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by alex.gellatly » Sat Apr 21, 2012 6:47 pm
This should be an easy question.... but I always seem to have problems with percents, can someone please help with this question, and percents in general. Thanks a lot!

Question:
A college admissions officer predicts that 20 percent of the students who are accepted will not attend the college. According to this prediction, how many students should be accepted to achieve a planned enrollment of x students?

(A) 1.05x
(B) 1.1x
(C) 1.2x
(D) 1.25x
(E) 1.8x
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by killer1387 » Sat Apr 21, 2012 7:21 pm
alex.gellatly wrote:This should be an easy question.... but I always seem to have problems with percents, can someone please help with this question, and percents in general. Thanks a lot!

Question:
A college admissions officer predicts that 20 percent of the students who are accepted will not attend the college. According to this prediction, how many students should be accepted to achieve a planned enrollment of x students?

(A) 1.05x
(B) 1.1x
(C) 1.2x
(D) 1.25x
(E) 1.8x
let students accepted = y
not attending one's = 0.2y
enrolled one's = 0.8 y = x
hence y = x/0.8 = 1.25x

hence D
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by Stuart@KaplanGMAT » Sat Apr 21, 2012 10:22 pm
killer1387 wrote: let students accepted = y
not attending one's = 0.2y
enrolled one's = 0.8 y = x
hence y = x/0.8 = 1.25x

hence D
Excellent solution!

My only advice would be to use fractions instead of decimals to minimize the calculations.

If x = 4/5(y), then y = 5/4(x)

(it's always just the reciprocal of the fraction on this type of question)

and we can quickly convert 5/4(x) to 1.25(x), since the answers are in decimal form.
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by Shalabh's Quants » Sat Apr 21, 2012 10:49 pm
alex.gellatly wrote:This should be an easy question.... but I always seem to have problems with percents, can someone please help with this question, and percents in general. Thanks a lot!

Question:
A college admissions officer predicts that 20 percent of the students who are accepted will not attend the college. According to this prediction, how many students should be accepted to achieve a planned enrollment of x students?

(A) 1.05x
(B) 1.1x
(C) 1.2x
(D) 1.25x
(E) 1.8x
If a quantity goes down by '1/a' then to to balance it, increase it by '1/a-1'.

In this problem, students not accepting is 20% = 1/5, hence to balance, increase intake by 1/(5-1) = 1/4. Which equals 25%. Ans. D.

Similarly if a quantity increases by '1/a' then to to balance it should go down by '1/a+1'..

'One can try with Petrol prices goes up by 10%, by what % should expense be reduced?'
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by GMATGuruNY » Sun Apr 22, 2012 3:03 am
alex.gellatly wrote: Question:
A college admissions officer predicts that 20 percent of the students who are accepted will not attend the college. According to this prediction, how many students should be accepted to achieve a planned enrollment of x students?

(A) 1.05x
(B) 1.1x
(C) 1.2x
(D) 1.25x
(E) 1.8x
Let the number of students accepted = 100.
Since 20% won't attend college, the number of students who attend college = x = 80.
Thus, our target here is 100: for a planned enrollment of x=80 students, the number who must be accepted = 100.

Now we plug x=80 into the answers to see which yields our target of 100.
Only answer choice D works:
1.25x = 1.25(80) = 100.

The correct answer is D.
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by [email protected] » Sun Apr 22, 2012 3:58 am
I have another way of giving the solution for this question:

Let the number of students be 100 (initially). Therefore, 20 percent i.e 20 students will not be able to attend the college. 80 will be able to attend the college. So now the question is that how many students should be accepted as to have all 100 students to attend the college. So the equation is as follows:

80 = 100
100 = x

i.e those 80 students who can attend the college should be 100 right. And the total no. of students who shall be accepted in the college will be 125.

Hence overall in the percent form it shall be 125/100 X x and hence 1.25x.

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