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A warehouse has \(n\) widgets to be packed in \(b\) boxes. Each box can hold \(x\) widgets. However, \(n\) is not evenly

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by Vincen » Tue Jun 02, 2020 7:10 am

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A warehouse has \(n\) widgets to be packed in \(b\) boxes. Each box can hold \(x\) widgets. However, \(n\) is not evenly divisible by \(x,\) so one of the boxes will contain fewer than \(x\) widgets. Which of the following expresses the number of widgets in that box, assuming all other boxes are filled to their capacity of \(x\) widgets?

A. \(n−bx\)

B. \(n−bx+x\)

C. \(n−bx−x\)

D. \(nx−bx\)

E. \(n−x\dfrac{bx}{b−1}\)

[spoiler]OA=B[/spoiler]

Source: Veritas Prep
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Re: A warehouse has \(n\) widgets to be packed in \(b\) boxes. Each box can hold \(x\) widgets. However, \(n\) is not ev

by Jay@ManhattanReview » Wed Jun 03, 2020 11:07 pm
Vincen wrote: ↑
Tue Jun 02, 2020 7:10 am
 A warehouse has \(n\) widgets to be packed in \(b\) boxes. Each box can hold \(x\) widgets. However, \(n\) is not evenly divisible by \(x,\) so one of the boxes will contain fewer than \(x\) widgets. Which of the following expresses the number of widgets in that box, assuming all other boxes are filled to their capacity of \(x\) widgets?

A. \(n−bx\)

B. \(n−bx+x\)

C. \(n−bx−x\)

D. \(nx−bx\)

E. \(n−x\dfrac{bx}{b−1}\)

[spoiler]OA=B[/spoiler]

Source: Veritas Prep
Given the information, we get that only \(b-1\) boxes would each have \(x\) widgets. Thus, \(b-1\) boxes would have \((b-1)x\) widgets.

No. of widgets left to be placed in the bth or the last box = \(n-(b-1)x=n-bx+x\) widgets.

The correct answer: B

Hope this helps!

-Jay
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Re: A warehouse has \(n\) widgets to be packed in \(b\) boxes. Each box can hold \(x\) widgets. However, \(n\) is not ev

by Scott@TargetTestPrep » Sat Jun 06, 2020 5:47 am
Vincen wrote: ↑
Tue Jun 02, 2020 7:10 am
 A warehouse has \(n\) widgets to be packed in \(b\) boxes. Each box can hold \(x\) widgets. However, \(n\) is not evenly divisible by \(x,\) so one of the boxes will contain fewer than \(x\) widgets. Which of the following expresses the number of widgets in that box, assuming all other boxes are filled to their capacity of \(x\) widgets?

A. \(n−bx\)

B. \(n−bx+x\)

C. \(n−bx−x\)

D. \(nx−bx\)

E. \(n−x\dfrac{bx}{b−1}\)

[spoiler]OA=B[/spoiler]

Solution:

We see that b - 1 boxes will be filled with x widgets each, and they together hold x(b - 1) = bx - x widgets. Therefore, the last box, i.e., the partially filled box, will hold n - (bx - x) = n - bx + x widgets.

Answer: B

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