Given Data:

Money becomes 216/125 times of itself in 3 years.

Concept:

where A is the amount,

P is the principal,

r is the rate of interest,

and n is the time in years.

Solution:

Using the formula,

taking cube-root on both sides,

⇒ 6/5 = 1 + r/100

⇒ 6/5 = (100 + r)/100

⇒ 120 = 100 + r

⇒ r = 120 - 100 = 20

Therefore, the rate of interest is 20 percent.

Given:

Radius of cylinder and cone = r

Height of cylinder and cone = h

Formulas used:

Volume of cylinder = πr²h

Volume of cone = 1/3 × πr²h

Calculation:

∴ Volume of cylinder/Volume of Cone = πr²h/1/3 × πr²h

⇒ 3/1 = 3 : 1

Given:

The profit earned when an article is sold for Rs. 800 is 20 times the loss incurred when it is sold for Rs. 275

Concept used:

Loss = Cost price – Selling price 

Profit = Selling price – Cost price 

S.P = C.P × (100 + P%)/100

Calculations:

Let the loss earned when an article is sold for Rs. 275 be x 

So, C.P of an article = 275 + x      ----(1)

According to the question 

The profit earned when an article is sold for Rs. 800 = 20x 

C.P of an article = 800 – 20x      ----(2)

From eq. (1) and eq. (2)

⇒ 275 + x = 800 – 20x

⇒ 21x = 800 – 275 

⇒ 21x = 525 

⇒ x = 25 

From eq. (1)

C.P of an article = 275 + 25 = 300 

S.P of an article when it is sold at 25% profit = 300 × 125/100 = 375

∴ S.P of an article when it is sold at 25% profit Rs. 375

Given:

The number of workers in the employment guarantee scheme increased = 15

Increase percentage of workers = 20%

Calculation:

Let the initial numbers of workers be x

Then, according to the question

⇒ 20% of x = 15

⇒ x = 75

∴ The initial number of workers is 75.

Given:

The ratio of the two numbers = 2 : 3

LCM of the two numbers = 48

Calculation:

Let the two numbers be 2y and 3y.

LCM (2y, 3y) = 6y

⇒ 6y = 48

⇒ y = 8

Now, The sum of numbers = (2y + 3y)

⇒ 5y

⇒ 5y = 5 × 8

⇒ 40

∴ The sum of the number is 40.

Part filled by A in 1 hr. = 1/6

Part filled by A in 12 hrs. = 12 × 1/6 = 2

Hence,

Part emptied by B = 2 – 1 = 1

Part emptied by B in 1 hr. = 1/8

∴ Time for which pipe B was opened = 1/(1/8) = 8 hrs.

Alternate Method

Total capacity of tank = LCM = 24 

The efficiency of A = 24/6 = 4

The efficiency of B = 24/8 = -3 (B can empty the tank so the efficiency of B is negative)

Tank fill by A and B together in one hr = 4 – 3 = 1

Let the A and B together fill the tank for x hrs 

Tank fill by A and B together in x hrs = x

Remaining = 24 – x 

Time taken by A to fill the remaining part = (24 – x)/4 

Total time = 12 

⇒ x + (24 – x)/4 = 12 

⇒ x = 8 

∴ Time for which pipe B was opened is 8 hrs

Given:

7-digit number 51032kk is divisible by 12

Concept used:

Divisibility rule for 3 = A number is completely divisible by 3 if the sum of its digits is divisible by 3.

Divisibility rule for 4 = If the last two digits of a number are divisible by 4, the number is divisible by 4

Calculation:

12 = 4 × 3

According to the concept,

Possible values of k = 0, 4, 8

5103200, 5103244, 510388 all are divisible by 4 as their last 2 digits are divisible by 4

Now,

5 + 1 + 0 + 3 + 2 + k + k = 11 + 2k should also be divisible by 3

For k = 0

11 + 0 = 11 not divisible by 3

For k = 4

11 + 8 = 19 not divisible by 3

For k = 8

11 + 16 = 27 divisible by 3

∴ Requird value of k is 8.

Given:

ΔABC is inscribed in a circle with centre O.

∠BAC = 2x and ∠BCO = 2y

Concept used:

The angle subtends at the centre is twice the angle subtended from the same arc to any point on its circumference.

The sum of all internal angles for a triangle is 180^∘.

Calculation:

According to the question, the required figure is:

Since the angle subtends at the centre is twice the angle subtended from the same arc to any point on its circumference.

Therefore, 

∠BOC = 2 × ∠BAC

⇒ ∠BOC = 2 × (2x)

⇒ ∠BOC = 4x

In ΔBOC,

Since OB = OC = r, 

∴ ∠BOC = ∠OCB = 2y

Now, 

In ΔBOC,

∠OCB + ∠OBC + ∠BOC = 180^∘ 

⇒ 2y + 2y + 4x = 180∘ 

⇒ 4y + 4x = 180∘ 

⇒ y + x = 45∘ 

∴ y + x = 45∘ is the required relation.

Calculation:

Range = Maximum score - Minimum score

⇒ Range = 63 - 10

⇒ Range = 53

∴ The range of his scores is 53.

Given:

The volume of cuboid 12250 cm³

Radius of cylinder 35 cm

Formula used:

CSA of cylinder = 2πrh

Volume of cylinder = π r²h

Solution:

Volume of cuboid = volume of cylinder 

⇒ 12250 = π r²h

⇒ π h = 12250/1225 

⇒ π h = 10 cm

CSA of cylinder = 2 π rh

⇒ 2 × 10 × 35 = 700 cm²

∴ The curved surface area of the cylinder is 700 cm².