Harvey teaches a certain number of biology students in 2

k= 7n
l=6p+6
1) is insufficient as the eq k=7n & l=6p +1 cannot be solved
2) insufficient as 7n=6p+6 cannot be resolvd
but if we combine these two then we get p=6
hence C

Hi,

In option (B) I was unable to prove equation 7N= 6P+6 Insufficient

As per my calculation only N=6 and P=6 can satisfy this equation.

Giving us K= 42 and L= 37 (6P+1 = 6(6) +1)

Could someone please elaborate on why option B in insufficient.

Thank you !

There are several values that can equate 7N to 6P+6 - they would generate multiples of 42 (because 6 and 7 are consecutive values, they share no common factors besides 1. And since the question doesn't restrict us in any way, all positive multiples of 42 are fair game)

7(6) = 6(6) + 6
42 = 42

7(12) = 6(13) + 6
84 = 84

7(18) = 6(20) + 6
126 = 126

And so on

Bunuel, ScottTargetTestPrep, ThatDudeKnows would you mind explaining with SII is sufficient by itself, please?

The second statement is no sufficient by itself. The OA is C, not B, which you can check under the spoiler in the original post.

Harvey teaches a certain number of biology students in 2 classes, K and L. He can divide the students in class K into 7 groups of n students each. He can divide the students in class L into 6 groups of p students each with 1 student left over. How many students are in class L ?

The stem gives that K = 7n and L = 6p + 1, and asks to find the value of L.

(1) n = p

The above is clearly insufficient.

(2) There are 5 more students in class K than in class L.

K = L + 5;
7n = (6p + 1) + 5;
7n = 6p + 6.

Infinitely many values of n and p satisfy this. For example, (n, p) can be (6, 6), (12, 13), (18, 20), and so on.

(1)+(2) Since from (1) n = p, from (2) we get 7p = 6p + 6, giving p = 6. Consequently, L = 6p + 1 = 37. Sufficient.

Answer: C.
_________________

AntonioGalindo

Statement two, by itself, is not sufficient.

Could I come up with a more elegant solution on a question like this? Sure, but my approach is typically to take a quick shot at seeing whether brute force is going to be manageable. In this case, it is, so why bother with anything else? There are no style points on the GMAT, only right/wrong.

Possible values for the number of students in K:
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, ...

Possible values for the number of students in L:
7, 13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, ...

(Notice that the first terms of the lists are equal, the second terms differ by 1, the third terms differ by 2, the fourth terms differ by 3, the fifth terms differ by 4, etc.)

Looking at statement 1 alone:
This just tells us that we need to pick the same numbered term from each list. We could have K=7 and L=7, K=14 and L=13, K=21 and L=19, K=28 and L=25, etc. Do we have enough information to say how many students are in L? No. BCE.

Looking at statement 2 alone:
We are looking for opportunities to select a number from the K list that is 5 more than some number from the L list. We could have K=42 and L=37, K=84 and L=79, etc. Do we have enough information to say how many students are in L? No. CE.

Looking at both statements together:
As we keep going to the right in both lists, the difference between them keeps increasing. There is only one spot where the same numbered terms are going to be separated by 5: K=42 and L=37. Do we have enough information to say how many students are in L? Yes. C.

Answer choice C.

Bunuel, ThatDudeKnows thank you to both of you for such detailed explanations!