Largest Integer

[rd_eastbay]

The average of 5 disitinct single digit integers is 5. If 2 of the integers are discarded, the new average is 5. What is the largest of the 5 integers.

 

(1) Exactly 3 of the integers are consecutive primes

(2) The smallest intger is 3

 

 

OA is A. I don't agree with it - I get B.

 

[arjmen]

Yep, I get B too with 7 as the largest integer.

Can't see how statement I would give a unique solution.

[rd_eastbay]

Thanks Arj.

 

I think the OA assumes (1) no negative integers and (2) all primes are consecutive (rather than the longest sequence is that of 3 primes).

[karmaholic]

Can we expect this question on the GMAT? I see the assumptions RD has made. I'm still not sure if we can make these assumptions while takin the test. Would you suggest doing that (making assumptions) if u found this question on the test, irrespective of what u chose as the answer? Just curious... and a bit confused.

[rd_eastbay]

Frankly, the assumptions should not be made in a quesiton like this. B is the correct answer.

[forshomide]

Actually, A by itselfis wrong.

 

First, all five numbers are single digits: 0,1,2,3,4,5,6,7,8,9

Also, we know that the sum of the 5 numbers gives 25(5*5) and the sum of 3 of the numbers gives 15(3*5). Now if 3 of the numbers are consecutive prime #s, the following are possible

 

0 + 1 + 3 = 3, which mean the other two numbers must add to 25-3=22. From above none of the remaining numbers will add to 22

 

Likewise, for 1+2+3 = 5, none of the two numbers left will add to 20

 

for 2+3+5 =10, two numbers, 8 and 7 will add to 15 => 10 + 15 = 25 . This makes 8 the largest integer

 

for, 3+5+7 = 15, 9 and 1 add to 10 to make 9 the largest number.

 

Sine A gives two possible largest numbers, it should not be the right ans.

 

Now, lets look at (2) and Let 3 be the smallest #, the following are possible:

3+4+5+6+7 = 25 => 7 = largest

3+6+7+9 = 25=> 9 = largest

Just like (1), we also have more than one possible ans.

 

For these reasons, I pick E

[alex_cute]

Guys, please explain how you arrived at B

[Vimo]

Hi guys i pick E as well, because like Foshomide stated the 2 statements will give us more than one value for the largest integer.

[forshomide]

Correction from my proevious explanation: 0 and 1 are not prime numbers. However, I still believe the ans. should be E

[arangarajan]

Guys,

 

The answer should be B

 

1st Statement:Exactly 3 of the integers are consecutive primes

 

1st statement rules out the fact that there can be no negative integers because we need a total of 25 to get the average of 5 for 5 numbers.

 

3 consecutive primes can be 2,3,5 or 3,5,7

 

Case 1 : 2,3,5 -- the other 2 numbers should be 9 and 6

 

they cannot be 7 and 8 because only 3 of the integers are consecutive primes which are already 2,3 and 5.

there fore the numbers can be 2,3,5,6,9.

 

But we cannot remove 2 numbers to arrive at a sum of 15 for the other 3 and maintain an average of 5.

 

therefore the 3 consecutive primes should be 3,5,7

 

if the 3 primes are 3,5,7 then the other 2 can be 1,9 or 6 and 4.

 

This gives 2 solutions for the largest integer namely 9 or 7

 

therefore statement 1 is insufficient ,

 

where as Statement 2 categorically mentions that 3 is the smallest integer and therefore only one set of solution exists to arrive at the sum of 25 and 15

 

Which is 3,4,5,6,7 .

 

Therefore the answer is B

 

Alternately statement 2 implies : 5/2(3+L) = 25 giving the highest value for L as 7. using the AP formula for the sum of 5 numbers starting with 3.

 

Hope this clarifies.

[satyendgmail.com]

i agree with the solution of arangarajan.

[vappana]

Ans sould be 'D'. In 'A' the 3 consequtive prime numbers can be 2,3,5 or 3,5,7 that result in an avg of 5. but when we take 2,3,5 we need '7' to get an avg of '5'. '7' is another prime number and we are resticted to only 3 prime numbers. thus, we will have 3,4,5,6,7. Ans (I) is suuff. (II) is suff as per others.

 

 

what is the OA.

[arangarajan]

statement 1 does not rule out the possibility of having 3,5,7,6,9 which would give a result of 9