avaneeshvyas wrote:

EvaJager wrote:

avaneeshvyas wrote:

The average of the first 5 terms of a set of numbers(S) is 6. If S has 12 terms and the average of the Set S is 7.5, find the average of the first 7 terms.

a)8.5

b)8.4

c)8.6

d)8.3

e)None Of these

The numbers in the given set must be ordered increasingly, otherwise the question makes no sense.

The sum of all the numbers in the set is 7.5 * 12 = 90, and the sum of the last 7 numbers is 90 - 5*6 = 60.

Therefore, the average of the last 7 numbers is 60/7. It follows that the 6th and the 7th number in the set must be no greater than 60/7.

Then the average of the first 7 numbers is at most (30 + 2*60/7)/7 = 330/49 < 8.

Answer E.

The question is worded the way it is typed. As far as the solution goes can we think of it as follows:

the Avg. of the last

8 terms is 60/7(As mentioned in the post above..)

Since the Avg. of the first 5 terms is 6, that means there is a deficit of 1.5 per number or (1.5*5)7.5 in total.

so the last 8 numbers should be not less than 60/7+7.5/7 = 67.5/7

Is the assumption in my solution wrong in someway...please point it out

I guess you meant 7 instead of 8 above.

The question is about the average of the first 7 numbers. What are you trying to deduce about the last 8 numbers? Or should be 7?

Sum of the first 5 numbers is 30. Sum of the last 7 numbers is 60 and their average is 60/7.

The 6th and 7th numbers in the set cannot be larger than the average of the last 7 numbers - we assumed that the numbers are ordered increasingly.

The average of a set of numbers is always between the smallest and the largest number of the set.

In order to estimate the average of the first 7 numbers, we have to get some information about the 6th and the 7th numbers.

And we found that neither can be greater than 60/7, which is enough to compare the average of the first 7 numbers to those given in the answers.

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PhD in Applied Mathematics

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