Question

out of 100 members in a club who play bridge,brilliards or basketball,80 play at least two of three games & 90 play at least one of the three games then how many numbers play exactly one of the three games?

Answer 1

Answer:

i think the question is wrong

Answer 2

Answer:

10

Step-by-step explanation:

bridge = a, brilliards = b, basketball = c

Total Number of members = 100

80 play at least two of three games - ab , bc, ac, abc combinations

ab + bc + ac + abc = 80 ---- (1)

90 play at least one of the three games - a , b, c, ab , bc, ac, abc combinations

a + b + c + (ab + bc + ac + abc) = 90 ---- (2)

So, players who play only a, b, c = a + b+ c

Solving (1) and (2), we get a+ b+ c = 10