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Shortcut Trick

Time taken by Pipe C and Pipe D to fill 3,000 L together = (6 × 4) ÷ (6 + 4) = 2.4 hr.

Since the new tank capacity (6,000 L) is exactly double the initial capacity (3,000 L), the time taken will also be doubled.

Required time = 2.4 × 2 = 4.8 hr.

∴ The correct answer is 4.8 hr.

Alternate Method

Given:

Capacity of initial tank = 3,000 L

Time for Pipe C = 6 hr | Time for Pipe D = 4 hr

Capacity of new tank = 6,000 L

Calculations:

⇒ Efficiency (Rate) of Pipe C = 3,000 ÷ 6 = 500 L/hr

⇒ Efficiency (Rate) of Pipe D = 3,000 ÷ 4 = 750 L/hr

⇒ Combined Efficiency of (C + D) = 500 + 750 = 1,250 L/hr

⇒ Time taken to fill 6,000 L tank = Total Capacity ÷ Combined Efficiency

⇒ Required Time = 6,000 ÷ 1,250

⇒ Required Time = 4.8 hr

∴ The correct answer is 4.8 hr.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to time for a constant volume of work. Formula: Work = Efficiency × Time.

Pipes and Cisterns Rule

If an inlet pipe fills in x hours and an outlet pipe empties in y hours, the net work in 1 hour = (1/x) − (1/y).

Direct Proportion in Work

If efficiency remains constant, the time required is directly proportional to the amount of work (tank capacity). If capacity doubles, time doubles.

Shortcut Trick

For a tower height h, when the angle changes from θ₁ to θ₂ and shadow increases by d:

Shortcut formula: h = d ÷ (cot θ₂ − cot θ₁)

Here, d = 20 m, θ₁ = 60°, and θ₂ = 45°

h = 20 ÷ (cot 45° − cot 60°) = 20 ÷ (1 − 1/1.7) = 20 ÷ (0.7/1.7) = 34 ÷ 0.7 ≈ 48.6 m

∴ The correct answer is 48.6 m.

Alternate Method

Let AB be the tower of height h. Let C and D be the points where the sun's altitude is 60° and 45° respectively.

Given: CD = 20 m, ∠ACB = 60°, ∠ADB = 45°, √3 = 1.7

In right-angled triangle ABC:

⇒ tan 60° = AB / BC

⇒ 1.7 = h / BC

⇒ BC = h / 1.7

In right-angled triangle ABD:

⇒ tan 45° = AB / BD

⇒ 1 = h / BD

⇒ BD = h

According to the question:

⇒ BD − BC = CD

⇒ h − h / 1.7 = 20

⇒ (1.7h − h) / 1.7 = 20

⇒ 0.7h / 1.7 = 20

⇒ 0.7h = 20 × 1.7

⇒ 0.7h = 34

⇒ h = 34 / 0.7 ≈ 48.5714

⇒ h ≈ 48.6 m

∴ The correct answer is 48.6 m.

Additional Information

Trigonometric Ratios

Commonly used ratios: tan θ = Perpendicular/Base and cot θ = Base/Perpendicular.

Angle of Elevation

As the object (Sun) moves lower in the sky (altitude decreases), the shadow of a vertical object becomes longer.

General Distance Formula

The distance between two points on the same side of a tower is given by d = h(cot θ₂ − cot θ₁), where θ₁ > θ₂.

Shortcut Trick

Use the distance relation: D = [S × (S + a) / a] × t₁ = [S × (S − b) / b] × t₂

⇒ [(S + 14) / 14] × 35 = [(S − 10) / 10] × 30

⇒ 5(S + 14) = 6(S − 10) ⇒ 5S + 70 = 6S − 60 ⇒ S = 130 km/h

⇒ D = [130 × (130 + 14) / 14] × (35 / 60) = 780 km

∴ The correct answer is 780 km.

