Given data:

Circumference of cylinder = 62.8 cm

Volume of cylinder = 8792 cm³

Concept used:

We would use the relationship between the circumference of the cylinder, volume, and radius to find the value of the curved surface area. 

Formula used:

C = 2 × π × r

V = π × r2 × h

Where "C" is the circumference of the circular base of the cylinder,  "r" is the radius, "V" is the volume of cylinder, and "h" is the Height.

Calculation:

From circumference of circular base formula-

The curved surface area of the cylinder is 1758.4 cm².

Alternate Method

Given:

The lengths of a pair of parallel sides of a trapezium = 18 cm and 15 cm

The perpendicular distance between these two sides = 12 cm

Formula used:

Area of trapezium = 1/2 × sum of its parallel sides × perpendicular distance between parallel sides

Calculation:

According to the question

Area of trapezium = 1/2 × sum of its parallel sides × perpendicular distance between parallel sides

⇒ 1/2 × (18 + 15) × 12 cm²

⇒ (1/2 × 33 × 12) cm²

⇒ (33 × 6) cm²

⇒ 198 cm²

∴ The area of the trapezium is 198 cm²

Given:

A person travels from P to Q at a speed of 50 km/h and returns by increasing his speed by 60%.

Concept used:

Average speed = 2S₁S₂/(S₁ + S₂)

Here S1, S2 are the speed

Calculation:

Speed at time of returning = 50 × 160%

⇒ 80 km/h

Average speed = (2 × 50 × 80)/(50 + 80)

⇒ 8000/130

⇒ 61.53

∴ The average speed for both trips is 61.53 km/h.

Concept used:

BODMAS consists of the first letter of the operations involved in a mathematical equation.

B- Bracket (), [], {}

O- Order, Square root, exponents, and powers

D- Division, ÷ /

M- Multiplication × *

A – Addition +

S – Subtraction –

Calculation:

Change the mixed fractions into fractions

Given:

Principal (P) = Rs. 12000, rate = 30% per annum, time = 1 year.

Compounded half-yearly.

Concept used:

When compound interest (C.I.) is compounded half yearly its rate becomes half and time becomes twice.

Amount (A) = P (1 + r / 100)^t

Solution:

As C.I. is compounded half yearly rate becomes 15% and time becomes two years.

Putting P = Rs. 12000, r = 15% and t = 2 years.

⇒ A = 12000 (1 + 15 / 100)²

⇒ A = 12000 × (23 / 20)²

⇒ A = 12000 × (23 / 20) × (23 / 20)

⇒ A = 15870.

∴ Amount to be paid after 1 year is Rs. 15870.

Given:

Ratio of number of coins:

25 paise:50 paise:Rs. 2:Rs. 5 = 5:4:3:1

Total amount of coins = Rs. 285 = 28500 paise

Solution:

Assume the common ratio to be x.

The number of 25 paise, 50 paise, Rs. 2, and Rs. 5 coins can be expressed,

⇒ 5x, 4x, 3x, and x respectively.

⇒ The total value in paise becomes (5x25) for 25 paise coins,

⇒ (4x50) for 50 paise coins,

⇒ (3x200) for Rs. 2 coins, and

⇒ (x500) for Rs. 5 coins.

Summing up all values, 

⇒ 125x + 200x + 600x + 500x = 1425x = Total amount in paise = 28500

⇒ x = 28500 / 1425 = 20

⇒ 5x = 100 and Rs. 5 coins = x = 20

The difference between the 25 paise and Rs. 5 coins

⇒ 100 - 20 = 80

∴ The difference between the number of 25 paise and Rs. 5 coins is 80.

Given:

The ratio of time taken in completing a work by a man and woman is 2 ∶ 3.

The man and the woman can complete the work together in 24 days.

Concepts used:

Time = Work/Efficiency

Time is inversely proportional to efficiency.

Calculation:

Time Ratio = 2 : 3

Efficiency Ratio = 3 : 2

Total work = (3  + 2)24

Time take by women = (5 × 24)/2 = 60

∴ A woman alone can complete the work in 60 days.

Given :

1) Alloy 1 = metal 1 : metal 2 = 1 : 3

2) Alloy 2 = metal 1 : metal 2 = 2 : 3

3) Alloy 3 = metal 1 : metal 2 = 4 : 11

4) All alloys are in equal quantity

Calculations :

Let quantity of each alloy be X

After mixing all the alloys quantity of metal 1 is mentioned below

⇒ ((1/4) + (2/5) + (4/15)) × a

LCM of 4, 5 and 15 is 60

Proportion of metal 1 in new mixture = ((1 × 15) + (2 × 12) + (4 × 4) ) / 60

⇒ 55 / 60

Now, after mixing all the alloys quantity of metal 2 is mentioned below

⇒ ((3/4) + (3/5) + (11/15)) × a

LCM of 4, 5 and 15 is 60

Proportion of metal  in new mixture = ((15 × 3) + (3 × 12) + (11 × 4))/60

⇒ 125/60

Now ratio of metal 1 and metal 2 in new mixture = 55 ∶ 125 = 11 ∶ 25

∴ The required ratio is 11 ∶ 25.

Given:

Sum to divided = Rs. 3900

R receives two-fifths of P's share and Q receives three times R's share.

Concept used:

Ratio Proportion and Partnership

Calculation:

Let the share of P is Rs. M.

Thus, the share of R and Q are Rs. (2M/5) and Rs. (6M/5) respectively.

Ratio between three of them = M : (2M/5) : (6M/5) = 5 : 2 : 6

Hence, Q's share = 6/13 × 3900 = 1800

∴ Q's share is Rs. 1800.