SOLUTION: A survey conducted among 150 high school students revealed that * 68 students like Math * 85 students like English * 55 students like History * 20 students like both Math

Question 1210179: A survey conducted among 150 high school students revealed that

* 68 students like Math
* 85 students like English
* 55 students like History
* 20 students like both Math and English
* 15 students like both Math and History
* 22 students like both English and History
* 8 students like all three subjects

How many students like at least one subject?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let $M$ be the set of students who like Math, $E$ be the set of students who like English, and $H$ be the set of students who like History. We are given the following information:
\begin{itemize}
\item $|M| = 68$
\item $|E| = 85$
\item $|H| = 55$
\item $|M \cap E| = 20$
\item $|M \cap H| = 15$
\item $|E \cap H| = 22$
\item $|M \cap E \cap H| = 8$
\end{itemize}
We want to find the number of students who like at least one subject, which is $|M \cup E \cup H|$. We can use the Principle of Inclusion-Exclusion to find this value:
$$|M \cup E \cup H| = |M| + |E| + |H| - |M \cap E| - |M \cap H| - |E \cap H| + |M \cap E \cap H|$$
Plugging in the given values, we get:
$$|M \cup E \cup H| = 68 + 85 + 55 - 20 - 15 - 22 + 8$$
$$|M \cup E \cup H| = 208 - 57 + 8$$
$$|M \cup E \cup H| = 151 + 8$$
$$|M \cup E \cup H| = 159$$
Thus, 159 students like at least one subject.
However, since there are only 150 students, this result is impossible. There must be an error in the given numbers. Let's recalculate:
$$|M \cup E \cup H| = 68 + 85 + 55 - 20 - 15 - 22 + 8$$
$$|M \cup E \cup H| = (68 + 85 + 55) - (20 + 15 + 22) + 8$$
$$|M \cup E \cup H| = 208 - 57 + 8$$
$$|M \cup E \cup H| = 151 + 8$$
$$|M \cup E \cup H| = 159$$
Let's verify the individual sets:
$|M \text{ only}| = |M| - |M \cap E| - |M \cap H| + |M \cap E \cap H| = 68 - 20 - 15 + 8 = 41$
$|E \text{ only}| = |E| - |M \cap E| - |E \cap H| + |M \cap E \cap H| = 85 - 20 - 22 + 8 = 51$
$|H \text{ only}| = |H| - |M \cap H| - |E \cap H| + |M \cap E \cap H| = 55 - 15 - 22 + 8 = 26$
$|M \cap E \text{ only}| = |M \cap E| - |M \cap E \cap H| = 20 - 8 = 12$
$|M \cap H \text{ only}| = |M \cap H| - |M \cap E \cap H| = 15 - 8 = 7$
$|E \cap H \text{ only}| = |E \cap H| - |M \cap E \cap H| = 22 - 8 = 14$
$|M \cap E \cap H| = 8$
$|M \cup E \cup H| = 41 + 51 + 26 + 12 + 7 + 14 + 8 = 159$
Since the result is 159, which is greater than the total number of students surveyed (150), there must be an error in the given data. However, assuming the given numbers are correct, the number of students who like at least one subject is 159.

Answer by ikleyn(52777) (Show Source): You can put this solution on YOUR website!
.

Where do you get wrong problems ?

Do you compose them on your own ?

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