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

Can someone provide a simple solution?

Here is a formula for exponential growth:

Final amount = original amount * multiplier^(number of changes).

Plug the following values into the formula above:
Original amount = 1.
Final amount = 2. (Since every 6 years the original amount doubles.)
Multiplier = x. (The factor by which the original amount will be multiplied every 2 years.)
Number of changes = 3. (Since over 6 years the original amount will be multiplied by x three times.)

2 = 1 * x^3
x = 2^(1/3).

Hi Aishwarya,

Such a series is called Geometric Progression

If you think about the terms in standard for considering a as First term and r as common ratio then terms will be

1st Term -- in Start of the year year --- a
2nd Term -- after 1 year --- ar
3rd Term -- after 2 year --- ar^2
4th Term -- after 3 year --- ar^3
5th Term -- after 4 year --- ar^4
6th Term -- after 5 year --- ar^5
7th Term -- after 6 year --- ar^6

As per the condition given

ar^6 = 2a [the sum become double in six years]

Therefore r^6 = 2

But the factor by which the sum changes in 2 years
ar^2/a = r^2 = Cube Root (r^6) = Cube root of 2 = 2^(1/3)

You can read more about the progressions from the links given below

https://www.beatthegmat.com/mba/2009/09/ ... th-section

https://www.questionbank.4gmat.com/mba_p ... 0504.shtml