Alternate Method

Given:

Increase in speed (a) = 14 km/hr, Reduction in time (t₁) = 35 min = 35/60 hr

Decrease in speed (b) = 10 km/hr, Increase in time (t₂) = 30 min = 30/60 hr

Formula Used:

Distance (D) = [S × (S + ΔS) / ΔS] × Δt

Calculations:

⇒ Let original speed be S km/hr and distance be D km.

⇒ According to the first condition: D = [S(S + 14) / 14] × (35 / 60) --- (i)

⇒ According to the second condition: D = [S(S − 10) / 10] × (30 / 60) --- (ii)

⇒ Equating (i) and (ii): [S(S + 14) / 14] × (35 / 60) = [S(S − 10) / 10] × (30 / 60)

⇒ [(S + 14) / 14] × 35 = [(S − 10) / 10] × 30

⇒ [(S + 14) / 2] × 5 = (S − 10) × 3

⇒ 5S + 70 = 6S − 60

⇒ S = 130 km/hr

⇒ Substituting S = 130 in equation (i):

⇒ D = [130 × (130 + 14) / 14] × (35 / 60)

⇒ D = [130 × 144 / 14] × (7 / 12)

⇒ D = (130 / 2) × (144 / 12) = 65 × 12 = 780 km

∴ The correct answer is 780 km.

Additional Information

Relationship between Speed, Time, and Distance

Distance is directly proportional to speed when time is constant (D = S × T). For a fixed distance, speed and time are inversely proportional.

Distance Formula with Time Difference

When a person travels at two different speeds S₁ and S₂ and the difference between the time taken is Δt, the distance D = [S₁ × S₂ / (S₁ − S₂)] × Δt.

Average Speed Formula

If a certain distance is covered at x km/hr and the same distance is covered at y km/hr, the Average Speed for the whole journey is 2xy / (x + y) km/hr.

Shortcut Trick

Let Cost Price (CP) = 100 units.

Initial Selling Price (SP) = 120 units (20% profit).

New SP = 120 − (10% of 120) = 120 − 12 = 108 units.

New Profit % = 108 − 100 = 8%.

∴ The correct answer is 8%.

Alternate Method

Given:

Initial Profit = 20%

Reduction in Selling Price = 10%

Formula used:

Profit Percentage = ((SP − CP) ÷ CP) × 100

Calculations:

⇒ Let Cost Price (CP) = ₹100

⇒ Initial Selling Price (SP₁) = CP + 20% of CP = 100 + 20 = ₹120

⇒ New Selling Price (SP₂) = SP₁ − 10% of SP₁

⇒ SP₂ = 120 − (10 ÷ 100) × 120 = 120 − 12 = ₹108

⇒ New Profit = SP₂ − CP = 108 − 100 = ₹8

⇒ New Profit Percentage = (8 ÷ 100) × 100 = 8%

∴ The correct answer is 8%.

Additional Information

Profit and Loss Basics

Profit occurs when Selling Price (SP) > Cost Price (CP). Profit % is always calculated on CP unless stated otherwise.

Successive Change Effect

When SP is changed, the new profit depends on the original CP, which remains constant unless specifically mentioned.

Marked Price and Discount

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

Shortcut Trick

Convert Interest Rate to fraction: 12.5% = 1/8. For 3 months (1/4 year), Rate = 1/8 × 1/4 = 1/32.

Principal to Amount Ratio: (8 × 8 × 32) : (9 × 9 × 33) ⇒ 2048 : 2673.

Principal 2048 units = ₹25,600 ⇒ 1 unit = ₹12.5.

Interest units = 2673 − 2048 = 625 units.

Compound Interest = 625 × 12.5 = ₹7,812.50.

∴ The correct answer is ₹7,812.50.

Alternate Method

Given: Principal (P) = ₹25,600, Rate (R) = 12.5% per annum, Time (T) = 2 years 3 months.

Formula Used: Amount (A) = P(1 + R/100)ⁿ × (1 + (R × t)/100) for fractional years.

