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Question

If A={x:x=2k,kāˆˆN and kā‰¤100}, B={x:x=2k,kāˆˆW and k<11} and C={x:x=k3,kāˆˆN and k<11}, then the value of n(Aā–³B)+n(Bā–³C)+n(Aā–³C) is

A
220
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B
216
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C
203
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D
207
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Solution

The correct option is B 216
A={x:x=2k,kN and k100}A={2,4,6,8,,200}n(A)=100
B={x:x=2k,kW and k<11}
B={20,21,22,23,,210}n(B)=11
C={x:x=k3,kN and k<11}C={13,23,33,,103}n(C)=10

Now, AB={21,22,23,24,25,26,27}
n(AB)=7n(AB)=n(A)+n(B)2n(AB)n(AB)=100+1114n(AB)=97 (1)

BC={1,23,43,83}n(BC)=4n(BC)=n(B)+n(C)2n(BC)n(BC)=11+108n(BC)=13 (2)

AC={23,43}n(AC)=2n(AC)=n(A)+n(C)2n(AC)n(AC)=100+104n(AC)=106 (3)

From equations (1),(2) and (3),
n(AB)+n(BC)+n(AC)=97+13+106=216

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