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!
