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Manager
Joined: 01 Feb 2018
Posts: 99
Location: India
Schools: ISB '20, NUS '21, NTU '20
GMAT 1: 700 Q47 V38
WE: Consulting (Consulting)

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Updated on: 30 Sep 2018, 04:52
00:00

Difficulty:

45% (medium)

Question Stats:

75% (02:36) correct 25% (02:31) wrong based on 32 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, 04:44.
Last edited by Bunuel on 30 Sep 2018, 04:52, edited 1 time in total.
Renamed the topic and edited the question.
Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 952
WE: Supply Chain Management (Energy and Utilities)

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30 Sep 2018, 05:56
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

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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Manager
Joined: 05 Oct 2017
Posts: 65
GMAT 1: 560 Q44 V23

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30 Sep 2018, 07:18
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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