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.


Which is prefer way to solve the above question? I was having a hard time using Venn diagram.

thanks in advance!!

Please be sure to post the first 6-8 words of the problem as your title. I'll let it slide this time b/c a lot of people haven't been doing this but, next time, we won't answer the question unless it has been posted properly!

That's b/c this one is a double-set matrix. :) (Are you in our class or have you used our Word Translations book? I'm going to assume you have and you're aware of the double-set matrix technique, but let me know if you haven't.)

This is a 2x2 problem: people can eithe rpass or not pass the written test and people can either pass or not pass the practical test.

---------PassW-----FailW-----Tot
PassP---
FallP----
Tot-----

So there's your grid.

They tell us nobody fails both, so put a zero in that square. Also, 30% of the ones who pass the written test don't pass the practical. I don't know how many pass the written, so put an x there. Then put 0.3x in the "pass written, fail practical" square. That's all the question stem tells me. I want to know how many are in the PassW-PassP square. Let's put a Y there. Y also equals 0.7x (because 0.7x + 0.3x = x).

---------PassW-----FailW-----Tot
PassP----Y=0.7x
FallP------0.3x-------0-------0.3x
Tot--------x------

The above is all I know to start.

Statement 1: total = 188. Add to chart.
---------PassW-----FailW-----Tot
PassP----Y=0.7x
FallP------0.3x-------0-------0.3x
Tot--------x--------------------188

Can I use this to find Y? Nope. NS - elim A and D.

Statement 2: those who passed practical (don't know this - call it z) - 20% failed written, so 80% didn't fail written. Add to chart.

---------PassW---------------FailW-----Tot
PassP----Y=0.7x=0.8z-------0.2z-------z
FallP------0.3x------------------0-------0.3x
Tot--------x-------------------0.2z

Can I use this to find Y, or 0.7x, or 0.8z? I can say that 0.7x = 0.8z. But that's not good enough. Elim B.

Statements 1 and 2

---------PassW---------------FailW-----Tot
PassP----Y=0.7x=0.8z-------0.2z-------z
FallP------0.3x------------------0-------0.3x
Tot--------x--------------------0.2z-----188

So now I have:
z+0.3x = 188
x + 0.2z = 188
and (from before) 0.7x = 0.8z
I can solve these. (But don't actually do it! Remember, it's data sufficiency!). Answer = C.

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

Firstly, I must say excellent explaination.

I understand that it is a 2X2 matrix problem, but ,at times, I tend to pick the wrong categories for the matrix.
As in this case, I picked :

Written Practical
Passed
Failed
Total

Is there a reasonable logic behind choosing the category :(

Thanks
DCE

test test

yes.

when you label the COLUMNS, the labels must refer to MUTUALLY EXCLUSIVE ALTERNATIVES or OPTIONS.
when you label the ROWS, the labels must also refer to MUTUALLY EXCLUSIVE ALTERNATIVES or OPTIONS.

this is why your labels don't work: the written and practical aren't mutually exclusive. in other words, you aren't choosing between the written and the practical.

if you examine stacey's correct version, you'll notice that the columns (pass written / fail written) are mutually exclusive alternatives, and that the rows (pass practical / fail practical) are also mutually exclusive alternatives.

in case you're wondering, the chart has to work like this in order to ensure that the numbers in the rows and columns will be additive.
in other words, if you write the categories the way you wrote them, then two things go wrong: (1) the rows/columns don't add anymore, and (2) elements may belong to more than one combination (and, therefore, you may be double-counting certain elements).