Problem Solving

can someone explain how to solve questions like these. I am looking both for a general strategy and a solution.

The answer is root 3(2)

A quantity increases in such a manner that the ratio of its values in a two consecutive years is constant. If the quantity doubles every 6 years by what factor does it increase in two years

Jinglander wrote:can someone explain how to solve questions like these. I am looking both for a general strategy and a solution.

The answer is root 3(2)

A quantity increases in such a manner that the ratio of its values in a two consecutive years is constant. If the quantity doubles every 6 years by what factor does it increase in two years

so assume in first year , it was C ..since the ratio of its values in a two consecutive years is constant - say it is X

so 2nd year = CX
3rd year = CX^2
4th year = CX^3
5th year = CX^4
6th year = CX^5

Now it is given , If the quantity doubles every 6 years
so , CX^5 = 2 C
X = 5th root of 2
So,2nd year = CX = C *[ 5th root of 2]

Another way of looking at it is that the quantity is increasing in a Geometric Progression . So if A where the starting / first term in the Sequence AND R is the common ratio,
then the rth term = A*R ^(n-1)

The original quatity is 1, we call the constant increasing rate each year is "x". For the year
Beginning year: quantity is 1
after 1 year: x
2 years: x^2
3 years: x^3
...................
6 years: x^6 = 2

General for "a" years

x^6 =2, so x = sqrt6 (2)
x^a = (sqrt6(2)) ^ a = 2^(a/6)
and for 2 years x^2 = 2^(2/6) = 2^(1/3) = sqrt3(2)