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Difficulty:
55%
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Question Stats:
67% (01:59) correct
33%
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wrong
based on 2118
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An investment of $1000 was made in a certain account and earned interest that was compounded annually. The annual interest rate was fixed for the duration of the investment, and after 12 years the $1000 increased to $4000 by earning interest. In how many years after the initial investment was made the $1000 have increased to $8000 by earning interest at that rate?
A. 16
B. 18
C. 20
D. 24
E. 30
A. 16
B. 18
C. 20
D. 24
E. 30
I have the solution presented by GMAT prep but it is not the way I like to do it. My strategy would simply be to find the interest rate. And then work backwards from the solutions to get the answer (an interest of 7000, or total of 8000).
But I'm having trouble calculating the rate here?
Can anyone help?
in 12 years he made 3000 in interest. So shouldn't it be 1000(r/100)^12 = 3000 ?? then solve for r ?
I keep getting a number in the 30s, or 36.... ugh.
But I'm having trouble calculating the rate here?
Can anyone help?
in 12 years he made 3000 in interest. So shouldn't it be 1000(r/100)^12 = 3000 ?? then solve for r ?
I keep getting a number in the 30s, or 36.... ugh.
anon1
Its actually an easier question with a simpler approach.
Amount 1000 is invested for 12 yrs and yields 4000. Amount is compounded annually.
we need number of years it will take to yield 8000.
from 1000 to 4000 = 4 times
from 4000 to 8000 = 2 times.
if compounding is happening at same rate throughout, then
since, time required for 4 times = 12 yrs
=>time required for 2 times = 6 yrs
Total time from initial = 12+6 = 18 yrs
Ans B.
shanmugamgsn
An investment of $1000 was made in a certain account and earned interest that was compounded annually. The annual interest rate was fixed for the duration of the investment, and after 12 years the $1000 increased to $4000 by earning interest. In how many years after the initial investment was made the $1000 have increased to $8000 by earning interest at that rate?
A. 16
B. 18
C. 20
D. 24
E. 30
In 12 years the investment quadrupled (from $1,000 to $4,000).
Thus, at the same rate compounded annually, it would need additional 12/2=6 years to double (from $4,000 to $8,000).
Therefore, 12+6=18 years are needed $1,000 to increase to $8,000.
Answer: B.
Hope it's clear.
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Posts: 2
HI anon1,I think you miss the point " compound interest" , the formula must be
1000* (1+r)^12 = 4000 (1)
1000* (1+r)^x =8000 (2) ( x : is the number of years we must find)
(1): --> 1+r= (12 root of 4)
(2)--> (1+r)^x=( 12 root of 4)^x = 8
---> 4^(x/12) =8
2^(x/6)=2^3
x/6=3 >>>x=18
hope this help ^^
1000* (1+r)^12 = 4000 (1)
1000* (1+r)^x =8000 (2) ( x : is the number of years we must find)
(1): --> 1+r= (12 root of 4)
(2)--> (1+r)^x=( 12 root of 4)^x = 8
---> 4^(x/12) =8
2^(x/6)=2^3
x/6=3 >>>x=18
hope this help ^^
General Discussion
I'm sorry, I'm losing you.
I'm getting
(1) (1+r)^12 = 4
(2) (1+r)^x = 8
I don't know what goes on after this... not understanding your work
But why can't we just find out what the r first and then work backwards from the solutions to find the amount of years necessary to get an interest of 7000 / total of 8000
I'm getting
(1) (1+r)^12 = 4
(2) (1+r)^x = 8
I don't know what goes on after this... not understanding your work
But why can't we just find out what the r first and then work backwards from the solutions to find the amount of years necessary to get an interest of 7000 / total of 8000
Posts: 2
Updated on: 24 Nov 2012, 20:17
Originally posted by tiennguyen1701 on 24 Nov 2012, 20:04.
Last edited by tiennguyen1701 on 24 Nov 2012, 20:17, edited 2 times in total.
Last edited by tiennguyen1701 on 24 Nov 2012, 20:17, edited 2 times in total.
anon1
(1): --> 1+r = √(12&4) (get 12 root of 4) then apply this to (2)
(2)--> (1+r)^x=√(12&4) ^x = 8
---> 4^(x/12) =8
2^(x/6)=2^3
x/6=3 >>>x=18
if you find r it hard for you to calculate !
shanmugamgsn
Joined: 04 Oct 2011
Last visit: 31 Dec 2014
Posts: 140
Location: India
Concentration: Entrepreneurship, International Business
GMAT 1: 440 Q33 V13
GPA: 3
GMAT 1: 440 Q33 V13
Posts: 140
Vips0000
Hey Vips,
I tried to figure out
from 1000 to 4000 = 4 times
from 4000 to 8000 = 2 times.
