Question

Box A can hold 10 marbles. Box B is double the height of Box A, four times the length of Box A, and half the width of box A. How many marbles can Box B hold?

Answer 1

40 marbles.

Explanation:

Given:

Box A can hold 10 marbles.

Box B is double the height of Box A, four times the length of Box A, and half the width of box A.

Question asked:

How many marbles can Box B hold?

Solution:

Let volume of box A =
`V_(a)`

Let volume of box B =
`V_(b)`

Now, suppose:

Length of box A =
`l`

Breadth =
`b`

Height =
`h`

So, volume of box A,
`V_(a)`
`=lbh`

As given, Box B is double the height of Box A, four times the length of Box A, and half the width of box A.

So, Length of box B =
`4l`

Breadth =
`(1)/(2) b`

Height =
`2h`

Volume of box B,
`V_(b)` =
`4l*(1)/(2) b*2h=4lbh`

By dividing volume of box A with volume of box B:

`(V_(a) )/(V_(b)) =(lbh)/(4lbh) \\\\ (V_(a) )/(V_(b))=(1)/(4) \\\ (lbh \ canceled\ by\ lbh)`

By cross multiplication:

`V_(b) =4V_(a)`

That means volume of box B is 4 times volume of box A;

∵ Box A can hold = 10 marbles

∴ Box B ( 4 times of box A ) can hold =
`10*4=40\ marbles`

Therefore, Box B can hold 40 marbles.