Search Previous Year Questions

Shortcut Trick

Using the Net Deviation Method:

Ratio of Junior to Senior employees = 30 : 45 = 2 : 3

Junior group deviation from overall average = 1.2 − 1.8 = −0.6 lakh

Total deviation of Junior parts = 2 × (−0.6) = −1.2 lakh

To balance the mean, total deviation of 3 Senior parts must be +1.2 lakh ⇒ per part = +0.4 lakh

Average sales of Senior employees = Overall average + Senior deviation = 1.8 + 0.4 = 2.2 lakh

∴ The correct answer is ₹2.2 lakh.

Alternate Method

Given:

Total employees = 75

Overall average monthly sales = ₹1.8 lakh

Junior employees = 30

Average sales of junior employees = ₹1.2 lakh

Formula Used:

Total Value = Average × Number of Entities

1.2 Junior (₹ lakh)

x Senior (₹ lakh)

1.8 Overall Avg

x − 1.8 Emp: 30 (Ratio 2)

0.6 Emp: 45 (Ratio 3)

Calculations:

⇒ Number of senior employees = 75 − 30 = 45

⇒ Total sales of all employees = 75 × 1.8 = 135 lakh

⇒ Total sales of junior employees = 30 × 1.2 = 36 lakh

⇒ Total sales of senior employees = 135 − 36 = 99 lakh

⇒ Average sales of senior employees = 99 ÷ 45

⇒ Average sales of senior employees = 2.2 lakh

∴ The correct answer is ₹2.2 lakh.

Additional Information

Weighted Average

When combining two different groups, the overall average is calculated using the formula: A = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂), where 'n' represents the number of elements and 'x' represents their respective averages.

Method of Deviation

This method assumes a central mean value and calculates the difference (deviation) of all group components from it. For any perfectly balanced average, the net sum of all weighted deviations must always be zero.

Alligation Cross Rule

A visual shortcut applied when two differently priced or valued sets are mixed to get a final average. The ratio of the quantities mixed is always given by (Higher Average − Target Average) : (Target Average − Lower Average).

Shortcut Trick

Total fuel consumed = Total distance ÷ Mileage = 3045 ÷ 15 = 203 litres

Increase in price per litre = 5% of ₹82 = ₹4.1

Total increase in expenditure = 203 × 4.1 = ₹832.3 ≈ ₹832

∴ The correct answer is ₹832.

Alternate Method

Given:

Initial price of petrol = ₹82 per litre

Price hike = 5%

Monthly distance travelled = 3045 km

Car mileage = 15 km/l

Petrol Consumed 3045 km ÷ 15 km/l = 203 litres

Price Increase 5% of ₹82 = ₹4.1 / litre

Total Increase 203 × 4.1 = ₹832.3 ≈ ₹832

Calculations:

⇒ Total petrol consumed per month = Total Distance ÷ Mileage

⇒ Total petrol consumed = 3045 ÷ 15 = 203 litres

⇒ Increase in petrol price = 5% of 82 = (5 ÷ 100) × 82 = ₹4.1 per litre

⇒ Total increase in monthly expenditure = Total petrol consumed × Increase per litre

⇒ Total increase = 203 × 4.1 = ₹832.3

⇒ Rounding to the nearest rupee, the increase is ₹832.

∴ The correct answer is ₹832.

Additional Information

Percentage Change

The formula to calculate the absolute change is: Change = Original Value × (Percentage Change ÷ 100).

Fuel and Distance Relationship

Total fuel required = Total Distance ÷ Fuel Efficiency (Mileage). Total cost is evaluated by multiplying the fuel required by the cost per unit of fuel.

Rounding Values

If the decimal part is less than 0.5, round down to the nearest integer. If it is 0.5 or greater, round up to the next integer.

Shortcut Trick

Let desktop price = 1, laptop price = 5, and the number of desktops ordered = x.

Original bill = 20 × 5 + x × 1 = 100 + x

New interchanged bill = x × 5 + 20 × 1 = 5x + 20

Given the new bill is 80% more, so (5x + 20) = 1.8 × (100 + x)

⇒ 5x + 20 = 180 + 1.8x ⇒ 3.2x = 160 ⇒ x = 50

∴ The correct answer is 2:5.

