Quantitative Aptitude - Arithmetic Ability Questions

Total number of possible ways = 9C3 = 84 ways

Required atleast one girl in the group of three = total possible ways - ways in which none is girl

None of the members in the group is girl = 6C3 = 20

Therefore, number of ways that at least one member is a girl in the group of three

= 84 - 20

= 64 ways.

When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36

Required, the sum of the two numbers that turn up is less than 12

That can be done as n(E)

= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5) }

= 35

Hence, required probability = n(E)/n(S) = 35/36.

A : B = 100 : 80.

A : C = 100 : 72.

B:C = B/A*A/C = 80/10*100/72 = 10/9 = 100/90 = 100:90

 B can give C 10 points

Let the Time be 'N' and Rate be 'R' 

J = (K x NR)/100 K = (L x NR)/100 

J/K = NR/100 K/L = NR/100 

J/K = K/L 

K x K = JL

P.W= Amount – (T.D)

Time = 100×TDPW×R

Let the present ages of the mother and daughter be 2x and x years respectively.
Then, (2x - 18) = 3(x - 18) => x = 36
Required sum = (2x + x) = 108 years.

Total Combination of getting a card from 52 cards = C152

Because there is no replacement, so number of cards after getting first card= 51

Now, Combination of getting an another card= C151

Total combination of getting 2 cards from 52 cards without replacement= (C152×C151) 

There are total 4 Ace in stack. Combination of getting 1 Ace is = C14

Because there is no replacement, So number of cards after getting first Ace = 3

Combination of getting an another Ace = C13

Total Combination of getting 2 Ace without replacement=C14×C13

Now,Probability of getting 2 cards which are Ace =C14×C13C152×C151 = 1/221.