Search Previous Year Questions

Shortcut Trick

In Buy X get Y free type questions, the Discount % is calculated on the total number of items received.

Total items received = 10

Items paid for = 6

Free items (Discount) = 10 − 6 = 4

Discount % = (Free items ÷ Total items) × 100 = (4 ÷ 10) × 100 = 40%

∴ The correct answer is 40%.

Alternate Method

Given: Total items handed over to the customer = 10; Price charged for items = 6.

Formula Used: Discount % = [(Marked Price − Selling Price) ÷ Marked Price] × 100

⇒ Let the Marked Price (MP) of 1 item be ₹1.

⇒ Total MP of 10 items = 10 × 1 = ₹10.

⇒ The customer pays for only 6 items, so Selling Price (SP) = 6 × 1 = ₹6.

⇒ Discount Amount = MP − SP = 10 − 6 = ₹4.

⇒ Discount % = (Discount ÷ Total MP) × 100

⇒ Discount % = (4 ÷ 10) × 100 = 40%.

∴ The correct answer is 40%.

Additional Information

Successive Discounts

When two discounts x% and y% are given, the single equivalent discount is (x + y − xy/100)%.

Relationship between CP and MP

The ratio of MP to CP is given by MP/CP = (100 + Profit%)/(100 − Discount%).

Markup Percentage

Markup % = [(Marked Price − Cost Price) ÷ Cost Price] × 100.

Shortcut Trick

When the distance traveled is the same for both parts of the journey, use the formula:

Average Speed = (2 × S₁ × S₂) / (S₁ + S₂)

Given: S₁ = 110 km/hr and S₂ = 140 km/hr

Average Speed = (2 × 110 × 140) / (110 + 140) = 30800 / 250 = 123.2 km/hr

∴ The correct answer is 123.2 km/hr.

Alternate Method

Given:

Speed from City A to City B (S₁) = 110 km/hr

Speed from City B to City A (S₂) = 140 km/hr

Formula Used:

Average Speed = Total Distance ÷ Total Time

Let the distance between City A and City B be D km.

⇒ Time taken to travel from A to B (t₁) = D / 110 hr

⇒ Time taken to travel from B to A (t₂) = D / 140 hr

⇒ Total Distance = D + D = 2D km

⇒ Total Time = D/110 + D/140 = (140D + 110D) / (110 × 140) = 250D / 15400

⇒ Average Speed = 2D ÷ (250D / 15400) = (2D × 15400) / 250D

⇒ Average Speed = 30800 / 250 = 123.2 km/hr

∴ The correct answer is 123.2 km/hr.

Additional Information

Average Speed for Different Distances

If distances d₁, d₂, d₃ are covered at speeds s₁, s₂, s₃, Average Speed = (d₁+d₂+d₃) / (d₁/s₁ + d₂/s₂ + d₃/s₃).

Average Speed for Three Equal Distances

When a journey is divided into three equal parts with speeds S₁, S₂, and S₃, Average Speed = (3 × S₁ × S₂ × S₃) / (S₁S₂ + S₂S₃ + S₃S₁).

Unit Conversion

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Shortcut Trick

If (1/a) of First = (1/b) of Second = (1/c) of Third, then the ratio of the parts is a : b : c.

Ratio of three parts = 2 : 4 : 8 = 1 : 2 : 4

Total units = 1 + 2 + 4 = 7 units

Value of 1 unit (First part) = 3920 ÷ 7 = 560

∴ The correct answer is ₹560.

Alternate Method

Given:

Total amount to be divided = ₹3,920

Condition: 1/2 of First Part = 1/4 of Second Part = 1/8 of Third Part

Formula Used:

If A/x = B/y = C/z, then the ratio A : B : C = x : y : z.

⇒ Let the three parts be A, B, and C.

⇒ (1/2) × A = (1/4) × B = (1/8) × C = k

⇒ A = 2k, B = 4k, C = 8k

⇒ A + B + C = 3920

⇒ 2k + 4k + 8k = 3920

⇒ 14k = 3920

⇒ k = 3920 ÷ 14 = 280

⇒ First part (A) = 2k = 2 × 280 = 560

∴ The correct answer is ₹560.

