Search Previous Year Questions

Shortcut Trick

Net Discount % = a + b − (ab/100) = 10 + 20 − (10 × 20)/100 = 28%

Selling Price = (100% − 28%) of Marked Price

SP = 72% of ₹3,000 = 0.72 × 3,000 = ₹2,160

∴ The correct answer is ₹2,160.

Alternate Method

Given: Marked Price (MP) = ₹3,000; First Discount = 10%; Second Discount = 20%.

Formula: Final SP = MP × (100 − d₁)/100 × (100 − d₂)/100

⇒ Price after 10% discount = 3,000 − (10% of 3,000) = 3,000 − 300 = ₹2,700

⇒ Price after 20% discount = 2,700 − (20% of 2,700) = 2,700 − 540 = ₹2,160

⇒ The final Selling Price is ₹2,160, which matches Option (4).

∴ The correct answer is ₹2,160.

Additional Information

Successive Discount Rule

When two successive discounts d₁% and d₂% are given, the equivalent single discount is (d₁ + d₂ − d₁d₂/100)%.

Selling Price vs Marked Price

Selling Price (SP) = Marked Price (MP) − Discount. If multiple discounts are applied, they are calculated on the diminishing balance.

Profit and Loss Relation

If CP is the cost price, then SP = CP + Profit or SP = CP − Loss. The markup is always calculated on the CP.

Shortcut Trick

Effective Simple Interest percentage for 2 years = Rate × Time = 16% × 2 = 32%.

According to the question, 32% of Principal = ₹1,440.

⇒ 1% of Principal = 1,440 ÷ 32 = 45.

⇒ 100% of Principal (Sum) = 45 × 100 = 4,500.

∴ The correct answer is ₹4,500.

Alternate Method

Given: Simple Interest (SI) = ₹1,440, Rate (R) = 16% per year, Time (T) = 2 years.

Formula Used: SI = (P × R × T) ÷ 100, where P is the Principal amount.

⇒ 1,440 = (P × 16 × 2) ÷ 100

⇒ 1,440 = 32P ÷ 100

⇒ 1,440 × 100 = 32P

⇒ 1,44,000 = 32P

⇒ P = 1,44,000 ÷ 32

⇒ P = 4,500

∴ The correct answer is ₹4,500.

Additional Information

Simple Interest vs Compound Interest

For the first year, Simple Interest and Compound Interest are equal. From the second year onwards, CI includes interest on the previous year's interest.

Amount Formula

Total Amount (A) = Principal (P) + Interest (I). For Simple Interest, A = P(1 + RT/100).

Rate of Interest Calculation

The annual Rate of Interest can be found using the formula R = (SI × 100) ÷ (P × T).

Shortcut Trick

Assume the Mean (40) as the balance point where the sum of deviations is zero.

Net deviation: ∑f(x − 40) = 2(−30) + 3(−20) + 1(−10) + 4(0) + f(10) + 4(20) = 0

−60 − 60 − 10 + 0 + 10f + 80 = 0 ⇒ 10f − 50 = 0

10f = 50 ⇒ f = 5

∴ The correct answer is 5.

Alternate Method

Given: Mean (&bar;x) = 40. Data points (x) and frequencies (f) are mapped below.

Formula: Mean (&bar;x) = ∑f_ix_i ÷ ∑f_i

⇒ ∑f_ix_i = (10 × 2) + (20 × 3) + (30 × 1) + (40 × 4) + (50 × f) + (60 × 4)

⇒ ∑f_ix_i = 20 + 60 + 30 + 160 + 50f + 240 = 510 + 50f

⇒ ∑f_i = 2 + 3 + 1 + 4 + f + 4 = 14 + f

⇒ 40 = (510 + 50f) ÷ (14 + f)

⇒ 40(14 + f) = 510 + 50f

⇒ 560 + 40f = 510 + 50f

⇒ 560 − 510 = 50f − 40f

⇒ 50 = 10f ⇒ f = 5

∴ The correct answer is 5.

Additional Information

Arithmetic Mean of Discrete Series

Calculated as the sum of products of observations and their frequencies divided by the total frequency: &bar;x = ∑fx / ∑f.

Properties of Mean

The algebraic sum of deviations of a set of observations from their arithmetic mean is always zero, i.e., ∑f_i(x_i − &bar;x) = 0.

Assumed Mean Method

A shortcut formula where Mean = A + (∑fd / ∑f), where A is the assumed mean and d = x − A is the deviation.

Shortcut Trick

Net difference in interest rate per annum = 6.5% − 3.5% = 3%

Total interest difference for 2 years = 3% × 2 = 6%

Positive difference in Simple Interest = 6% of ₹3,40,000

Difference = 3,40,000 × (6 / 100) = 3,400 × 6 = ₹20,400

∴ The correct answer is ₹20,400.

