Question

Two containers designed to hold water are side by side, both in the shape of a cylinder. Container A has a diameter of 6 feet and a height of 18 feet. Container B has a diameter of 8 feet and a height of 17 feet. Container A is full of water and the water is pumped into Container B until Container A is empty. To the nearest tenth, what is the percent of Container B that is empty after the pumping is complete?​

Answer 1

The percent of container B that is left after pumping the water of Container A full in it is 40 per cent .

Step by step explanation :

Given -

• Two containers ( Container A & Container B ) in shape of cylinders.
• Diameter of Container A = 6 feet
• Height of Container A = 18 feet
• Diameter of container B = 8 feet
• Height of Container B = 17 feet

To find -

The percent of container B that is left after pumping the water of Container A full in it.

Solution -

Firstly, we have to find how much water each container can hold i.e. volume of containers .

We know that -

$\boxed{ \mathfrak \purple{volume \: of \:cylinder = \pi {r}^{2} h} }$

where, r is the radius of the cylinder & h is the height of the cylinder.

Now,

For Container A

Given, diameter = 6 feet

.•. Radius = $\frac{6}{2}$

$= 3 \: \mathfrak{feet}$

Height = 18 feet

•.• Volume of Container A = $\frac{22}{7} \times {3}^{2} \times 18$

$= \frac{22}{7} \times 3 \times 3 \times 18$

$= \frac{3564}{7}$

$= 509.142857 \:{ft.}^{3} \mathfrak{(approximately)}$

Rounding off to the nearest tenth . ..

$=> 510 \: \mathfrak{ {feet}^{3} }$

For Container B

Given, Diameter = 8 feet

.•. Radius = $\frac{8}{2}$

$= 4 \: \mathfrak{feet}$

Height = 17 feet

•.• Volume of Container B = $\frac{22}{7} \times {4}^{2} \times 17$

$= \frac{22}{7} \times 4 \times 4 \times 17$

$= \frac{5984}{7}$

$= 854.857143 \: {ft.}^{3} \mathfrak{(approximately)}$

Rounding off to the nearest tenth. ..

$=> 850 \: \mathfrak{ {feet}^{3} }$

Now,

Volume container B that is left after pumping the water of Container A in it = Volume of Container B - Volume of Container A

= ( 850 - 510 ) ft^3

= 340 ft^3

Now ,

The percent of container B that is left after pumping the water of Container A in it = Volume container B that is left after pumping the water of Container A /Total volume of Container B × 100

$= \frac{340}{850} \times 100$

$= \boxed{ 40 \: \%}$