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.

In each term second two numbers are multiples of the first number. But in 4-15-24 15 is not multiple of 4.

The given series follows the pattern that,

×5+5,   ×5+5,   ×5+5,   ×5+5,   ×5+5

38 x 5 + 5  = 195

195 x 5 + 5 = 980

980 x 5 + 5 = 4905

4905 x 5 + 5 = 24530

24530 x 5 + 5 = 122655

So the missing number is 980 .

When R and W are the first and last letters of all the words then we can arrange them in 5!ways. Similarly When W and R are the first and last letters of the words then the remaining letters can be arrange in 5! ways.

Thus the total number of permutations = 2 x 5!  = 2 x 120 = 240