Alternate Method

Original OrderLaptops = 20Desktops = xBill = 100 + x+80%New OrderLaptops = xDesktops = 20Bill = 5x + 20Required Ratio = 20 : x

Given:

⇒ Number of high-end laptops originally ordered = 20

⇒ Let the number of standard desktops originally ordered = x

⇒ Price of a high-end laptop = 5 × Price of a standard desktop

Formula Used:

⇒ Total Bill = (Quantity of Laptops × Price of Laptop) + (Quantity of Desktops × Price of Desktop)

⇒ New Bill = Original Bill × (100% + Increase%)

Calculations:

⇒ Let the price of a standard desktop be 1 unit. Then the price of a high-end laptop is 5 units.

⇒ Original Bill = (20 × 5) + (x × 1) = 100 + x

⇒ Due to the mistake, the new quantity of laptops = x, and desktops = 20.

⇒ New Bill = (x × 5) + (20 × 1) = 5x + 20

⇒ It is given that the new bill is 80% higher than the original bill. This means the new bill is 180% of the original bill.

⇒ 5x + 20 = (100 + x) × (180 ÷ 100)

⇒ 5x + 20 = (100 + x) × 1.8

⇒ 5x + 20 = 180 + 1.8x

⇒ 5x − 1.8x = 180 − 20

⇒ 3.2x = 160

⇒ x = 160 ÷ 3.2

⇒ x = 50

⇒ Number of standard desktops originally ordered = 50

⇒ Required Ratio = (Number of High-End Laptops) : (Number of Standard Desktops) = 20 : 50 = 2 : 5

∴ The correct answer is 2:5.

Additional Information

Interchange of Quantities Rule

When the quantities of two items are interchanged, the change in total value is equal to the product of the difference in their unit prices and the difference in their original quantities: Change in Value = (P₁ − P₂) × (Q₂ − Q₁).

Percentage Increase Formula

The percentage increase between two values is calculated as: % Increase = [(New Value − Original Value) ÷ Original Value] × 100.

Base Value in Percentages

Whenever a value increases by R%, the new value becomes Original Value × (1 + R/100). Identifying the correct base value (the original order bill) is crucial to avoid calculation errors.

Shortcut Trick

Using the multiplying factor method for Compound Interest:

Rate for 3 full years = 12% (Multiplier = 1.12)

Rate for next 4 months = 12% × (4/12) = 4% (Multiplier = 1.04)

Amount = P × (1.12)³ × (1.04)

Amount = 8000 × 1.404928 × 1.04 ≈ 11689

Compound Interest = 11689 − 8000 = 3689

∴ The correct answer is ₹3,689.

Alternate Method

Given:

Principal (P) = ₹8,000

Rate (R) = 12% per annum

Time (T) = 3 years 4 months = 3 + 4/12 years = 10/3 years

Formula Used:

A = P × (1 + R/100)ⁿ × (1 + r/100)

Where n is the full years and r is the rate for the fractional year.

Start ₹8000

+12% (1Yr)

Year 1 ₹8960

+12% (1Yr)

Year 2 ₹10035.2

+12% (1Yr)

Year 3 ₹11239.4

+4% (4Mo)

3 Yr 4 Mo ₹11689

Total CI = ₹11689 − ₹8000 = ₹3689

Calculations:

⇒ Rate for the fractional 4 months (1/3 year) = 12% × 1/3 = 4%

⇒ Amount (A) = 8000 × (1 + 12/100)³ × (1 + 4/100)

⇒ A = 8000 × (112/100)³ × (104/100)

⇒ A = 8000 × 1.12 × 1.12 × 1.12 × 1.04

⇒ A = 8000 × 1.404928 × 1.04

⇒ A = 11239.424 × 1.04 = 11689.00

⇒ Compound Interest (CI) = A − P

⇒ CI = 11689 − 8000 = 3689

∴ The correct answer is ₹3,689.

