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A shopkeeper purchases birdfeeders for $10 each and s

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A shopkeeper purchases birdfeeders for $10 each and s [#permalink] New post 27 Jul 2017, 09:59
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A shopkeeper purchases birdfeeders for $10 each and sells them for $18 each. If the cost of the feeders increases by 50% for two months in a row, what is the smallest percent increase the shopkeeper can apply to the selling price in order to avoid selling at a loss?

Enter your value as %

[Reveal] Spoiler:
25 %

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Re: A shopkeeper purchases birdfeeders for $10 each and s [#permalink] New post 25 Sep 2017, 06:15
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If the cost increases by 50% for two consecutive months, it becomes (1+0.5)*(1+0.5)*20 = 22.5. Then, in order to avoid a loss, the selling cost must be equal to 22.5. Then, the percent increase in the selling cost should be [(22.5-18)/18]*100 = 25%
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Re: A shopkeeper purchases birdfeeders for $10 each and s [#permalink] New post 25 Sep 2017, 07:59
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Carcass wrote:
A shopkeeper purchases birdfeeders for $10 each and sells them for $18 each. If the cost of the feeders increases by 50% for two months in a row, what is the smallest percent increase the shopkeeper can apply to the selling price in order to avoid selling at a loss?

Enter your value as %

[Reveal] Spoiler:
25 %


We have two times increase in birdfeed price:

for 1st $10 we have \(10+50*\frac{10}{100}\) = 15

For 2nd price increase we have \(15+15*\frac{50}{100}\) = 22.5

Now smallest % increase in Selling price should be = \(\frac{(22.5 -18)}{18} *100\) = 25%
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Re: A shopkeeper purchases birdfeeders for $10 each and s [#permalink] New post 18 Dec 2017, 19:46
pranab01 wrote:
Carcass wrote:
A shopkeeper purchases birdfeeders for $10 each and sells them for $18 each. If the cost of the feeders increases by 50% for two months in a row, what is the smallest percent increase the shopkeeper can apply to the selling price in order to avoid selling at a loss?

Enter your value as %

[Reveal] Spoiler:
25 %


We have two times increase in birdfeed price:

for 1st $10 we have \(10+50*\frac{10}{100}\) = 15

For 2nd price increase we have \(15+15*\frac{50}{100}\) = 22.5

Now smallest % increase in Selling price should be = \(\frac{(22.5 -18)}{18} *100\) = 25%


I AM a little confused since the cost rises in 2 month in a row, don't we have to consider how much percent increase in ONE month?
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Re: A shopkeeper purchases birdfeeders for $10 each and s [#permalink] New post 19 Dec 2017, 10:26
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No. The stem says just this: 50% the first-month increase, then the second month a more 50% upon the first increase. Pretty straight.

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Re: A shopkeeper purchases birdfeeders for $10 each and s [#permalink] New post 19 Dec 2017, 17:10
Carcass wrote:
No. The stem says just this: 50% the first-month increase, then the second month a more 50% upon the first increase. Pretty straight.

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i know what the stem means; problem is that the asnwer should be considered the percent increase in one month or 2?
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Re: A shopkeeper purchases birdfeeders for $10 each and s [#permalink] New post 20 Dec 2017, 14:55
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Peter wrote:
Carcass wrote:
No. The stem says just this: 50% the first-month increase, then the second month a more 50% upon the first increase. Pretty straight.

Regards


i know what the stem means; problem is that the asnwer should be considered the percent increase in one month or 2?


2.

Moreover, the question implies that you do have a threshold over which you must have a gain and not a loss. So, if the shop sells at $18 and in the first month have the increase at $15 , if being this way you have a loss. No matter what.

\(\frac{15-18}{18} = \frac{-3}{18}\) not possible

Considering the 2 you do have a certain amount (in this case 22.5) minus 18, you can obtain the % for not having a loss but a gain.

The gist of the problem is to find the second amount.

Hope this helps.

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Re: A shopkeeper purchases birdfeeders for $10 each and s   [#permalink] 20 Dec 2017, 14:55
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A shopkeeper purchases birdfeeders for $10 each and s

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