⇒ Convert 3 months to years: 3/12 = 1/4 year.

⇒ Annual rate in fraction: 12.5/100 = 1/8.

⇒ Rate for 3 months: (1/8) × (1/4) = 1/32.

⇒ Amount (A) = 25600 × (1 + 1/8)² × (1 + 1/32)

⇒ A = 25600 × (9/8) × (9/8) × (33/32)

⇒ A = 25600 × (81/64) × (33/32)

⇒ A = (25600 ÷ 2048) × (81 × 33)

⇒ A = 12.5 × 2673 = ₹33,412.50

⇒ Compound Interest (CI) = Amount − Principal

⇒ CI = ₹33,412.50 − ₹25,600 = ₹7,812.50

∴ The correct answer is ₹7,812.50.

Additional Information

Fractional Time Period in CI

When time is in the form T years and m months, calculate for T years normally and apply rate (m/12) × R for the remaining fractional part.

Difference between CI and SI

For a 2-year period, the difference between compound and simple interest is D = P(R/100)².

Effective Rate of Interest

The effective rate for two successive years with rates r₁ and r₂ is r₁ + r₂ + (r₁ × r₂)/100.

Shortcut Trick

Speed in km/h = Distance ÷ Time = 360 ÷ 8 = 45 km/h.

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

Speed = 45 × (5/18) = 2.5 × 5 = 12.5 m/s.

∴ The correct answer is 12.5 m/s.

Alternate Method

Given:

Total Distance = 360 km

Total Time = 8 hr

Formula Used:

Speed = Distance ÷ Time

Conversion factor: 1 km/h = 5/18 m/s

⇒ Speed in km/h = 360 ÷ 8

⇒ Speed = 45 km/h

⇒ Speed in m/s = 45 × (5 ÷ 18)

⇒ Speed = (5 × 5) ÷ 2

⇒ Speed = 25 ÷ 2 = 12.5 m/s

∴ The correct answer is 12.5 m/s.

Additional Information

Unit Conversion Rules

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

Speed-Distance-Time Formula

Distance = Speed × Time. Ensure all units are in the same system (SI or Metric) before calculating.

Average Speed

Average Speed = (Total Distance) ÷ (Total Time). For two equal distances at speeds x and y, Average Speed = 2xy ÷ (x + y).

Shortcut Trick

Observe the common difference between the divisors and their respective remainders:

3 − 2 = 1

4 − 3 = 1

5 − 4 = 1

Smallest Number = LCM(3, 4, 5) − Common Difference

Required Number = 60 − 1 = 59.

∴ The correct answer is 59.

Alternate Method

Given:

Divisors: 3, 4, 5

Remainders: 2, 3, 4

Formula Used:

When the difference between the divisor (d) and remainder (r) is constant (k = d − r), the smallest number is given by: LCM(d₁, d₂, d₃, ...) − k.

Calculations:

⇒ Difference 1 = 3 − 2 = 1

⇒ Difference 2 = 4 − 3 = 1

⇒ Difference 3 = 5 − 4 = 1

⇒ Constant difference (k) = 1

⇒ LCM of 3, 4, 5:

⇒ 3 = 3¹

⇒ 4 = 2²

⇒ 5 = 5¹

⇒ LCM = 2² × 3¹ × 5¹ = 4 × 3 × 5 = 60

⇒ Smallest number = 60 − 1 = 59

∴ The correct answer is 59.

Additional Information

LCM and Remainder Concept

If a number leaves the same remainder 'r' in each case, the number is LCM(divisors) + r.

Successive Division

In successive division, the quotient of the first division becomes the dividend for the second division.

Chinese Remainder Theorem

A general method to find a number satisfying multiple linear congruences when divisors are coprime.

Shortcut Trick

Buy 2 Get 1 Free means the customer gets 3 items but pays for only 2 items.

Total Selling Price (SP) of 3 items = 2 × 2,700 = ₹ 5,400.

