1.Set Theory
medium

Of the members of three athletic teams in a school $21$ are in the cricket team, $26$ are in the hockey team and $29$ are in the football team. Among them, $14$ play hockey and cricket, $15$ play hockey and football, and $12$ play football and cricket. Eight play all the three games. The total number of members in the three athletic teams is

A

$43$

B

$76$

C

$49$

D

None of these

Solution

(a) Let $B, H, F$ denote the sets of members who are on the basketball team, hockey team and football team respectively.

Then we are given $n\,(B) = 21,\,n\,(H) = 26,n\,(F) = 29$

$n\,(H \cap B) = 14$, $n\,(H \cap F) = 15$, $n\,(F \cap B) = 12$
and $n\,(B \cap H \cap F) = 8$.

We have to find $n\,(B \cup H \cup F)$.

To find this, we use the formula

$n\,(B \cup H \cup F) = n\,(B) + n\,(H) + n\,(F)$

$ – n\,(B \cap H) – n\,(H \cap F) – n\,(F \cap B) + n\,(B \cap H \cap F)$

Hence,$n\,(B \cup H \cup F) = (21 + 26 + 29) – (14 + 15 + 12) + 8 = 43$

Thus these are $43$ members in all.

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.