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# Sarah operated her lemonade stand Monday through

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Manager
Joined: 11 Aug 2012
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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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95% (hard)

Question Stats:

50% (03:27) correct 50% (03:39) wrong based on 226 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. $3.25 The question is easy; I already solved it. However, it takes time to do it and many steps. Do you know a method to solve it in less tan 1.75 minutes? [Reveal] Spoiler: OA Kudos [?]: 135 [0], given: 16 Intern Joined: 05 Jun 2012 Posts: 24 Kudos [?]: 13 [0], given: 0 WE: Marketing (Retail) Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 29 Dec 2012, 03:20 danzig wrote: Sarah operated her lemonade stand Monday through Friday over a two week period and made a total profit of 276 dollars. On hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days. Each cup she sold had a total cost of 75 cents and Sarah did not incur any other costs. If every day she sold exactly 32 cups and 3 of the days were hot, then what was the price of 1 cup on a hot day? A.$ 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

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

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28 May 2013, 10:02
Ok I am not getting the answer ... 1.50 .. How many days are you cobsidering 10 or 13 ?

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

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28 May 2013, 10:31
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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. _________________ Kudos [?]: 132517 [3], given: 12324 Intern Joined: 27 Dec 2012 Posts: 14 Kudos [?]: 1 [1], given: 2 Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 28 May 2013, 11:42 1 This post received KUDOS But the OA says A .. Is it wrong ? Posted from GMAT ToolKit Kudos [?]: 1 [1], given: 2 Math Expert Joined: 02 Sep 2009 Posts: 42248 Kudos [?]: 132517 [1], given: 12324 Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 28 May 2013, 12:26 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. _________________ Kudos [?]: 132517 [1], given: 12324 Intern Joined: 27 Dec 2012 Posts: 14 Kudos [?]: 1 [1], given: 2 Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 28 May 2013, 12:31 1 This post received KUDOS Thanks ... I also got the same answer ... Posted from GMAT ToolKit Kudos [?]: 1 [1], given: 2 Intern Joined: 15 Dec 2007 Posts: 13 Kudos [?]: 28 [2], given: 138 Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 03 Nov 2013, 07:50 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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Re: Sarah operated her lemonade stand Monday through [#permalink]

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14 Jan 2015, 10:20
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Re: Sarah operated her lemonade stand Monday through [#permalink]

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15 Jan 2015, 20:43
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 Kudos [?]: 40 [0], given: 224 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7736 Kudos [?]: 17766 [1], given: 235 Location: Pune, India Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 16 Jan 2015, 00:08 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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Re: Sarah operated her lemonade stand Monday through [#permalink]

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16 Jan 2015, 01:18
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17766 [1], given: 235 Senior Manager Status: DONE! Joined: 05 Sep 2016 Posts: 408 Kudos [?]: 25 [0], given: 283 Re: Sarah operated her lemonade stand Monday through [#permalink] ### Show Tags 27 Jan 2017, 15:43 Mon-Fri = 5 day work week Hot Days: PH = PR +0.25PR = 1.25PR Solution: = 3{32(1.25PR)}+2{32(PR)} =$276
--> PR = 276/184 = $1.5 --> PH = 1.25PR --> PH =$1.88

B.

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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.

So, 32*1.25*p*3 +32*p * 7 - 32*10 *0.75 = 276
-> 32p(3.75 + 7) = 276 + 320*0.75 = 276 +240 = 516
-> p = 516/32/10.75 = 129/8*10.75= 129/86 = 3/2 = 1.5

So, price of 1 cup on hot day = 1.5 *1.25 ~ 1.88

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Re: Sarah operated her lemonade stand Monday through   [#permalink] 14 Aug 2017, 02:50
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# Sarah operated her lemonade stand Monday through

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