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An investment of $1000 was made in a certain account and ear
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24 Nov 2012, 13:28
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65% (02:00) correct 35% (02:17) wrong based on 729 sessions
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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 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.
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Re: An investment of $1000 was made in a certain account and ear
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26 Nov 2012, 01:36
shanmugamgsn wrote: Heyy 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... !!! 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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Re: An investment of $1000 was made in a certain account and ear
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24 Nov 2012, 22:18
anon1 wrote: 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
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. 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.
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Re: An investment of $1000 was made in a certain account and ear
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24 Nov 2012, 17:57
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 ^^



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Re: An investment of $1000 was made in a certain account and ear
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24 Nov 2012, 19:48
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



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Re: An investment of $1000 was made in a certain account and ear
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Updated on: 24 Nov 2012, 21:17
anon1 wrote: 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 (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 !



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Re: An investment of $1000 was made in a certain account and ear
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25 Nov 2012, 17:50
Vips0000 wrote: anon1 wrote: 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
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. 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. 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... !!!
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Re: An investment of $1000 was made in a certain account and ear
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06 Jun 2014, 10:10
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



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Re: An investment of $1000 was made in a certain account and ear
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07 Dec 2014, 14:51
Bunuel wrote: shanmugamgsn wrote: Heyy 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... !!! 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. Hi Bunuel, Two questions regarding your approach: 1) Since interest is compounded, isn't the amount that increases after every annual period change? For example, if I invest $100 at a rate of 20% compounded annual  after year 1, I have 120. After year 2, I have 144 and so on. It rises a lot quicker because they are exponents  doesn't it? By the same token, how can we assume that it will take the same amount of time to double from $1000 to $2000 as $4000 to $800. I would assume that the latter would double a lot quicker? 2) If I was to follow the math, I get 4 = (1+r)^12. What exactly can we do to simplify this further. I tried to multiply both sides by (1/12) but that didn't really get me anywhere? If I simplified it further, I ended up getting: (2^6) = (100 + R)/(100) R = 6200? Thanks



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Re: An investment of $1000 was made in a certain account and ear
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07 Dec 2014, 15:37
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?



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Re: An investment of $1000 was made in a certain account and ear
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08 Dec 2014, 03:37
russ9 wrote: Bunuel wrote: shanmugamgsn wrote: Heyy 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... !!! 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. Hi Bunuel, Two questions regarding your approach: 1) Since interest is compounded, isn't the amount that increases after every annual period change? For example, if I invest $100 at a rate of 20% compounded annual  after year 1, I have 120. After year 2, I have 144 and so on. It rises a lot quicker because they are exponents  doesn't it? By the same token, how can we assume that it will take the same amount of time to double from $1000 to $2000 as $4000 to $800. I would assume that the latter would double a lot quicker? 2) If I was to follow the math, I get 4 = (1+r)^12. What exactly can we do to simplify this further. I tried to multiply both sides by (1/12) but that didn't really get me anywhere? If I simplified it further, I ended up getting:
(2^6) = (100 + R)/(100) R = 6200?Thanks 1. You can answer this yourself if you try simpler numbers. 2. Your simplification there is not correct. Do not understand what you've done. Also, why do you need to solve for r?
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Re: An investment of $1000 was made in a certain account and ear
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08 Dec 2014, 04:01
bankerboy30 wrote: 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? 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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Re: An investment of $1000 was made in a certain account and ear
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21 Mar 2015, 06:04
shanmugamgsn wrote: Vips0000 wrote: anon1 wrote: 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
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. 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. 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... !!! 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!



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Re: An investment of $1000 was made in a certain account and ear
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07 Jan 2017, 05:18
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



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Re: An investment of $1000 was made in a certain account and ear
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19 Dec 2017, 00:22
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



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An investment of $1000 was made in a certain account and ear
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19 Dec 2017, 12:31
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



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