In a class of 80 students, 60% students play carrom, 45% play chess and 10% student neither play carrom nor chess. What is the number of student who play chess only?

Explanation

No. of students play carrom in class = $80 \times \frac{60}{100} = 48$ and no. of students play chess in class = $80 \times \frac{45}{100} = 36$ Students who neither play carrom nor chess = $80 \times \frac{10}{100} = 8$ No. of such students play atleast one game. = $80 - 8 = 72$ Venn diagram showing intersection of Carrom and Chess sets Carrom Chess 8 36 12 24 No. of students who play both game = $(48 + 36) - 72$ = $84 - 72 = 12$ No. of students who play only chess = $36 - 12 = 24$