Question

A school awarded 58 medals for Honesty, 20 for Punctuality and 25 for Obedience. If theseedals were agged by a total of 78 students and only 5 students got meadals for all the three values, findthe number of students twho received medals for exactly two of the three values.e the sets of students who bagged medals in Honesty Punetuality

Answer 1

There are 15 students two received medals for exactly two of the three values.

Step-by-step explanation:

We are given that a school awarded 58 medals for Honesty, 20 for Punctuality, and 25 for Obedience.

If these medals were tagged by a total of 78 students and only 5 students got medals for all the three values.

Let the total number of students who got the medals = n(T) = 78

Number of students who got medals for Honesty = n(H) = 58

Number of students who got medals for Punctuality = n(P) = 20

Number of students who got medals for Obedience = n(O) = 25

Number of students who got medals for all the three values = $\text{n} (\text{H} \bigcap\text{P} \bigcap \text{O} )$ = 5

Now as we know that;

n(T) = n(H) + n(P) + n(O) - n(H $\bigcap$ P) - n(P $\bigcap$ O) - n(O $\bigcap$ H) + $\text{n} (\text{H} \bigcap\text{P} \bigcap \text{O} )$

78 = 58 + 20 + 25 - n(H $\bigcap$ P) - n(P $\bigcap$ O) - n(O $\bigcap$ H) + 5

78 = 108 - n(H $\bigcap$ P) - n(P $\bigcap$ O) - n(O $\bigcap$ H)

n(H $\bigcap$ P) + n(P $\bigcap$ O) + n(O $\bigcap$ H) = 108 - 78 = 30

This means that there are 30 students who received two medals but in this also there may be some students who have received all three medals.

So, to get the number of students two received medals for exactly two of the three values is given by = 30 - ($3 \times \text{n} (\text{H} \bigcap \text{P} \bigcap \text{O} )$)

                                                = 30 - 15 = 15.

Hence, there are 15 students two received medals for exactly two of the three values.