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19 May 2019, 21:33
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
65% (02:04) correct
35%
(02:20)
wrong
based on 156
sessions
History
Date
Time
Result
Not Attempted Yet
A group of ten candy bars has an average cost of $0.89 per candy bar. How many additional bars must be bought at the cost of $0.72 a piece to bring the average below $0.80?
A. 8
B. 9
C. 10
D. 11
E. 12
A. 8
B. 9
C. 10
D. 11
E. 12
Archit3110
Major Poster
20 May 2019, 02:11
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Bunuel
solve
0.79=8.9+.72*x / 10+x
x = ~11 so value has to be >11
IMO E 12
21 May 2019, 00:53
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The average cost for 10 candies = $0.89
Let the additional number of bars that must be bought to bring the average below $0.80 to be "\(x\)".
The average cost of the candies after x number of candies have been bought = \(\frac{(0.89*10 + 0.72*x)}{10 + x}\)
= \(\frac{(8.9 + 0.72x)}{10+x}\)
As per the question, the new average cost per candy should be less than $0.80
Hence, \((8.9 + 0.72x)10+x < 0.8\)
\(8.9 + 0.72x < 8 + 0.8x\)
\(0.9 < 0.08x\)
\(0.08x > 0.90\)
\(8x > 90\)
\(x > 90/8\)
\(x > 11.25\)
Since the number of candies will be a natural number, the least possible value of \(x\) is 12.
The correct answer is E.
Let the additional number of bars that must be bought to bring the average below $0.80 to be "\(x\)".
The average cost of the candies after x number of candies have been bought = \(\frac{(0.89*10 + 0.72*x)}{10 + x}\)
= \(\frac{(8.9 + 0.72x)}{10+x}\)
As per the question, the new average cost per candy should be less than $0.80
Hence, \((8.9 + 0.72x)10+x < 0.8\)
\(8.9 + 0.72x < 8 + 0.8x\)
\(0.9 < 0.08x\)
\(0.08x > 0.90\)
\(8x > 90\)
\(x > 90/8\)
\(x > 11.25\)
Since the number of candies will be a natural number, the least possible value of \(x\) is 12.
The correct answer is E.