But couldn't
You mean 1000 to 4000 = 4*1000
4000 to 8000 = 2 *2000
If so how u took first as 1000 and next as 2000 ..... Or any other logic?
Can you brief it plzzz... !!!
I used a logical approach to solve this:
In 12 years, the investment doubled twice. Therefore, in the first 6 years, the investment doubled from 1000 to 2000, and in the next 6 years the investment doubled again from 2000 to 4000. Since the rate of return remains the same, the investment will take another 6 years to double from 4000 to 8000.
Total time is 12+6=18
In 12 years, the investment doubled twice. Therefore, in the first 6 years, the investment doubled from 1000 to 2000, and in the next 6 years the investment doubled again from 2000 to 4000. Since the rate of return remains the same, the investment will take another 6 years to double from 4000 to 8000.
Total time is 12+6=18
Posts: 71
Bunuel following that same logic: if I put 1000 into an account with 100% interest in year 1 I will get 2000, year 2 4000 year 3 8000. So I increased my investment eight fold in three years. Now if I want my investment to go to double to 16,000 it would take one more year. But according to what has been said above 3/4 would be the answer. Are you counting the initial deposit as year 1?
bankerboy30
At the rate of 100% (*2) the investment increases 8 times in 3 years.
In 3 years the investment increases 2*2*2 = 8 times (from $1,000 to $8,000). The doubling period is 1 year.
Thus, at the same rate compounded annually, it would need additional 3/3 = 1 year to double (from $8,000 to $16,000).
Therefore, 3+1=4 years are needed $1,000 to increase to $16,000.
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Posts: 4
shanmugamgsn
We can solve this another way-
The amount 1000 quadrupled in 12 years- that means the investment increased by 1/3rd (4/12) each year.
Thus to increase the investment by 2 more times ( 1/3 a year + 1/3+ 1/3+ 1/3 + 1/3 + 1/3) it will take 6 years.
12+ 6 = 18 years!
DeD
Joined: 30 Aug 2016
Last visit: 20 Jan 2026
Posts: 3
Location: Italy
Schools: HEC Sept19 intake (A$$)
GMAT 1: 700 Q47 V41
GPA: 3.83
The math is not that bad if you just use the exponential notation for roots. We have:
1000 (1+r) ^12 = 4000 ----> (1+r) = 4 ^(1/12) = 2 ^ (1/6)
1000 (1+r) ^x = 8000 ----> (1+r) = 8 ^ (1/x) = 2 ^ (3/x)
Find x:
1 / 6 = 3 / x -----> x = 18
1000 (1+r) ^12 = 4000 ----> (1+r) = 4 ^(1/12) = 2 ^ (1/6)
1000 (1+r) ^x = 8000 ----> (1+r) = 8 ^ (1/x) = 2 ^ (3/x)
Find x:
1 / 6 = 3 / x -----> x = 18
Posts: 5
Can solve it like this...
Taking simple numbers: Invested $1 and it grew $4 in 12 years
4 = (1+R)^12
Taking square root both the side, we get: 2 = (1+R)^6
To solve for $8:
Multiply both equation we get: 4*2 = (1+R)^(12+6)
8 = (1+R)^18
Answer: 18 years
Taking simple numbers: Invested $1 and it grew $4 in 12 years
4 = (1+R)^12
Taking square root both the side, we get: 2 = (1+R)^6
To solve for $8:
Multiply both equation we get: 4*2 = (1+R)^(12+6)
8 = (1+R)^18
Answer: 18 years
Posts: 34
start=1000
year1= 1000x
year2=1000x^2
...
year 12=1000x^12
so 1000x^12=4000
x^12=4
1000(x^12)^?=8000
(x^12)^?=8
since x^12=4, 4^?=8
2^2y=2^3
2y=3 so y=3/2
1000(x^12)^(3/2)=8000
Using the power property, we obtain 1000(x^18)=8000
so 18 years
year1= 1000x
year2=1000x^2
...
year 12=1000x^12
so 1000x^12=4000
x^12=4
1000(x^12)^?=8000
(x^12)^?=8
since x^12=4, 4^?=8
2^2y=2^3
2y=3 so y=3/2
1000(x^12)^(3/2)=8000
Using the power property, we obtain 1000(x^18)=8000
so 18 years
Posts: 292
Not the best way but made an educated guess by estimating with simple interest:
-- If it took 12 years to get from 1000 to 4000, it should take considerably less than +12 to get to 8000 since we will be increasing the amount of interest added each year.
-- If we used simple interest the answer would be 28 years (3000/12 so $1000 per 4 years).
