m, n , k

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m, n , k

by clar » Wed Aug 19, 2009 1:42 am
If m+n+k = 15 where m , n , k are positive integers and
m<n<k what is n-m?

a) k is greater than 10

b) k is a prime number

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by Mia09 » Wed Aug 19, 2009 3:19 am
IMO C

1)K>10

if k =11 n can be 3 and m can be 1 (m<n<k)(n-m=2)
if k =12 n cab be 2 and m =1(n-m=1)

Not Sufficient

2)K is a prime number

k=7;n=6;m=2 (n-m=4)
k=7;n=5;m=3 (n-m=2)

Not Sufficient

Taking both together the answer will be k=11
k=12 aint prime and in k=13 n and m both =1 which does not satisfy m<n<k

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by real2008 » Wed Aug 19, 2009 3:41 am
it is C

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Re: m, n , k

by shahdevine » Wed Aug 19, 2009 4:15 am
clar wrote:If m+n+k = 15 where m , n , k are positive integers and
m<n<k what is n-m?

a) k is greater than 10

b) k is a prime number
OA is C

s1) k is at least equal to 11 or greater. So set k=11 --> m+n+11=15
m+n=4
since m<n we have m and n can only equal 1 and 3
n-m=3-1=2
However what if k=12 m and n could equal 1 and 2 and n-m=1

Insufficient

s2) k is prime so could equal 2,3,7,11,13 but because m<n<k and m+n+k=15 eliminate 2,3 and 13. if k=11, m and n could equal 1 and 3 so n-m=2. However if k=7, m and could equal 5 and 3 or 6 and 2, where in both cases n-m has varying sums. Insufficient

s1+s2) k greater than 10 and prime and has to equal 15-m-n. Sufficient

you got this man

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by abhigyanc » Thu Aug 20, 2009 7:58 am
why are you not taking 13 into consideration? its a prime number?!?!

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by ankitns » Thu Aug 20, 2009 1:41 pm
abhigyanc wrote:why are you not taking 13 into consideration? its a prime number?!?!
If we take K = 13, then the only way we could get the sum = 15 if m = 1 and n = 1. but since it is given that m < n < k, thus the minimum value for m can be 1 and n can be 2. This gives us the minimum value for (m + n) = 3. If minimum value for (m + n) = 3, then maximum value of k = 15 - 3 = 12.

Hope this helps.
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