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The ACME company manufactured x brooms per month from Januar [#permalink]
18 Dec 2012, 00:33

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Question Stats:

65% (03:12) correct
35% (01:47) wrong based on 142 sessions

The ACME company manufactured x brooms per month from January to April, inclusive. On the first of each month, during the following May to December, inclusive, it sold x/2 brooms. At the beginning of production on January 1st, the ACME company had no brooms in its inventory. If storage costs were $1 per month per broom, approximately how much, in terms of x, did the ACME company pay for storage from May 2nd to December 31st, inclusive?

Re: The ACME company manufactured x brooms per month from Januar [#permalink]
18 Dec 2012, 01:33

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Expert's post

dcastan2 wrote:

The ACME company manufactured x brooms per month from January to April, inclusive. On the first of each month, during the following May to December, inclusive, it sold x/2 brooms. At the beginning of production on January 1st, the ACME company had no brooms in its inventory. If storage costs were $1 per month per broom, approximately how much, in terms of x, did the ACME company pay for storage from May 2nd to December 31st, inclusive?

A. $x B. $3x C. $4x D. $5x E $14x

Pick some smart number for \(x\), let \(x=2\) (I chose \(x=2\) as in this case monthly shipments would be \(\frac{x}{2}=1\)).

From January to April, inclusive \(4x=8\) brooms were produced and in May the company paid for storage of 8-1=7 brooms, in next month for storage of 6 and so on.

So the total storage cost would be: \(1*(7+6+5+4+3+2+1+0)=28\) --> as \(x=2\), then \(28=14x\).

Re: The ACME company manufactured x brooms per month from Januar [#permalink]
21 Dec 2012, 02:50

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Expert's post

skamal7 wrote:

In this case if x=4 then how the equation will turn out to be can you please explain

Sure. Say \(x=4\), in this case monthly shipments would be \(\frac{x}{2}=2\). Then from January to April, inclusive \(4x=16\) brooms were produced and in May the company paid for storage of 16-2=14 brooms, in next month for storage of 12 and so on.

So the total storage cost would be: \(1*(14+12+10+8+6+4+2+0)=56\) --> as \(x=4\), then \(56=14x\).

Re: The ACME company manufactured x brooms per month from Januar [#permalink]
23 Nov 2013, 05:29

Bunuel wrote:

dcastan2 wrote:

The ACME company manufactured x brooms per month from January to April, inclusive. On the first of each month, during the following May to December, inclusive, it sold x/2 brooms. At the beginning of production on January 1st, the ACME company had no brooms in its inventory. If storage costs were $1 per month per broom, approximately how much, in terms of x, did the ACME company pay for storage from May 2nd to December 31st, inclusive?

A. $x B. $3x C. $4x D. $5x E $14x

Pick some smart number for \(x\), let \(x=2\) (I chose \(x=2\) as in this case monthly shipments would be \(\frac{x}{2}=1\)).

From January to April, inclusive \(4x=8\) brooms were produced and in May the company paid for storage of 8-1=7 brooms, in next month for storage of 6 and so on.

So the total storage cost would be: \(1*(7+6+5+4+3+2+1+0)=28\) --> as \(x=2\), then \(28=14x\).

Re: The ACME company manufactured x brooms per month from Januar [#permalink]
23 Nov 2013, 10:14

ronr34 wrote:

Can you please post an algebraic solution to this problem? I solved it plugging numbers, but I can't seem to do so algebraically. Thanks

Hello ron

We are given that on 1st of each month x brooms are made from Jan to Apr. that gives us \(4x\) brooms in inventory by end of April. On first of each month from May to December \(\frac{x}{2}\) brooms are sold. Therefore we will have

\(4x-\frac{x}{2} = \frac{7x}{2}\) brooms on May 2nd ---> storage cost is $1 per broom per month so in may the company pays \(\frac{7x}{2}\) Similarly we have \(\frac{7x}{2}-\frac{x}{2} = \frac{6x}{2}\) brooms on June 2 ---> storage cost is \(\frac{6x}{2}\). July 2nd \(\frac{5x}{2}\) Continuing this we have \(\frac{x}{2}\) brooms left by Nov 2nd which are sold on Dec 1st, so no more brooms are left on Dec 2nd and no storage costs in december. By adding storage costs as derived above we get

