Mark has a box that is $3 \frac{1}{2}$ $3 1/2$ feet high, $4 \frac{2}{7}$ $4 2/7$ feet wide and 5 feet long. Jeff has a box that measures 4 feet along each edge. Whose box has the greater volume? By how much?

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Mark has a box that is $3 \frac{1}{2}$ $3 1/2$ feet high, $4 \frac{2}{7}$ $4 2/7$ feet wide and 5 feet long. Jeff has a box that measures 4 feet along each edge. Whose box has the greater volume? By how much?

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Answer 1 · 2018-03-15T01:03:59

Mark's box is larger by $11$ $11$ cubic feet.

Explanation:

Determining the volume of the square is relatively simple so we'll start with that:
$V_{J} = s^{3}$ $V_J = s^3$
$V_{J} = 4^{3}$ $V_J = 4^3$
$V_{J} = 64 f t^{3}$ $V_J = 64 ft^3$
To calculate the volume of a rectangular prism, we multiply length by width by height:
$V_{M} = l w h$ $V_M = lwh$
$V_{M} = 5 \cdot 4 \frac{2}{7} \cdot 3 \frac{1}{2}$ $V_M = 5*4 2/7*3 1/2$
$V_{M} = 5 \cdot \frac{30}{7} \cdot \frac{7}{2}$ $V_M = 5*30/7*7/2$
$V_{M} = \frac{1050}{14}$ $V_M = 1050/14$
$V_{M} = 75 f t^{3}$ $V_M = 75 ft^3$
Find the difference:
$V_{D} = V_{M} - V_{D}$ $V_D = V_M - V_D$
$V_{D} = 75 - 64$ $V_D = 75-64$
$V_{D} = 11 f t^{3}$ $V_D = 11 ft^3$
Therefore, Mark's box is larger by $11$ $11$ cubic feet.

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Mark has a box that is $3 \frac{1}{2}$ $3 1/2$ feet high, $4 \frac{2}{7}$ $4 2/7$ feet wide and 5 feet long. Jeff has a box that measures 4 feet along each edge. Whose box has the greater volume? By how much?

1 Answer

Explanation:

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Mark has a box that is $3 \frac{1}{2}$ $3 1/2$ feet high, $4 \frac{2}{7}$ $4 2/7$ feet wide and 5 feet long. Jeff has a box that measures 4 feet along each edge. Whose box has the greater volume? By how much?