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Shortcut Trick

Effective CP = 400 + 50 = 450 (in thousands).

SP = 555 (in thousands).

Profit = 555 − 450 = 105.

Gain % = (105 / 450) × 100 = (7 / 30) × 100 = 23.33%.

∴ The correct answer is 23.33%.

Alternate Method

Given:

Purchase price of Tractor = ₹4,00,000

Amount spent on Repairs = ₹50,000

Selling Price (SP) = ₹5,55,000

Formula Used:

Total Cost Price (CP) = Purchase Price + Overhead Expenses

Gain = SP − CP

Gain Percentage = (Gain / CP) × 100

Calculations:

⇒ Total Cost Price (CP) = 4,00,000 + 50,000 = ₹4,50,000

⇒ Gain = 5,55,000 − 4,50,000 = ₹1,05,000

⇒ Gain % = (1,05,000 / 4,50,000) × 100

⇒ Gain % = (105 / 450) × 100

⇒ Gain % = (7 / 30) × 100

⇒ Gain % = 70 / 3 = 23.3333...

Rounding off to two decimal places, we get 23.33%.

∴ The correct answer is 23.33%.

Additional Information

Total Cost Price

When extra expenses like Repairs, Transportation, or Installation are incurred on an item, they must be added to the purchase price to find the actual Effective Cost Price.

Profit and Loss Percentage

Profit or Loss is always calculated on the Cost Price (CP) unless stated otherwise. Gain % = (SP − CP) / CP × 100.

Rounding Rule

For Rounding off to two decimal places, if the third decimal digit is 5 or more, increase the second digit by 1; otherwise, leave it unchanged.

Shortcut Trick

Price after 1^st discount = 3,000 × (100 − 25)/100 = 3,000 × 0.75 = ₹2,250.

Required Net Price = ₹1,800.

Additional Discount needed = 2,250 − 1,800 = ₹450.

Additional Discount % = (450 / 2,250) × 100 = 1/5 × 100 = 20%.

∴ The correct answer is 20%.

Alternate Method

Given:

Marked Price (MP) = ₹3,000

First Discount (D₁) = 25%

Final Selling Price (SP) = ₹1,800

Calculations:

⇒ Price after 1^st discount = 3,000 − (25% of 3,000)

⇒ Price after 1^st discount = 3,000 − 750 = ₹2,250

⇒ Let the additional discount be x%.

⇒ Final Price = 2,250 × (100 − x) / 100 = 1,800

⇒ (100 − x) / 100 = 1,800 / 2,250

⇒ (100 − x) / 100 = 180 / 225 = 4 / 5

⇒ 100 − x = (4 / 5) × 100

⇒ 100 − x = 80

⇒ x = 100 − 80 = 20

∴ The correct answer is 20%.

Additional Information

Successive Discount Formula

When two discounts a% and b% are given successively, the single equivalent discount is (a + b − (a × b)/100)%.

Marked Price and Selling Price

Selling Price = Marked Price × (100 − Discount%)/100. Discounts are always calculated on the Marked Price.

Net Discount

The net discount is the difference between the original Marked Price and the final Selling Price after all successive discounts.

Shortcut Trick

Relative Speed (same direction) = 66 − 48 = 18 km/hr

Convert to m/s: 18 × (5/18) = 5 m/s

Total distance covered (sum of lengths) = 5 m/s × 35 sec = 175 m

Since lengths are equal, length of each train = 175 ÷ 2 = 87.5 m

∴ The correct answer is 87.5 m.

Alternate Method

Given:

Speed of faster train (v₁) = 66 km/hr

Speed of slower train (v₂) = 48 km/hr

Time taken to pass (t) = 35 sec

Length of each train = L (Equal length)

Formula Used:

Relative Speed (Same Direction) = v₁ − v₂

Total Distance = Speed × Time

Calculations:

⇒ Relative Speed = 66 − 48 = 18 km/hr

⇒ Converting to m/s = 18 × (5 ÷ 18) = 5 m/s

⇒ Total distance to cross = Length of Train 1 + Length of Train 2

⇒ Distance = L + L = 2L

⇒ 2L = Relative Speed × Time

⇒ 2L = 5 × 35

⇒ 2L = 175

⇒ L = 175 ÷ 2 = 87.5 m

∴ The correct answer is 87.5 m.

Additional Information

Relative Speed in Opposite Direction

When two objects move in opposite directions, their relative speed is the sum of their individual speeds (v₁ + v₂).

Unit Conversion Shortcut

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

Train Crossing Concepts

When a train crosses a stationary object with length (like a platform or bridge), total distance = Length of train + Length of object.

Shortcut Trick

Total distance covered = 50 + 70 = 120 km

Time taken for first part = 50 ÷ 25 = 2 hours

Time taken for second part = 70 ÷ 17.5 = 4 hours

Total time taken = 2 + 4 = 6 hours

Average speed = Total Distance ÷ Total Time = 120 ÷ 6 = 20 km/h

∴ The correct answer is 20 km/h.

