Question

Of the final grades received by the students in a certain math course, \[\dfrac{1}{5}\] are A’s, $\dfrac{1}{4}$ are B’s, $\dfrac{1}{2}$ are C’s and the remaining 10 grades are D’s. What is the number of students in the course?

Answer 1

Hint: Multiply the ratio of A, B and C with the total number of students, that is T and add the remaining 10 to obtain one equation in T. By using this methodology, we can easily find the total number of students.

Complete step-by-step solution -
Let, T be the total number of the students. We are given that the in-math course, \[\dfrac{1}{5}\] are A’s, $\dfrac{1}{4}$ are B’s, $\dfrac{1}{2}$ are C’s and the remaining 10 grades are D’s.
When T is the total number of students, so we get
Number of students getting A grade $=\dfrac{1}{5}\times T$
Number of students getting B grade $=\dfrac{1}{4}\times T$
Number of students getting C grade $=\dfrac{1}{2}\times T$
Number of students getting D grade =10.
Total number of students in the class = Number of students getting A grade + Number of students getting B grade + Number of students getting C grade + Number of students getting D grade. Therefore, it can be expressed as:
$T=\dfrac{T}{5}+\dfrac{T}{4}+\dfrac{T}{2}+10$
By taking L.C.M (Lowest common multiple of 5, 4 and 2) 20, we get:
$\begin{align}
  & T=\dfrac{4T+5T+10T+200}{20} \\
 & \Rightarrow 20T=19T+200 \\
 & \Rightarrow 20T-19T=200 \\
 & \Rightarrow T=200 \\
\end{align}$
Hence, the total number of students in math courses are 200.

Note: This is a direct problem in which one variable is involved and one equation is formed by manipulating the problem statement. Students must be careful while taking the LCM to obtain a common denominator.