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Shortcut Trick

Let Total Work = LCM(6, 30, 2) = 30 units.

Efficiency: Gopal = 5, Akshay = 1, (Gopal + Akshay + Atul) = 15.

Atul's Efficiency = 15 − (5 + 1) = 9 units/hr.

Time for thrice the work = (30 × 3) ÷ 9 = 90 ÷ 9 = 10 hr.

∴ The correct answer is 10 hr.

Alternate Method

Given:

Time taken by Gopal = 6 hr

Time taken by Akshay = 30 hr

Time taken by Gopal, Akshay, and Atul together = 2 hr

Formula Used:

Total Work = Time × Efficiency

Calculations:

⇒ Total Efficiency of Gopal + Akshay + Atul = 15

⇒ Efficiency of Atul = 15 − (Efficiency of Gopal + Efficiency of Akshay)

⇒ Efficiency of Atul = 15 − (5 + 1) = 9 units/hr

⇒ Thrice the work = 3 × 30 = 90 units

⇒ Time taken by Atul = 90 ÷ 9 = 10 hr

∴ The correct answer is 10 hr.

Additional Information

LCM Method in Time & Work

This method assumes the Total Work as the Least Common Multiple of the individual time durations to avoid complex fractions.

Efficiency Concept

Efficiency is inversely proportional to the time taken to complete a task. Efficiency = Work ÷ Time.

Work & Wages

When multiple people work together, the wages are distributed in the ratio of their respective efficiencies or the work done by each.

Shortcut Trick

Identify that the journey is divided into four equal distance parts: 1/4, 1/4, 1/4, and 1/4.

Assume each part is the LCM of (30, 40, 50, 60) = 600 km.

Total Distance = 600 × 4 = 2400 km.

Total Time = (600 ÷ 30) + (600 ÷ 40) + (600 ÷ 50) + (600 ÷ 60) = 20 + 15 + 12 + 10 = 57 hours.

Average Speed = 2400 ÷ 57 ≈ 42.11 km/h.

∴ The correct answer is 42.11 km/h.

Alternate Method

Given:

Speeds for different parts: s₁ = 30 km/h, s₂ = 40 km/h, s₃ = 50 km/h, s₄ = 60 km/h.

Formula Used:

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Let Total Distance = D km.

⇒ First part (d₁) = 1/4 × D = D/4

⇒ Second part (d₂) = 1/3 × (D − D/4) = 1/3 × 3/4 D = D/4

⇒ Third part (d₃) = 1/2 × (D − D/4 − D/4) = 1/2 × D/2 = D/4

⇒ Fourth part (d₄) = D − (D/4 + D/4 + D/4) = D/4

⇒ Total Time = (d₁/s₁) + (d₂/s₂) + (d₃/s₃) + (d₄/s₄)

⇒ Total Time = (D/4)/30 + (D/4)/40 + (D/4)/50 + (D/4)/60

⇒ Total Time = D/120 + D/160 + D/200 + D/240

⇒ Total Time = (20D + 15D + 12D + 10D) ÷ 2400 = 57D ÷ 2400

⇒ Average Speed = D ÷ (57D/2400) = 2400 ÷ 57 ≈ 42.1052 km/h.

∴ The correct answer is 42.11 km/h.

Additional Information

Average Speed for Two Equal Distances

When a person covers two equal distances at speeds x and y, the Average Speed = 2xy ÷ (x + y).

Average Speed for Three Equal Distances

When a person covers three equal distances at speeds x, y, and z, the Average Speed = 3xyz ÷ (xy + yz + zx).

Unit Conversion

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

Shortcut Trick

Identify the Pythagorean triplet: (7, 24, 25) scaled by 2 → (14, 48, 50).

So, Slant Height (l) = 50 m.

Total Cost = Area × Rate = (π × r × l) × 5

⇒ (22/7 × 14 × 50) × 5 = 2200 × 5 = 11,000.

∴ The correct answer is ₹11,000.

Alternate Method

Given:

⇒ Radius (r) = 14 m

⇒ Height (h) = 48 m

⇒ Cost Rate = ₹5 per m²

Formula Used:

⇒ Slant height (l) = √(r² + h²)

⇒ Curved Surface Area (CSA) = πrl

Calculations:

⇒ l = √(14² + 48²)

⇒ l = √(196 + 2304) = √2500

⇒ Slant Height (l) = 50 m

⇒ Curved Surface Area = (22/7) × 14 × 50

⇒ Curved Surface Area = 22 × 2 × 50 = 2200 m²

⇒ Total Cost = Area × Rate

⇒ Total Cost = 2200 × 5

⇒ Total Cost = ₹11,000

∴ The correct answer is ₹11,000.

Additional Information

Total Surface Area of Cone

It includes the base area and curved area: TSA = πr(l + r).