Additional Information

Fractional Year Compounding

When time is in fractions (e.g., n years and m months), apply the full compounding rate for complete years and a proportionally reduced rate for the remaining fractional year.

Effective Compound Interest Rate

For two consecutive periods at rates x% and y%, the overall effective compound interest percentage can be calculated using the successive percentage formula: x + y + (x × y)/100.

Difference Between CI and SI

For 2 years, the difference between compound and simple interest is CI − SI = P(R/100)². For 3 years, the difference is CI − SI = P(R/100)²(3 + R/100).

Shortcut Trick

CP per candy: Red = 15 ÷ 3 = ₹5, Blue = 18 ÷ 4 = ₹4.5, Green = 22 ÷ 5 = ₹4.4

Total CP for 4 mixed candies (in ratio 1:1:2) = 5 + 4.5 + (2 × 4.4) = ₹18.3

SP for 4 mixed candies = 4 × (10 ÷ 2) = ₹20

Profit = 20 − 18.3 = ₹1.7

Profit % = (1.7 ÷ 18.3) × 100 ≈ 9.29%

∴ The correct answer is Profit of 9.29%.

Alternate Method

Candy Mixture (1 : 1 : 2 Ratio)

1 Red @ ₹5

1 Blue @ ₹4.5

2 Green @ ₹4.4

Total CP = ₹18.3

• Profit ₹1.7

Total SP = ₹20

Profit % = (1.7 ÷ 18.3) × 100 ≈ 9.29%

Given:

Red candies bought at 3 for ₹15

Blue candies bought at 4 for ₹18

Green candies bought at 5 for ₹22

Mixture ratio = 1 : 1 : 2

Selling price = 2 for ₹10

Calculations:

⇒ Let the number of Red, Blue, and Green candies mixed be 1, 1, and 2 respectively.

⇒ Total candies mixed = 1 + 1 + 2 = 4 candies

⇒ Cost Price (CP) of 1 Red candy = 15 ÷ 3 = ₹5

⇒ Cost Price (CP) of 1 Blue candy = 18 ÷ 4 = ₹4.5

⇒ Cost Price (CP) of 1 Green candy = 22 ÷ 5 = ₹4.4

⇒ Total Cost Price of 4 mixed candies = (1 × 5) + (1 × 4.5) + (2 × 4.4)

⇒ Total CP = 5 + 4.5 + 8.8 = ₹18.3

⇒ Selling Price (SP) of 1 candy = 10 ÷ 2 = ₹5

⇒ Total SP of 4 mixed candies = 4 × 5 = ₹20

⇒ Since SP is greater than CP, there is a profit.

⇒ Profit = SP − CP = 20 − 18.3 = ₹1.7

⇒ Profit % = (Profit ÷ Total CP) × 100

⇒ Profit % = (1.7 ÷ 18.3) × 100

⇒ Profit % = 170 ÷ 18.3 ≈ 9.289% ≈ 9.29%

∴ The correct answer is Profit of 9.29%.

Additional Information

Profit & Loss Formulas

Profit occurs when Selling Price (SP) > Cost Price (CP). Profit = SP − CP. Profit Percentage is calculated on the total Cost Price: (Profit ÷ CP) × 100.

Mixture Average Cost

When different quantities of items are mixed, the Average Cost Price is simply the total cost of all components combined, divided by the total physical quantity.

Ratio Multipliers Trick

Instead of finding a large common multiple (LCM) to equate quantities, directly taking sample quantities equal to the given ratio (such as 1, 1, and 2 units) makes percentage-based calculations perfectly proportional and much faster.

Shortcut Trick

Initial Ratio of Boys : Girls = 7 : 5 and Final Ratio = 7 : 6

Since the number of boys remains the same, the ratio part for boys (7) is unchanged.

Increase in girls' ratio = 6 − 5 = 1 unit.

This 1 unit increase corresponds to the 20 girls who joined.

Total initial students = 7 + 5 = 12 units = 12 × 20 = 240.

∴ The correct answer is 240 students.

Alternate Method

Given:

Initial ratio of Boys : Girls = 7 : 5

Number of girls added = 20

Final ratio of Boys : Girls = 7 : 6

Formula Used:

If a ratio is given as a:b, the actual values can be taken as ax and bx.

