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On Monday, a person mailed 8 packages weighing an average
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13 Aug 2012, 05:57
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Re: On Monday, a person mailed 8 packages weighing an average
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13 Aug 2012, 05:57




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Re: On Monday, a person mailed 8 packages weighing an average
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07 Sep 2012, 10:47
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.
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Re: On Monday, a person mailed 8 packages weighing an average
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13 Aug 2012, 07:46
a person mailed 8 packages=99 pounds (total of all 8 packages)
and on Tuesday, 4 packages=61 pounds (total of all 4 packages)
total weight =99+61 =160 pounds
the average weight =160/(8+4) =option a



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Re: On Monday, a person mailed 8 packages weighing an average
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16 Aug 2012, 10:10
Bunuel wrote: RESERVED FOR A SOLUTION. Bunuel, had an offtopic 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"
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Re: On Monday, a person mailed 8 packages weighing an average
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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?
Thanks a lot for your help !



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Re: On Monday, a person mailed 8 packages weighing an average
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07 Sep 2012, 09:15



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Re: On Monday, a person mailed 8 packages weighing an average
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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 :
12(3/8) = 12+3/8 and not 12x(3/8).
Can be confusing.



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Re: On Monday, a person mailed 8 packages weighing an average
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24 Aug 2014, 23:54
I used weightedaverages to solve this as follows:
Let 12.375 be the lowest weight, therefore, (15.25  12.375) = 2.875, the difference in weights between the light and heavy packages.
Hence, the amount we need to add to 12.375 is:
(8x0 + 4x2.875)/(8+12) = 11.50 / 12 ~ 12/12 = 1 (Need to round final answer down a bit since I rounded numerator up)
Therefore, the average weight of all packages is approximately 12.375 + 1 = 13.375, since I need to round the numerator down, ans = 13.333, or A



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Re: On Monday, a person mailed 8 packages weighing an average
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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



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Re: On Monday, a person mailed 8 packages weighing an average
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20 Nov 2014, 02:08
Total weight on Monday \(= 8*12 + 8 * \frac{3}{8} = 96 + 3\) Total weight on Tuesday \(= 4*15 + 4 * \frac{1}{4} = 60 + 1\) Average of all days \(= \frac{96 + 60 + 4}{12} = 8 + 5 + \frac{4}{12} = 13\frac{1}{3}\) Answer = A
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Re: On Monday, a person mailed 8 packages weighing an average
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20 May 2016, 03:29
Bunuel wrote: 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) \(16\frac{1}{2}\) Solution: To solve this question we can use the weighted average equation. Weighted Average = (Sum of Weighted Terms) / (Total Number of Items) We'll first determine the sum (numerator). We see that on the first day we had 8 items that averaged 12 3/8 pounds. We don't know the weights of the individual packages, but we can determine that the sum of all 8 packages is: Sum of first day's packages = 8 x 12 3/8 = 99 pounds Similarly, the sum of the second day's packages is: Sum of second day's packages = 4 x 15 ¼ = 61 We now can use the weighted average equation to find the average weight of the 12 packages: Weighted Average = (99 + 61) / 12 Weighted Average = 160 /12 Weighted Average = 13 1/3 Answer: A
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Re: On Monday, a person mailed 8 packages weighing an average
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20 Jul 2016, 11:55
The total weight of packages on Monday = 12 \(\frac{3}{8}\) * 8 = 99 pounds
The total weight of packages on tuesday = 15 \(\frac{1}{4}\) * 4 = 61 pounds
Average = Total weight / Number of packages = \(\frac{(99+61)}{12}\) = 13 \(\frac{1}{3}\) pounds Correct Answer  A



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Re: On Monday, a person mailed 8 packages weighing an average
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20 Jul 2016, 14:25
(2*12 3/8)+(1*15 1/4)=40 pounds 40/3=13 1/3 pounds average



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On Monday, a person mailed 8 packages weighing an average
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15 Dec 2016, 14:15
This Question is giving us mean of two data and asking us for the combined average.
Using \(Mean = Sum/#\)
Sum(8) = \(\frac{99}{8}*8=99\) Sum(4)=\(\frac{61}{4}*4=61\)
Hence combined average =\(\frac{99+61}{12}= \frac{40}{3}\) Hence A
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Re: On Monday, a person mailed 8 packages weighing an average
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28 Jul 2017, 15:02
Using the weighted average formula = weight of x (x) + weight of y (y) = weight of x and y (x+y)
In this situation, x =8, weighting of x = 12 3/8; y = 4, weighting of y = 15 1/4
substitute in the formula to find 99 + 61 = weight of x and y (8 +4) 99 + 61 = weight of x and y (12) (99+61)/12 = weight of x and y = 13 1/3



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Re: On Monday, a person mailed 8 packages weighing an average
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09 Sep 2018, 16:40
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