Suppose A1,A2,....,A30 are thirty sets, each having 5 elements and B1,B2,.....,Bn are n sets, each with 3 elements.
Let ⋃30i=1Ai=⋃ni=1Bj=S. Each element of S belongs to exactly 10 of the Ais and exactly 9 of the Bis. Then n is equal to
A
15
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
45
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
50
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C 45 Let m be the number of elements in S. Given that each of the Ais have 5 elements.
If no element in Ais is repeated, then the number of elements in A1∪A2∪A3.....∪A30=(30×5).
But each element of the Ais is used 10 times. ∴m=30×510=15.
If the elements in B1,B2,.....,Bn, are not repeated, then the total number of elements in B1∪B2∪B3.....∪Bn=3n.
But each element in Bis is repeated 9 times. ∴m=3n9⇒3n9=15⇒n=45