Alternate Method

Given:

Principal (P) = ₹3,40,000

Rate for Bank A (R₁) = 3.5% p.a.

Rate for Bank B (R₂) = 6.5% p.a.

Time (T) = 2 years

Formula Used:

Simple Interest (SI) = (P × R × T) / 100

Calculations:

⇒ Interest paid to Bank A (SI₁) = (3,40,000 × 3.5 × 2) / 100

⇒ SI₁ = 3,400 × 7 = ₹23,800

⇒ Interest paid to Bank B (SI₂) = (3,40,000 × 6.5 × 2) / 100

⇒ SI₂ = 3,400 × 13 = ₹44,200

⇒ Positive difference = SI₂ − SI₁

⇒ Difference = 44,200 − 23,800 = ₹20,400

∴ The correct answer is ₹20,400.

Additional Information

Simple Interest Formula

SI = (P × R × T) / 100, where P is Principal, R is Rate, and T is Time.

Rate Difference Rule

When P and T are constant, the difference in interests is P × T × (R₂ − R₁) / 100.

Amount Formula

Total Amount (A) = Principal (P) + Simple Interest (SI).

Shortcut Trick

LCM of (12, 15, 20) = 60 units (Total Capacity).

Efficiencies: A = 5, B = 4, C = −3 (emptying).

Work done by Pipe A and Pipe B in 2 hr = (5 + 4) × 2 = 18 units.

Remaining Work = 60 − 18 = 42 units.

Time for remaining part = 42 ÷ (5 + 4 − 3) = 42 ÷ 6 = 7 hr.

Total time = 2 + 7 = 9 hr.

∴ The correct answer is 9 hr.

Alternate Method

Given: Time for Pipe A = 12 hr, Pipe B = 15 hr, Pipe C = 20 hr.

Formula: Total Work = LCM of individual times; Efficiency = Total Work ÷ Time.

⇒ Tank filled by Pipe A and Pipe B in the first 2 hours = (5 + 4) × 2 = 18 units

⇒ Remaining capacity to be filled = 60 − 18 = 42 units

⇒ Combined efficiency of A, B, and C = 5 + 4 − 3 = 6 units/hr

⇒ Time taken to fill the remaining part = 42 ÷ 6 = 7 hr

⇒ Total time taken to fill the tank = 2 hr + 7 hr = 9 hr

∴ The correct answer is 9 hr.

Additional Information

Pipes and Cisterns Efficiency

An inlet pipe (filling) has positive efficiency, while an outlet pipe (emptying) has negative efficiency.

Net Work Done

When multiple pipes operate, the net rate is the algebraic sum of their individual efficiencies.

LCM Method for Total Capacity

The Total Capacity of a tank is best assumed as the LCM of the individual times to simplify calculations without fractions.

Shortcut Trick

Sum of ages 2 years ago = 46 − (2 × 2) = 42 years.

Ratio of P to Q was 2 : 1.

Total units = 2 + 1 = 3 units → 42 years.

1 unit = 14 years.

P's age 2 years ago = 2 × 14 = 28 years.

Present age of P = 28 + 2 = 30 years.

∴ The correct answer is 30 years.

Alternate Method

Given:

Present combined age of P and Q = 46 years

Condition: 2 years ago, P = 2 × Q

Let the present age of P be P and Q be Q.

⇒ P + Q = 46 (Equation 1)

Two years ago, their ages were (P − 2) and (Q − 2).

⇒ P − 2 = 2 × (Q − 2)

⇒ P − 2 = 2Q − 4

⇒ P − 2Q = −2 (Equation 2)

Subtracting Equation 2 from Equation 1:

⇒ (P + Q) − (P − 2Q) = 46 − (−2)

⇒ 3Q = 48

⇒ Q = 16 years

Substitute the value of Q in Equation 1:

⇒ P + 16 = 46

⇒ P = 46 − 16

⇒ P = 30 years

∴ The correct answer is 30 years.

Additional Information

Constant Age Difference

The numerical difference between the ages of two individuals always remains constant, regardless of how many years pass or are subtracted.

Sum Adjustment for Time

When calculating the sum of ages 'n' years ago for 'k' people, subtract (n × k) from the present total sum.

Ratio and Proportions in Ages

If the ratio of ages at time 't' is a : b, and the total sum at that same time is S, then individual ages are [a/(a+b)] × S and [b/(a+b)] × S.

Shortcut Trick

The mean proportional of two numbers a and b is calculated as √(a × b).

Given a = 16 and b = 25.

New side = √(16 × 25) = 4 × 5 = 20 m.

∴ The correct answer is 20 m.