Re: The ACME company manufactured x brooms per month from Januar [#permalink]
23 Nov 2013, 12:15

gmatprav wrote:

ronr34 wrote:

Can you please post an algebraic solution to this problem? I solved it plugging numbers, but I can't seem to do so algebraically. Thanks

Hello ron

We are given that on 1st of each month x brooms are made from Jan to Apr. that gives us \(4x\) brooms in inventory by end of April. On first of each month from May to December \(\frac{x}{2}\) brooms are sold. Therefore we will have

\(4x-\frac{x}{2} = \frac{7x}{2}\) brooms on May 2nd ---> storage cost is $1 per broom per month so in may the company pays \(\frac{7x}{2}\) Similarly we have \(\frac{7x}{2}-\frac{x}{2} = \frac{6x}{2}\) brooms on June 2 ---> storage cost is \(\frac{6x}{2}\). July 2nd \(\frac{5x}{2}\) Continuing this we have \(\frac{x}{2}\) brooms left by Nov 2nd which are sold on Dec 1st, so no more brooms are left on Dec 2nd and no storage costs in december. By adding storage costs as derived above we get

Re: The ACME company manufactured x brooms per month from Januar [#permalink]
24 Nov 2013, 05:58

ronr34 wrote:

gmatprav wrote:

ronr34 wrote:

Can you please post an algebraic solution to this problem? I solved it plugging numbers, but I can't seem to do so algebraically.

Hello ron

We are given that on 1st of each month x brooms are made from Jan to Apr. that gives us \(4x\) brooms in inventory by end of April. On first of each month from May to December \(\frac{x}{2}\) brooms are sold. Therefore we will have

\(4x-\frac{x}{2} = \frac{7x}{2}\) brooms on May 2nd ---> storage cost is $1 per broom per month so in may the company pays \(\frac{7x}{2}\) Similarly we have \(\frac{7x}{2}-\frac{x}{2} = \frac{6x}{2}\) brooms on June 2 ---> storage cost is \(\frac{6x}{2}\). July 2nd \(\frac{5x}{2}\) Continuing this we have \(\frac{x}{2}\) brooms left by Nov 2nd which are sold on Dec 1st, so no more brooms are left on Dec 2nd and no storage costs in december. By adding storage costs as derived above we get

I was able to do this calculation but I am looking for a general formula for cases like this.... Is there anything of the sort?

This problem is not a generic problem that warrants a formula. If you solved it like this then you are on right track. You mentioned how to do it without plugging in numbers Note that I did not plug in numbers. We can create a formula for similar problems, but in the end it will be harder to remember the formula than to solve it directly. _________________

Re: The ACME company manufactured x brooms per month from Januar [#permalink]
03 Apr 2014, 23:49

gmatprav wrote:

ronr34 wrote:

Can you please post an algebraic solution to this problem? I solved it plugging numbers, but I can't seem to do so algebraically. Thanks

Hello ron

We are given that on 1st of each month x brooms are made from Jan to Apr. that gives us \(4x\) brooms in inventory by end of April. On first of each month from May to December \(\frac{x}{2}\) brooms are sold. Therefore we will have

\(4x-\frac{x}{2} = \frac{7x}{2}\) brooms on May 2nd ---> storage cost is $1 per broom per month so in may the company pays \(\frac{7x}{2}\) Similarly we have \(\frac{7x}{2}-\frac{x}{2} = \frac{6x}{2}\) brooms on June 2 ---> storage cost is \(\frac{6x}{2}\). July 2nd \(\frac{5x}{2}\) Continuing this we have \(\frac{x}{2}\) brooms left by Nov 2nd which are sold on Dec 1st, so no more brooms are left on Dec 2nd and no storage costs in december. By adding storage costs as derived above we get

Re: The ACME company manufactured x brooms per month from Januar [#permalink]
17 Jul 2015, 06:56

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