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Planned enrolment of x students

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by enriqueta26 » Sat Jul 17, 2010 9:32 am
Could someone explain the steps for this one, please? (source: Gmat practice test)

A college admissions officer predicts that 20% of the students who are acepted will not attend college. Acording to this, How many students should be acepted to achieve a planned enrollment of x students?

[spoiler]answer: 1,25x[/spoiler]
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by Patrick_GMATFix » Sat Jul 17, 2010 9:48 am
I would setup a proportion to solve.

20% (1 in 5) of the people admitted does not attend. So 4 of every 5 admitted student attend. Thus the ratio of the number admitted to the number who attend will be 5:4. For x people to attend, z people will be admitted such that z/x = 5/4 >> z = (5/4)x = 1.25x. For x to attend, admit 1.25x people.

-Patrick
Last edited by Patrick_GMATFix on Sat Jul 17, 2010 9:59 am, edited 1 time in total.
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by barcebal » Sat Jul 17, 2010 9:53 am
Another way to think of it....

a=# of students accepted

(% of students that enroll)*A=X

We know that 80% will enroll if 20% do not.

Thus, 0.8(A) = X

Solve for A

(I convert to fractions)

(4/5)*(a)=x

a=(5/4)*x

a=1.25x
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by singhpreet1 » Sat Jul 17, 2010 9:54 am
enriqueta26 wrote:Could someone explain the steps for this one, please? (source: Gmat practice test)

A college admissions officer predicts that 20% of the students who are acepted will not attend college. Acording to this, How many students should be acepted to achieve a planned enrollment of x students?

[spoiler]answer: 1,25x[/spoiler]
this is a ratio you will play with a lot in the GMAT questions...make it a thumb rule, when you want to scale down by 20%, to keep it equal you want to go up by 25% and vice versa. please correct me if im wrong on this one guys.

ok to your question.

if i take 100 people: 80 will enroll and 20 wont. so now to fill the rest of the 20 seats i need to get another 25 people, so that 20 enroll which is the ratio of 4/5 as already taken up by Patrick. therefore total people u want for 100% enrollment is 100+25=125

hope thats helpful.

Preet
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by enriqueta26 » Sat Jul 17, 2010 10:47 am
thank you all!!
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