Volume of a Cone

The capacity of a conical structure is given by V = (1/3)πr²h.

Pythagorean Property in Cones

The radius, vertical height, and slant height always satisfy l² = r² + h².

Shortcut Trick

Radius (r) = Circumference ÷ 2π = 352 ÷ (44/7) = 56 cm.

Volume (V) = (Curved Surface Area × Radius) ÷ 2

⇒ V = (8800 × 56) ÷ 2 = 4400 × 56 = 2,46,400 cm³.

∴ The correct answer is 2,46,400 cm³.

Alternate Method

Given:

Curved Surface Area (CSA) = 8800 cm²

Circumference of base = 352 cm

Formula Used:

Circumference = 2πr

CSA = 2πrh

Volume = πr²h

Calculations:

⇒ Circumference = 2 × (22/7) × r = 352

⇒ (44/7) × r = 352

⇒ r = (352 × 7) ÷ 44 = 8 × 7 = 56 cm

⇒ CSA = (2πr) × h = 8800

⇒ 352 × h = 8800

⇒ h = 8800 ÷ 352 = 25 cm

⇒ Volume = (22/7) × 56 × 56 × 25

⇒ Volume = 22 × 8 × 56 × 25

⇒ Volume = 176 × 1400 = 2,46,400 cm³

∴ The correct answer is 2,46,400 cm³.

Additional Information

Total Surface Area of Cylinder

It is the sum of the curved surface area and the areas of the two circular bases: TSA = 2πr(r + h).

Base Area of Cylinder

The base of a cylinder is a circle, so its area is calculated as Area = πr².

Volume of a Cone

The volume of a cone with the same radius and height as a cylinder is exactly one-third of the cylinder's volume: V = (1/3)πr²h.

Shortcut Trick

Difference in Profit Percentage = 45% − 30% = 15%.

This 15% difference is equivalent to the price increase of ₹5.

Initial Selling Price (SP₁) represents 100% + 30% = 130% of CP.

Value of 130% = (5 ÷ 15) × 130 = 130 ÷ 3 = ₹43.33.

∴ The correct answer is ₹43.33.

Alternate Method

Given: Initial Profit = 30%, Final Profit = 45%, Change in SP = ₹5.

Formula: CP = (Difference in SP ÷ Difference in Profit%) × 100

⇒ Let the Cost Price (CP) be 100x.

⇒ Initial Selling Price (SP₁) = 130% of 100x = 130x.

⇒ New Selling Price (SP₂) = 145% of 100x = 145x.

⇒ Difference = SP₂ − SP₁ = 145x − 130x = 15x.

⇒ 15x = ₹5

⇒ x = 5 ÷ 15 = 1 ÷ 3.

⇒ Initial Selling Price = 130x = 130 × (1 ÷ 3)

⇒ SP₁ = 130 ÷ 3 = ₹43.333...

∴ The correct answer is ₹43.33.

Additional Information

Profit and Loss Basics

Profit or Loss percentage is always calculated on the Cost Price (CP) unless stated otherwise. Profit = SP − CP.

Direct SP Calculation

To find SP directly from CP and profit%: SP = CP × (100 + Profit%) ÷ 100.

Change in Profit%

When the same object is sold at two different profit percentages, the difference in Selling Price is directly proportional to the difference in profit percentages of the CP.

Shortcut Trick

Ratio of Boys to Girls = 26 : 25

Total units = 26 + 25 = 51 units

51 units → 2448 students

1 unit → 2448 ÷ 51 = 48

Number of boys = 26 units = 26 × 48 = 1248

∴ The correct answer is 1248 boys.

Alternate Method

Given:

Total number of students = 2448

Ratio (Boys : Girls) = 26 : 25

Formula Used:

Value of a part = (Ratio of that part ÷ Sum of ratio parts) × Total value

Calculations:

⇒ Let the number of boys be 26x and the number of girls be 25x.

⇒ 26x + 25x = 2448

⇒ 51x = 2448

⇒ x = 2448 ÷ 51

⇒ x = 48

⇒ Number of boys = 26x = 26 × 48

⇒ Number of boys = 1248

∴ The correct answer is 1248 boys.

Additional Information

Sum of Ratio Parts

To divide a total quantity into a given ratio a:b, the total parts are (a + b). Each part is calculated as [a/(a+b)] × Total and [b/(a+b)] × Total.

Ratio and Percentage

A ratio can be converted to percentage by (Part ÷ Total) × 100. For example, the percentage of boys here is (26 ÷ 51) × 100 ≈ 50.98%.

Simplification Tip

When multiplying 26 × 48, use 26 × (50 − 2) = 1300 − 52 = 1248 for faster calculation.

