Quantitative Aptitude - Arithmetic Ability Questions

Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.

Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.

P(E) = n(E) / n(S)

      = 4/8= 1/2

By replacing the “?” with “x”

(12.5×14÷8.75) x (100/100) + 42 = 50 + X

(125*140)/875 + 42 = 50 + X

140/7 + 42 = 50 + X

20 + 42 = 50 + X

62 = 50 + X

X = 62 - 50

X = 12

Given Number series is 18     37     76     155     314     635

follows a pattern that, 

18

18 x 2 +1 = 37

37 x 2 +2 = 76

76 x 2 +3 = 155

155 x 2 +4 = 314

314 x 2 +5 = 633 not 635

Hence, the odd man is 635.

100 cm is read as 102 cm. 

A1 = (100*100)Sq.cm 

A2 = (102*102)Sq.cm

(A2 - A1) = 1022-1002 = (102 + 100) x (102 - 100) = 404 sq.cm.

Here the given series 1, 6,  ?, 15, 45, 66, 91 follows a

Pattern in the series is, +5, +9, +13,...,.. +21, +25

So +4 is getting increased at every term addition.

Missing Number in the series will be 15+(9+4) = 15 + 13 = 28.

M + Tu + W + Th = 4 x 48 = 192
Tu + W + Th + F = 4 x 46 = 184
M = 42
Tu + W + Th = 192 - 42 = 150
F = 184 – 150 = 34

H.C.F of co-prime numbers is 1. So, L.C.M = 117/1 =117