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Sarah operated her lemonade stand Monday through [#permalink]

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28 Dec 2012, 09:19

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50% (03:39) wrong based on 225 sessions

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Sarah operated her lemonade stand Monday through Friday over a two week period and made a total profit of 276 dollars. On hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days. Each cup she sold had a total cost of 75 cents and Sarah did not incur any other costs. If every day she sold exactly 32 cups and 3 of the days were hot, then what was the price of 1 cup on a hot day?

A. $ 1.50 B. $ 1.88 C. $ 2.25 D. $ 2.50 E. $ 3.25

The question is easy; I already solved it. However, it takes time to do it and many steps. Do you know a method to solve it in less tan 1.75 minutes?

Re: Sarah operated her lemonade stand Monday through [#permalink]

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29 Dec 2012, 03:20

danzig wrote:

Sarah operated her lemonade stand Monday through Friday over a two week period and made a total profit of 276 dollars. On hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days. Each cup she sold had a total cost of 75 cents and Sarah did not incur any other costs. If every day she sold exactly 32 cups and 3 of the days were hot, then what was the price of 1 cup on a hot day?

A. $ 1.50 B. $ 1.88 C. $ 2.25 D. $ 2.50 E. $ 3.25

The question is easy; I already solved it. However, it takes time to do it and many steps. Do you know a method to solve it in less tan 1.75 minutes?

The steps are time taking , thought the calculations are easy..It took me also close to 2 minutes

Sarah operated the stand Monday through Friday (5 days) over a two week period, thus she operated the stand for total of 10 days.

Sarah operated her lemonade stand Monday through Friday over a two week period and made a total profit of 276 dollars. On hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days. Each cup she sold had a total cost of 75 cents and Sarah did not incur any other costs. If every day she sold exactly 32 cups and 3 of the days were hot, then what was the price of 1 cup on a hot day?

A. $ 1.50 B. $ 1.88 C. $ 2.25 D. $ 2.50 E. $ 3.25

7 regular days --> sales = 7*32*x = 224x; 3 hot days --> sales = 3*32*(1.25x) = 120x;

Total sales = 224x+120x = 344x. Total cost = 10*32*0.75 = 240.

Re: Sarah operated her lemonade stand Monday through [#permalink]

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03 Nov 2013, 07:50

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This post received KUDOS

You can put it all in one formula: 276 = (32*7*p+32*3*1.25*p) - 75*10*32 276 = (224p + 120p) - 24000 (thats in cents, or $240) 276+240 = 344p 1.5=p p*1.25=~1.8

Re: Sarah operated her lemonade stand Monday through [#permalink]

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14 Jan 2015, 10:20

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Sarah operated the stand Monday through Friday (5 days) over a two week period, thus she operated the stand for total of 10 days.

Sarah operated her lemonade stand Monday through Friday over a two week period and made a total profit of 276 dollars. On hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days. Each cup she sold had a total cost of 75 cents and Sarah did not incur any other costs. If every day she sold exactly 32 cups and 3 of the days were hot, then what was the price of 1 cup on a hot day?

A. $ 1.50 B. $ 1.88 C. $ 2.25 D. $ 2.50 E. $ 3.25

7 regular days --> sales = 7*32*x = 224x; 3 hot days --> sales = 3*32*(1.25x) = 120x;

Total sales = 224x+120x = 344x. Total cost = 10*32*0.75 = 240.

Profit = 344x - 240 = 276 --> x=1.5. 1.25x=~1.88.

Answer: B.

Hope it's clear.

Hi,

Is there a quick way to calculate the following without a calculator:

Is there a quick way to calculate the following without a calculator:

(516/344)*1.25

TO

\(\frac{516}{344} * \frac{5}{4}\)

516 is divisible by 4 so cancel it out with 4 in the denominator: \(\frac{129}{344} * 5\)

Even now, the numbers are big. Try to see if you have any other common factors: 129 = 3 * 43 We see that 3 is not a factor of 344 (since 3+4+4= 11 which is not a factor of 3). Next check if 43 is a factor of 344. Note that 43*8 = 344 - brilliant!

Sarah operated her lemonade stand Monday through Friday over a two week period and made a total profit of 276 dollars. On hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days. Each cup she sold had a total cost of 75 cents and Sarah did not incur any other costs. If every day she sold exactly 32 cups and 3 of the days were hot, then what was the price of 1 cup on a hot day?

A. $ 1.50 B. $ 1.88 C. $ 2.25 D. $ 2.50 E. $ 3.25

The question is easy; I already solved it. However, it takes time to do it and many steps. Do you know a method to solve it in less tan 1.75 minutes?

Total 10 days of sale - 3 hot, 7 regular

Total cost price = $(3/4) * 32 cups * 10 days = $240 Total profit = $276 Total selling price = 240 + 276 = 516

Say, x is the selling price on regular day

516 = 7*x*32 + 3*(5x/4)*32 = Revenue on Regular day + Revenue on Hot day 516/32 = 7x + 15x/4 129/8 = 43x/4

129/2*43 = x

3/2 = x

Selling price on hot day = (5/4)*(3/2) = 15/8 = 1.88

Concentration: General Management, Entrepreneurship

GPA: 3.8

WE: Engineering (Energy and Utilities)

Re: Sarah operated her lemonade stand Monday through [#permalink]

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14 Aug 2017, 02:50

danzig wrote:

Sarah operated her lemonade stand Monday through Friday over a two week period and made a total profit of 276 dollars. On hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days. Each cup she sold had a total cost of 75 cents and Sarah did not incur any other costs. If every day she sold exactly 32 cups and 3 of the days were hot, then what was the price of 1 cup on a hot day?

A. $ 1.50 B. $ 1.88 C. $ 2.25 D. $ 2.50 E. $ 3.25

The question is easy; I already solved it. However, it takes time to do it and many steps. Do you know a method to solve it in less tan 1.75 minutes?

In the two week period, Sarah operated lemonade for total 10 days out of which 3 were hot days and 7 were regular days. Let p be the selling price on regular day and 1.25 p be the selling price on hot days.

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