E) is definitely too big
D) is out too because 24 is too close to 28 (compounding should make the amount of time much less)
A) is definitely too small (no way to make up that much in only 4 years even with compounding)
Between B and C, I thought B was more likely to be correct since it's closer to 12 and 20 is exactly 8 away from 12 and 28.
-- If it took 12 years to get from 1000 to 4000, it should take considerably less than +12 to get to 8000 since we will be increasing the amount of interest added each year.
-- If we used simple interest the answer would be 28 years (3000/12 so $1000 per 4 years).
E) is definitely too big
D) is out too because 24 is too close to 28 (compounding should make the amount of time much less)
A) is definitely too small (no way to make up that much in only 4 years even with compounding)
Between B and C, I thought B was more likely to be correct since it's closer to 12 and 20 is exactly 8 away from 12 and 28.
\(1000* \frac{(100+r)}{100}^{12} = 4000\)
let 100+r/100 be t
t^12=4
t= 4^1/12
to find:
1000*t^n=8000
substituting value of t
4^n/12=8
2^2n/12=2^3
n/6=3
n=18
let 100+r/100 be t
t^12=4
t= 4^1/12
to find:
1000*t^n=8000
substituting value of t
4^n/12=8
2^2n/12=2^3
n/6=3
n=18
Let me give a simplified approach to such questions
use percentages to solve this ...
Increase of 300% (from 1k to 4k) in 12 Years....
Next 12 Years, again 300%.. which comes out to be 16k, but we need 8k.... so divide time/2 ... ie. 6 more Years...
Total = 12 Years + 6 Years = 18 Years
use percentages to solve this ...
Increase of 300% (from 1k to 4k) in 12 Years....
Next 12 Years, again 300%.. which comes out to be 16k, but we need 8k.... so divide time/2 ... ie. 6 more Years...
Total = 12 Years + 6 Years = 18 Years
We're told : \(1,000 (1 + \frac{r}{100})^{12} = 4,000\)
= \( (1 + \frac{r}{100})^{12} = 4\)
= \( (1 + \frac{r}{100})^{6} = 2\)
= \( (1 + \frac{r}{100})^{18} = 2^3 = 8\)
After 18 years, the initial investment would have increased to $8,000.
= \( (1 + \frac{r}{100})^{12} = 4\)
= \( (1 + \frac{r}{100})^{6} = 2\)
= \( (1 + \frac{r}{100})^{18} = 2^3 = 8\)
After 18 years, the initial investment would have increased to $8,000.
Posts: 125
Bunuel
Bunuel Could you explain why you took 3/3? Based on the original question posted by someone else, I also arrived at 3/4.
for 8 times -> it took 3 years. to double -> it takes 3/4th of year using proportions.
I thought you followed the same pattern with original question:
for 4 times -> it took 12 years. To double -> it takes 6 years using proportions.
where are we missing ?
GMAT Focus 1: 745 Q86 V90 DI85
Posts: 3,004
The tricky thing here is that the method folks are (correctly) using to determine doubling periods does NOT automatically tell us the annual rate of change. When we're dealing with exponential growth, it's not correct to say that something that quadruples in 12 years is growing by 4/12 = 1/3 of the initial investment per year. If that were the case, the growth would not be exponential.
The helpful thing about doubling periods is that they give us a nice stable number to work with. Once our exponential growth leads to a doubling, that same time span will keep producing another doubling. So we're working specifically with number of DOUBLINGS, not total growth. In the original example, there were 2 doublings and we just needed one, so it would take 1/2 as long. In the follow-up question, there were 3 doublings and we just wanted one, so it would take 1/3 as long.
If we don't have a clean number like 2 or 3, you can see the trouble we get into. What if our investment has grown by 60% after 3 years? Has it grown by 20% each year? No, that's just the average growth. We've multiplied by some factor 3 separate times to reach 1.6, so x^3 = 1.6. That's not going to be too fun without a calculator, although you can predict that the annual rate will be a bit less than 20% or the compounding would cause the growth to exceed 60%.
_________________
The helpful thing about doubling periods is that they give us a nice stable number to work with. Once our exponential growth leads to a doubling, that same time span will keep producing another doubling. So we're working specifically with number of DOUBLINGS, not total growth. In the original example, there were 2 doublings and we just needed one, so it would take 1/2 as long. In the follow-up question, there were 3 doublings and we just wanted one, so it would take 1/3 as long.
If we don't have a clean number like 2 or 3, you can see the trouble we get into. What if our investment has grown by 60% after 3 years? Has it grown by 20% each year? No, that's just the average growth. We've multiplied by some factor 3 separate times to reach 1.6, so x^3 = 1.6. That's not going to be too fun without a calculator, although you can predict that the annual rate will be a bit less than 20% or the compounding would cause the growth to exceed 60%.
sriharsha4444
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