Given:

Length of the arc = 15 cm

Angle subtended by the arc at the centre = 24°

Formula used:

Where, 2πr = circumference and θ = angle subtended by the arc at the centre.

Calculations:

According to the question,

⇒ 2πr = 15 × 15

⇒ 2πr = 225 cm

∴ The circumference is of length 225 cm.

Concept:

Obtain the remainder by dividing the given number from the divisor

The remainder, so obtained, when subtracted from the divisor gives the required value.

Calculation:

⇒ 56789 = 56580 + 209 = 345 × 164 + 209

Remainder = 209

Now, 209 + x = 345

⇒ x = 345 – 209

⇒ x = 136

∴ Sum of the digits of 136 = (1 + 3 + 6) = 10

Alternate Method

Given:

Number = 56789

Divisor = 345

Formula Used:

Smallest number to be added = Divisor - (Number % Divisor)

Calculation:

Remainder when 56789 is divided by 345:

⇒ 56789 % 345

⇒ 56789 - (56789 / 345) × 345

⇒ 56789 - 164 × 345

⇒ 56789 - 56580

⇒ 209

Smallest number to be added = 345 - 209

⇒ x = 136

Sum of the digits of x:

⇒ 1 + 3 + 6

⇒ 10

The sum of the digits of x is 10.

Given:

Two successive discounts of 20% and 10% on the marked price of an article.

Formula used:

Then, a 20% discount on Rs. 90 is Rs 72.

This means that the article marked as Rs. 100 is sold at Rs 72 which is equal to a discount of 

∴ A single discount equivalent to two successive discounts of 20% and 10% on the marked price of an article is 28%.

Shortcut Trick

⇒ - x - y + (x × y)/100

⇒ - 20 - 10 + (20 × 10)/100

⇒ - 30 + 2

⇒ - 28

∴ The Discount is 28%.

Given:

Let x men can do a job in 18 days

6 of them absent so (x - 6) can complete the job in 20 days

Formula used:

M₁D₁ = M₂D₂

Calculation:

x × 18 = (x - 6) × 20

⇒ 18x = 20x - 120

⇒ 18x - 20x = -120

⇒ -2x = -120

⇒ x = 120/2

⇒ x = 60

∴ original number of men = 60

Given:

y ∝ x

And y = 32 when x = 16

Calculation:

According to the question,

⇒ y ∝ x

By removing constant,

⇒ y = kx

Now putting values given in question,

⇒ k × 16 = 32

⇒ k = 2

Now,

⇒ y = 2 × 24

⇒ y = 48

∴ The correct answer is 48.

Given:

12 - [24 - (72 ÷ 3 - (30 - 50 ÷ 5) ÷ 20)]

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

⇒ 12 - [24 - (72 ÷ 3 - (30 - 50 ÷ 5) ÷ 20)]

⇒ 12 - [24 - (72 ÷ 3 - (30 - 10) ÷ 20)]

⇒ 12 - [24 - (72 ÷ 3 - 20 ÷ 20)]

⇒ 12 - [24 - (24 - 1)]

⇒ 12 - [24 - 24 + 1]

⇒ 12 - 1

⇒ 11

∴ Option 1 is the correct answer.

Given:

Sum of a number and its reciprocal = –2

Concept Used:

Quadratic equation form: ax² + bx + c = 0

Roots are calculated using the middle-term split method to form factors.

Calculation:

Let the number be x.

According to the question,

Therefore, the required quadratic equation is x2 + 2x + 1 = 0 and the number is –1.

Given:

Discounts = 12% and 8%

Concept Used:

Single Equivalent Discount = (x + y) – xy/100

Calculation:

Single equivalent discount of 12% and 8%

= 12 + 8 – (12 × 8)/100

Single equivalent discount = 20 – 96/100

= 20 – 0.96 = 19.04%

Therefore, the single discount equivalent to two successive discounts of 12% and 8% is 19.04%.

Given:

Speed of the two trains is 80 km/h and 90 km/h

They cross each other in 3 minutes

Concept used:

Time = Distance/speed

Relative speed = When two bodies moving in the same direction then their relative speed of them will be different of the individual speed

Calculation:

3 min = 180 sec

The relative speed of the trains = 90 - 80

⇒ 10 km/h

Speed in m/sec = 10 × (5/18)

⇒ 25/9

Let the length of the other train be x m

Now,

[(230 + x)/(25/9)] = 180

⇒ (230 + x) = 180 × (25/9)

⇒ 230 + x = 500

⇒ x = 270

So, length of the other train = 270 m

∴ The length (in m) of the other train is 270.

Given:

Dimension of the tank = 15 × 12 × 5 (in meters)

Width of the steel sheet = 1.5 m

Concept used:

Surface area of a cuboid = 2 × [Length × Breadth + Length × Height + Breadth × Height]

TSA of the open tank = 2 × [Length × Breadth + Length × Height + Breadth × Height] - Length x Breadth.   

Calculation:

The total surface area of the open tank = Total surface area of the tank - Area of the upper cover

Total surface area of the open tank = 2 × [15 × 12 + 12 × 5 + 15 × 5] - 15 × 12 = 450 m²

Hence, requires length of sheet = 450 ÷ 1.5 = 300 m

∴ 300 m of the sheet will be required.

Important Points

Total surface area of the open tank = Total surface area of the tank - Area of the upper cover