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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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

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

Source: Beat The GMAT — Problem Solving | Tue Jun 02, 2020 7:10 am

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

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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Scott@TargetTestPrep: GMAT Instructor; Posts: 8246; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

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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