MGMAT - sets

Combining both sentences, you get 50% passed only 1 of 2 exams i.e. either the Written or the practical. Since no one failed both, it means 50% passed both

If there are 188 students, then 94 got their Driver's license.

Good one Neo.

OA in indeed C

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.

Let people who passed written be x
Let people who passed practical be y

.3x did not pass practical (Hence .7x passed practical)
We have to find .7x (people who passed both written and practical..

Statement I: 188 = x+y...but it gives no other clue....Hence it is alone insufficient to find .7x

Statement II: Clearly insufficient; we have no numbers; we only have percentages

Statement I and II: Accordint to Statement II .2y who passed practical failed written...thus .8y passed both written and practical

Now from I and II .... .8y = .7x (Students who passed both)
So y = .7x/.8 = 7x/8

x+y = 188; we can solve for x and y using these 2 statements...

Hence answer should be C

Cybermusings wrote: .3x did not pass practical (Hence .7x passed practical)
We have to find .7x (people who passed both written and practical..

Statement I and II: Accordint to Statement II .2y who passed practical failed written...thus .8y passed both written and practical
Now from I and II .... .8y = .7x (Students who passed both)
So y = .7x/.8 = 7x/8
x+y = 188; we can solve for x and y using these 2 statements...
Hence answer should be C

No offence but this is rather a long approach. Plus it may have been easier had you considered a Venn Diagram(atleast it worked for me that way)

Neo2000 wrote: No offence but this is rather a long approach. Plus it may have been easier had you considered a Venn Diagram(atleast it worked for me that way)

I agree with Neo. Actually I ended up solving it like Cybermusings
(and ofcourse wasted time). Later found out that it could be solved
in a rather simple way.

Hence this question on the forum

jayhawk2001 wrote: I agree with Neo. Actually I ended up solving it like Cybermusings
(and ofcourse wasted time). Later found out that it could be solved
in a rather simple way.
Hence this question on the forum

What was your approach? The easier one