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On a certain date, Pat invested $10,000 at x percent annual
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Updated on: 02 Jul 2014, 10:15
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On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000? A. 15 B. 16 C. 18 D. 20 E. 24
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Originally posted by arjtryarjtry on 22 Jul 2008, 18:59.
Last edited by Bunuel on 02 Jul 2014, 10:15, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.




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Re: On a certain date, Pat invested $10,000 at x percent annual
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03 Sep 2008, 12:17
$40,000 is 4 times the original amount.
Therefore it doubled twice in 12 years.
Therefore its doubling time is 6.
$80,000 is double $40,000 so another 6 years will get us to $80,000.
Therefore 12+6=18 is the necessary amount of time.




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Re: On a certain date, Pat invested $10,000 at x percent annual
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22 Jul 2008, 21:13
arjtryarjtry wrote: On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000? A. 15 B. 16 C. 18 D. 20 E. 24 x interest rate 80.000= 10.000 (1+x)^year => 8=(1+x)^year 40.000=10.000 . (1+x)^12 => 4= (1+x)^12 =>2= (1+x)^6 => 8 = (1+x)^ 18 So, after 18 years, the total value of the investment plus interest will increase to $80,000.



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Re: On a certain date, Pat invested $10,000 at x percent annual
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02 Sep 2008, 19:50
On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000? A. 15 B. 16 C. 18 D. 20 E. 24
for compound interest problems, when the time is large, then how does one approach???
dont just give the steps... also mention, for variations in the problems, how does one approach???



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Re: On a certain date, Pat invested $10,000 at x percent annual
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02 Sep 2008, 23:24
arjtryarjtry wrote: On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000? A. 15 B. 16 C. 18 D. 20 E. 24
for compound interest problems, when the time is large, then how does one approach???
dont just give the steps... also mention, for variations in the problems, how does one approach??? your second post blocked my post. i cannot do it without calculator or computer and also beleieve this is not real gmat type question cuz its very difficult to get the value without those machines. if i were to choose during the test, would go for 18/ or 20. probably 18 cuz not it wont take too long to get the value doubled....
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Re: On a certain date, Pat invested $10,000 at x percent annual
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03 Sep 2008, 01:22
given 40000=10000(1+x/100)^12 or 4 ^1/12 = (1+x/100) or x/100 = 4^1/121........A
asked 80000=10000(1+x/100)^n..............b n=?
substitute value of x/100 from A in B. 8 = (1+4^1/121)^n or 8= 4^n/12 2^3=2^2n/12 n=18
ans = C



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Re: On a certain date, Pat invested $10,000 at x percent annual
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03 Sep 2008, 06:31
arjtryarjtry wrote: On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000? A. 15 B. 16 C. 18 D. 20 E. 24
for compound interest problems, when the time is large, then how does one approach???
dont just give the steps... also mention, for variations in the problems, how does one approach??? 40k = 10k(1+x/100)^12 > 4=(1+x/100)^12 > (1+x/100)^6=2 80k=40k(1+x/100)^n > (1+x/100)^n =2 =(1+x/100)^6 n=6 Total years = 12+6=18
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Re: On a certain date, Pat invested $10,000 at x percent annual
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03 Sep 2008, 12:47
arjtryarjtry wrote: On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000? A. 15 B. 16 C. 18 D. 20 E. 24
for compound interest problems, when the time is large, then how does one approach???
dont just give the steps... also mention, for variations in the problems, how does one approach??? Let's simplify this question. First equation; 40,000 = 10,000(1 + x)^12 Given that (1 + x) = a and rearrange the equation above we get 4 = a^12 = 2^2 So now we know that a^12 = (a^6)^2 = 2^2 Therefore, (a^6) = 2 Second equation; 80,000 = 10,000(1 + x)^n Or 8 = a^n 2^3 = a^n (a^6)^3 = a^n = a^18  (since a^6 = 2) therefore, n = 18. C is the answer



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Re: On a certain date, Pat invested $10,000 at x percent annual
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03 Sep 2008, 13:42
Even though it looks like a tough question, it is actually not. 4 times in 12 yesrs = > 2 times in 6 years ( from 40K to 80K ). So total year = 12 +6 =18



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Re: On a certain date, Pat invested $10,000 at x percent annual
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02 Feb 2009, 02:35
$40,000 is 4 times the original amount.
Therefore it doubled twice in 12 years.
Therefore its doubling time is 6.
$80,000 is double $40,000 so another 6 years will get us to $80,000.
Therefore 12+6=18 is the necessary amount of time. ________________________
is this the right method???



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Re: On a certain date, Pat invested $10,000 at x percent annual
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02 Feb 2009, 08:51
GMAT TIGER wrote: arjtryarjtry wrote: On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000? A. 15 B. 16 C. 18 D. 20 E. 24
for compound interest problems, when the time is large, then how does one approach???
dont just give the steps... also mention, for variations in the problems, how does one approach??? your second post blocked my post. i cannot do it without calculator or computer and also beleieve this is not real gmat type question cuz its very difficult to get the value without those machines. if i were to choose during the test, would go for 18 or 20. probably 18 cuz it wont take too long to get the value doubled.... I found a method: Rule of 72. Given an x% return, it takes 10,000 to quadralope 12 years. So according to the rule: 72/x is the no of years 10,000.00 took to double 20,000.00. Again, 20,000.00 took to double 40,000.00 same (72/x) no of years. 72/x+ 72/x = 12 x = 12% (though rate here is not very much required). Again, 40,000.00 takes the same (72/x) no of years to double 80,000.00. 72/x = 6 years. So altogather: 10,000  20,000 = 6 years 20,000  40,000 = 6 years 40,000  80,000 = 6 years total 18 years.
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Re: On a certain date, Pat invested $10,000 at x percent annual
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26 Jun 2015, 23:29
The key is to identify how quickly the money gets doubled..
for the current problem money gets doubled in every 6 years..
t=0 Investment = 10000 t=6 Investment= 20000 t=12 Investment = 40000..
Therefore for money to double from 40,000 to 80,000 it will take another 6 years. i.e 18years



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Re: On a certain date, Pat invested $10,000 at x percent annual
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12 Dec 2015, 11:33
i tried different approaches, but I believe this is the most straight forward one. in 12y, the amount is 4x, where x is the initial amount. it means that if we start with 40k in year 12, in another 12 years, we'll have 160k. we can eliminate right away 24 years. now, if the amount 4x in 12y, it might be true that the amount will be 2x in 6y. ok, so we have 12y initial, then another 6 years, so total 18 years to reach 80k. C.



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Re: On a certain date, Pat invested $10,000 at x percent annual
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01 Dec 2016, 20:46
Tricky little bugger...
Compound Interest Formula: P[1+(r/n)]^(nt)
10,000(1+x)^12 = 40,000 (1+x)^12 = 4 (1+x) = 4^(1/12)
10,000(1+x)^t = 80,000 (4^(1/12))^t = 8 (2^2)^(1/12) = 2^3 (2^(1/6)^t = 2^3
Bases are equal, so we can set the exponents equal to each other > (1/6)t = 3 > t = 18
C.



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Re: On a certain date, Pat invested $10,000 at x percent annual
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