Shortcut Trick

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

Speed = (Distance ÷ Time) × (18 ÷ 5)

Speed = (1200 ÷ 40) × (18 ÷ 5) = 30 × 3.6 = 108 km/h.

∴ The correct answer is 108 km/h.

Alternate Method

Given:

Distance = 1200 m

Time = 40 sec

Formula Used:

Speed = Distance ÷ Time

Calculations:

⇒ Speed in m/s = 1200 ÷ 40

⇒ Speed in m/s = 30 m/s

⇒ We know that 1 m/s = 18/5 km/h

⇒ Speed in km/h = 30 × (18 ÷ 5)

⇒ Speed in km/h = 6 × 18

⇒ Speed in km/h = 108 km/h

∴ The correct answer is 108 km/h.

Additional Information

Unit Conversion Rules

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

Average Speed

Average Speed = Total Distance Traveled ÷ Total Time Taken.

Relative Speed

When two objects move in the opposite direction, relative speed = S₁ + S₂. In the same direction, relative speed = S₁ − S₂.

Shortcut Trick

To find the ratio A:C when A:B and B:C are given, multiply the ratios directly.

A/C = (A/B) × (B/C)

A/C = (7/8) × (2/2.5) = 14 / 20

Simplify 14/20 by dividing by 2 ⇒ 7/10.

∴ The correct answer is 7 : 10.

Alternate Method

Given: Ratio A:B = 7:8 and B:C = 2:2.5.

Method: Make the common term B equal in both ratios by multiplying B:C by 4.

⇒ A : B = 7 : 8

⇒ B : C = 2 : 2.5

⇒ Multiply B : C by 4 ⇒ (2 × 4) : (2.5 × 4) = 8 : 10

⇒ Since B is now 8 in both ratios, A : B : C = 7 : 8 : 10

⇒ A : C = 7 : 10

∴ The correct answer is 7 : 10.

Additional Information

Compound Ratio

When two or more ratios are multiplied together (a/b × c/d), the result is called a compound ratio.

Duplicate and Sub-duplicate Ratio

The duplicate ratio of a:b is a²:b² and the sub-duplicate ratio is √a : √b.

Proportion Rule

If a:b = c:d, then the product of extremes (a × d) must equal the product of means (b × c).

Shortcut Trick

Identify the Pythagorean triplet for the given sides 15 and 17.

The standard triplet is (8, 15, 17), where 17 is the hypotenuse.

The missing side (perpendicular) = 8 cm.

Perimeter = Sum of all sides = 15 + 17 + 8 = 40 cm.

∴ The correct answer is 40 cm.

Alternate Method

Given: Hypotenuse (h) = 17 cm, Base (b) = 15 cm.

Formula: According to Pythagoras Theorem, p² + b² = h², where p is the perpendicular.

⇒ p² + 15² = 17²

⇒ p² + 225 = 289

⇒ p² = 289 − 225

⇒ p² = 64

⇒ p = √64 = 8 cm

⇒ Perimeter of triangle = Sum of all sides = p + b + h

⇒ Perimeter = 8 + 15 + 17 = 40 cm

∴ The correct answer is 40 cm.

Additional Information

Pythagoras Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: a² + b² = c².

Common Pythagorean Triplets

Standard sets of integers that satisfy the theorem include (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), and (9, 40, 41).

Area of Right Triangle

The area is calculated as 1/2 × Base × Height. For this triangle, Area = 1/2 × 15 × 8 = 60 cm².

Shortcut Trick

For two pipes filling a tank, the combined time taken is calculated using the formula: (A × B) ÷ (A + B).

Given Pipe A = 9 hr and Pipe B = 5 hr.

Combined Time = (9 × 5) ÷ (9 + 5) = 45 ÷ 14 ≈ 3.21 hr.

Rounding off to one decimal place, we get 3.2 hr.

∴ The correct answer is 3.2 hr.

Alternate Method

Given:

Time taken by Pipe A = 9 hr

Time taken by Pipe B = 5 hr

Let the Total Work be the LCM of 9 and 5, which is 45 units.

Calculations:

⇒ Total Efficiency of (A + B) = 5 + 9 = 14 units/hr

⇒ Time Taken to fill the tank = Total Work ÷ Total Efficiency

⇒ Time Taken = 45 ÷ 14

⇒ Time Taken = 3.2142... hr

⇒ Rounding to one decimal place = 3.2 hr

∴ The correct answer is 3.2 hr.

Additional Information

Net Efficiency with Outlet Pipe

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

Time and Efficiency Relationship

The time taken to complete a task is inversely proportional to the efficiency of the pipe (Time = Work ÷ Efficiency).

Rule for N Pipes

For multiple pipes, the combined efficiency is the sum of individual efficiencies: 1/T_total = 1/T₁ + 1/T₂ + ... + 1/T_n.