Initial (7:5) Final (7:6)

Boys (7 units)

Girls (5 units)

Boys (7 units)

Girls (5 units)

+1

• 20 Girls

Calculations:

⇒ Let the initial number of boys be 7x and girls be 5x.

⇒ Total initial students = 7x + 5x = 12x.

⇒ After 20 girls join, the new number of girls = 5x + 20.

⇒ According to the given condition, 7x ÷ (5x + 20) = 7 ÷ 6

⇒ Cross-multiplying: 6 × 7x = 7 × (5x + 20)

⇒ 42x = 35x + 140

⇒ 42x − 35x = 140

⇒ 7x = 140

⇒ x = 20

⇒ Total initial students = 12x = 12 × 20 = 240

∴ The correct answer is 240 students.

Additional Information

Ratio Balancing Method

When one quantity changes while the other remains constant, ensure the ratio parts of the constant quantity are made equal before directly equating the change in units.

Compound Ratio

The compound ratio of two independent ratios a:b and c:d is evaluated as (a × c):(b × d).

Basic Proportion Property

If four quantities are in proportion such that a:b = c:d, then their cross products are equal (a × d = b × c).

Shortcut Trick

Total work = LCM(30, 45, 90) = 90 units.

Efficiencies: A = 3, B = 2, C = 1.

Work done in a 5-day cycle = (3 × 5) + 2 + 1 = 18 units.

Number of cycles = Total work ÷ Work per cycle = 90 ÷ 18 = 5 cycles.

Total days = 5 cycles × 5 days/cycle = 25 days.

∴ The correct answer is 25 days.

Alternate Method

Given:

Time taken by A = 30 days

Time taken by B = 45 days

Time taken by C = 90 days

Calculations:

⇒ In 1 cycle of 5 days, A works for all 5 days, while B and C work only on the 5^th day.

⇒ Work done by A in 5 days = 5 × 3 = 15 units

⇒ Work done by B and C on the 5^th day = 2 + 1 = 3 units

⇒ Total work done in 5 days = 15 + 3 = 18 units

⇒ Number of 5-day cycles required to complete 90 units = 90 ÷ 18 = 5 cycles

⇒ Total time taken = 5 cycles × 5 days/cycle = 25 days

∴ The correct answer is 25 days.

Additional Information

Total Work Concept

Total work is considered as the Least Common Multiple (LCM) of the days taken by individuals. Efficiency is the work done per unit time (Total Work ÷ Days).

Alternate or Cyclic Work

Find the work done in one complete cycle. Divide the total work by the work per cycle to find the number of cycles, then multiply by days per cycle to find the total time required.

Pipes and Cisterns

Works exactly on the same principle as Time and Work, but pipes that empty the tank are assigned a negative efficiency value.

Shortcut Trick

Apply the alligation method on the water concentration of the mixtures.

Initial mixture water = 40%, Added pure acid water = 0%, Final mixture water = 30%.

Ratio of remaining mixture to added pure acid = (30 − 0) : (40 − 30) = 30 : 10 = 3 : 1.

This means 1 part out of a total of (3 + 1) = 4 parts was replaced.

Replaced quantity = 1 / 4 of total volume = (1 / 4) × 100 = 25 litres.

∴ The correct answer is 25 litres.

Alternate Method

Given:

Initial volume of the solution = 100 litres

Initial ratio of acid to water = 3 : 2

Final ratio of acid to water = 7 : 3

Calculations:

⇒ Quantity of water in the initial solution = (2 / 5) × 100 = 40 litres

⇒ Let x litres of the solution be removed and replaced with pure acid.

⇒ Quantity of water removed = (2 / 5) × x = 0.4x litres

⇒ Remaining water in the solution = 40 − 0.4x litres

⇒ Since only pure acid is added back, the overall quantity of water in the final mixture does not increase.

⇒ Final volume of the solution is maintained at 100 litres.

