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by manik11 » Sat Feb 20, 2016 9:46 pm
All numbers in a list of six positive integers are less than 8. What is the smallest integer that if added to the list would guarantee that the average becomes greater than 8?

A) 14
B) 35
C) 42
D) 50
E) 51

OA : E

Hi experts!
I thought that to find the smallest integer to add to the list to bump the average above 8 I should maximize all the existing 6 numbers in the list.Hence, I chose 7 as the value of all the numbers in the list.
So, my equation looked like this (7*6 + X)/7 > 8. This gave me X>14 in which case B could be the answer. Could you guys please tell me what I'm doing wrong?
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by MartyMurray » Sat Feb 20, 2016 11:11 pm
manik11 wrote:Hi experts!
I thought that to find the smallest integer to add to the list to bump the average above 8 I should maximize all the existing 6 numbers in the list.Hence, I chose 7 as the value of all the numbers in the list.
So, my equation looked like this (7*6 + X)/7 > 8. This gave me X>14 in which case B could be the answer. Could you guys please tell me what I'm doing wrong?
Hmm.

You are seeking to guarantee that adding to the list the number that you choose will make the average become greater than 8.

You maximized all of the existing numbers in the list. Is it guaranteed that the existing numbers will be the maximum value?

What if they were smaller?
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by manik11 » Sat Feb 20, 2016 11:23 pm
Marty Murray wrote:
You maximized all of the existing numbers in the list. Is it guaranteed that the existing numbers will be the maximum value?

What if they were smaller?
Thanks Marty!
I have a follow up question. If its not guaranteed that all the existing numbers will have the max value, how is it correct to assume that all the numbers will have minimum value of 1?

Maybe I'm missing something really basic here. Could you please point that out?
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by MartyMurray » Sat Feb 20, 2016 11:41 pm
manik11 wrote:I have a follow up question. If its not guaranteed that all the existing numbers will have the max value, how is it correct to assume that all the numbers will have minimum value of 1?

Maybe I'm missing something really basic here. Could you please point that out?
You are not assuming that the numbers all have the minimum value of 1.

The point is that in order to answer the question, you need to find a number the use of which guarantees that the new average will be greater than 8.

So the number you choose has to have a value such that given any possible values of the six existing integers in the set, the new average will be greater than 8. While the values of the six existing integers may not all be 1, in order to guarantee that adding the new number will make the average become greater than 8, the number has to work in all cases, including the case in which the value of all six existing integers is 1.
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by GMATGuruNY » Sun Feb 21, 2016 2:46 am
manik11 wrote:All numbers in a list of six positive integers are less than 8. What is the smallest integer that if added to the list would guarantee that the average becomes greater than 8?

A) 14
B) 35
C) 42
D) 50
E) 51
After an integer is added to the list, the total number of integers = 7.
If the average of the 7 integers = 8, then the sum of the 7 integers = (number)(average) = 7*8 = 56.
Since the actual new average must be GREATER THAN 8, the actual new sum must be GREATER THAN 56.
In other words:
new sum ≥ 57.

To GUARANTEE that the new sum ≥ 57, we must consider the WORST-CASE SCENARIO.
Here, the worst-case scenario occurs when the original 6 positive integers are all as SMALL AS POSSIBLE:
{1, 1, 1, 1, 1, 1}.
In the worst-case scenario, the sum of the 6 original integers = 1+1+1+1+1+1 = 6.
Of the answer choices, only E will yield a sum ≥ 57:
51+6 = 57.

The correct answer is E.
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by sandipgumtya » Sun Feb 21, 2016 8:38 am
Hi Mitch,
I have a doubt here.We are asked to find the min value of the integer.Can we take all six value to be 7?then we can get the min value of the integer.is it correct?.Pl explain.
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by DavidG@VeritasPrep » Sun Feb 21, 2016 9:40 am
I have a doubt here.We are asked to find the min value of the integer.Can we take all six value to be 7?then we can get the min value of the integer.is it correct?.Pl explain.
Read the question stem carefully. It says: What is the smallest integer that if added to the list would GUARANTEE that the average becomes greater than 8?

Take the scenario you mentioned: {7, 7, 7, 7, 7, 7, x} In this case, if x = 15, the average would exceed 8. So it's possible that the average could exceed 8 with a final term of 15, but we don't know what the first six integers are, so it isn't true that an additional term of 15 will guarantee an average greater than 8.

If, for example, the set were {1, 1, 1, 1, 1, 1, 15} the average would clearly not be greater than 8, so a final term of 15 does not guarantee that the average will be greater than 8.
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by [email protected] » Sun Feb 21, 2016 12:21 pm
Hi sandipgumtya,

Since we don't know what the original 6 numbers are, we have to consider ALL of the possible outcomes and choose the ONE answer that will ALWAYS raise the average of the group above 8. Since it's not practical to 'test' every possible group of 6 numbers, we have to consider the 'extreme'/lowest set of 6 numbers.

In this case, that group would be {1, 1, 1, 1, 1, 1}. Adding a 51 to that group will raise the average to 57/7 = 8 1/7. If we change the 6 numbers to any other possible combination of positive integers, then the 51 will still raise the average above 8 (the newer average will just end up being higher than 8 1/7).

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by sandipgumtya » Mon Feb 22, 2016 10:08 pm
Hi David/Rich
Thanks a lot.Got it clear now.Thanks.
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by Matt@VeritasPrep » Thu Mar 03, 2016 12:55 am
Another way to catch this is to think of the sum of the set. Seven terms with an average of 8 would have a set sum of 56, so whatever the sum of our set is, it must be > 56.

From there, we could think of the integers we already have. Since they're positive, they must each be at least 1. So the six of those six must be at least 6*1, or 6. To go from 6 to >56, we need >50, so the answer is E.
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by Jeff@TargetTestPrep » Mon Jan 29, 2018 10:18 am
manik11 wrote:All numbers in a list of six positive integers are less than 8. What is the smallest integer that if added to the list would guarantee that the average becomes greater than 8?

A) 14
B) 35
C) 42
D) 50
E) 51
Let's say each integer is equal to 1. Then, the sum of the six integers is 6(1) = 6. Let the unknown integer = x. Then we have:

(6 + x)/7 > 8

6 + x > 56

x > 50

So the minimum value of x is 51.

You might wonder why we chose the smallest possible value of 1 for each of the six integers. This was to fulfill the requirement that the added number would guarantee that the average of the 7 numbers would be greater than 8.

Answer: E

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