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MGMAT - sets

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MGMAT - sets

by jayhawk2001 » Sun May 13, 2007 2:56 pm
Another interesting DS. OA coming after a few reply.

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.
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Source: — Data Sufficiency |
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by Neo2000 » Tue May 15, 2007 7:09 pm
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.
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by jayhawk2001 » Tue May 15, 2007 9:12 pm
Good one Neo.

OA in indeed C
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by Cybermusings » Wed May 16, 2007 8:41 am
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
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by Neo2000 » Wed May 16, 2007 12:10 pm
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)
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by jayhawk2001 » Wed May 16, 2007 6:30 pm
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 :-)
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by Neo2000 » Wed May 16, 2007 6:42 pm
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 :)
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