Last semester, Professor K taught two classes, A and B. Each student

Hi,

In many cases, there is built in pattern that allow you solve without any tedious work. I solved this equation in less than 30 seconds.

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"

We should consider any possible cases from any statements in data sufficiency problem especially these type of question which sometime can be really tricky.
Let say if the statement (1) is changed a little into "The students in both classes combined handed in a total of 40 assignments."
Considered from the given info that 7A + 5B = 40 ---> the only possible combination is A=5 and B=1 which this statement alone is sufficient to answer the question.

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.

there are 2 kinds of 2 variable equation on gmat. if the numbers are big, 2 variables need 2 equation to solve. if numbers are small, one equation is enough to find 2 variable .
this is the lession. gmat is a little unfair when playing this trick

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

Is it fine if i am able to solve orally rather than solving like you did or Should i make a habit of solving each and every question?

I don't know why people are using such complex methods to solve this problem.

For the GMAT, you have to know the N variables rule - to solve for n variables, you need n distinct equations.
N variables rule.
Exceptions:
1. When there is a combination of variables and you can substitute for that, you do not need as many equations to solve.
2. If the variables cancels out – you do not need as many equations.
3. If the equations are identical – you cannot solve.

Now, what do we need?
Number of students in class A, let's call it X.

What are we given?
7 assignments doe students in class A and 5 assignments for students in class B.

(1) 7x+5y=85 (equation number 1) Not Sufficient bc we have 1 equation and 2 variables
(2) y=10 (equation number 2) No sufficient bc again we have 1 equation and 2 variables

(1) and (2) together: we have 2 equations and 2 variables. Sufficient

Thanks! But what made you decide to test cases?

Is there a pattern to learn (maybe prime numbers or the fact that A & B must be whole numbers?)

Cheers

Answer: Option C

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total students who submitted assignment
class a =7x and class b = 5y
#1
7x+5y = 85---(1)
x= 35, 70 ; y 50 , 15
#2
There were 10 students in class B.

what no info about class A students
insufficient
from 1 &2
students in class are 5 and that in B are 10
sufficient
7*5+5*10 =85
option C

Solution:

We need to determine the number of students in class A.

Statement One Alone:

Even though we know a total of 85 assignments were handed in by the students in the two classes, we still can’t determine the number of students in class A (or class B). It’s possible that class A has 5 students and class B has 10 students (notice that 5 x 7 + 10 x 5 = 85). However, it’s also possible that class A has 10 students and class B has 3 students (notice that 10 x 7 + 3 x 5 = 85). Therefore, statement one alone is not sufficient.

Statement Two Alone:

Knowing only the number of students in class B does not allow us to determine the number of students in class A. We know that the 10 students in class B handed in a total of 10 x 5 = 50 assignments, but without knowing the combined number of assignments for both classes, we can’t determine the number of students in class A. Statement two alone is not sufficient.

Statements One and Two Together:

From statement two, we know that the students in class B handed in a total of 10 x 5 = 50 assignments. Since we are told in statement one that the two classes handed in a total of 85 assignments, we see that the students in class A handed in a total of 85 - 50 = 35 assignments. Since each student in class A handed in 7 assignments, there must be 35/7 = 5 students in class A.

Answer: C

I made a silly error here and gave this question more respect than it is worth. Since I found that 7x5 + 5x10 altogether yield 85, I thought the GMAT is playing a trick with me that I needed both statements.

People need to remember that sometimes the rule that i need '2 separate equations since I have 2 separate variables' does not work

Because sometimes even within that 1 equation (with the two variables) there is only 1 combination of numbers that work.

Correct me if I am wrong.