# In a group of children 35 play football out of which 20 play football only, 22 play hockey; 25 play cricket out of which 11 play cricket only. Out of these 7 play cricket and football but not hockey, 3 play football and hockey but not cricket and 12 play football and cricket both. How many play all the three games?

A. 5

B. 5

C. 12

D. 7

Answer

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Hint: At first jot down the given data in the question and then try to make Venn diagram and from that Venn diagram use the formula that \[n\left( F\cap C \right)=n\left( F\cap C \right)\left\{ not\text{ hockey} \right\}+n\left( F\cap C\cap H \right)\] to get the answer.

In the question we are told that there are a total 35 children out of which 20 play football only, 22 play hockey and 25 play cricket out of which 11 play cricket only. Out of these 7 play cricket and football but not hockey, 3 play football and hockey but not cricket and 12 play football and cricket both and now from all the data given we will find a number of children who play all these games.

Let football be considered as F, hockey be as H and cricket as C. So according to the data given we can write,

\[\begin{align}

& n\left( F\text{ only} \right)=20 \\

& n\left( H \right)=22 \\

& n\left( C \right)=25 \\

& n\left( C\text{ only} \right)=11 \\

& n\left( C\cap F \right)\left\{ not\text{ hockey} \right\}=7 \\

& n\left( F\cap H \right)\left\{ not\text{ cricket} \right\}=3 \\

& n\left( F\cap C \right)=12 \\

& n\left( F\cap H\cap C \right)=x \\

\end{align}\]

Let’s represent the above written data in Venn diagram:

As we were given that \[n\left( F\cap C \right)=12\]

Here are ordering to Venn diagram,

\[n\left( F\cap C \right)=n\left( F\cap C \right)\left\{ not\text{ hockey} \right\}+n\left( F\cap C\cap H \right)\]

So, as we know the values we can substitute them so we get,

\[12=7+x\Rightarrow x=5\]

Note: While solving these kinds of questions students have questions regarding \[n\left( F \right)\text{ and }n\left( F\text{ only} \right)\] as they are both different values. This also goes for \[n\left( C \right)\text{ and }n\left( H \right)\] too.

__Complete step-by-step answer:__In the question we are told that there are a total 35 children out of which 20 play football only, 22 play hockey and 25 play cricket out of which 11 play cricket only. Out of these 7 play cricket and football but not hockey, 3 play football and hockey but not cricket and 12 play football and cricket both and now from all the data given we will find a number of children who play all these games.

Let football be considered as F, hockey be as H and cricket as C. So according to the data given we can write,

\[\begin{align}

& n\left( F\text{ only} \right)=20 \\

& n\left( H \right)=22 \\

& n\left( C \right)=25 \\

& n\left( C\text{ only} \right)=11 \\

& n\left( C\cap F \right)\left\{ not\text{ hockey} \right\}=7 \\

& n\left( F\cap H \right)\left\{ not\text{ cricket} \right\}=3 \\

& n\left( F\cap C \right)=12 \\

& n\left( F\cap H\cap C \right)=x \\

\end{align}\]

Let’s represent the above written data in Venn diagram:

As we were given that \[n\left( F\cap C \right)=12\]

Here are ordering to Venn diagram,

\[n\left( F\cap C \right)=n\left( F\cap C \right)\left\{ not\text{ hockey} \right\}+n\left( F\cap C\cap H \right)\]

So, as we know the values we can substitute them so we get,

\[12=7+x\Rightarrow x=5\]

**So, the correct answer is “Option A”.**Note: While solving these kinds of questions students have questions regarding \[n\left( F \right)\text{ and }n\left( F\text{ only} \right)\] as they are both different values. This also goes for \[n\left( C \right)\text{ and }n\left( H \right)\] too.

Last updated date: 23rd Sep 2023

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