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Group A, which is a set of positive integers, has an average (arithmet

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Group A, which is a set of positive integers, has an average (arithmet  [#permalink]

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New post 04 Dec 2019, 02:36
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Group A, which is a set of positive integers, has an average (arithmetic mean) of 8 while group B, a different set of positive integers, has an average of 10. In which group is the sum of the numbers greater?

(1) The average of all numbers in both groups combined is \(8 \frac{2}{3}\)
(2) The sum of all numbers in both groups is 52


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Re: Group A, which is a set of positive integers, has an average (arithmet  [#permalink]

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New post 04 Dec 2019, 05:54
Let the number of elements in set A be m and that in set B be n.
So, Sum of elements in A = 8m and Sum of elements in B = 10n

We need to find whether 8m>10n which is possible when the values of m and n are known.

(1) The average of all numbers in both groups combined is \(8\frac{2}{3}\)

\(\frac{8m+10n}{m+n}=8\frac{2}{3}\)

At first glance we can see that this is a weighted average set up which is less than 9 (the mean of 8 & 10) and so m>n.

On simplifying, we can find that m=2n

When n is 1 and m is 2, 8m=16, 10n=10 (8m>10n)

When n is 2 and m is 4, 8m=32, 10n=20 (8m>10n)

When n is 3 and m is 6, 8m=48, 10n=30 (8m>10n)

The difference only keeps growing as we go further and in all cases 8m>10n

1 is sufficient

(2) The sum of all numbers in both groups is 52

8m+10n=52

Since m and n are integers, this is only possible when m=4, n=2

The values are m and n are now known

2 is sufficient

Answer is (D)
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Re: Group A, which is a set of positive integers, has an average (arithmet   [#permalink] 04 Dec 2019, 05:54
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