Respuesta :
Answer:
40 marbles.
Step-by-step 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 = [tex]V_{a}[/tex]
Let volume of box B = [tex]V_{b}[/tex]
Now, suppose:
Length of box A = [tex]l[/tex]
Breadth = [tex]b[/tex]
Height = [tex]h[/tex]
So, volume of box A, [tex]V_{a}[/tex] [tex]=lbh[/tex]
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 = [tex]4l[/tex]
Breadth = [tex]\frac{1}{2} b[/tex]
Height = [tex]2h[/tex]
Volume of box B, [tex]V_{b}[/tex] = [tex]4l\times\frac{1}{2} b\times2h=4lbh[/tex]
By dividing volume of box A with volume of box B:
[tex]\frac{V_{a} }{V_{b}} =\frac{lbh}{4lbh} \\\\ \frac{V_{a} }{V_{b}}=\frac{1}{4} \\\ (lbh \ canceled\ by\ lbh)[/tex]
By cross multiplication:
[tex]V_{b} =4V_{a}[/tex]
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 = [tex]10\times4=40\ marbles[/tex]
Therefore, Box B can hold 40 marbles.
