A quantity increases in a manner such that the ratio of its values in

Question

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?

A. 4

B. 2

C.  \sqrt{2}

D.  \sqrt {2}

E.  \frac{1}{2}

LeoN88 · Accepted Answer

a, ak, ak^2, ak^3, ak^4, ak^5 and ak^6
Now, ak^6 = 2*a
So, k^6 = 2,

Now required= ak^2/a = k^2
k^6 = 2
=> k^2 = (2)^(1/3)

Balkrishna · Answer

As per question stem, a quantity increases in a manner such that the  ratio of its values in any two consecutive years is constant .

Let's check whether ratio of every two consecutive number in your solution is same or not

Year 1 = 100
Year 2 = 120
Year 3 = 140

Ratio of year 2 to 1 = (value in year 2) / (value in year 1)
= 120/100 = 1.2

Ration of year 3 to 2 = (value in year 3) / (value in year 2)
= 140/120 = 1.6667

Look, ratio of quantity increase in every two consecutive number is not same in the value taken by you; that is because you assumed that increase in the quantity is same every year. It is not anywhere mentioned in the question stem that quantity equally increases every year.

Here, the concept of compounding is used.

Here is the correct answer.
Suppose 1 becomes 2 after 6 years.
let x be the ratio by which 1 increases every year.

value of 1 after 6 years = 1 * (x)^6
so, 2 = 1 *(x)^6
Taking 6th root both side,
2^1/6 = x

So, x = 2^1/6

Value of 1 after 2 years = 1 * (X)^2
Value of 1 after 2 years = 1 * 2^1/6^2
Value of 1 after 2 years = 1 *2^1/3
Value of 1 after 2 years = 2^1/3 (which is third root of 2)

I hope this will be help to you in understanding.

Regards,
Balkrushna Vaghasia

SPatel1992 · Answer

Hi!
I used a number to gauge:

Year 1= 100, Year 2= 120, Year 3= 140, Year 4=160, Year 5= 180 and Year 6= 200 (double of Year 1).

120/100= 1.2. I thought the answer was root 2.

Is this the correct method? How did you arrive at cube root 2?

arorni · Answer

Hi,
Can someone help me with this question? Shouldn't we calculate the value of the geometric sequence term 6 as a1*r^5. So the answer should be 5th root of 2. But none of the answer choice is matching.

Posted from my mobile device

Gauravji21 · Answer

Question says the ratio of any two consecutive years is constant. 
So, if we start with first year as "a" i.e. the first term of GP..we will miss the ratio between the starting value (zeroth year) and the first year value.

Let the starting value (zeroth year) be "a"
By First year, value will be "aK".. where K= constant rate at which the value increases or will increase
 Second year = "aK" x "K" = "ak2" .... third year = ak3.....sixth year = ak6

ak6 = 2a => k = 2^ (1/6)
After two years.. value will be ak2 = ak2= 2^ (1/6 *2) =  2^(1/3)

altairahmad · Answer

Can you please elaborate on the highlighted statement. If we start with 'a' as first year, I don't quite get how the question stem puts a constraint on it. It just says that ratio between two consecutive years should be constant. It should be independent of the starting point, no ?

Bunuel   chetan2u

chetan2u · Answer

You start in any way but ensure you have six years..

start point -a
End of first year -ak
.......
End of 6th year -ak^6

Start point -1
End of first year -k
....
End of 6th year - k^6

So you can see a to ak is in one year, AK to ak^2 in two years and so on

chrtpmdr · Answer

What I don't get:

if we start at T1 = x 
 T2 = x*k 
 ...
 T6 = x*k^5

In general, for a geometric sequence xn = x*k^(n-1) 
How do we read out of the problem that in year 6 we increased by 6 and not by 5?

Sabby · Answer

Let's say our first term is 1 and as the 6th years it doubles, our 6th term is 2

We know that our ratio, let's call it x, is constant for 2 years then changes. As it changes every two years on a 6 year period, the total number of changes is 3.

As per exponential growth formula :

Final amount = original amount * multiplier^(number of changes).
2=1*x^3
2=x^3
x= 2^1/3

Answer D)

Elite097 · Answer

Bunuel   avigutman   ThatDudeKnows  how can we tell for sure that this series starts from a to ak^6 and not a to ak^5? Technically it should be latter because we are not given even whether a was the amount in the beginning or not and it doesnt make sense to assume randomly that we re measuring for end of year or so. Pls clarify how to identify such patterns

ThatDudeKnows · Answer

Elite097

I'll answer the question you asked, but then please answer mine.

Here's the answer to yours. 
We aren't given an "a" or a "k" at all. We are simply told that after six years, the amount will be double what we started with. If we start with $1, at the end of the first year, we have 1*k^1. At the end of the second year, we have 1*k^2. ... At the end of the sixth year, we have 1*k^6 = $2.

Here's mine.
Why are we messing around with "a" and/or "k" at all?
We start with $1. In six years, we have $2. We are asked what we have after two years. 
Is it 2 or greater than 2? No. A and B are out.
Is it less than 1? No. E is out. 
We are down to C and D. At this point, D sure   feels   like the better option, but we can prove it by just testing C or D. 
C looks easier to work with. If we have  \sqrt{2}  at the end of two years, we will have  \sqrt{2}\sqrt{2}  at the end of four years. That's just equal to 2. But we have 2 at the end of six years, so this isn't right and we can eliminate C.

Answer choice D.

Use the answer choices. You do not need to solve for the correct answer if you can prove that four are wrong. Plug In The Answers (PITA).
ThatDudeKnowsPITA

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