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List of 5 unknown numbers - 700+ Q

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by rohit_gmat » Sun Sep 11, 2011 8:03 am
Q. The average of 5 distinct single digit integers is 5. If two of the integers are discarded, the new average is 4. What is the largest of the 5 integers?

1) Exactly 3 of the integers are consecutive primes

2) The smallest integer is 3

OA : A
IMO : D
Source : TPR Math drills

Plz help explain why my answer is wrong...
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Source: — Data Sufficiency |
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by cans » Sun Sep 11, 2011 8:50 am
sum of 5 integers = 25. sum of 3 = 12... thus sum of 2 = 13.
b) smallest = 3. if that's discarded, the other discarded = 10. but all are single digit, thus not possible.
but if 3's not discarded, sum of 2 = 9.
Sum of 2 = 13 (4,9) (5,8) (6,7)
sum of 2 =9 (1,8) (2,7) (3,6) (4,5)
only possibility = (3,4,5,6,7)
Sufficient
a)consecutive single digit primes: (2,3,5) (3,5,7)
if (2,3,5), sum of other 2 = 15. which can be made by (6,9) or (7,8). (7,8) not possible because then 4 integers will be consecutive prime. (6,9) not possible because of sum of 3 not 12.
if (3,5,7), then sum of other 2 = 10 which can be made by: (1,9) , (2,8 not possible: 4 consecutive primes), (4,6).
if (1,9), sum of 3 numbers is not 12.
if (4,6), highest=7.
only case: (3,4,5,6,7)
Sufficient..
IMO d
Last edited by cans on Sun Sep 11, 2011 10:06 am, edited 1 time in total.
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by rohit_gmat » Sun Sep 11, 2011 9:46 am
cans wrote:sum of 5 integers = 25. sum of 3 = 12... thus sum of 2 = 13.
b) smallest = 3. if that's discarded, the other discarded = 10. but all are single digit, thus not possible.
but if 3's not discarded, sum of 2 = 9.
Sum of 2 = 13 (4,9) (5,8) (6,7)
sum of 2 =9 (1,8) (2,7) (3,6) (4,5)
if we take (4,9)(1,8) highest=9 but if we take (2,7)&(5,8) highest=8. Insufficient.
a)consecutive single digit primes: (2,3,5) (3,5,7)
if (2,3,5), sum of other 2 = 15. which can be made by (6,9) or (7,8). (7,8) not possible because then 4 integers will be consecutive prime. (6,9) not possible because of sum of 3 not 12.
if (3,5,7), then sum of other 2 = 10 which can be made by: (1,9) , (2,8 not possible: 4 consecutive primes), (4,6).
if (1,9), sum of 3 numbers is not 12.
if (4,6), highest=7.
only case: (3,4,5,6,7)
Sufficient..
IMO A

ok cans.. i hate to take this back up again..
but there is smth that you missed in ur logic which makes stmnt 2 suff... that the numbers have to be unique/distinct no matter wat... and the other thing is that 3 has to be the smallest number...

so here...
sum of 2 =9 (1,8) (2,7) (3,6) (4,5)
the only possible choice is ... sum of 2 (kept) = (4,5) (since the rest either repeat 3 or have numbers smaller than 3)

So ...
Sum of 2 = 13 (4,9) (5,8) (6,7)
can only be ... Sum of 2 (discarded) = 13 = (6,7) ... the other repeat the digits 4 & 5 which we confirmed above do exist in the list...
so there is only one largest possible value in the list ... 7 ..


my way to approach stmnt 2 was... if the avg is 5 and the smallest possible number in the list is 3.. the difference between the avg (5) & 3 is 2.. so the largest possible number (to keep the avg still equal to 5) can be avg (5) + 2 = 7 ...

plz help
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by cans » Sun Sep 11, 2011 10:03 am
Sorry again for wrong explanation. IMO D
if smallest is 3. then numbers have to be 3,4,5,6,7. no other possibility.....
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by AbhiJ » Sun Sep 11, 2011 11:05 am
TPR Math means med school prep material ?
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by rohit_gmat » Sun Sep 11, 2011 6:39 pm
TPR = The Princeton Review ...
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