Last semester, Professor K taught two classes, A and B. Each student in class A handed in 7 assignments, and each student in class B handed in 5 assignments. How many students were in class A ?

(1) The students in both classes combined handed in a total of 85 assignments.
(2) There were 10 students in class B.

NEW question from GMAT® Official Guide 2019

(DS12062)
_________________

Say class A has 'a' students and class B has 'b' students. We need to find a = ?

(1) The students in both classes combined handed in a total of 85 assignments.
7a + 5b = 85
Note that a, b are pos. integers.


There are only two possible solutions (which I got by putting numbers in the equation; tedious, I know)
a = 5, b = 10
a = 10, b = 3
Since we have two possible values of a, Statement 1 is not sufficient.

(2) There were 10 students in class B.
b = 10
Does not help with a.
Insufficient.

Combining two gives us a unique answer.
Answer: C

From the information given, let A = the number of student in class A and B = the number of student in class B
(1) The students in both classes combined handed in a total of 85 assignments.
So we got the equation 7A + 5B = 85
let consider some possible combination between A and B which are
1. A = 5 and B = 10
2. A = 10 and B = 3
Since there are 2 possible combinations. (1) is insufficient

(2) There were 10 students in class B.
This statement alone just tell us that the total assignments from class B is 10*5=50 which is insufficient to answer what is the value of A

But when we combine both statements. It is sufficient to answer the question.
So the answer is "C"

Let A = # of student in class A & B = # of student in class B

Total Assignment of class A = 7A

Total Assignment of class B = 5B

(1) The students in both classes combined handed in a total of 85 assignments.

Equation: 7A + 5B = 85
Let's look how we can solve in less with minimum effort. Look to number properties (in same cases unit digit plays great role too but not here)

'85' is multiple of 5 & '5B' is multiple of 5......Hence 7A MUST be multiple of 5.

Therefore 'A' can take values 5, 10, 15...etc

Let A=5.......Then B=10

Let A=10.......Then B=3

We can't proceed more as it will exceed 85

Insufficient

(2) There were 10 students in class B.

No info about A.

Insufficient

Combine 1 & 2
We have only one clear answer when B=10 so A = 5

Answer : C.

I would say its a tricky question.

We need, How many students were in class A?

Statement 1: Total assignment combined = 85
Because each student of Class A handed in 7 assignments, and class B handed in 5 assignments.
Class A student's number would be multiple of 7 and the student number of class B would be multiple of 5.

Here possible combinations are A(X7) B
7 78
14 71
21 64
28 57
35 50
42 43
49 36
56 29
63 22
70 15
77 8
84 1
Here two combinations are possible
So, Stat 1 Insufficient. (If there was one combination Stat 1 would have been sufficient)

Statement 2: B=10 clearly Insufficient

Combined, Sufficient

Answer : C