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# A lemonade stand sold only small and large cups of lemonade

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25 Nov 2011, 15:09
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A lemonade stand sold only small and large cups of lemonade on Tuesday. 3/5 of the cups sold were small and the rest were large. If the large cups were sold for 7/6 as much as the small cups, what fraction of Tuesday's total revenue was from the sale of large cups?

A. 7/16
B. 7/15
C. 10/21
D. 17/35
E. 1/2
[Reveal] Spoiler: OA

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Last edited by Bunuel on 05 Feb 2012, 15:17, edited 1 time in total.
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05 Feb 2012, 15:27
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manalq8 wrote:
A lemonade stand sold only small and large cups of lemonade on Tuesday. 3/5 of the cups sold were small and the rest were large. If the large cups were sold for 7/6 as much as the small cups, what fraction of Tuesday's total revenue was from the sale of large cups?

A. 7/16
B. 7/15
C. 10/21
D. 17/35
E. 1/2

This problem is very good example how easy and quick plug-in method might be.

Let the total # of cups be 10.
# of small cups sold 3/5*10=6;
# of large cups sold 10-6=4;

Let the price of small cup be $6, then the price of larges cup would be 7/6*6=$7;

Revenue from small cups: 6*$6=$36;
Revenue from large cups cups: 4*$7=$28;

Fraction of total revenue from large cups: 28/(36+28)=7/16.

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03 Oct 2014, 09:57
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03 Oct 2014, 22:33
......................... Small .................... Large ............ Total

Sale .................... $$\frac{3x}{5}$$ ..................... $$\frac{2x}{5}$$ ................. x (Assume)

Price ..................... a .......................... $$\frac{7a}{6}$$

Required Fraction$$= \frac{\frac{2x}{5} * \frac{7a}{6}}{\frac{3xa}{5} + \frac{14xa}{6}} = \frac{14xa}{30} * \frac{30}{32xa} = \frac{7}{16}$$
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12 Nov 2014, 02:15
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Let total no. of cups be x
Small cups ( $$\frac{3}{5} x$$) $$+$$ Large Cups ($$\frac{2}{5} x$$) $$= x$$

Let the price of small cups be $$y$$
then price of large cups be $$\frac{7}{6} y$$

Therefore required fraction = (Price of large cups * no. of large cups) / (Price of large cups * no. of large cups + Price of small cups * no. of small cups)

= $$\frac{\frac{2}{5} x * y}{\frac{2}{5}x * \frac{7}{6}y + \frac{3}{5} x * y}$$

= $$\frac{\frac{7}{15} xy}{\frac{7}{15} xy + \frac{3}{5} xy}$$

= $$\frac{xy * \frac{7}{15} }{xy * \frac{7}{15} + \frac{9}{15}}$$

= $$\frac{7}{16}$$

Last edited by AayushGMAT on 26 Oct 2015, 01:45, edited 2 times in total.

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21 Oct 2015, 04:09
PareshGmat wrote:
......................... Small .................... Large ............ Total

Sale .................... $$\frac{3x}{5}$$ ..................... $$\frac{2x}{5}$$ ................. x (Assume)

Price ..................... a .......................... $$\frac{7a}{6}$$

Required Fraction$$= \frac{\frac{2x}{5} * \frac{7a}{6}}{\frac{3xa}{5} + \frac{14xa}{6}} = \frac{14xa}{30} * \frac{30}{32xa} = \frac{7}{16}$$

I understand that the Fraction we need is: (Large Cup Sale * Large Cup Price) / (Small Cup Sale * Small Cup Price) + (Large Cup Sale * Large Cup Price).
Now I don't understand, why the Large Cup term in the denominator is $$\frac{14xa}{6}$$ and not $$\frac{7xa}{15}$$ or $$\frac{14xa}{30}$$.
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13 Mar 2017, 03:51
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15 Mar 2017, 16:45
manalq8 wrote:
A lemonade stand sold only small and large cups of lemonade on Tuesday. 3/5 of the cups sold were small and the rest were large. If the large cups were sold for 7/6 as much as the small cups, what fraction of Tuesday's total revenue was from the sale of large cups?

A. 7/16
B. 7/15
C. 10/21
D. 17/35
E. 1/2

We can let the total number of cups = n. Since 3/5 of the cups that were sold were small, (3/5)n = the number of small cups sold and (2/5)n = the number of large cups sold.

Also, if we let the price of each small cup = p, then the price for each large cup = (7/6)p.

The revenue from the sale of large cups is (7/6)p * (2/5)n = (7/15)np and the revenue from the sale of small cups is p * (3/5)n = (3/5)np. Therefore, the total revenue from the sale of all the cups is (7/15)np + (3/5)np = (7/15)np + (9/15)np = (16/15)np.

We can now determine what fraction of Tuesday's total revenue was from the sale of large cups.

[(7/15)np]/[(16/15)np] = (7/15)/(16/15) = 7/16

Alternate Solution:

Let’s denote the total number of cups sold by 5n in which n is an integer. Since 3/5 of the number of cups sold were small, the number of small cups that were sold is (5n)(3/5) = 3n. Similarly, since 1 - 3/5 = 2/5 of the number of cups sold were large, the number of large cups that were sold is (5n)(2/5) = 2n.

Also, let’s denote the price of each small cup as6p. Then, the price of a large cup was (6p)(7/6) = 7p.

Now, the revenue earned from small cups was (3n)(6p) = 18np and the revenue earned from large cups was (2n)(7p) = 14np. In total, a revenue of 18np + 14np = 32np was earned. The large cups constituted 14np/32np = 7/16 of this revenue.

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Re: A lemonade stand sold only small and large cups of lemonade   [#permalink] 15 Mar 2017, 16:45
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