GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 27 Jan 2020, 15:22

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 60687
Group A, which is a set of positive integers, has an average (arithmet  [#permalink]

### Show Tags

04 Dec 2019, 02:36
00:00

Difficulty:

65% (hard)

Question Stats:

52% (02:23) correct 48% (01:59) wrong based on 28 sessions

### HideShow timer Statistics

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

Are You Up For the Challenge: 700 Level Questions

_________________
Director
Joined: 16 Jan 2019
Posts: 544
Location: India
Concentration: General Management
WE: Sales (Other)
Re: Group A, which is a set of positive integers, has an average (arithmet  [#permalink]

### Show Tags

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

Re: Group A, which is a set of positive integers, has an average (arithmet   [#permalink] 04 Dec 2019, 05:54
Display posts from previous: Sort by