Additional Information

Ratio of Equality Rule

When portions of quantities are equal (e.g., A/x = B/y), the quantities are in direct proportion to their denominators (A : B = x : y).

Division into Ratios

To divide a total sum S into parts with ratio a : b : c, the value of the first part is [a ÷ (a + b + c)] × S.

Inverse Proportion

If a × A = b × B, then the ratio A : B is the inverse of the coefficients, i.e., A : B = 1/a : 1/b = b : a.

Shortcut Trick

When combined time (T_XY) and one individual time (T_Y) are known, the other time (T_X) is:

T_X = (T_Y × T_XY) ÷ (T_Y − T_XY)

T_X = (15 × 10) ÷ (15 − 10)

T_X = 150 ÷ 5 = 30 min

∴ The correct answer is 30 min.

Alternate Method

Given:

Time taken by Pipe X and Pipe Y together = 10 min

Time taken by Pipe Y alone = 15 min

Calculations:

⇒ Total capacity of the tank = LCM of 10 and 15 = 30 units

⇒ Efficiency of (X + Y) = 30 ÷ 10 = 3 units/min

⇒ Efficiency of Y = 30 ÷ 15 = 2 units/min

⇒ Efficiency of X = Efficiency of (X + Y) − Efficiency of Y

⇒ Efficiency of X = 3 − 2 = 1 unit/min

⇒ Time taken by Pipe X alone = Total Capacity ÷ Efficiency of X

⇒ Time = 30 ÷ 1 = 30 min

∴ The correct answer is 30 min.

Additional Information

Pipes and Cisterns Basics

If a pipe can fill a tank in 'x' hours, then the part filled in 1 hour = 1/x. This is known as the efficiency of the pipe.

Work done by Outlet Pipe

An outlet pipe (leak) performs negative work. If Pipe A fills in 'x' hours and Pipe B empties in 'y' hours, net work in 1 hour = (1/x − 1/y).

Combined Time Formula

If two pipes fill a tank in 'a' and 'b' hours respectively, then the time taken to fill the tank together = (a × b) ÷ (a + b).

Shortcut Trick

Efficiency: 5 Men = 7 Women.

Work remaining after 5 days for 5 men = 8 − 5 = 3 days of work.

Remaining work in terms of man-days = 5 men × 3 days = 15 man-days.

Convert man-days to woman-days: Since 5M = 7W, then 15M = 3 × 7W = 21 woman-days.

Number of women required to finish in 7 days = 21 woman-days ÷ 7 days = 3.

∴ The correct answer is 3 women.

Alternate Method

Given: 5 men or 7 women can do a job in 8 days. 5 men work for 5 days.

Formula Used: Total Work = Efficiency × Time | M₁D₁ = M₂D₂

⇒ Efficiency Relation: 5M = 7W

⇒ Total Work in woman-days = 7 women × 8 days = 56 units

⇒ Total Work in man-days = 5 men × 8 days = 40 units

⇒ Work done by 5 men in 5 days = 5 × 5 = 25 units (man-days)

⇒ Remaining Work = 40 − 25 = 15 units (man-days)

⇒ Convert units to woman-days: Since 5 units (M) = 7 units (W)

⇒ 15 man-days = (15 ÷ 5) × 7 = 3 × 7 = 21 woman-days

⇒ Number of women needed to finish 21 woman-days in 7 days = 21 ÷ 7

⇒ Number of women = 3

∴ The correct answer is 3 women.

Additional Information

Work Equivalency Formula

When M₁ persons do a work in D₁ days and M₂ persons do it in D₂ days, then M₁ × D₁ = M₂ × D₂. If efficiency varies, multiply by efficiency (E₁M₁D₁ = E₂M₂D₂).

Or/And Case in Time & Work

The word 'OR' (e.g., 5 men or 7 women) implies their total work capacities are equal (5M = 7W). 'AND' implies the sum of their individual capacities (5M + 7W).

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a task. If the ratio of efficiency is x:y, the ratio of time taken is y:x.

Shortcut Trick

Let initial Cost Price (CP₁) = 100 units → Initial SP₁ = 90 units (10% loss).

New CP₂ = 100 units − ₹100. New SP₂ = 120% of (100 units − ₹100) = 120 units − ₹120.

Difference in Selling Prices: (120 units − ₹120) − 90 units = ₹180.

