A cardboard box has a length of 3 feet, height of \(2\frac{\mathrm{1} }{\mathrm{2}}\) feet, and depth of 2 feet. If the length and depth are doubled, by what percent does the volume of the box change?
Get all 1000+ ASVAB exam-like questions with our mobile apps!
Detailed Explanation
The volume of the original box is 3 × \(2\frac{\mathrm{1} }{\mathrm{2}}\) × 2=15.
The volume of the box with the length and depth doubled is 6 × \(2\frac{\mathrm{1} }{\mathrm{2}}\) × 4=60.
The amount of change in volume is 60 – 15 = 45. The percent change is the amount of change in volume divided by the original volume: \(\frac{\mathrm{45} }{\mathrm{15}}\) × 100% = 300%.
The volume of the box with the length and depth doubled is 6 × \(2\frac{\mathrm{1} }{\mathrm{2}}\) × 4=60.
The amount of change in volume is 60 – 15 = 45. The percent change is the amount of change in volume divided by the original volume: \(\frac{\mathrm{45} }{\mathrm{15}}\) × 100% = 300%.
Take more free practice tests for other ASVAB topics with our ASVAB practice test now!