Quantitative Aptitude - Arithmetic Ability Questions

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = 15C3  =15*14*133*2*1= 455.  

Let E = event of getting all the 3 red balls.

 n(E) = 5C3 = 5*42*1 = 10.

 => P(E) = n(E)/n(S) = 10/455 = 2/91.

$$\begin{gathered} ?2 = 2×78-21+8+49-148 \\ = 57 - 21 \\ =36 \\ => ? = ±6 \end{gathered}$$

Here the series follow the rule that

12+22 = 5

32+42 = 25

52+62 = 61

72+82 = 113

SO NEXT IS, 92+102 = 181

Here the given number series follows a pattern of two alternate series.

1. 7 19 31 ...(Difference of 12)

2. 9 18 27 ...(Difference of 9)

Here the odd number is 29 instead of 27

And the next number is 31 + 12 = 43 & 27 + 9 = 36

Terms at odd places are 5, 6, 7, 8 etc. and each term at even place is 16.

So, 9 is wrong.

Given the number series is 4, 6, 12, 30, 81, 315.

Here it follows a pattern that

4

4 x 1.5 = 6

6 x 2 = 12

12 x 2.5 = 30

30 x 3 = 90 not equals to 81

90 x 3.5 = 315

Hence the odd one in the series is 81.