30 units − ₹120 = ₹180 ⇒ 30 units = ₹300 ⇒ 1 unit = ₹10.

Actual CP = 100 units = 100 × 10 = ₹1000.

∴ The correct answer is ₹1000.

Alternate Method

Given: Initial Loss = 10%, reduction in CP = ₹100, new Profit = 20%, increase in SP = ₹180.

Formula Used: Selling Price (SP) = Cost Price (CP) × (100 ± P/L)%

Let the original Cost Price (CP) of the Table be ₹100x.

⇒ Initial Selling Price (SP₁) = 100x × 90/100 = 90x

⇒ New Cost Price (CP₂) = 100x − 100

⇒ New Selling Price (SP₂) = (100x − 100) × 120/100

⇒ SP₂ = 1.2 × (100x − 100) = 120x − 120

According to the question, New SP is ₹180 more than Initial SP:

⇒ (120x − 120) − 90x = 180

⇒ 30x − 120 = 180

⇒ 30x = 300

⇒ x = 10

⇒ Actual Cost Price = 100x = 100 × 10 = ₹1000

∴ The correct answer is ₹1000.

Additional Information

Profit and Loss Percentage

Profit % = (SP − CP)/CP × 100 and Loss % = (CP − SP)/CP × 100. Both are always calculated on the Cost Price unless stated otherwise.

Successive Changes in CP and SP

When CP and Profit/Loss % both change, the final SP is found by applying the new percentage on the modified CP base.

Marked Price (MP) and Discount

Discount is the reduction offered on the Marked Price. SP = MP − Discount = MP × (100 − D%)/100.

Shortcut Trick

Ratio of investments (Zakir : Yamini) = 1560 : 2210 = 156 : 221 = 12 : 17.

Since the time period is same, Profit Ratio = Investment Ratio = 12 : 17.

Zakir's share (12 units) = ₹720 ⇒ 1 unit = ₹60.

Total profit = (12 + 17) units = 29 units = 29 × 60 = ₹1740.

∴ The correct answer is ₹1740.

Alternate Method

Given: Zakir's investment = ₹1560, Yamini's investment = ₹2210, Zakir's profit share = ₹720.

Formula: Ratio of Profit = Ratio of (Investment × Time)

⇒ Ratio of Profit (Zakir : Yamini) = 1560 : 2210

⇒ Ratio = 156 ÷ 13 : 221 ÷ 13 = 12 : 17

⇒ Let Zakir's profit be 12x and Yamini's profit be 17x.

⇒ 12x = 720

⇒ x = 720 ÷ 12 = 60

⇒ Total Profit = 12x + 17x = 29x

⇒ Total Profit = 29 × 60 = 1740

∴ The correct answer is ₹1740.

Additional Information

Simple Partnership

When partners invest capital for the same time period, the profit is divided in the ratio of their investments: P₁ : P₂ = I₁ : I₂.

Compound Partnership

When partners invest for different time periods, the profit ratio is the product of investment and time: P₁ : P₂ = (I₁ × T₁) : (I₂ × T₂).

Working vs Sleeping Partner

A working partner manages the business and may receive a fixed percentage of profit as a salary before the remaining profit is distributed.

Shortcut Trick

Relative Speed = 78 − 66 = 12 km/hr (same direction).

Lead Distance of first train at 9 a.m. = 66 × 2 hours = 132 km.

Time to meet = Lead Distance ÷ Relative Speed = 132 ÷ 12 = 11 hours.

Meeting time = 9:00 a.m. + 11 hours = 8:00 p.m.

∴ The correct answer is 8 p.m.

Alternate Method

Given:

Speed of Train 1 (S₁) = 66 km/hr

Speed of Train 2 (S₂) = 78 km/hr

Start time of Train 1 = 7 a.m. | Start time of Train 2 = 9 a.m.

Formula Used:

Distance = Speed × Time

Time to catch up = (Initial distance between them) ÷ (Relative Speed)

Calculations:

⇒ Time duration for which Train 1 traveled alone = 9 a.m. − 7 a.m. = 2 hours.

⇒ Distance covered by Train 1 in 2 hours = 66 × 2 = 132 km.

⇒ Since both trains move from Station A to Station B, they are in the same direction.

