A warehouse has n widgets to be packed in b boxes. Each

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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?

\(A. \ n−bx\)
\(B. \ n−bx+x\)
\(C. \ n−bx−x\)
\(D. \ nx−bx\)
\(E. \ n−x\dfrac{bx}{b−1}\)

The OA is the option _B_

Source: Veritas Prep

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by Scott@TargetTestPrep » Wed Jul 10, 2019 4:21 pm
M7MBA wrote: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?

\(A. \ n−bx\)
\(B. \ n−bx+x\)
\(C. \ n−bx−x\)
\(D. \ nx−bx\)
\(E. \ n−x\dfrac{bx}{b−1}\)

The OA is the option _B_

Source: Veritas Prep
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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