Lemonades of two sizes

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Lemonades of two sizes

by sivaspurthybtg » Fri Jun 14, 2013 8:12 pm
Julie is selling lemonades in two sizes, small and large. Small lemonades cost $0.52 and large lemonades cost $0.58. How many small lemonades did Julie sell?
(1) Julie sold a total of 9 lemonades.
(2) Julie's total revenue from the sale of lemonades was $4.92
Source: — Data Sufficiency |

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by GMATGuruNY » Sat Jun 15, 2013 3:17 am
sivaspurthybtg wrote:Julie is selling lemonades in two sizes, small and large. Small lemonades cost $0.52 and large lemonades cost $0.58. How many small lemonades did Julie sell?
(1) Julie sold a total of 9 lemonades.
(2) Julie's total revenue from the sale of lemonades was $4.92
Statement 1: Julie sold a total of 9 lemonades.
No way the to determine the number of small lemondades.
INSUFFICIENT.

Statement 2: Julie's total revenue from the sale of lemonades was $4.92.
52S + 58L = 492.

The total number sold will be MAXIMIZED if all of the revenue is yielded by the SMALL lemonades.
If 10 cups of S are sold, the total revenue = 52*10 = 520.
Too MUCH revenue,
Thus, the total number sold must be LESS THAN 10.

The total number sold will be MINIMIZED if all of the revenue is yielded by the LARGE lemonades.
If 8 cups of L are sold, the total revenue = 58*8 = 464.
Too LITTLE revenue.
Thus, the total number sold must be GREATER THAN 8.

Thus, to yield $492 in revenue, exactly 9 lemonades must be sold:
S + L = 9.

Since we have two variables and two distinct linear equations (52S + 58L = 492 and S+L=9), we can solve for each variable.
SUFFICIENT.

The correct answer is B.
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by aaggar7 » Sat Jun 15, 2013 9:07 pm
Hi Mitch.

The below part from your response is not clear.Please explain again.

The total number sold will be MAXIMIZED if all of the revenue is yielded by the SMALL lemonades.
If 10 cups of S are sold, the total revenue = 52*10 = 520.
Too MUCH revenue,
Thus, the total number sold must be LESS THAN 10.

The total number sold will be MINIMIZED if all of the revenue is yielded by the LARGE lemonades.
If 8 cups of L are sold, the total revenue = 58*8 = 464.
Too LITTLE revenue.
Thus, the total number sold must be GREATER THAN 8.

Thus, to yield $492 in revenue, exactly 9 lemonades must be sold:
S + L = 9.

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by GMATGuruNY » Sun Jun 16, 2013 2:46 am
aaggar7 wrote:Hi Mitch.

The below part from your response is not clear.Please explain again.

The total number sold will be MAXIMIZED if all of the revenue is yielded by the SMALL lemonades.
If 10 cups of S are sold, the total revenue = 52*10 = 520.
Too MUCH revenue,
Thus, the total number sold must be LESS THAN 10.

The total number sold will be MINIMIZED if all of the revenue is yielded by the LARGE lemonades.
If 8 cups of L are sold, the total revenue = 58*8 = 464.
Too LITTLE revenue.
Thus, the total number sold must be GREATER THAN 8.

Thus, to yield $492 in revenue, exactly 9 lemonades must be sold:
S + L = 9.
An analogy:
John has exactly $492. The price of each small shirt is $52. The price of each large shirt is $58. If John spends all of his money, how many shirts does he buy?

John will buy the greatest number of shirts possible if he buys only small shirts, which are cheaper.
If John buys 10 small shirts, the total cost = 10*52 = 520.
Since the amount that John spends ($492) is less than $520, John must buy fewer than 10 shirts.

John will buy the least number of shirts possible if he buys only large shirts, which are more expensive.
If John buys 8 large shirts, the total cost = 8*58 = 464.
Since the amount that John spends ($492) is greater than $464, John must buy more than 8 shirts.

Since John must buy fewer than 10 shirts and more than 8 shirts, he must buy exactly 9 shirts:
S + L = 9.

The same reasoning holds true for the lemonade problem.
To yield exactly $492 in revenue, exactly 9 lemonades must be sold:
S + L = 9.
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