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A certain manufacturer increased its gross profit on a

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A certain manufacturer increased its gross profit on a product from 10 percent of the cost of the product to 15 percent of the cost by changing the selling price. If the new selling price was $92.00 and the cost of the product remained the same, what was the old selling price?

A. $77.40
B. $80.00
C. $83.64
D. $87.40
E. $88.00

Last edited by Bunuel on 27 Nov 2012, 03:20, edited 1 time in total.
Renamed the topic and edited the tags.

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Re: A certain manufacturer increased its gross profit on a [#permalink]

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New post 27 Nov 2012, 03:19
saxenarahul021 wrote:
A certain manufacturer increased its gross profit on a product from 10 percent of the cost of the product to 15 percent of the cost by changing the selling price. If the new selling price was $92.00 and the cost of the product remained the same, what was the old selling price?

A. $77.40
B. $80.00
C. $83.64
D. $87.40
E. $88.00


Given that {cost of the product}*1.15=$92 --> {cost of the product}=$80.

The old price was $80*1.1=$88.

Answer: E.
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Re: A certain manufacturer increased its gross profit on a [#permalink]

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A certain manufacturer increased its gross profit on a [#permalink]

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New post 08 Sep 2017, 09:14
I am not sure I am following this question (or rather the solution that Bunuel responded with). Here is my thought:


Selling Price - Cost Price = Profit

Let CP = x
Old Prof = \(\frac{110}{100}\)(x)
New Prof =\(\frac{115}{100}\)(x)

Old SP = Old Prof + CP [CP is the same in both conditions as given in the prompt]
=> Old SP = \(\frac{110}{100}\)(x) + x
=> Old SP = \(\frac{210}{100}\)(x)

Also, given

New SP = 92

New SP = New Prof + CP [CP is the same in both conditions as given in the prompt]

=> 92 = \(\frac{115}{100}\)(x) + x

Solving,

x = \(\frac{92*100}{215}\)

So, the CP = \(\frac{92*100}{215}\)

Now, plug this back into the Old SP equation above.

Old SP = \(\frac{92*100}{215}\) * \(\frac{210}{100}\)

This yields a value of 89.86. Where am I going wrong? This, of course, is a loooong-winded math.

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A certain manufacturer increased its gross profit on a [#permalink]

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New post 08 Sep 2017, 14:34
saxenarahul021 wrote:
A certain manufacturer increased its gross profit on a product from 10 percent of the cost of the product to 15 percent of the cost by changing the selling price. If the new selling price was $92.00 and the cost of the product remained the same, what was the old selling price?

A. $77.40
B. $80.00
C. $83.64
D. $87.40
E. $88.00

Work backwards. Find cost price, which has not changed, from NEW Sell Price. Then take cost price * 1.10 to get old sell price.

1) COST PRICE?

NEW Sell Price is 15% of (more than) cost price, meaning, cost price plus 15% of cost price

(.15CP as a sell price won't work -- he would be selling at a price 85 percent lower than CP. Huge loss. No gross profit.)

NEW Sell Price = 1.15CP = $92

$92 = 1.15CP

\(\frac{92}{1.15}\) = $80

CP = $80

2) OLD Selling Price? Use CP. Old sell price was 10 percent "of" (more than) CP.

Old SP = 1.10CP

We now have value for CP, which is 80

Old SP is thus 1.1 * 80 = $88
Old SP = $88

ANSWER E

Bunuel , given that you posted first and announced the answer, perhaps this one ought to have an official answer? :-) Not sure.

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A certain manufacturer increased its gross profit on a [#permalink]

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New post 08 Sep 2017, 14:43
Blackbox wrote:
I am not sure I am following this question (or rather the solution that Bunuel responded with). Here is my thought:

Selling Price - Cost Price = Profit

Let CP = x
Old Prof = \(\frac{110}{100}\)(x)
New Prof =\(\frac{115}{100}\)(x)

Old SP = Old Prof + CP [CP is the same in both conditions as given in the prompt]
=> Old SP = \(\frac{110}{100}\)(x) + x
=> Old SP = \(\frac{210}{100}\)(x)

Also, given

New SP = 92

New SP = New Prof + CP [CP is the same in both conditions as given in the prompt]

=> 92 = \(\frac{115}{100}\)(x) + x

Solving,

x = \(\frac{92*100}{215}\)

So, the CP = \(\frac{92*100}{215}\)

Now, plug this back into the Old SP equation above.

Old SP = \(\frac{92*100}{215}\) * \(\frac{210}{100}\)

This yields a value of 89.86. Where am I going wrong? This, of course, is a loooong-winded math.

Blackbox, I think you added in CP twice. See, e.g., partial quote of your post below.

You define CP as x, but then also define old SP, it seems, as

\(\frac{110}{100}\)(x) + x =
1.10x + x

Your original "100" base of the percentage of CP gets used twice, once invisibly in the second term, x. Given your direction, I think your equation should have been:

\(\frac{10}{100}\)(x) + \(\frac{100}{100}\)(x) =

\(\frac{110}{100}\)(x) = Old SP

Old SP is not \(\frac{210}{100}\)(x), or 2.10x

Old SP is 1.10x. See my solution above.

