Quantitative Aptitude - Arithmetic Ability Questions

Relative speed of the thief and policeman  =  (11 – 10) km/hr = 1 km/hr 

Distance covered in 6 minutes  = (1/60) x 6 km   = 1/10 km = 100 m

Therefore, Distance between the thief and policeman = (200 – 100) m = 100 m.

Here the given number series is 6, 3, 3, 4.5, 10

The odd one in the given series is 10

The pattern it follows is

6

6 x 0.5 = 3

3 x  1  = 3

3 x 1.5 = 4.5

4.5 x 2 = 9

Hence, the odd number is 10.

Here the given series is 1, 3, 6, 11, 20, 37, ??,

and the logic behind the series is

1 x 2 + 1 = 3

3 x 2 + 0 = 6

6 x 2 + (-1) = 11

11 x 2 + (-2) = 20

20 x 2 + (-3) = 37

37 x 2 + (-4) = 70

Hence, the next number in the given series is 70.

The terms of the series are products of two consecutive prime numbers i.e.

(2 x 3), (3 x 5), (5 x 7), (7 x 11),.....

So, 165 is wrong and must be replaced by (11 x 13) i.e. 143.

The next number in the series is

143 143 136 110

      0    7      26 (Differences)

 $$\begin{gathered} 13 - 1 , 23 - 1, 33 - 1, 43 - 1,... \\ Now the next number is given by \\ 110 - (43 - 1) = 110 - 63 = 47. \end{gathered}$$

Given are the two AP'S:

15,12,9.... in which a=15, d=-3.............(1) 

-15,-13,-11..... in which a'=-15 ,d'=2.....(2)

now using the nth term's formula,we get

a+(n-1)d = a'+(n-1)d'

substituting the value obtained in eq. 1 and 2,

15+(n-1) x (-3) = -15+(n-1) x 2

=> 15 - 3n + 3 = -15 + 2n - 2

=> 12 - 3n = -17 + 2n

=> 12+17 = 2n+3n

=> 29=5n

=> n= 29/5

ANS:      log₅512 = log512/log5  =  log29log10/2  =9log2log10-log2 =9*0.30101-0.3010 =2.709/0.699 =2709/699 =3.876