Question

In a class $A$ of \[25\] students, $20$ passed with $60\% $ or more marks; in another class $B$ of $30$ students, $24$ passed with $60\% $ or more marks. In which class is a greater fraction of students getting $60\% $ or more marks?

Answer 1

Hint: We are asked to find which class has a greater fraction in number of students passed with $60\% $ or above marks. So, we have to compare the fractions of students with those marks out of the total number of students in both classes. Thus, we can find the greater fraction.

Formula used:
If there are $n$ persons/objects in total and $x$ with a certain feature fraction of persons/objects with that feature is $\dfrac{x}{n}$.

Complete step-by-step answer:
Given that out of \[25\] students in class $A$, $20$ passed with $60\% $ or more marks and in another class $B$ of $30$ students, $24$ passed with $60\% $ or more marks.
We have to find which class has a greater fraction of students with $60\% $ or above marks.
Total number of students in class $A$, ${T_A} = 25$
Number of students with $60\% $ or above marks, ${N_A} = 20$
If there are $n$ persons/objects in total and $x$ with a certain feature fraction of persons/objects with that feature is $\dfrac{x}{n}$.
So here, fraction of students $60\% $ or above marks is $\dfrac{{{N_A}}}{{{T_A}}} = \dfrac{{20}}{{25}}$
Dividing both numerator and denominator by $5$, we get $\dfrac{{{N_A}}}{{{T_A}}} = \dfrac{4}{5}$
Total number of students in class $B$, ${T_B} = 30$
Number of students with $60\% $ or above marks, ${N_B} = 24$
If there are $n$ persons/objects in total and $x$ with a certain feature fraction of persons/objects with that feature is $\dfrac{x}{n}$.
So here, fraction of students $60\% $ or above marks is $\dfrac{{{N_B}}}{{{T_B}}} = \dfrac{{24}}{{30}}$
Dividing both numerator and denominator by $6$, we get $\dfrac{{{N_A}}}{{{T_A}}} = \dfrac{4}{5}$
This gives the two fractions are equal.
$\therefore $ Both classes have the same fraction for the students with $60\% $ or above marks.

Note: Here, when simplified by cancelling the common factors from numerator and denominator, we get two equal fractions. But this may not be the case always. If we get two different fractions try to convert them into a common denominator (by means of LCM) and find the greatest.