Quantitative Aptitude - Arithmetic Ability Questions

Probability of getting a tail when a single coin is tossed =12
Probability of getting 4 when a die is thrown =16

Required probability  =(12)×(16)
= 1/12

Volume of cylinder / Volume of cone = 3/1

Given total applicants = 135
Given graduates are G=60
Otherthan graduates=135-60 = 75
Given experienced candidates = 80
1) For maximum number of graduates have experience
    Total graduates to have experience = 60

2) For minimum number of graduates have experience
    Remaining after taking other than graduates in experience= 80-75 = 5

$$\begin{gathered} 10+25+108+154+225 \\ 10+25+108+154+15 \\ 10+25+108+13 \\ 10+25+11 \\ 10+6 \\ 16 \\ =4 \end{gathered}$$

Maximum earning will be only when he will won on the maximum yielding table.

     A ----> 10:1

     B ----> 20:1

     C ----> 30:1

 i.e, he won on B and C but lost on A

 20 x 200 + 30 x 200 -1 x 200 = 9800

 minimum earning will be when he won on  table A and B and lose on that table 3.

Therefore, 10 x 200 + 20 x 200 - 1 x 200  =  6000-200 = 5800

Therefore, Difference= 9800 - 5800 = 4000

3300 ----- 3% ↑ (1st time interval, 99) → 3399.

Here, time interval is given as half-yearly i.e. 6 months.