Quantitative Aptitude - Arithmetic Ability Questions

Here the LCM of 54, 60 is

=> 2 x 3 x 3 x 3 x 10

= 6 x 9 x 10

= 54 x 10

= 540.

K x 3/8 = 141 => K = 376

376 x 32.08/100 = 120

S = { 1, 2, 3, 4, .....18 } 

=> n(S) = 18

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

 E3 =E1∩E2={6, 12, 18} 

=> n(E3) = 3

∴E=E1 ∪ E2 = E1+E2-E3 

=> n(E) = 9 + 6 - 3 =12

 where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

∴P(E)=n(E)n(S)=1218=23

Since,each member of the league must meet every other member of the league.If they only played each other once,there would be 8C2 games.Since,each pairing of teams will occur three times,the answer will be triple.

Therfore, 8C2*3=84

6P4 = 6! / (6-4)! = 360

Probability of left handed = 4/5
total student = 40
total left handed = 40 x(4/5)
= 32

The question requires you to find number of the outcomes in which at most 3 coins turn up as heads.

i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.

The number of outcomes in which 0 coins turn heads is 6C0=1 

The number of outcomes in which 1 coin turns head is =6C1=6 

The number of outcomes in which 2 coins turn heads is6C2=15 

The number of outcomes in which 3 coins turn heads is6C3=20

Therefore, total number of outcomes =1+6+15+20= 42 outcomes