Complex DS problem

[RoadtoIVY]

If Carlos, Eric, Jeff, and Nigel together have a total of 100 guitars, and each guitar has only one owner, then Nigel has how many more guitars than Eric does?

(1) Carlos has two more guitars than Jeff does: Eric has twelve fewer guitars than Jeff does.
(2) If Nigel had three times as many guitars as he does, he would have two fewer guitars than Carlos and Jeff together have now.

The source I have says B . But as per my calculations, answer should be C

[thephoenix]

imo b

here we need to have an eqn having N and E

with B and info given in q we can develop that relation

i.e N=1/4(98-E)

hence suff

[heshamelaziry]

imo b

here we need to have an eqn having N and E

with B and info given in q we can develop that relation

i.e N=1/4(98-E)

hence suff

So, what is N - E ? this equation does not solve.

[gmatmachoman]

If Carlos, Eric, Jeff, and Nigel together have a total of 100 guitars, and each guitar has only one owner, then Nigel has how many more guitars than Eric does?

(1) Carlos has two more guitars than Jeff does: Eric has twelve fewer guitars than Jeff does.
(2) If Nigel had three times as many guitars as he does, he would have two fewer guitars than Carlos and Jeff together have now.

The source I have says B . But as per my calculations, answer should be C

From the question I could understand:

E= 100-(C+J+N)

Fr\om Statement no 2:

3N=(C+J)-2

-----> C+J = 3N+2

SUBSTITUTE THE VALUE OF C+J IN EQUATION 1.

E= 100-(3N +2+N)
-----> 98-4N

N=(98-E)/4...

SO BY THE QUESTION STATEMENT 2 ALONE IS NOT SUFFICENT TO SOLVE THE PROBLEM.


ON FURTHER SOLVING, C SEEMS TO THE RIGHT ANSWER USING BOTH THE GIVEN STATEMENTS.

[Stuart@KaplanGMAT]

If Carlos, Eric, Jeff, and Nigel together have a total of 100 guitars, and each guitar has only one owner, then Nigel has how many more guitars than Eric does?

(1) Carlos has two more guitars than Jeff does: Eric has twelve fewer guitars than Jeff does.
(2) If Nigel had three times as many guitars as he does, he would have two fewer guitars than Carlos and Jeff together have now.

The source I have says B . But as per my calculations, answer should be C

Let's turn the words into algebra:

Q: if C+E+J+N = 100, what's the value of N-E?

We think: 4 variables, 1 equation so far. To solve, we need either 3 more distinct linear equations or, since we're asked for the value of a relationship rather than an actual variable, a special equation that gives us the exact relationship we desire.

(1) C - J = 2; J - E = 12

Two more equations, so unless we can get the exact relationship we want, insufficient.

We can rewrite the equations as:

C = J + 2 and J = E + 12 and then sub in for J to get C = E + 14

We can now turn "C" and "J" into "E" in our first equation:

E + 14 + E + E + 12 + N = 100

At this point we should recognize that we're not getting "N - E" out of this, so we're done: insufficient.

(If you had recognized that earlier, you should have stopped earlier.)

(2) 3N = C + J - 2

Only 1 new equation, so unless we can plug in to get the exact relationship we desire, it will be insufficient.

Let's rewrite as:

C + J = 3N - 2

and rewrite our original equation as:

C + J + E + N = 100

then sub in for C + J to get:

3N - 2 + E + N = 100

or

4N + E = 102

Again, no way to get N - E out of that mess: insufficient.

When we combine, we have 4 equations and 4 unknowns - we can solve any question related to this system: sufficient, choose (C).