Quantitative Aptitude - Arithmetic Ability Questions

Given equation can be written as

$$\begin{gathered} 215 x 103 + ? = 24 \\ => 14 + ? = 24 \\ => ? = 24 - 14 \\ => ? = 10 \end{gathered}$$

132 = 4 x 3 x 11, So if the number is divisible by all three numbers 4,3 and 11,then the number is divisible by 132 also.
264   => 4,3,11(/)
396   => 4,3,11(/)
462   => 11,3
792   => 4,3,11(/)
968   => 11,4
2178 => 11,3
5184 => 3,4
6336 => 4,3,11(/)
Required number of numbers=4.

There are two series, beginning respectively with 3 and 7. In one 3 is added and in another 2 is subtracted.

The next number is 1 - 2 = -1.

Total number of outcomes possible, n(S) = 10 + 25 = 35

Total number of prizes, n(E) = 10

 P(E)=n(E)n(S)=1035=27

a+b=5   ...(1)    and    3a+2b=20 ...(2)

Multiplying (1) by 2 and subtracting from (2), we get : a=10.

Putting a=10 in (1), we get : b=-5

Therefore,   (3a+b) = 3 x 10+(-5)=30-5=25.