Alternate Method

Given:

First distance (d₁) = 50 km, Speed (s₁) = 25 km/h

Second distance (d₂) = 70 km, Speed (s₂) = 17.5 km/h

Formula Used:

Average Speed = Total Distance ÷ Total Time

Time = Distance ÷ Speed

Calculations:

⇒ Total distance = d₁ + d₂ = 50 + 70 = 120 km

⇒ Time for first part (t₁) = 50 ÷ 25 = 2 hours

⇒ Time for second part (t₂) = 70 ÷ 17.5 = 4 hours

⇒ Total time = t₁ + t₂ = 2 + 4 = 6 hours

⇒ Average speed = 120 ÷ 6

⇒ Average speed = 20 km/h

∴ The correct answer is 20 km/h.

Additional Information

Average Speed (Equal Distances)

If an object covers two equal distances at speeds x and y, then the average speed is given by 2xy ÷ (x + y).

Unit Conversion

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

Relative Speed

When two objects move in the same direction, relative speed is the difference of their speeds. If they move in opposite directions, it is the sum of their speeds.

Shortcut Trick

Time taken by Inlet Pipe (A) = 28 hr

Time taken with Leak (A + L) = 28 + 14 = 42 hr

Time to empty by Leak = (Time₁ × Time₂) / (Time₂ − Time₁)

Time to empty = (28 × 42) / (42 − 28) = (28 × 42) / 14 = 2 × 42 = 84 hr

∴ The correct answer is 84 hr.

Alternate Method

Given:

Filling time of Inlet Pipe = 28 hr

Extra time due to Leak = 14 hr

Total time with Leak = 28 + 14 = 42 hr

Formula Used:

Total Work = LCM of individual times

Efficiency = Total Work ÷ Time

Calculations:

⇒ Efficiency of Inlet Pipe (A) = 3 units/hr

⇒ Combined Efficiency of Pipe and Leak (A + L) = 2 units/hr

⇒ Efficiency of Leak (L) = (A + L) − A

⇒ Efficiency of Leak = 2 − 3 = −1 unit/hr (Negative sign indicates emptying)

⇒ Time taken by Leak to empty the tank = Total Work ÷ |Efficiency of Leak|

⇒ Time = 84 ÷ 1 = 84 hr

∴ The correct answer is 84 hr.

Additional Information

Inlet and Outlet Pipes

An Inlet Pipe fills the tank (positive work), while an Outlet Pipe or Leak empties the tank (negative work).

Net Efficiency

When both pipes are open, the Net Efficiency is the algebraic sum of their individual efficiencies: Net E = E_Inlet − E_Outlet.

Tank Capacity Irrelevance

If the question asks for time and provides relative filling rates, the actual volume (e.g., 5000 liters) is not required for the calculation.

Shortcut Trick

Total time from A to B = 240 ÷ 60 = 4 hours.

Since return time is also 4 hours, the Average Speed for return journey = 240 ÷ 4 = 60 km/h.

Using Alligation on return speeds: (Car Speed 80 − Average 60) : (Average 60 − Bus Speed 40) = 20 : 20 = 1 : 1.

Time ratio (Bus : Car) = 1 : 1 ∴ Bus time = (1/2) × 4 = 2 hours.

Distance by bus = Speed × Time = 40 × 2 = 80 km.

∴ The correct answer is 80 km.

Alternate Method

Given:

Distance between town A and town B = 240 km.

Speed (A to B) = 60 km/h.

Return Bus speed = 40 km/h | Return Car speed = 80 km/h.

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Time taken (A to B) = 240 ÷ 60 = 4 hours.

⇒ Let the distance travelled by bus on the return journey be 'd' km.

⇒ Then, distance travelled by car = (240 − d) km.

⇒ Total return time = (d ÷ 40) + ((240 − d) ÷ 80) = 4 hours.

⇒ Multiply by 80: 2d + (240 − d) = 4 × 80

⇒ 2d + 240 − d = 320

⇒ d + 240 = 320

⇒ d = 320 − 240 = 80 km.

∴ The correct answer is 80 km.

Additional Information

Average Speed for Entire Journey

If distances are equal, Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂).

Time and Speed Relation

For a fixed distance, Speed is inversely proportional to Time (S₁/S₂ = T₂/T₁).

Alligation in Speed-Distance

When Alligation is applied to speeds, the resulting ratio is always the ratio of the time taken for each segment.

Shortcut Trick

Semi-perimeter (Length + Breadth) = Total Perimeter ÷ 2 = 50 ÷ 2 = 25 cm.

Since Length is 5 cm more than Breadth, use the sum and difference logic:

Length = (Sum + Difference) ÷ 2 = (25 + 5) ÷ 2 = 15 cm.

∴ The correct answer is 15 cm.

Alternate Method

Given:

Perimeter of the Rectangle = 50 cm

Length (L) = Breadth (B) + 5 cm

Formula Used:

Perimeter = 2 × (Length + Breadth)

Calculations:

⇒ Let the Breadth of the rectangle be B cm.

