks the wall).

Work done in first 4 days by A = 4 × 3 = +12 units.

Work done in next 2 days by A and B together = 2 × (3 − 8) = −10 units.

Total work completed = 12 − 10 = 2 units. Remaining work = 24 − 2 = 22 units.

Time for A to build the remaining work = 22 ÷ 3 = 7¹/₃ days.

∴ The correct answer is 7¹/₃ days.

Alternate Method

Given:

Time taken by A to build the wall = 8 days

Time taken by B to break the wall = 3 days

Calculations:

⇒ Work completed by A alone in the first 4 days = 4 × 3 = 12 units

⇒ Combined efficiency of A and B when working together = 3 − 8 = −5 units/day

⇒ Work done by A and B together in 2 days = 2 × (−5) = −10 units

⇒ Total work completed by the end of 6 days = 12 − 10 = 2 units

⇒ Remaining work to be completed = 24 − 2 = 22 units

⇒ Time required by A alone to build the remaining wall = Remaining work ÷ Efficiency of A

⇒ Time required = 22 ÷ 3 = 7¹/₃ days

∴ The correct answer is 7¹/₃ days.

Additional Information

Pipes & Cisterns Analogy (Negative Work)

When a person breaks a wall or a pipe empties a tank, their work is considered negative. Combined work = Positive Efficiency − Negative Efficiency.

Alternate Day Work

When people work on alternate days, calculate the work done in a complete cycle of days, then find the remaining work and assign it sequentially to the person whose turn it is next.

Work & Wages

Wages are strictly distributed in the ratio of the total work done by each individual. If the time worked by all is the same, the wages are simply distributed in the ratio of their efficiencies.

Shortcut Trick

When distance is constant, speed is inversely proportional to time.

Time ratio = 5 : 1

Speed ratio = 1 : 5

Given 1 unit of speed = 240 km/h

Required speed = 5 units = 5 × 240 = 1200 km/h.

∴ The correct answer is 1200 km/h.

Alternate Method

Case 1: 5 Hours

Distance = D Speed = 240 km/h | Time = 5 h

Case 2: 1 Hour

Distance = D Speed = ? | Time = 1 h

Distance is constant in both cases

Given:

Initial speed (S₁) = 240 km/h

Initial time (T₁) = 5 hours

Required time (T₂) = 1 hour

Formula Used:

Distance = Speed × Time

Speed = Distance ÷ Time

Calculations:

⇒ Total distance covered = S₁ × T₁

⇒ Distance = 240 × 5 = 1200 km

⇒ To cover the same distance in 1 hour, required speed (S₂) = Distance ÷ T₂

⇒ S₂ = 1200 ÷ 1 = 1200 km/h

∴ The correct answer is 1200 km/h.

Additional Information

Unit Conversion

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

Average Speed

Average Speed = Total Distance ÷ Total Time. If two equal distances are covered at speeds a and b respectively, then the average speed is given by (2 × a × b) ÷ (a + b).

Relative Speed

When two objects move in the same direction, their relative speed is S₁ − S₂. When they move in opposite directions towards each other, their relative speed is S₁ + S₂.

Shortcut Trick

Distance = [ (S₁ × S₂) / |S₁ − S₂| ] × Δt

Here, S₁ = 50 km/h, S₂ = 30 km/h.

Time difference Δt = 1 hour early + 1 hour late = 2 hours.

⇒ Distance = [ (50 × 30) / (50 − 30) ] × 2

⇒ Distance = [ 1500 / 20 ] × 2 = 150 km.

∴ The correct answer is 150 km.

Alternate Method

City A City B

Distance = d km

50 km/h → Arrives 1 hr early

30 km/h → Arrives 1 hr late

Given:

Speed 1 (S₁) = 50 km/h, Time 1 (t₁) = t − 1 hours

Speed 2 (S₂) = 30 km/h, Time 2 (t₂) = t + 1 hours

Formula Used:

Distance = Speed × Time

Calculations:

Let the usual time taken to cover distance d be t hours.

⇒ d = 50 × (t − 1) --- (Equation 1)

⇒ d = 30 × (t + 1) --- (Equation 2)

Equating both the distances since the actual distance is constant:

⇒ 50 × (t − 1) = 30 × (t + 1)

⇒ 5t − 5 = 3t + 3

⇒ 2t = 8

⇒ t = 4 hours

Substituting the value of t in Equation 1:

⇒ d = 50 × (4 − 1)

⇒ d = 50 × 3 = 150 km

∴ The correct answer is 150 km.

Additional Information

Average Speed

When a person travels equal distances at two different speeds S₁ and S₂, the average speed for the entire journey is given by (2 × S₁ × S₂) / (S₁ + S₂).

Relative Speed

If two bodies are moving in the same direction, relative speed = |S₁ − S₂|. If they are moving in opposite directions, relative speed = S₁ + S₂.

Time Difference Formula

When travelling the same distance with varying speeds resulting in early or late arrivals, the direct distance is always (S₁ × S₂) / |S₁ − S₂| × Total Time Difference.