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A lemonade stand sells cups of Yellow Lemonade and Pink Lemonade. Pink

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A lemonade stand sells cups of Yellow Lemonade and Pink Lemonade. Pink  [#permalink]

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New post Updated on: 30 Sep 2018, 05:52
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  45% (medium)

Question Stats:

76% (02:19) correct 24% (02:39) wrong based on 21 sessions

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A lemonade stand sells cups of Yellow Lemonade and Pink Lemonade. Pink Lemonade sells for 50% more than Yellow Lemonade. On a certain day, 1/x of all the cups of lemonade sold was Yellow Lemonade. In terms of x, what fraction of the total revenue from lemonade sales came from sales of Yellow Lemonade?


A. \(\frac{x − 1}{x}\)

B. \(\frac{2}{3x − 1}\)

C. \(\frac{3x − 3}{3x − 1}\)

D. \(\frac{3}{2x + 1}\)

E. \(\frac{2x − 1}{2x + 1}\)

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Originally posted by Sreyoshi007 on 30 Sep 2018, 05:44.
Last edited by Bunuel on 30 Sep 2018, 05:52, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A lemonade stand sells cups of Yellow Lemonade and Pink Lemonade. Pink  [#permalink]

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New post 30 Sep 2018, 06:56
Sreyoshi007 wrote:
A lemonade stand sells cups of Yellow Lemonade and Pink Lemonade. Pink Lemonade sells for 50% more than Yellow Lemonade. On a certain day, 1/x of all the cups of lemonade sold was Yellow Lemonade. In terms of x, what fraction of the total revenue from lemonade sales came from sales of Yellow Lemonade?


A. \(\frac{x − 1}{x}\)

B. \(\frac{2}{3x − 1}\)

C. \(\frac{3x − 3}{3x − 1}\)

D. \(\frac{3}{2x + 1}\)

E. \(\frac{2x − 1}{2x + 1}\)


Let total number of lemonade cups be 100.
Let Price of a cup of yellow lemonade be $y.

1) Given, Pink Lemonade sells for 50% more than Yellow Lemonade.
Or, Price of a cup of Pink lemonade=$1.5y

2) Given, 1/x of all the cups of lemonade sold was Yellow Lemonade
Or, Number of cups of yellow lemonade sold=\(\frac{1}{x}\)*100
So, Number of cups of yellow lemonade sold=(1-\(\frac{1}{x}\))*100=\(\frac{x-1}{x}\)*100

Total revenue from sales of 100 nos of lemonade cups=Revenue from yellow lemonades+Revenue from Pink Lemonades=1/x*100*y+(1-1/x)*100*1.5y

Now, fraction of the total revenue from lemonade sales came from sales of Yellow Lemonade=\(\frac{\frac{1}{x}*100*y}{\frac{1}{x}*100*y+(\frac{x-1}{x})*100*1.5y}\)
factoring out 1/x, y, and 100 from the numerator and denominator , we have,

Required fraction=\(\frac{1}{1+(x-1)*1.5}\)
Multiplying both numerator and denominator by 2, we have
Required fraction=\(\frac{2}{2+(x-1)*3}\)=\(\frac{2}{3x-1}\)

Ans. (B)
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A lemonade stand sells cups of Yellow Lemonade and Pink Lemonade. Pink  [#permalink]

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New post 30 Sep 2018, 08:18
Let the price of YL=m
so the price of PL= 1.5m

Let the number of cups he sold on that day = C
so revenue from YL=C*1/x*m ......eq(1)
revenue from PL= C*(1-1/x)*1.5m .......eq(2)

fraction of revenue from YL = eq(1)/{eq(1)+eq(2)}

on solving we get the answer= 2/(3x-1)
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A lemonade stand sells cups of Yellow Lemonade and Pink Lemonade. Pink &nbs [#permalink] 30 Sep 2018, 08:18
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A lemonade stand sells cups of Yellow Lemonade and Pink Lemonade. Pink

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