Similarly, you're right on track when you define new SP as \(\frac{115}{100}\)(x). So you could just write

\(\frac{115}{100}\)(x) = 92, then solve. You would get cost price of 80.

I didn't run your math without the "extra" CP. But that's at least one issue. Easy mistake to make. Hope that helps. :-)
Quote:
Old SP = Old Prof + CP [CP is the same in both conditions as given in the prompt]
=> Old SP = \(\frac{110}{100}\)(x) + x
=> Old SP = \(\frac{210}{100}\)(x)

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A certain manufacturer increased its gross profit on a [#permalink]

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New post 08 Sep 2017, 15:36
Dear genxer123,

Thank you for responding. My calculations are based on my assumption Selling Price - Cost Price = Profit. Is this formula incorrect? What I understood from the prompt is that the Gross Profit was affected as per this condition -- "... increased its gross profit on a product from 10 percent of the cost of the product ...". What I do not understand is why that condition is equated to Selling Price. For e.g., NEW Sell Price = 1.15CP = $92. I am clearly confused :?

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A certain manufacturer increased its gross profit on a [#permalink]

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New post 08 Sep 2017, 21:17
Blackbox wrote:
Dear genxer123,

Thank you for responding. My calculations are based on my assumption Selling Price - Cost Price = Profit. Is this formula incorrect? What I understood from the prompt is that the Gross Profit was affected as per this condition -- "... increased its gross profit on a product from 10 percent of the cost of the product ...". What I do not understand is why that condition is equated to Selling Price. For e.g., NEW Sell Price = 1.15CP = $92. I am clearly confused :?

Dear Blackbox ,

[EDITED TO DELETE DISCUSSION OF "CONDITION." I don't think that is the problem, because your own formula, with the PROFIT recalculated, yielded the answer.]

Of course the formula is correct! I've seen your posts, including the recent one on that horrible problem about a gasoline price increase (and a guy, Ron, who could afford part of the increase, such that we had to calculate by what percentage he must decrease consumption). You nailed it. I solved it. Barely. In FOUR minutes. I got stuck.

So I think you are just a little stuck on this one. You are not calculating profit correctly in the equation "Selling Price - Cost Price = Profit"!

You are forgetting to subtract 100 percent of the cost price from the equation!

I would PM you this whole thing, but math symbols and formulas are not translated any longer on PMs.
Incorrect figures are in red. Strike through will not work on fractions.
Correct figures are in blue.
I beg you: at some point, use multipliers for percentages? :dazen

Selling Price - Cost Price = Profit

Let CP = x
Old Prof = \(\frac{110}{100}\)(x)
New Prof =\(\frac{115}{100}\)(x)

Old Prof = \(\frac{110}{100}\)(x) - 1x

Old Prof = \(\frac{110}{100}\) (x) - \(\frac{100}{100}\) (x) = \(\frac{10}{100}\)(x)

New Prof = \(\frac{115}{100}\)(x) - 1x

New Prof = \(\frac{115}{100}(x) - \frac{100}{100}(x) = \frac{15}{100}(x)\)


Old SP = Old Prof + CP [CP is the same in both conditions as given in the prompt]

=> Old SP =\(\frac{110}{100}\)(x) + x
=> Old SP =\(\frac{210}{100}\)(x) + x

Old SP = Old Prof + CP [CP is the same in both conditions as given in the prompt]

=> Old SP= \(\frac{10}{100}\)(x) + x

=> Old SP \(\frac{10}{100}\)(x) +\(\frac{100}{100}\)(x)

=> Old SP = \(\frac{110}{100}\)(x)


Also, given

New SP = 92

New SP = New Prof + CP [CP is the same in both conditions as given in the prompt]

=> 92 = \(\frac{115}{100}\)(x) + x

=> 92 = \(\frac{15}{100}\)(x) + x

=>92 = \(\frac{115}{100}\)(x)


Solving, REPEAT THE EQUATION FROM ABOVE

92 = \(\frac{115}{100}\)(x)

x = \(\frac{92*115}{100}\)

x = 92 * \(\frac{100}{115}\)

x = 80

So, the CP = \(\frac{92*100}{215}\)

So, the CP = 80

Now, plug this back into the Old SP equation above.

Old SP = \(\frac{92*100}{215}\) * \(\frac{210}{100}\)

Old SP = (92 * \(\frac{100}{115}\) ) * \(\frac{110}{100}\) = $88

I so hope that helps. Your formula was fine. (Hard, but fine.) I just redid your math, and voila, the answer! :-)

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Re: A certain manufacturer increased its gross profit on a [#permalink]

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New post 08 Sep 2017, 22:54
Oh.My.God. I can totally see where I went wrong. I feel so stupid right now :-). I guess that's what you get when you do 60 Verbal and Math questions back to back.

Thank you, genxer123. I really appreciate your taking time to lay the math out.

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Re: A certain manufacturer increased its gross profit on a   [#permalink] 08 Sep 2017, 22:54
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