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The ACME company manufactured x brooms per month from Januar

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The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post Updated on: 10 Dec 2017, 09:51
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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

Originally posted by dcastan2 on 18 Dec 2012, 01:33.
Last edited by Bunuel on 10 Dec 2017, 09:51, edited 2 times in total.
Renamed the topic.
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 18 Dec 2012, 02:33
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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 \(28=14x\).

Answer: E.

Identical question from GMAT Prep to practice: a-certain-business-produced-x-rakes-each-month-form-november-101738.html
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 23 Nov 2013, 11:14
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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

\(\frac{7x}{2}+\frac{6x}{2}+\frac{5x}{2}+\frac{4x}{2}+\frac{3x}{2}+\frac{2x}{2}+\frac{x}{2} = \frac{28x}{2} = 14x\)

Hope this helps!
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 20 Dec 2012, 11:04
In this case if x=4 then how the equation will turn out to be can you please explain
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 20 Dec 2012, 11:36
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January - April : 4 months , so 4x brooms produced.

From May its an A.P. with
a = 3.5x (since we are counting from May 2nd, so 0.5x has already been sold on May 1st)
d = - 0.5 x

So 8th Term (December ) is - a + 7d = 0

Sum of AP is (first term + last term) * n/2

= 3.5x * 8 / 2 = 14x

Hence E
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 21 Dec 2012, 03:50
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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 \(56=14x\).

Answer: E.
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 21 Dec 2012, 04:12
your reply helped me a lot in understanding this question.
Are there any 700 + level questions on geomenrty??
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 23 Nov 2013, 06:29
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 \(28=14x\).

Answer: E.

Identical question from GMAT Prep to practice: a-certain-business-produced-x-rakes-each-month-form-november-101738.html

Hi Bunuel,
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
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 23 Nov 2013, 13:15
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

\(\frac{7x}{2}+\frac{6x}{2}+\frac{5x}{2}+\frac{4x}{2}+\frac{3x}{2}+\frac{2x}{2}+\frac{x}{2} = \frac{28x}{2} = 14x\)

Hope this helps!

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?
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 24 Nov 2013, 06:58
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

\(\frac{7x}{2}+\frac{6x}{2}+\frac{5x}{2}+\frac{4x}{2}+\frac{3x}{2}+\frac{2x}{2}+\frac{x}{2} = \frac{28x}{2} = 14x\)

Hope this helps!

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.
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 04 Apr 2014, 00:49
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

\(\frac{7x}{2}+\frac{6x}{2}+\frac{5x}{2}+\frac{4x}{2}+\frac{3x}{2}+\frac{2x}{2}+\frac{x}{2} = \frac{28x}{2} = 14x\)

Hope this helps!



Did in the same way ; with just a addition

Wrote \(4x = \frac{8x}{2}\) for the simplicity of calculation & proceeded

As denominator is same, just add the numerator & then divide by 2

Addition of 1 to 7; used formula \(\frac{n(n+1)}{2} = 7 * \frac{8}{2} = 28\)

Dividing by 2

28/2 = 14
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 16 Nov 2016, 09:54
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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


first, let's see how many x were manufactured.
January X
February X
March X
April X
total manufactured: 4x
May 1st sold 0.5x = remained till end of the month 3.5x - paid 3.5x for storage
June 1st sold 0.5x = remained till end of the month 3x - paid 3x for storage
July 1st sold 0.5x = remained till end of the month 2.5x - paid 2.5x for storage
August 1st sold 0.5x = remained till end of the month 2x - paid 2x for storage
September 1st sold 0.5x = remained till end of the month 1.5x - paid 1.5x for storage
October 1st sold 0.5x = remained till end of the month 1x - paid 1x for storage
November 1st sold 0.5x = remained till end of the month 0.5x - paid 0.5x for storage
December 1st sold 0.5x = remained till end of the month 0x - paid 0.

so total paid:
3.5x + 3x + 2.5x + 2x + 1.5x + 1x + 0.5x
3.5x + 2.5x = 6x
3x + 2x + 1x = 6x
1.5x + 0.5x = 2x

6x+6x+2x=14x

answer is E.
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Re: The ACME company manufactured x brooms per month from Januar  [#permalink]

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New post 04 Sep 2019, 03:39
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Best way to do is not calculating, see the differences in the answer values.
At may itself the cost would be 4x, so it makes sense the answer will be higher than 5x, so 14 only
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Re: The ACME company manufactured x brooms per month from Januar   [#permalink] 04 Sep 2019, 03:39
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