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Given data:

Length = 20 cm

Breadth = 10 cm

Height = 8 cm

Formula:

TSA = 2 × (l × b + b × h + l × h)

Where-

TSA = Total Surface Area

l = Length

b = Breadth

h = Height

Calculations:

TSA = 2 × (20 × 10 + 10 × 8 + 20 × 8) = 880 cm²

∴ The total cardboard area needed to make a closed box is 880 cm².

Shortcut Trick

Radius (r) = Curved Surface Area ÷ (2 × π × height)

r = 126π ÷ (2 × π × 14) = 126 ÷ 28 = 4.5 cm

Volume of cylinder = Curved Surface Area × (r ÷ 2)

Volume = 126π × (4.5 ÷ 2) = 126π × 2.25 = 283.5π cm³

∴ The correct answer is 283½ π cm³.

Alternate Method

Given:

Curved Surface Area (CSA) = 126π cm²

Height (h) = 14 cm

Formula Used:

CSA of cylinder = 2πrh

Volume of cylinder = πr²h

Calculations:

⇒ 2 × π × r × 14 = 126 × π

⇒ 28 × r = 126

⇒ r = 126 ÷ 28 = 4.5 cm

⇒ Volume = π × (4.5)² × 14

⇒ Volume = π × 20.25 × 14

⇒ Volume = 283.5π cm³

⇒ Volume = 283½ π cm³

∴ The correct answer is 283½ π cm³.

Additional Information

Total Surface Area (TSA) of Cylinder

The total surface area includes the curved surface and the two circular bases: TSA = 2πr(h + r).

Volume of a Cone

A cone with the same radius and height as a cylinder has exactly one-third of its volume: V = (1/3)πr²h.

Relation between Volume and CSA

For any cylinder, the ratio of its volume to its curved surface area is always r/2, simplified as V = CSA × (r/2).

Formula:

M₁ × D₁ = M₂ × D₂, where

M1 and M₂ = Number of men working on D1 and D₂.

D1 and D₂ = Days for which M1 and M₂ worked respectively.

Calculations:

Total work days = 20 × 30 = 600 man-days

Total work done in 10 days = 20 × 10 = 200 man-days

Total work remaining = 600 - 200 = 400 man-days

Total time needed to complete the remaining work = 400 ÷ 16 = 25

Total time needed to complete the whole work = 25 + 10 = 35 days

∴ The work would be completed in 35 days.

Given:

$\rm x^4+\frac{1}{x^4} = 727,x>1$

Formulas used:

(a + b)² = a² + b² + 2ab

(a - b)² = a2 + b2 - 2ab

Calculation:

$\rm x^4+\frac{1}{x^4} = 727$ 

By adding 2 on both sides 

x⁴ + 1/x⁴ + 2 = 727 + 2 

⇒ (x² + 1/x²)² = 729

⇒ (x2 + 1/x2) = √729  

⇒  (x2 + 1/x2) = 27

By subtracting 2 from both sides

⇒ (x2 + 1/x2) - 2 = 27 - 2

⇒ (x - 1/x)² = 25

⇒  (x - 1/x) = √25

∴ (x - 1/x) = 5

Given 

Speed of train = 72 km/hr

Time taken to  cross platform = 30 seconds 

Formula Used 

Distance = Speed × Time

Distance = Length of train + Length of Platform 

Calculation 

According to the question 

Length of train = 72 × (5/18) × 18 = 360 m 

Length of train + Length of platform = 72 × (5/18) × 30 

⇒ 360 + length of platform = 600 

⇒ Length of Platform = 600 - 360 = 240 meters 

∴ The correct answer is 240 meters

Given:

Selling price (SP) = Rs. 55

Profit = 10%

New SP = 58

Formula used:

SP = CP × (100 + P%)/100

SP = CP × (100 - L%)/100

Calculations:

⇒ CP × (100 + 10/100) = 55

⇒ CP × (11/10) = 55

⇒ CP  = 55 × (10/11)

⇒ CP = Rs. 50

Profit% = (58 - 50)/50 × 100

Profit% = 8 × 2 = 16%

The answer is 16%.

Calculation:

Let the two numbers be 5x and 4x

As we know,

LCM of 5x and 4x= 20x

HCF of 5x and 4x = x = 4

According to the question

20x = 20 × 4 = 80

∴ The LCM of the two numbers is 80.

Given:

x - y = 5 and xy = 150

Calculation:

So from the equation we can assume the possible values of x and y

So x = 15 and y = 10

So X + Y = 25

∴ The correct answer is 2

Shortcut Trick

Total Work (LCM of 20, 30, 60) = 60 units. Efficiency: A = 3, B = 2, C = 1.

Work done by A (7 am to 9 am) = 3 × 2 = 6 units.

Work done by B (8 am to 9 am) = 2 × 1 = 2 units.

Remaining work at 9 am = 60 − (6 + 2) = 52 units.

Time after 9 am = 52 ÷ (3 + 2 + 1) = 52 ÷ 6 = 8 hours 40 minutes.

∴ The correct answer is 5 : 40 p.m.

Alternate Method

Given: Pipe A fills in 20 hrs, B in 30 hrs, C in 60 hrs. Start times: A at 7 a.m., B at 8 a.m., C at 9 a.m.

⇒ Work done by A till 9 a.m. (2 hours) = 3 × 2 = 6 units

⇒ Work done by B till 9 a.m. (1 hour) = 2 × 1 = 2 units

⇒ Total work completed till 9 a.m. = 6 + 2 = 8 units

⇒ Remaining work to be filled = 60 − 8 = 52 units

⇒ Combined efficiency of A, B, and C = 3 + 2 + 1 = 6 units/hour

⇒ Time required after 9 a.m. = 52 ÷ 6 = 8 hours 40 minutes

⇒ Final filling time = 9:00 a.m. + 8 hours 40 minutes = 5:40 p.m.

∴ The correct answer is 5 : 40 p.m.

Additional Information

Pipes and Cisterns Logic

If an inlet pipe fills in X hours and an outlet pipe empties in Y hours, the net work in 1 hour is (1/X − 1/Y).

Fraction to Time Conversion

To convert hours in fraction to minutes, multiply the fraction part by 60. E.g., 2/3 hours = 2/3 × 60 = 40 minutes.

Combined Efficiency

When multiple pipes operate together, their efficiencies are additive. Total time = Total Work ÷ Sum of individual efficiencies.

Shortcut Trick

Ratio of Cost Price (CP) : Selling Price (SP) = 20 : 21

Profit = 21 − 20 = 1 unit

Gain % = (Profit ÷ CP) × 100

Gain % = (1 ÷ 20) × 100 = 5%

∴ The correct answer is 5%.

Alternate Method

Given:

CP : SP = 20 : 21

Formula Used:

Gain % = [(SP − CP) ÷ CP] × 100

⇒ Let CP = 20x and SP = 21x

⇒ Gain = SP − CP

⇒ Gain = 21x − 20x = x

⇒ Gain % = (x ÷ 20x) × 100

⇒ Gain % = (1 ÷ 20) × 100

⇒ Gain % = 5%

∴ The correct answer is 5%.

Additional Information

Profit and Loss Percentage

Profit % and Loss % are always calculated on the Cost Price (CP) unless stated otherwise. Formula: Loss % = (Loss ÷ CP) × 100.

Relationship between SP and CP

Selling Price can be directly found using CP and Gain%: SP = CP × (100 + Gain%) ÷ 100.

Selling Price for Loss

When there is a loss: SP = CP × (100 − Loss%) ÷ 100.