I think this can be solved as follows:

There are 3 groups of students since no one failed both tests:
1) pass w and fail p = 30% (given)
2) fail w and pass p = a
3) pass w and pass p = b

Statement 1 is not sufficient because you can only compute the number of students in group 1 with it --> answer B),C) or E) left.

Statement 2 alone is not sufficient because you can work out the % of students in group 3 but not the asked quantity. And for that you need info given in statement 1, so answer is C).

Cheers, Joe.

Well my opinion differs a bit though I amstill figuring out how to solve this problem. The problem statement says that 30% of the students who passed written couldnt pass practical.

So we can not assume that the number of students who passed written test and could not clear the practical is 30% of the total student headcount as mentioned in solution above.

Thanks

This seems to be a really decent problem, hence reopening this post.....

gmatjoe wrote:I think this can be solved as follows:

There are 3 groups of students since no one failed both tests:
1) pass w and fail p = 30% (given)
2) fail w and pass p = a
3) pass w and pass p = b

Thanks Joe for the explanation, but still not entirely clear. Agree with the three groups but cannot understand 30% of what..

There are 3 groups of students since no one failed both tests:
1) pass w and fail p = 30% of pass w
2) fail w and pass p = 20% of pass p
3) pass w and pass p = b (what we need to find).

I cant get on to Step2..what next.

Joe/Stuart help..
Engin...can you please let us know the source of this question.

Regards,
Ash

the questions is testing to a firm understanding on overlapping. TO fix the problem, one can use ven diagram or just a normal diagram with accompanying solution. This question comes from manhattan review.

Enginpasa1 wrote:To receive a driver license, sixteen year-olds at Culliver High School have to pass both a written and a practical driving test. Everyone has to take the tests, and no one failed both tests. If 30% of the 16 year-olds who passed the written test did not pass the practical, how many sixteen year-olds at Culliver High School received their driver license?

(1) There are 188 sixteen year-olds at Culliver High School.

(2) 20% of the sixteen year-olds who passed the practical test failed the written test.

No one failed both, so there are indeed 3 categories:

Pass written, fail practical
Pass practical, fail written
Pass both

One way to solve this is to use the overlapping sets formula:

Total # = Group 1 + Group 2 + neither - both

Which in this case means:

Total # of students = #pass practical + #pass written - #passboth

Let's let # of students = s
# of pass practical = p
# of pass written = w
# of pass both = b

so:

s = p + w - b

we also know that 30% of the students who passed the written did not pass the practical: in other words, 70% of students who passed the written DID pass the practical. So:

b = 70%(w)

To start, we have 2 equations and 4 unknowns, so unless we get something special we need 2 more equations.

(As an aside, the # of equations/# of unknowns rule is THE most powerful tool in data sufficiency - knowing the rule and its exceptions (and of course when to use it) will get you a LOT of points on test day.)

(1) s=188

Well, we now know that:

188 = p + s - b

and

b = 70%(w)

Sadly, not enough to solve for b: insufficient.

(2) 20% of those who passed the practical failed the written.

In other words, 80% of those who passed the pracitcal passed the written. So:

b = 80%(p)

We now have:

b = 80%(p)
b = 70%(w)
s = p + w - b

Still no way to solve for b: insufficient.

Combined, however, we have 4 equations and 4 unknowns!

s = p + w - b
s = 188
b = 70%(w)
b = 80%(p)

So, we can reduce our equation to b and some numbers: sufficient, choose (c).

Note: in my explanation, I actually set out all the equations, which is WAY more work than you actually want to do on test day. If you understand the concepts, you should be able to do DS questions much quicker than PS questions, because we don't actually need to set up the math. If we can see how many equations we have and how many variables we can eliminate, we can get through DS questions much more quickly.

Hope that helps!

awesome explanation. thanks