45 students are enrolled in a digital marketing course. To t

Question

45 students are enrolled in a digital marketing course. To the nearest integer, p percent of the students have a degree in computer science. Which of the following cannot be the value of p?

A. 11
B. 24
C. 33
D. 50
E. 89

kaustubhkohli12 · Answer

Can anyone explain what exactly is this question testing 
?

kevincan · Answer

Think about the factors of 45

kevincan · Answer

45 is a multiple of 9, so the fraction with a degree in CS could be 1/9, 1/3, or 8/9. Thus p could be 11,33, or 89.

45 is an odd number, so it is impossible that exactly half have a degree in CS. 
22/45 =1/2 - 1/90 ——- p would be 49
23/45 = 1/2 + 1/90——- p would be 51
Thus p cannot be 50

11/45 =1/4 - 1/180——- p would be 24

sarkarsarbani · Answer

I couldn't understand the solution  kevincan . I was solving like this :45*p/100=9p/20

kevincan · Answer

That would be a nice way to proceed if p could be a decimal. However ,only if the number of students with a degree in CS is a multiple of 9 would that be the case

9/45 = 20 percent 
18/45 =40 percent etc

However , 11% of 45 is close to 1/9 of 45 . 
5/45 =0.111... close to 11%

Likewise 40/45=0.8888... close to 89%

15/45 =1/3 = 33 1/3 %

The tricky ones are 24 and 50

50% of 45 = 45/2 =22.5

22/45 is 1/2 -1/90, close to 49%
23/45 is 1/2 + 1/90 , close to 51%

FYI 11/45 is 1/4 - 1/180 , close to 24%

suntprovident · Answer

I did not understand the solution

Only 33 percent would be a whole number i.e. 15 for all remaining solution the number of students would be a fraction or decimal

a. 11 percent of 45 = 4.95
b. 24 percent of 45 = 10.80
d 50 percent of 45 = 22.5 
e. 89 percent of 45 = 40.05

sarkarsarbani · Answer

kevincan  I was also solving how  suntprovident  is doing.My answer is also 33 %.What is wrong in my understanding?

yashsharma1 · Answer

sarkarsarbani   suntprovident  I understood the question and tried to solve in my own way. This way is good for understanding but took me almost 5 mins and hence not the way to go for GMAT probably.

My understanding of the question is that there exists a percentage p which is the rounded form of the actual % of students that have a degree in CS. That means the actual percentage can be +- 0.5% of p. For example, if I go by the answer choices, if answer was 11%, that means p=11%, actual % of students that have CS degree can be 10.5% to 11.49% of 45. The only condition is that the resulting number has to be an integer. So, I simply calculated 10.5% and 11.49% of 45 and saw if there is at least 1 integer in that range. Only option D did not have an integer in the range and hence that was the answer. Refer below for detailed calculations for each option.

A. 11%: Range = 10.5% to 11.49%: No of students = 4.725 to 5.175 students (Since No of students can be 5 when actual % is ~11.112%, this is a possible solution)
B. 24%: Range = 23.5% to 24.49%: No of students = 10.575 to 11.0205 students (Again, no of students can be 11)
C. 33%: Range = 32.5% to 33.49%: No of students = 14.625 to 15.0705 students (No of students can be 15)
D. 50%: Range = 49.5% to 50.49%: No of students = 22.275 to 22.7205 students (No integer in the range and hence the answer)
E. 89%: Range = 88.5% to 89.49%: No of students = 39.825 to 40.2705 students (No of students can be 40)

kevincan · Answer

This question is a lot less time consuming if we know the following :

Fraction Percent rounded to the nearest integer

1/3 ........................33%
1/9 ........................11%
8/9.........................89% 
1/8..........................13% not needed but nice to know 
1/6...........................17%
2/3...........................67%

kevincan · Answer

33% of 45 is not an integer , but p can be 33.

I wonder whether some of you have misread the question. We are looking for an impossible value of integer p

5/45 = 1/9 ................... p= 11
11/45 = 1/4 - 1/180 ...... p=24
15/45 =1/3.................... p=33
22/45 =1/2- 1/90 ........... p=49
23/45 =1/2 + 1/90 ...........p= 51
40/45 = 8/9.................. p= 89

50 cannot be the value of integer p

culpavoluptatum · Answer

I think the easiest solution to this question could be the PITA method. Just plug in the answers. By just using the first 2 options a and b, you will realise that Option D is the answer as it falls in the middle of the decimal number which 22.5. As it is difficult to state which is the nearest integer as it could be either 23 or 24.

Answer D- 50%

Answer D- 50%

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=89267 I realized that given that my "c" is an integer, my "a" (unrounded % value, which when rounded to integer gives "p") is essentially some multiple of 20 divided by 9. Knowing that 1/9 is 11%, choices like 11, 33, and 89 can be intuitively seen to be possible "p" values. In any case, by some quick calculations, we can see that 50% is not possible. The 20-multiple numerator is either 440 or 460. - At 440, we get 48.8% which rounds to 49%. - At 460, we get 51.11%, which rounds to 51%. 50% is not possible. Hope this helps! --- Harsha GMAT-Club-Forum-b5uuf6kx.png

45 students are enrolled in a digital marketing course. To t

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