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A warehouse has n widgets to be packed in b boxes. Each box can hold x

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A warehouse has n widgets to be packed in b boxes. Each box can hold x  [#permalink]

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New post 16 Oct 2016, 04:17
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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 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. \(n-x\frac{(bx)}{(b-1)}\)



Cheers ! :punk


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A warehouse has n widgets to be packed in b boxes. Each box can hold x  [#permalink]

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New post Updated on: 17 Oct 2016, 05:37
Taking simple numbers for n,b,x
Assume Total 51(n) widgets to be packed. We have 6(b) boxes, of which 5(b-1) boxes can have 10(x) widgets inside it.

1 widget cannot be packed and has to be separately packed in a new box(b).

This can be got by the expression n(51) - b-1(5)*x(10) (Option B)

Corrected solution. Thanks for pointing out the error.
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Originally posted by pushpitkc on 16 Oct 2016, 05:21.
Last edited by pushpitkc on 17 Oct 2016, 05:37, edited 1 time in total.
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Re: A warehouse has n widgets to be packed in b boxes. Each box can hold x  [#permalink]

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New post 17 Oct 2016, 02:16
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pushpitkc wrote:
Taking simple numbers for n,b,x
Assume Total 51(n) widgets to be packed. We have 5(b) boxes and each box can have 10(x) widgets inside it.

1 widget cannot be packed and has to be separately packed in a new box. This can be got by the expression

n(51) - b(5)*x(10) (Option A)


You have missed something here. Question says out of b boxes, b-1 boxes are filled to their capacity whereas last box is not filled to its capacity.

So, if I say (b-1) boxes have x widgets. Total widgets in (b-1) boxes = (b-1) * x

We have total n widgets. So leftover widgets = n - (b-1)*x = n - bx + x. Hence, Answer B.
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Re: A warehouse has n widgets to be packed in b boxes. Each box can hold x  [#permalink]

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New post 17 Oct 2016, 03:22
XavierAlexander 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 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. \(n-x\frac{(bx)}{(b-1)}\)

Cheers ! :punk



Total Boxes = b
i.e. Boxes will to the full capacity = (b-1)
Number of widgets in completely filled boxes = (b-1)*x

i.e. Total Widgets, n = (b-1)*x + Remaining widgets

i.e. Remaining Widgets = n - (b-1)*x = n-bx +x

Answer: Option B
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A warehouse has n widgets to be packed in b boxes. Each box can hold x  [#permalink]

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New post 13 Feb 2017, 07:05
XavierAlexander 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 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. \(n-x\frac{(bx)}{(b-1)}\)

Cheers ! :punk



Official solution from Veritas Prep.

This word problem is essentially asking about the relationship between dividend, divisor, quotient, and remainder. The total number of widgets, \(n\), is the dividend (the number you start with and then divide). The number of widgets per box \((x)\) serves as the divisor (the number you're dividing by), and the number of FULL boxes will be the quotient. Since the last box will not be full with x widgets, that box is not part of the quotient, so the quotient will be \(b−1\) (the total number of boxes, minus the one box that isn't full). Then the remaining widgets, to go in the last box, will be the remainder.

Algebraically, you can set this up using the Quotient/Remainder relationship (or formula):

Dividend = Divisor * Quotient + Remainder

Since you know which variables correspond to each value other than remainder, you can plug those in:

\(n=(b−1)(x)+R\)

And then solve for \(R\):

\(R=n−(b−1)(x)\)

Since this form does not match any answer choices, distribute the multiplication:

\(R=n−(bx−x)\)

Which then becomes (as you distribute the negative across parentheses):

\(R=n−bx+x\)

, which is answer choice B. Alternatively since this problem involves variables in all answer choices, you can simply pick numbers to test the relationship. If you were to say that there were 91 widgets to start, being packed in 10 boxes of 10 widgets each. then your variables are:

\(n=91\);\(x=10\);\(b=10\)

And you know that that will give you 9 boxes of 10 with one remaining for the last box.

Plug those in to the answer choices and you will see that only answer choice B will give you the correct remainder of \(1\), so \(B\) must be correct.
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Re: A warehouse has n widgets to be packed in b boxes. Each box can hold x  [#permalink]

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Re: A warehouse has n widgets to be packed in b boxes. Each box can hold x &nbs [#permalink] 17 Sep 2018, 04:16
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