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87% (01:55) correct
13%
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On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of \(12\frac{3}{8}\) pounds, and on Tuesday, 4 packages weighing an average of \(15\frac{1}{4}\) pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?
(A) \(13\frac{1}{3}\)
(B) \(13\frac{13}{16}\)
(C) \(15\frac{1}{2}\)
(D) \(15\frac{15}{16}\)
(E) \(16\frac{1}{2}\)
(A) \(13\frac{1}{3}\)
(B) \(13\frac{13}{16}\)
(C) \(15\frac{1}{2}\)
(D) \(15\frac{15}{16}\)
(E) \(16\frac{1}{2}\)
07 Sep 2012, 11:47
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Since the final average cannot be greater than \(15\frac{1}{4}\), answers C, D and E are out.
We can use the property of weighted averages.
\(15\frac{1}{4}=15\frac{2}{8}\), the distance between the two initial averages is almost 3.
Since the number of packages are in a ratio of 8:4 = 2:1, the differences between the final average and the initial averages are in a ratio 1:2.
So, the distance between \(12\frac{3}{8}\) and the final average is almost 1, close to \(12\frac{3}{8}+1\approx{13}\frac{1}{4}\).
The final answer should be close to \(13\frac{1}{4}\).
Answer A.
We can use the property of weighted averages.
\(15\frac{1}{4}=15\frac{2}{8}\), the distance between the two initial averages is almost 3.
Since the number of packages are in a ratio of 8:4 = 2:1, the differences between the final average and the initial averages are in a ratio 1:2.
So, the distance between \(12\frac{3}{8}\) and the final average is almost 1, close to \(12\frac{3}{8}+1\approx{13}\frac{1}{4}\).
The final answer should be close to \(13\frac{1}{4}\).
Answer A.
13 Aug 2012, 06:57
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SOLUTION
On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of \(12\frac{3}{8}\) pounds, and on Tuesday, 4 packages weighing an average of \(15\frac{1}{4}\) pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?
(A) \(13\frac{1}{3}\)
(B) \(13\frac{13}{16}\)
(C) \(15\frac{1}{2}\)
(D) \(15\frac{15}{16}\)
(E) \(16\frac{1}{2}\)
The total weight of 8 packages is \(8*12\frac{3}{8}=99\) pounds;
The total weight of 4 packages is \(4*15\frac{1}{4}=61\) pounds;
The average weight of all 12 packages is \(\frac{total \ weight}{# \ of \ packages}=\frac{99+61}{12}=13\frac{1}{3}\).
Answer: A.
On Monday, a person mailed 8 packages weighing an average (arithmetic mean) of \(12\frac{3}{8}\) pounds, and on Tuesday, 4 packages weighing an average of \(15\frac{1}{4}\) pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?
(A) \(13\frac{1}{3}\)
(B) \(13\frac{13}{16}\)
(C) \(15\frac{1}{2}\)
(D) \(15\frac{15}{16}\)
(E) \(16\frac{1}{2}\)
The total weight of 8 packages is \(8*12\frac{3}{8}=99\) pounds;
The total weight of 4 packages is \(4*15\frac{1}{4}=61\) pounds;
The average weight of all 12 packages is \(\frac{total \ weight}{# \ of \ packages}=\frac{99+61}{12}=13\frac{1}{3}\).
Answer: A.
