Question

A Sharbat maker prepares Sharbat in a cylindrical vessel of diameter and height equal to 54 cm, up to its brim. He sells the sharbat in small cylindrical glasses of radius 3 cm and height 9 cm. Find the total money earned by the sharbat maker at the end of the day if he charges Rs.15 for each glass.
1. Rs. 9,160
2. Rs. 9,060
3. Rs. 7,290
4. Rs. 7,920

Answer 1

Correct Answer - Option 3 : Rs. 7,290

Given:

Diameter of the cylindrical vessel = 54 cm

Height of the cylindrical vessel = 54 cm

Radius of the cylindrical glass = 3 cm

Height of the cylindrical glass = 9 cm

Selling Price of one glass of Sharbat = Rs. 15

Formula Used:

Volume of the Cylinder = πr²h

where, 

r = Radius of the cylinder

h = Height of the cylinder

Calculation:

Let the radius and height of the cylindrical vessel be 'R' and 'H' respectively.

And let the radius and height of the cylindrical glass be 'r' and 'h' respectively.

According to the question,

R = Diameter/2

⇒ 54/2

⇒ 27 cm

Now, Number of glasses of Sharbat sold in a day = (Volume of the cylindrical vessel)/(Volume of the cylindrical glass)

\(⇒ \frac{πR^2H}{πr^2h}\)

\(⇒ \frac{R^2H}{r^2h}\)

\(⇒ \frac{[(27)^2 × 54]}{[3^2 × 9]} \)

⇒ 9 × 9 × 6

⇒ 81 × 6

⇒ 486 glasses

So, Total money earned by the Sharbat maker at the end of the day = Number of glasses × Selling price of one glass of Sharbat

⇒ 486 × 15

⇒ 7,290

∴ The total money earned by the Sharbat maker at the end of the day Rs. 7,290.