Quantitative Aptitude - Arithmetic Ability Questions

Total weight increased = (8 x 2.5) kg = 20 kg.

Weight of new person = (65 + 20) kg = 85 kg.

Let the sum be Rs.P.then
P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)
On dividing,we get (1+R/100)^3=10025/6690=3/2.
Substituting this value in (i),we get:
P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.

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.

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.

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