Number of students who study both Hindi and English

Question

In a survey of $100$ students, the number of students studying the various languages is found as: English only $18; $ English but not Hindi $23; $ English and Chinese $8;$ Chinese and Hindi $8;$ English $26;$ Chinese $48$ and no language $24.$ Find

$(1)$ How many students are studying Hindi?

$(2)$ How many students are studying both English and Hindi?

Let $U$ be the universal set. Then $n(U)=100$

And $E: $English and $H: $Hindi and $C: $ Chinese

Then $\displaystyle n(E\cap H'\cap H')=n(E)-n(E\cap H)-n(E\cap C)+n(E\cap H\cap C)=18$

$\displaystyle n(E\cap H')=n(E\cap H'\cap C')+n(E\cap C)-n(E\cap C\cap H)=23$ and $n(E\cap C)=8$ and $n(C\cap H)=8$ and $n(E)=26 ,n(C)=48$ and

$\displaystyle n(E'\cap C'\cap H')=n(U)-n(E\cup H\cup C)=24$

From the above data, we get $n(E\cap H)=n(E\cap H\cap C)$ and $n(E\cup H \cup C)=76$

Now using $\displaystyle n(E\cup H\cup C)=n(E)+n(H)+n(C)-n(E\cap H)-n(E\cap C)-n(H\cap C)+n(E\cap H\cap C)$

$\displaystyle 76=26+48+n(H)-8-8\Longrightarrow n(H)=18$

Now how can I find part $(2)$?

Answer 1

There are three languages involved: Chinese, English and Hindi.

I would like to divide all the $100$ students into eight groups, any two of which do not overlap.

Suppose:

• $n(C' \cap E' \cap H') = a_0$.
• $n(C' \cap E' \cap H) = a_1$.
• $n(C' \cap E \cap H') = a_2$.
• $n(C' \cap E \cap H) = a_3$.
• $n(C \cap E' \cap H') = a_4$.
• $n(C \cap E' \cap H) = a_5$.
• $n(C \cap E \cap H') = a_6$.
• $n(C \cap E \cap H) = a_7$.

We learn:

• $a_2 = 18$.
• $a_2 + a_6 = 23$, which means that $a_6 = 5$.
• $a_6 + a_7 = 8$, which means that $a_7 = 3$.
• $a_5 + a_7 = 8$, which means that $a_5 = 5$.
• $a_2 + a_3 + a_6 + a_7 = 26$, which means that $a_3 = 0$.
• $a_4 + a_5 + a_6 + a_7 = 48$, which means that $a_4 = 35$.
• $a_0 = 24$.
• $a_0 + a_1 + \dots + a_7 = 100$, which means that $a_1 = 10$.

Hence

• $n(C' \cap E' \cap H') = 24$.
• $n(C' \cap E' \cap H) = 10$.
• $n(C' \cap E \cap H') = 18$.
• $n(C' \cap E \cap H) = 0$.
• $n(C \cap E' \cap H') = 35$.
• $n(C \cap E' \cap H) = 5$.
• $n(C \cap E \cap H') = 5$.
• $n(C \cap E \cap H) = 3$.

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(1) We learn from the list above that there are $10 + 0 + 5 + 3 = 18$ students who study Hindi.

(2) We learn from the list above that there are $0 + 3 = 3$ students who study both Chinese and Hindi.

Answer 2

Here is an image and slightly different explanation to Feather Wing's answer.

1. We have the amount of students who only study English being $18$,
   and the amount of students who study both Chinese and English being $8$.
   Combining those two groups result in the group with
   people who study English without studying Hindi combined with people who study all three languages, which is $18+8=26$ people.
2. Also, the amount of people who study English without studying Hindi is $23$.
3. This means that the amount of people who study all three languages is $26-23=3$.
4. This means that the amount of people who study both Chinese and English without studying Hindi is $8-3=5$.
5. Since the total number of people who study English is $26$, and the total number of people who study English without Hindi is $23$, we have the total number of people who study both English and Hindi being $\boldsymbol{26-23=3}$, which is the answer to the second part.
6. We know that the total amount of people who study English is $26$, and the total amount of people who study Chinese is $48$, and the amount of people who study both Chinese and English is $8$, so the amount of people who study both Chinese and English in total is $26+48-8=66$.
7. As there are $100$ people in total and $24$ people who do not study a language, $100-24=76$ people study a language.
8. This means there are $76-66=10$ people who study only Hindi.
9. Therefore, there is a total of $\boldsymbol{10+0+3+5=18}$ people who study Hindi, which is the answer to the first part.