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Volume

Expert replies
by Deepthi Subbu » Wed Jan 19, 2011 12:57 am
3 spherical chocolates are packed in a straight line in a rectangular box that encloses them flush on all sides. If the volume of each of the chocolates is increased by 50% and they are placed in a similar fashion in a new box, what is the ratio of the volume of the new box to that of the original box?

I dont have the answer choice to this question.
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Source: — Problem Solving |

by Anurag@Gurome » Wed Jan 19, 2011 1:16 am
Deepthi Subbu wrote:3 spherical chocolates are packed in a straight line in a rectangular box that encloses them flush on all sides. If the volume of each of the chocolates is increased by 50% and they are placed in a similar fashion in a new box, what is the ratio of the volume of the new box to that of the original box?
If you view the chocolates in the box from above, it'll look like as the following figure
Image

Now, say the radius of each chocolate = r
Hence, height and width of the box = 2r
And, length of the box = 6r
Thus, volume of the box = (2r)*(2r)*(6r) = 24r³

Now, volume of each of the chocolates = (4/3)πr³

If volume of each chocolate is increased by 50%, the new volume will be (3/2)*(4/3)πr³ = (4/3)πR³, where R is the new radius of each chocolate. Hence, R³ = (3/2)r³

Ratio of the volume of the new box to that of the original box = (24R³)/(24r³) = R³/r³ = 3/2
Last edited by Anurag@Gurome on Wed Jan 19, 2011 1:50 am, edited 1 time in total.
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by Deepthi Subbu » Wed Jan 19, 2011 1:43 am
Anurag@Gurome wrote:
Deepthi Subbu wrote:3 spherical chocolates are packed in a straight line in a rectangular box that encloses them flush on all sides. If the volume of each of the chocolates is increased by 50% and they are placed in a similar fashion in a new box, what is the ratio of the volume of the new box to that of the original box?
If you view the chocolates in the box from above, it'll look like as the following figure
Image

Now, say the radius of each chocolate = r
Hence, height and width of the box = 2r
And, length of the box = 6r
Thus, volume of the box = (r)*(2r)*(6r) = 12r³

Now, volume of each of the chocolates = (4/3)πr³

If volume of each chocolate is increased by 50%, the new volume will be (3/2)*(4/3)πr³ = (4/3)πR³, where R is the new radius of each chocolate. Hence, R³ = (3/2)r³

Ratio of the volume of the new box to that of the original box = (12R³)/(12r³) = R³/r³ = 3/2
How do you arrive at the height , width and the length of the box to be 2r and 6r?
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by Anurag@Gurome » Wed Jan 19, 2011 1:53 am
Deepthi Subbu wrote:How do you arrive at the height , width and the length of the box to be 2r and 6r?
Image

Refer to the figure above.
Consider it as a upper view of the chocolates in the box.

Length of the box will be equal to sum of the diameters of the three chocolates. Hence, l = 3*diameter = 3*2*radius = 6r

Similarly, width of the box = diameter of one chocolate = 2r

Now side view of the chocolates in the box will be same as the above figure with width replaced by height. Hence, height of the box = 2r
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