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.

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.

i=j/m

The monthly payment will be=PV*I

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

We know that,

$$\begin{gathered} S.I. = PTR100 \\ ⇒ P = S.I x 100T x R \\ And also, \\ C.I = P1 + R100n- 1 \end{gathered}$$

From given data, P = Rs. 8625

Now, C.I  = $$\begin{gathered} 86251+41002-1 \\ = 862526252-1 \\ = 8625676625-1 \\ = 8625 x 51625 \\ C.I = Rs. 703.80 \end{gathered}$$

We know that I = PTR/100

11 = (2x30xR)/100

R = 18.333 %

Since two of the eight first class petty officers are to fill two different offices, we write 8P2=56

Then, two of the six second class petty officers are to fill two different offices; thus, we write 6P2 =30

The principle of choice holds in this case; therefore, the members have 56 x 30 = 1680 ways to select the required office holders