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On Monday, a person mailed 8 packages weighing an average [#permalink]
13 Aug 2012, 05:57
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A
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C
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Difficulty:
5% (low)
Question Stats:
82% (03:25) correct
18% (01:47) wrong based on 687 sessions
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) \(16\frac{1}{2}\)
Practice Questions Question: 16 Page: 154 Difficulty: 600
Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
07 Sep 2012, 09:15
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Gmbrox wrote:
There must be something easy that i just don't get for me : 8*12(3/8) = 36 as you simplify the 8 between them. Therefore how do you manage to arrive at 99?
I guess it must be something different of spelling or something?
Thanks a lot for your help !
It's not 12 multiplied by 3/8. it's \(12\frac{3}{8}=\frac{12*8+3}{8}=\frac{99}{8}\) (the same way as \(1\frac{1}{2}=\frac{3}{2}\)). _________________
Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
13 Aug 2012, 05:57
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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 \(\frac{total \ weight}{# \ of \ packages}=\frac{99+61}{12}=13\frac{1}{3}\).
Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
07 Sep 2012, 10:47
2
This post received KUDOS
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. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
16 Aug 2012, 10:10
Bunuel wrote:
RESERVED FOR A SOLUTION.
Bunuel, had an off-topic request for you: Could you please post questions from non-OG sources as well? I'm not sure if that might breach a copyright arrangement bsaed on the source you use, and of course your comments on other's questions are supremely valuable for those of us subscribed to your daily updates - but if you could include occasional 700+ non-OG questions, would be much appreciated by your "followers"
_________________
How to improve your RC score, pls Kudo if helpful! http://gmatclub.com/forum/how-to-improve-my-rc-accuracy-117195.html Work experience (as of June 2012) 2.5 yrs (Currently employed) - Mckinsey & Co. (US Healthcare Analyst) 2 yrs - Advertising industry (client servicing)
Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
16 Aug 2012, 23:54
Expert's post
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 \(\frac{total \ weight}{# \ of \ packages}=\frac{99+61}{12}=13\frac{1}{3}\).
Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
07 Sep 2012, 09:09
There must be something easy that i just don't get for me : 8*12(3/8) = 36 as you simplify the 8 between them. Therefore how do you manage to arrive at 99?
I guess it must be something different of spelling or something?
Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
07 Sep 2012, 09:20
Thank you a lot bunuel, Ok after reviewing the official book, i now got it, it is a mix number, it does not exists in france so that's why. If anyone has difficulties to understand like me :
Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
26 Oct 2013, 03:38
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Re: On Monday, a person mailed 8 packages weighing an average [#permalink]
15 Nov 2014, 01:55
I used ratio of packages, which is 2:1. converted both to the same fractions, so 15 1/4 = 15 2/8
2x(12 3/8) + 1x( 15 2/8) = 24+15+ 6/8+2/8 = 39 and 8/8, 8/8 is also obviously 1. Could also be together 40 but that's not easily divisible with three and you know you're left with a remainder. Instead just: 39/3 + 1/3 = 13 and 1/3
gmatclubot
Re: On Monday, a person mailed 8 packages weighing an average
[#permalink]
15 Nov 2014, 01:55
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