Quantitative Aptitude - Arithmetic Ability Questions

r = (100 x I)/(P x T)

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.

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.

These two events cannot be disjoint because P(K) + P(L) > 1.

P(AꓴB) = P(A) + P(B) - P(AꓵB).

An event is disjoint if P(A ꓵ B) = 0. If K and L are disjoint P(K ꓴ L) = 0.8 + 0.6 = 1.4

And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.

To find the LCM of 24, 36 and 40

24 = 2 x 2 x 2 x 3

36 = 2 x 2 x 3 x 3

40 = 2 x 2 x 2 x 5 

Now, LCM of 24, 36 and 40 = 2 x 2 x 2 x 3 x 3 x 5

= 8 x 9 x 5

= 72 x 5

= 360

For these type of problems,

Quantity of Diesel remained = q1-pqn   

Here p = 12 , q = 120 

=> 1201-121203 = 120 x 0.9 x 0.9 x 0.9 = 87.48 lit.