NandishSS wrote:John purchased large bottles of water for $2 each and small bottles of water for $1.50 each. What percent of the bottles purchased were small bottles?
(1) John spent $33 on the bottles of water
(2) The average price of bottles purchased was $1.65
Statement 2:
To determine the ratio of L to S, use ALLIGATION -- a very efficient way to handle mixture problems.
Step 1: Plot the 3 values on a number line, with the prices for L and S on the ends and the average price in the middle.
L 200---------165----------150 S
Step 2: Calculate the distances between the percentages.
L 200----
35----165----
15----150 S
Step 3: Determine the ratio in the mixture.
The required ratio of L to S is equal to the RECIPROCAL of the distances in red.
L:S = 15:35 = 3:7.
Implication:
Of every 10 bottles, 3 were large and 7 were small.
Thus, the percentage of small bottles = 7/10 = 70%.
SUFFICIENT.
Statement 1:
Since the two statements cannot contradict each other, the percentage yielded by Statement 2 must also be possible in Statement 1.
Thus, it must be possible in Statement 1 that 70% of the bottles were small.
Equation implied by Statement 1:
2L + 1.5S = 33
4L + 3S = 66.
In the resulting equation, it's possible that S=2 and L=15, since 4(15) + 2(3) = 66.
In this case, 2 of every 17 bottles are small.
Since the percentage of small bottles can be different values, INSUFFICIENT.
The correct answer is
B.
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