In a class,
5
1
received an A grade,
4
1
received a B grade,
3
1
received a C grade and the rest received a D grade. What is the minimum number of people who received a D grade?
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I thought it based on number of students in the photo...and the answer is 5 #LOL but if we talk about how fractions work then it must be 13
Nope, lets do it like… Assuming there are 5 students in the class, 1 student got an A. And he is separated from the ones who get a B. So, now it is just 4, in which 1 gets a B. As it goes on, the rest receive a D, which is 2. P.S. In the question it is given to find the least no. who received a D!!
That is not how fractions work.
The problem in your reasoning is that you are separating people, and that's incorrect.
Your answer would work if the problem is rephrased as such:
"In a class, 5 1 received an A grade. After removing the those A grade students, the remaining 4 1 of the students received a B grade. After removing the those A grade students and B grade students, the remaining 3 1 received a C grade. And the rest received a D grade. What is the minimum number of people who received a D grade?"
Do you see the difference?
Let the number of students in the class = x
Then
The number of student who received a D grade =
x - (1/5 + 1/4 + 1/3)x = (13/60)x
The number (13/60)x to be an integer, x must be a multiple of 60
i.e.
x = 60, 120, 180, ...............
To get the minimum number, x must be 60
The minimum number of student who received a D grade = 13
You need to find the lowest common multiple of 5, 4, and 3 so you can get whole number solutions for each fraction of the whole class. The LCM of 5, 4 and 3 is 5 x 4 x 3 = 60. (Keep in mind LCM is not always all three numbers multiplied together, but it is in this case)
1/5 of 60 is 12. 1/4 of 60 is 15. 1/3 of 60 is 20.
20 + 15 + 12 = 47 60 - 47 = the rest of the class, people who received a D grade = 13.
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If there are N students, then since the numbers of students who receive each of the four grades must be integer-valued, we require that N be divisible by each of 3 , 4 and 5 . The least such value is LCM( 3 , 4 , 5 ) = 6 0 , which will in turn yield the minimum possible value for the number of students who receive a D grade. This value will be
6 0 − 5 6 0 − 4 6 0 − 3 6 0 = 6 0 − 1 2 − 1 5 − 2 0 = 1 3 .