Sarah operated her lemonade stand Monday through : GMAT Problem Solving (PS)
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# Sarah operated her lemonade stand Monday through

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28 Dec 2012, 08:19
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54% (04:17) correct 46% (03:57) wrong based on 149 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. $3.25 The question is easy; I already solved it. However, it takes time to do it and many steps. Do you know a method to solve it in less tan 1.75 minutes? [Reveal] Spoiler: OA Intern Joined: 05 Jun 2012 Posts: 24 WE: Marketing (Retail) Followers: 0 Kudos [?]: 12 [0], given: 0 Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 29 Dec 2012, 02:20 danzig wrote: Sarah operated her lemonade stand Monday through Friday over a two week period and made a total profit of 276 dollars. On hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days. Each cup she sold had a total cost of 75 cents and Sarah did not incur any other costs. If every day she sold exactly 32 cups and 3 of the days were hot, then what was the price of 1 cup on a hot day? A.$ 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
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28 May 2013, 09:02
Ok I am not getting the answer ... 1.50 .. How many days are you cobsidering 10 or 13 ?

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28 May 2013, 09:31
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Asishp wrote:
Ok I am not getting the answer ... 1.50 .. How many days are you cobsidering 10 or 13 ?

Posted from GMAT ToolKit

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. _________________ Intern Joined: 27 Dec 2012 Posts: 14 Followers: 0 Kudos [?]: 1 [1] , given: 2 Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 28 May 2013, 10:42 1 This post received KUDOS But the OA says A .. Is it wrong ? Posted from GMAT ToolKit Math Expert Joined: 02 Sep 2009 Posts: 36520 Followers: 7064 Kudos [?]: 92885 [1] , given: 10528 Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 28 May 2013, 11:26 1 This post received KUDOS Expert's post Asishp wrote: But the OA says A .. Is it wrong ? Posted from GMAT ToolKit Yes, OA is B. Edited. _________________ Intern Joined: 27 Dec 2012 Posts: 14 Followers: 0 Kudos [?]: 1 [1] , given: 2 Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 28 May 2013, 11:31 1 This post received KUDOS Thanks ... I also got the same answer ... Posted from GMAT ToolKit Intern Joined: 15 Dec 2007 Posts: 13 Followers: 3 Kudos [?]: 15 [2] , given: 138 Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 03 Nov 2013, 06:50 2 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
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14 Jan 2015, 09:20
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15 Jan 2015, 19:43
Bunuel wrote:
Asishp wrote:
Ok I am not getting the answer ... 1.50 .. How many days are you cobsidering 10 or 13 ?

Posted from GMAT ToolKit

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: (516/344)*1.25 TO Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7118 Location: Pune, India Followers: 2128 Kudos [?]: 13622 [1] , given: 222 Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 15 Jan 2015, 23:08 1 This post received KUDOS Expert's post thorinoakenshield wrote: Hi, 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! The numbers have reduced to $$\frac{3}{8} * 5 = \frac{15}{8}$$ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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16 Jan 2015, 00:18
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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. $3.25 The question is easy; I already solved it. However, it takes time to do it and many steps. Do you know a method to solve it in less tan 1.75 minutes? Total 10 days of sale - 3 hot, 7 regular Total cost price =$(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

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Re: Sarah operated her lemonade stand Monday through   [#permalink] 16 Jan 2015, 00:18
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