Problem Solving

Enrollment in City College in 1980 was 83 1/3 percent of enrollment in 1990. What was the percent increase in the college's enrollment from 1980 to 1990?

A. 10%
B. 16 2/3 %
C. 20%
D. 25%
E. 183 1/3%

can someone explain why the answer is c??? im getting b over and over again! what am i missing here or has the wrong answer been indicated?

mitzwillrockgmat wrote:Enrollment in City College in 1980 was 83 1/3 percent of enrollment in 1990. What was the percent increase in the college's enrollment from 1980 to 1990?

A. 10%
B. 16 2/3 %
C. 20%
D. 25%
E. 183 1/3%

can someone explain why the answer is c??? im getting b over and over again! what am i missing here or has the wrong answer been indicated?

Let 100 be the number of students in 1990

SO in 1980 : it will be (250/3)

Percentage increase : { (No of students in 1990 - No of Students in 1989)/No of Students in 1989 }* 100

{100 - (250/3)}/(250/3)
{50/250} * 100
20 %

pick C

mitzwillrockgmat wrote:Enrollment in City College in 1980 was 83 1/3 percent of enrollment in 1990. What was the percent increase in the college's enrollment from 1980 to 1990?

A. 10%
B. 16 2/3 %
C. 20%
D. 25%
E. 183 1/3%

can someone explain why the answer is c??? im getting b over and over again! what am i missing here or has the wrong answer been indicated?

Let population be 'x' in 1990
So, population in 1980 = 83 1/3 % of x = x * (250/3) / 100 = 5x / 6

percent increase in the college's enrollment from 1980 to 1990 =( [ x - (5x / 6) ] / ( 5x / 6) ) * 100 =[( x / 6)/( 5x / 6)] * 100

equates to 20%

I highly recommend memorizing the decimal equivalents of 1/2, 1/3, 1/4...1/10. Knowing these can save you a lot of time.

1/6 is 16.666%, so 83 1/3% (83.333%) is 5/6.

This problem tells us that enrollment in 1980 is 5/6 of enrollment in 1990, so you can think of 1980's value as 5x and 1990's as 6x. The increase from 5x to 6x is x, or 20% of 5x.

Answer is C.

For any two values A and B ,if A is a/b times less than B,B will be a/b-a times more than A

Here ,
enrollment in 1980 was 83 1/3 percent of enrollment in 1990
So enrollment in 1980 is (100-83 1/3)/100 or 1/6 times less than enrollment in 1990
So enrollment in 1990 is 1/6-1 or 1/5 th time more than enrollment in 1980

1/5th=20%

Ans option C

@gmatmachoman, U have beat me here buddy!!!!

liferocks wrote:For any two values A and B ,if A is a/b times less than B,B will be a/b-a times more than A

Here ,
enrollment in 1980 was 83 1/3 percent of enrollment in 1990
So enrollment in 1980 is (100-83 1/3)/100 or 1/6 times less than enrollment in 1990
So enrollment in 1990 is 1/6-1 or 1/5 th time more than enrollment in 1980

1/5th=20%

Ans option C

@gmatmachoman, U have beat me here buddy!!!!

yeah yeah... Becox of u I am forced to become Nocturnal bird...But the good news is i AM GETTING free in my office...