If 5 is a divisor of m

[sanjib]

If 5 is a divisor of m and 7 is a divisor of n, what is the value of n?
(1) mn is 140
(2) n is an even integer.


IMO E
OA is C
because Lets say m=5K
and n=7K' where K and K' are quotients.
from st.1 we gat that mn=5.7.K.K'=140=>K.K'=140/35=4
this 4 could be (2x2) or (4x1) or (1x4)
from this it is insufficient to tell n

from st.2 we get n=7K' is Even. we cant say any thing from it so it is insufficient.
By taking both st.-as 7 is odd so K'has to be even in oredr to n=7K' to be even. so K' could be 2 or 4
still insufficient.
It must be E

[Stuart@KaplanGMAT]

If 5 is a divisor of m and 7 is a divisor of n, what is the value of n?
(1) mn is 140
(2) n is an even integer.


IMO E
OA is C

Whenever you post a question that you doubt, the very first piece of information you should include is the source. For example, if this question came from Bob's Discount GMAT House, then there's a good chance that it's wrong. On the other hand, if the question came from a GMAT Prep Test (i.e. retired GMAT question), then there's a better chance that you're wrong.

Since you're right, I'm guessing that this question comes from an untrustworthy source. Let's take it apart:

Q: If m is a multiple of 5 and n is a multiple of 7, what's the value of n?

Not much to think about, let's head to the statements.

(1) mn=140

Breaking 140 into primes, we get 140 = 2*2*5*7.

m must be at least 5, so there are a number of different possibilities for n (7, 2*7, 2*2*7): insufficient.

(2) n is even. There are an infinite number of even multiples of 7 (14, 28, 42, ...): insufficient.

Together: out of the three possibilities from (1), 2 are even (2*7 and 2*2*7), so there are still 2 possible values for n: insufficient, choose (E).

[adityanarula]

I got C:

Statement 1: mn =140

Therefore the pairs can be: 1X140,2X70, 4X35, 5X28, 7X20

Given that m/5 & n/7, 5X28 or 7X20. Thus, Insuff

Statement 2: n is even

Insuff

Combining both:

Thus, n = 28, as it also has to be divisible by 7

Thus sufficient

Am I missing anything?

[cashie19]

lets take
m = 5 x a
n = 7 x b

from (1)
mn = 5 x 7 x ab = 140 = 5 x 7 x 4
=> ab = 4
=> a,b as 1,4 or 4,1 or 2,2
insufficient

from (2)
n is even => insufficient

comb (1) n (2)
a,b can be either 1,4 or 2,2
still insufficient

=> Ans is E

[ghacker]

Answer is E

m=5q and n =7p ; we only know that m is divisible by 5 and n is divisible by 7 , but they can also be divisible by other integers , we don't know that , so we look at the helping statements

Statement I : mn = 140 , we know that when we multiply m and n we get 35 as a factor so pq = 4 but we dont know what p is and q is we only know that pq = 4

statement insufficient

Statement II : clearly insufficient

So both together : we know that n is even so p can take either 4 or 2
hence both statement together ---> insufficient

Answer is E

[Stuart@KaplanGMAT]

I got C:

Statement 1: mn =140

Therefore the pairs can be: 1X140,2X70, 4X35, 5X28, 7X20

Given that m/5 & n/7, 5X28 or 7X20. Thus, Insuff

Statement 2: n is even

Insuff

Combining both:

Thus, n = 28, as it also has to be divisible by 7

Thus sufficient

Am I missing anything?

Yes - your'e missing that if mn is 140 it can also be 10 * 14, which also fits both statements, so n could also be 14.

[sanjib]

Thanks for vouching it with confidence Mr. Stuart.