⇒ Therefore, the Length (L) = (B + 5) cm.

⇒ Perimeter = 2 × (L + B)

⇒ 50 = 2 × ((B + 5) + B)

⇒ 50 = 2 × (2B + 5)

⇒ 25 = 2B + 5

⇒ 2B = 25 − 5

⇒ 2B = 20

⇒ B = 10 cm

⇒ Length = B + 5 = 10 + 5 = 15 cm.

∴ The correct answer is 15 cm.

Additional Information

Area of Rectangle

The total surface area enclosed is given by Area = Length × Breadth.

Diagonal of Rectangle

The distance between opposite vertices is calculated using Pythagoras theorem: d = √(L² + B²).

Relationship with Square

A Square is a special rectangle where Length = Breadth. Its Perimeter is 4 × Side.

Shortcut Trick

The difference between the Total Surface Area (TSA) and the Lateral Surface Area (LSA) of a cube is simply the area of the top and bottom faces.

Difference = 2 × (side)²

Difference = 2 × 18² = 2 × 324 = 648 cm²

∴ The correct answer is 648 cm².

Alternate Method

Given:

Side of the cube (a) = 18 cm

Formula Used:

Total Surface Area (TSA) = 6a²

Lateral Surface Area (LSA) = 4a²

Calculations:

⇒ TSA = 6 × (18)² = 6 × 324 = 1944 cm²

⇒ LSA = 4 × (18)² = 4 × 324 = 1296 cm²

⇒ Difference = TSA − LSA

⇒ Difference = 1944 − 1296 = 648 cm²

∴ The correct answer is 648 cm².

Additional Information

Volume of a Cube

The volume of a cube with side length 'a' is given by V = a³.

Diagonal of a Cube

The length of the longest diagonal that can fit inside a cube is d = a√3.

Surface Areas of a Cuboid

For a cuboid with length (l), breadth (b), and height (h): TSA = 2(lb + bh + hl) and LSA = 2h(l + b).

Shortcut Trick

Third Proportional of 3 and 11√3 = (11√3)² ÷ 3 = (121 × 3) ÷ 3 = 121.

Mean Proportional of 121 and 9 = √(121 × 9) = 11 × 3 = 33.

Required Ratio = 121 : 33 = 11 : 3 (by dividing both terms by 11).

∴ The correct answer is 11:3.

Alternate Method

Given:

First set: a = 3, b = 11√3

Second set: x = 121, y = 9

Formulas used:

1. Third Proportional (c) of a and b is given by: b² = a × c ⇒ c = b² ÷ a
2. Mean Proportional (m) of x and y is given by: m = √(x × y)

Calculations:

⇒ Third proportional = (11√3)² ÷ 3

⇒ Third proportional = (121 × 3) ÷ 3 = 121

⇒ Mean proportional = √(121 × 9)

⇒ Mean proportional = 11 × 3 = 33

⇒ Ratio = 121 : 33

⇒ Ratio = (11 × 11) : (11 × 3)

⇒ Ratio = 11 : 3

∴ The correct answer is 11:3.

Additional Information

Mean Proportional

If three numbers a, b, and c are in continued proportion, then a/b = b/c. Here, b is called the mean proportional and b = √(ac).

Third Proportional

In the continued proportion a : b :: b : c, the term c is known as the third proportional to a and b, calculated as c = b² ÷ a.

Fourth Proportional

If four numbers a, b, c, and d are in proportion (a : b :: c : d), then d is the fourth proportional and d = (b × c) ÷ a.

Shortcut Trick

Time taken by inlet pipe (T₁) = 4 hr

Time taken with leak (T₂) = 4 + 4 = 8 hr

Leakage time = (T₁ × T₂) ÷ (T₂ − T₁)

Time = (4 × 8) ÷ (8 − 4) = 32 ÷ 4 = 8 hr

∴ The correct answer is 8 hr.

Alternate Method

Given:

Tank Capacity = 600 liters

Inlet pipe filling time = 4 hr

Time taken with leak = 4 hr + 4 hr = 8 hr

Formula Used:

Efficiency = Total Work ÷ Time taken

⇒ Efficiency of Leak = Combined Efficiency − Inlet Efficiency

⇒ Efficiency of Leak = 1 − 2 = −1 unit/hr

⇒ Time taken by leak to empty = Total Work ÷ |Efficiency of Leak|

⇒ Time = 8 ÷ 1 = 8 hr

∴ The correct answer is 8 hr.

Additional Information

Negative Work (Leakage)

In pipes and cisterns, a leak or an outlet pipe performs negative work. Its efficiency is subtracted from the total filling efficiency.

Efficiency and Time Relationship

Efficiency is inversely proportional to Time. If the time taken to fill increases, the net efficiency has decreased due to an opposing force like a leak.

Combined Pipe Formula

If pipe A fills in x hours and pipe B empties in y hours, net work per hour = (1÷x) − (1÷y).