Math, asked by RUTVI1, 1 year ago

find the following questions

Attachments:

Answers

Answered by MADHANSCTS
1
U = { natural number between 25 and 45}
A = {even numbers} = { 26,28,30,32,34,36,38,40,42,44}
B = {multiples of 3} = { 27,30,33,36,39,42}

i)n(A) + n(B) = 10 + 6 = 16

ii) n(AUB) = n[{26,27,28,30,32,33,34,36,38,39,40,42,44}]
                  = 13
    n(A∩B) = n[{30,36,42}]
                  = 3
    n(AUB) + n(A∩B) = 16 
iii) A-B = { 26,28,32,34,38,40,44}
    n(A-B) = 7

RUTVI1: thanks
MADHANSCTS: mark as brainliest
RUTVI1: ok
MADHANSCTS: not mine
Anonymous: but the ques is the numbers b/w 25 and 45 know... why u included 25 and 45 also?
MADHANSCTS: ok
Answered by Anonymous
1
(i) n(A) +n(B) :
 no. of numbers b/w 25 and 45 = 18
no .of even numbers are= 9
no.of multiples of 3 are = 6
there fore n(A) +n(B) = 9+6 = 15
(ii) n(A U B) + n( A ∩ B) :
      A = 26, 28,30,32,34,36,38,40,42,44
      B = 27,30,33,36,39,42,
A U B = 26,27,28,30,32,33,34,36,38,39,40,42,44
n( A U B) = 13
A ∩B = 30,36,42
n(A U B) =3
therefore n(AUB) + n(A ∩ B) = 13 + 3 =16
(iii) n( A - B) :
     A -B = 26,28,32,34,38,40,44
     n( A -B) =7

Anonymous: NOT SURE
Anonymous: it's not fair choosing wrong answer as brainliest
Similar questions