Quantitative Aptitude - Arithmetic Ability Questions

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.

From the given data,

$$\begin{gathered} 40100 x 75 + 80100 x 25 = K100 x 250 \\ 30 + 20 = 5K2 \\ 50 x 2 = 5K \\ K = 1005 = 20 \end{gathered}$$

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.

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

Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.

Then  [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.

or 5x + 20 = 305 or x = 57.

So, required difference = (57 + 8) - 57  = 8

The Sum of the digits in each number, Except 324 is 10.

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.