⇒ Quantity of water in the final solution = (3 / 10) × 100 = 30 litres

⇒ Equating the remaining water to final water:

⇒ 40 − 0.4x = 30

⇒ 0.4x = 10

⇒ x = 10 ÷ 0.4 = 25 litres

∴ The correct answer is 25 litres.

Additional Information

Alligation Rule

It is a quick method used to find the ratio in which two ingredients at their given prices or concentrations must be mixed to produce a mixture of a desired average concentration.

Tracking the Untouched Component

When solving replacement questions, always frame your equation using the component that is NOT being added from the outside. In this case, since pure acid is added back, it is much faster to track the changing quantity of water.

Replacement Formula

If a container initially holds V units of pure liquid and x units are taken out and replaced by water n times sequentially, the final quantity of pure liquid remaining is given by V × (1 − x/V)ⁿ.

Shortcut Trick

Time and efficiency are inversely proportional to each other.

Ratio of Time taken by P and Q = 2 : 5

Here, 2 parts correspond to 12 days.

So, 1 part = 12 ÷ 2 = 6 days.

Time taken by Q (5 parts) = 5 × 6 = 30 days.

∴ The correct answer is 30 days.

Alternate Method

Given:

Ratio of the efficiencies of P and Q = 5 : 2

Time taken by P to complete the project = 12 days

Formula Used:

Total Work = Efficiency × Time

Time = Total Work ÷ Efficiency

Calculations:

⇒ Let the efficiency of P be 5 units/day and Q be 2 units/day.

⇒ Total Work = Efficiency of P × Time taken by P

⇒ Total Work = 5 × 12 = 60 units

⇒ Time taken by Q = Total Work ÷ Efficiency of Q

⇒ Time taken by Q = 60 ÷ 2 = 30 days

∴ The correct answer is 30 days.

Additional Information

Time and Efficiency Relationship

Efficiency is inversely proportional to the time taken to complete a work. If the ratio of efficiencies of A and B is x : y, then the ratio of time taken by them is y : x.

Combined Work Rule

If A can do a piece of work in x days and B can do it in y days, their combined efficiency is (1/x + 1/y). Together they complete the work in (x × y) ÷ (x + y) days.

Work and Wages

Wages are distributed in proportion to the total work done by each person. If workers work for the exact same number of days, wages are distributed in the direct ratio of their efficiencies.

Shortcut Trick

Relative speed of Car with respect to SUV = 15 − 10 = 5 m/s.

Time required to close the initial gap of 600 m = 600 ÷ 5 = 120 seconds.

Distance traveled by car in this time = Speed × Time = 15 × 120 = 1800 m.

∴ The correct answer is 1800 m.

Alternate Method

Point P Car (15 m/s)

Point Q SUV (10 m/s)

Meeting

Gap = 600 m

SUV dist = 1200 m

Car distance = 1800 m

Given:

⇒ Speed of car (S_C) = 15 m/s

⇒ Speed of SUV (S_S) = 10 m/s

⇒ Initial distance between them in direction of motion (d) = 600 m

⇒ Track circumference = 1.5 km = 1500 m (Extra information)

Formula Used:

⇒ Time taken to overtake = Distance ÷ Relative Speed

⇒ Distance = Speed × Time

Calculation:

⇒ Since both vehicles are moving in the same direction, Relative Speed = S_C − S_S

⇒ Relative Speed = 15 − 10 = 5 m/s

⇒ Time taken to overtake = 600 ÷ 5 = 120 seconds

⇒ Distance traveled by car = S_C × Time taken

⇒ Distance = 15 × 120 = 1800 m

∴ The correct answer is 1800 m.

Additional Information

Relative Speed (Same Direction)

When two bodies move in the same direction, their relative speed is the difference of their individual speeds: S_R = S₁ − S₂. The distance required to overtake is determined by the initial gap between them.

Relative Speed (Opposite Direction)

When two bodies move in opposite directions (e.g., towards each other), their relative speed is the sum of their individual speeds: S_R = S₁ + S₂.

Circular Track Properties

The circumference of the track is only needed when finding the time taken for the vehicles to meet again at the starting point, or for a subsequent overtake cycle (which requires covering an extra 1 full lap).