Alternate Method

Given:

Side of the first square garden (a) = 16 m

Side of the second square garden (b) = 25 m

Formula Used:

If x is the mean proportional between a and b, then a : x = x : b ⇒ x² = ab ⇒ x = √ab

Calculations:

⇒ Let the side of the third square garden be x.

⇒ According to the question, x is the mean proportional of 16 and 25.

⇒ x = √(16 × 25)

⇒ x = √400

⇒ x = 20 m

∴ The correct answer is 20 m.

Additional Information

Mean Proportional

For any two numbers a and b, the mean proportional x is the square root of their product, i.e., x = √ab.

Third Proportional

If a and b are two numbers, their third proportional c is such that a:b = b:c, which gives c = b²/a.

Fourth Proportional

For three numbers a, b, and c, the fourth proportional d satisfies a:b = c:d, implying d = (b × c) / a.

Shortcut Trick

Total volume remains constant when objects are melted and recast.

Volume of 16 identical cubes = 16 × (Side)³

Volume = 16 × (2 cm)³ = 16 × 8 cm³ = 128 cm³

∴ The correct answer is 128 cm³.

Alternate Method

Given: Sixteen identical cubes with side = 2 cm. For the cuboid, length (L) = 2 × breadth (B) and height (H) = breadth (B).

Formula Used: Volume of cube = a³ | Volume of cuboid = L × B × H

⇒ Volume of one solid cube = (2 cm)³ = 8 cm³

⇒ Total volume of 16 cubes = 16 × 8 cm³ = 128 cm³

⇒ Let the breadth of the cuboid be B.

⇒ Length L = 2B and Height H = B.

⇒ Volume of cuboid = L × B × H = (2B) × B × B = 2B³

⇒ Since volume is conserved: 2B³ = 128

⇒ B³ = 64 ⇒ B = 4 cm

⇒ Volume of cuboid = 2 × (4)³ = 2 × 64 = 128 cm³

∴ The correct answer is 128 cm³.

Additional Information

Volume Conservation Principle

When a physical object is melted and recast into one or more different shapes, the total volume remains unchanged, assuming no material is lost in the process.

Total Surface Area of Cuboid

The total surface area (TSA) of a cuboid is given by the formula: 2 × (LB + BH + HL). For the calculated dimensions (8, 4, 4), TSA = 2 × (32 + 16 + 32) = 160 cm².

Diagonal of a Cuboid

The space diagonal (longest distance between opposite corners) is calculated as √(L² + B² + H²).

Shortcut Trick

Convert 1 hr 20 min to total minutes: 60 + 20 = 80 minutes.

Printing rate per minute = 2400 ÷ 80 = 30 pages/min.

Pages printed in 45 minutes = 30 × 45 = 1350.

∴ The correct answer is 1350 pages.

Alternate Method

Given:

Total pages = 2400

Time taken = 1 hr 20 min

Required time = 45 min

Calculation:

⇒ 1 hr 20 min = (1 × 60) + 20 = 80 minutes

⇒ Let the pages printed in 45 min be x.

⇒ 2400 ÷ 80 = x ÷ 45

⇒ 30 = x ÷ 45

⇒ x = 30 × 45

⇒ x = 1350

∴ The correct answer is 1350 pages.

Additional Information

Unitary Method

Find the value of a single unit by dividing the total value by the number of units, then multiply to find the required value.

Steady Rate / Constant Speed

When a machine works at a steady rate, the work done is directly proportional to the time spent (Work = Rate × Time).

Time Conversion

Always ensure all time units are consistent (e.g., converting hours to minutes using 1 hr = 60 min) before performing division or multiplication.

Shortcut Trick

Net filling rate of all three taps = (1/4 + 1/8 − 1/8) = 1/4 of the cistern per hour.

The time required to fill the entire cistern is 4 hours.

To fill one-fourth of the empty cistern, time taken = 4 ÷ 4 = 1 hour.

∴ The correct answer is 1 hr.

Alternate Method

Given: Filling times are 4 hr and 8 hr; Emptying time is 8 hr.

⇒ Total Efficiency of all taps = 2 + 1 − 1 = 2 units/hr.

⇒ Target volume to fill (one-fourth) = 8 × (1/4) = 2 units.

⇒ Time required = Target Volume ÷ Total Efficiency

⇒ Time required = 2 ÷ 2 = 1 hr.

∴ The correct answer is 1 hr.

Additional Information

Net Filling Rate

The combined rate of multiple pipes is the algebraic sum of their individual rates, where inlet pipes are positive and outlet pipes are negative.

Work and Efficiency

In pipes and cisterns, the total capacity is treated as work. Time = Total Capacity ÷ Net Efficiency.

LCM Method

Assume the total capacity of the cistern to be the Least Common Multiple (LCM) of the individual time periods for easier calculation.