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In a class of 40 students, each one plays either Tennis or Badminton or both. If 28 play Tennis and 26 play Badminton, find

(i) how many play both the games;

(ii) how many play Tennis only;

(iii) how many play Badminton only.

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Answer

Given:

Total students in the class: n(T ∪ B) = 40

Students who play Tennis: n(T) = 28

Students who play Badminton: n(B) = 26

(i) how many play both the games

This represents the intersection of the two sets, n(T ∩ B).

We use the formula:

n(T ∩ B) = n(T) + n(B) - n(T ∪ B)

Substituting the values in above, we get:

n(T ∩ B) = 28 + 26 - 40

n(T ∩ B) = 54 - 40

n(T ∩ B) = 14

∴ 14 students play both the games.

(ii) how many play Tennis only

This represents the set T - B, consisting of students who play Tennis but do not play Badminton.

We use the formula:

n(T - B) = n(T) - n(T ∩ B)

Substituting the values in above, we get:

n(T - B) = 28 - 14

n(T - B) = 14

∴ 14 students play Tennis only.

(iii) how many play Badminton only

This represents the set B - T, consisting of students who play Badminton but do not play Tennis.

We use the formula:

n(B - T) = n(B) - n(T ∩ B)

Substituting the values in above, we get:

n(B - T) = 26 - 14

n(B - T) = 12

∴ 12 students play Badminton only.

The Venn diagram is shown below:

In a class of 40 students, each one plays either Tennis or Badminton or both. If 28 play Tennis and 26 play Badminton, find. R.S. Aggarwal Mathematics Solutions ICSE Class 7.

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