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
m, n , k
This topic has expert replies
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
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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OA is Cclar 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
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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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.abhigyanc wrote:why are you not taking 13 into consideration? its a prime number?!?!
Hope this helps.
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
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