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# Out of 30 students in a class, 6 like football, 12 like cricket, and the remaining like tennis. Find the ratio of:(a) Number of students liking football to the number of students liking tennis.(b) Number of students liking cricket to the total number of students.

Last updated date: 13th Jun 2024
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Hint: We have given the number of students liking each game. A simple subtraction will give us the number of students liking each game individually. But we are asked to calculate the certain ratios so we will have to find the common factors and then we simplify the ratios as simply as possible.

It is given that the class contains a total $30$ students who like different games.
It can be seen that the games are tennis, cricket and football.
Let us assume that everyone likes some game and no one likes more than one game.
Otherwise it will be impossible to find any of the required ratios.
It is given that $6$ students like football.
Also, it is given that $12$ students like cricket.
This takes care of a total of $12 + 6 = 18$ students.
It is given that remaining students like tennis.
As the total number of students is $30$, the students liking tennis are:
$30 - 18 = 12$
Thus, we established that $12$ students like tennis.
Now first we are asked to find the ratio of students liking football to the students liking tennis.
It is given that $6$ students like football and we have just established that $12$ students like tennis.
(a) Therefore, the required ratio is calculated as follows:
$\dfrac{6}{{12}} = \dfrac{1}{2}$
That is this ratio is $1:2$.
(b) Secondly, we are asked to find the ratio of students who like cricket to the total number of students.
It is given that $12$ students like cricket and the total number of students is $30$.
Therefore, the required ratio is calculated as follows:
$\dfrac{{12}}{{30}} = \dfrac{2}{5}$

Therefore, the answer to question (a) is $1:2$ and question (b) is $2:5$. Hence the correct answer is option (A).

Note:
We first listed the given data correctly and used it to calculate the required data by simple subtraction. Later we performed some divisions to calculate the final ratios as required. Note that here the problem is not a Venn diagram but just a simple ratio.