⇒ Relative Speed = 78 − 66 = 12 km/hr.

⇒ Time taken by Train 2 to catch Train 1 = 132 ÷ 12 = 11 hours.

⇒ Meeting time = 9:00 a.m. + 11 hours.

⇒ 9 a.m. + 3 hours = 12:00 p.m. (noon), and remaining 8 hours = 8:00 p.m.

∴ The correct answer is 8 p.m.

Additional Information

Relative Speed Concepts

When two objects move in the same direction, Relative Speed = S₁ − S₂. When they move in opposite directions, Relative Speed = S₁ + S₂.

Time of Meeting

Meeting Time = (Initial Gap between objects) ÷ (Relative Speed). Always ensure the start time is synchronized before calculating the gap.

Average Speed

If a journey is divided into two equal distances with speeds x and y, Average Speed = 2xy ÷ (x + y).

Shortcut Trick

For two successive discounts of a% and b%, the effective discount percentage is given by the formula:

Effective Discount = (a + b − (a × b) / 100)%

Here, a = 4 and b = 5.

⇒ Total Discount = 4 + 5 − (4 × 5) / 100

⇒ Total Discount = 9 − 20 / 100 = 9 − 0.2 = 8.8%

∴ The correct answer is 8.8%.

Alternate Method

Given:

First discount rate = 4%

Second discount rate = 5%

Calculations:

⇒ Let the Marked Price (MP) of the article be 100 units.

⇒ Price after the 1^st discount = 100 − (4% of 100) = 100 − 4 = 96 units.

⇒ Price after the 2^nd discount = 96 − (5% of 96) = 96 − 4.8 = 91.2 units.

⇒ This final price is the Selling Price (SP).

⇒ Total Discount = Marked Price − Selling Price = 100 − 91.2 = 8.8 units.

⇒ Total Discount % = (Total Discount / MP) × 100 = (8.8 / 100) × 100 = 8.8%.

∴ The correct answer is 8.8%.

Additional Information

Three Successive Discounts

For three discounts a%, b%, and c%, the final multiplier is (1 − a/100) × (1 − b/100) × (1 − c/100). The single equivalent discount is [1 − Final Multiplier] × 100%.

Marked Price and Selling Price Relation

Selling Price (SP) = Marked Price (MP) × (100 − D₁)/100 × (100 − D₂)/100, where D₁ and D₂ are successive discount rates.

Order of Discounts

The effective discount remains the same regardless of the order in which the successive discounts are applied (Commutative Property).

Shortcut Trick

Initial TSA ratio part (where r=1, h=2) = 1 × (1 + 2) = 3.

New TSA ratio part (where r=1.2335, h=2) = 1.2335 × (1.2335 + 2) = 3.98852225.

Percentage Increase = [(3.98852225 − 3) ÷ 3] × 100 = 32.9507...%.

Rounding to two decimal places, we get 32.95%.

∴ The correct answer is 32.95%.

Alternate Method

Given:

Initial height (h) = 2 × radius (r)

Increase in radius = 23.35%

Formula Used:

Total Surface Area (TSA) of a cylinder = 2πr(r + h)

Calculations:

⇒ Initial TSA = 2 × π × r × (r + 2r) = 2 × π × r × 3r = 6πr²

⇒ New radius (r') = r + 23.35% of r = 1.2335r

⇒ New height (h') = 2r (remains unchanged)

⇒ New TSA = 2 × π × r' × (r' + h')

⇒ New TSA = 2 × π × 1.2335r × (1.2335r + 2r) = 2 × π × 1.2335r × 3.2335r

⇒ New TSA = 7.9770445 × πr²

⇒ Increase in TSA = 7.9770445πr² − 6πr² = 1.9770445πr²

⇒ Percentage Increase = (Increase ÷ Initial TSA) × 100

⇒ Percentage Increase = (1.9770445 ÷ 6) × 100 = 32.9507...%

∴ The correct answer is 32.95%.

Additional Information

Curved Surface Area (CSA)

The area of the side surface of the cylinder is given by CSA = 2πrh.

Volume of Cylinder

The total space occupied by the cylinder is given by V = πr²h.

Successive Percentage Change

When multiple dimensions change, the net percentage change in area or volume can be calculated using the product of the individual multiplying factors.