Since the trader earns 20% profit, Total Cost Price (CP) = 5,400 ÷ 1.2 = ₹ 4,500.

Cost Price of one pair of trousers = 4,500 ÷ 3 = ₹ 1,500.

∴ The correct answer is ₹ 1,500.

Alternate Method

Given:

Marked Price (MP) per pair = ₹ 2,700

Offer: Buy 2, Get 1 Free (Total items received = 3)

Profit percentage = 20%

Calculations:

⇒ Total Selling Price (SP) for 3 items = 2 × 2,700 = ₹ 5,400

⇒ Let the Cost Price (CP) of one pair of trousers be ₹ x

⇒ Total Cost Price for 3 pairs = 3x

⇒ Profit = 20%, so Total SP = 120% of Total CP

⇒ 5,400 = 1.20 × (3x)

⇒ 5,400 = 3.6x

⇒ x = 5,400 ÷ 3.6

⇒ x = 1,500

∴ The correct answer is ₹ 1,500.

Additional Information

Discount % in 'Buy X Get Y Free' schemes

Discount % = [Free Items ÷ Total Items] × 100. In this case: [1 ÷ (2 + 1)] × 100 = 33.33%.

Relation between CP and MP

CP ÷ MP = (100 − Discount %) ÷ (100 + Profit %). This ratio helps find CP directly when MP, discount, and profit are known.

Profit and Loss Basics

Profit % = [(SP − CP) ÷ CP] × 100. Always ensure CP and SP are calculated for the same number of units.

Shortcut Trick

Net change in savings = (Increase in Income) − (Increase in Expenditure)

Increase in Income = 10% of 45,000 = ₹4,500

Increase in Expenditure = 20% of 36,000 = ₹7,200

Change in savings = 4,500 − 7,200 = −₹2,700 (Decrease)

New monthly savings = (45,000 − 36,000) − 2,700 = 9,000 − 2,700 = ₹6,300

∴ The correct answer is ₹6,300.

Alternate Method

Given: Original monthly income = ₹45,000, Original monthly expenditure = ₹36,000.

Formula Used: Savings = Income − Expenditure

⇒ New Income = 45,000 × (110 ÷ 100) = ₹49,500

⇒ New Expenditure = 36,000 × (120 ÷ 100) = ₹43,200

⇒ New Savings = New Income − New Expenditure

⇒ New Savings = 49,500 − 43,200

⇒ New Savings = ₹6,300

∴ The correct answer is ₹6,300.

Additional Information

Basic Budget Equation

Income = Expenditure + Savings. This relation holds true for both original and updated values.

Percentage Change in Savings

Percentage change in savings can be calculated as [(New Savings − Old Savings) ÷ Old Savings] × 100.

Income Increase vs Expenditure Increase

If the absolute increase in expenditure exceeds the absolute increase in income, the savings will always decrease.

Shortcut Trick

Identify factors of the product 30 that add up to 13.

Factors of 30: (1, 30), (2, 15), (3, 10), (5, 6).

Sum check: 10 + 3 = 13 (Correct).

Positive Difference: 10 − 3 = 7.

∴ The correct answer is 7.

Alternate Method

Given:

Sum of two positive numbers (x + y) = 13

Product of the numbers (xy) = 30

Formula Used:

(x − y)² = (x + y)² − 4xy

Calculations:

⇒ (x − y)² = (13)² − 4 × (30)

⇒ (x − y)² = 169 − 120

⇒ (x − y)² = 49

⇒ (x − y) = √49

⇒ x − y = 7

∴ The correct answer is 7.

Additional Information

Quadratic Roots Relation

Any two numbers can be treated as roots of a quadratic equation: z² − (Sum)z + (Product) = 0.

Algebraic Identities

Remember that (x + y)² = x² + y² + 2xy and (x − y)² = x² + y² − 2xy. Their difference is always 4xy.

Root Difference Formula

For a quadratic equation az² + bz + c = 0, the difference of roots is given by √D/|a|, where D = b² − 4ac.