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An investment of d dollars at k percent simple annual

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An investment of d dollars at k percent simple annual  [#permalink]

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New post 10 Dec 2010, 10:16
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An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 over a 3 year period?

A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3

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An investment of d dollars at k percent simple annual  [#permalink]

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An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period. In terms of d, what dollar amount invested at the same rate will yield $2,4,00 over a 3 year period?
A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3

Simple interest = principal * interest rate * time, where "principal" is the starting amount and "rate" is the interest rate at which the money grows per a given period of time (note: express the rate as a decimal in the formula). Time must be expressed in the same units used for time in the Rate.

An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period --> \(d*\frac{k}{100}*2=600\) --> \(k=\frac{30,000}{d}\)

What (x) dollar amount invested at the same rate will yield $2,400 over a 3 year period --> \(x*\frac{k}{100}*3=2,400\) --> \(x*\frac{30,000}{100d}*3=2,400\) --> \(x=\frac{8d}{3}\).

Answer: E.

Or another way: In order interest to be 4 times more in 1.5 times longer time period then investment must be 4d/1.5=8d/3.

Answer: E.
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Re: An investment of d dollars at k percent  [#permalink]

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New post 06 Feb 2012, 01:01
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Since you want to earn 4 times more interest in 1.5 times the current time period, the answer will be 4d/1.5 or 8d/3
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Re: Another simple one i guess...  [#permalink]

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New post 10 Dec 2010, 10:38
Thanks. Didnt know the Simple interest thing.

+1 from me for all ur help.

thanks.
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Re: an investment of d dollars at k percent  [#permalink]

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New post 18 Jun 2011, 06:58
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WishMasterUA wrote:
an investment of d dollars at k percent simple annual interest yield $600 interest over a 2-year period. in term of d, what dollar amount invested at the same rate will yild $2400 interest income over a 3-year period?
1) 2d/3
2) 3d/4
3) 4d/3
4) 3d/2
5) 8d/3

Please explain in detail how to solve this problem.

simple interest = PNR/100
dk2=600
d = 300/k
thus
(x*300/k)*k*3= 2400
X= 2400/900= 24/9 = 8/3

hence E
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Re: An investment of d dollars at k percent simple annual  [#permalink]

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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 09 Feb 2015, 11:27
Hi Bunuel,

If we take simple annual interest of 2% for $10,000 investment, for 2 years it will amount to $404.

How do we know "An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period" = d*k/100 *2
and not similar to $404 calc?

Thanks
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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 09 Feb 2015, 12:22
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Hi nitestr,

The GMAT will likely ask you to deal with an "interest formula" question once or twice, so you need to note the wording of the prompt and you need to know the formulas for simple interest and compound interest.

Simple Interest = (Principal)(1 + RT)

Compound Interest = (Principal)(1 + R)^T

R = interest rate written as a decimal
T = length of time (usually years)

So, if we calculate 2% simple interest on a $10,000 principal over 2 years, we'll have...

(10,000)(1 + (.02)(2))
(10,000)(1.04) = 10,400
The interest = 400 (not 404)

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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 25 May 2015, 10:54
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EMPOWERgmatRichC wrote:
Hi nitestr,

The GMAT will likely ask you to deal with an "interest formula" question once or twice, so you need to note the wording of the prompt and you need to know the formulas for simple interest and compound interest.

Simple Interest = (Principal)(1 + RT)

Compound Interest = (Principal)(1 + R)^T

R = interest rate written as a decimal
T = length of time (usually years)

So, if we calculate 2% simple interest on a $10,000 principal over 2 years, we'll have...

(10,000)(1 + (.02)(2))
(10,000)(1.04) = 10,400
The interest = 400 (not 404)

GMAT assassins aren't born, they're made,
Rich


Hi Rich,

The formula for the Simple interest is S.I = Principal (RT) no ?

Thanks for your clarification.

Regards
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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 25 May 2015, 21:31
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Hi SD007,

If you want to calculate the amount of simple interest generated, then YES, the calculation is...

(Principal)(RT)

If you're interested in the TOTAL amount of money (e.g. in a bank account) AFTER the simple interest is generated, then the calculation is....

(Principal)(1 + RT)

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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 12 Jul 2015, 19:01
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Simple way to solve it -
600 dollars in 2 years means 300 dollars in 1 year.
To get 2400 dollars' it will take 8 years.
To get 2400 in 3 years, we need 8/3 times money. Answer is E.
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An investment of d dollars at k percent simple annual  [#permalink]

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New post 15 Dec 2015, 04:49
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144144 wrote:
An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 over a 3 year period?

A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3


Hi guys, I just did my first CAT and this one was the first question that I answered correctly.
We don't need to involve any intetrest formulas at all here: the ratio of interest earnings per year = \(\frac{800}{300} = \frac{8}{3}\) so the ratio of the investment that generates 2400 (let it be X) to D is the same (interest rate remains the same)
\(\frac{8}{3}\)=\(\frac{x}{d}\) --> \(x=\frac{8}{3}*d\)
Answer E
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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 13 Jan 2018, 06:43
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144144 wrote:
An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 over a 3 year period?

A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3


An investment of d dollars at k percent simple annual interest yields $600 over a 2-year period.
In other words, an investment of d dollars yields $300 in interest each year.

What dollar amount invested at the same rate will yield $2,400 interest over a 3-year period?
In other words, how much money must we invest to earn $800 in interest EACH YEAR?

If a d dollar investment yields $300 in interest EACH YEAR, then:
- a 2d dollar investment would yield $600 (2 times $300) in interest EACH YEAR
- a 3d dollar investment would yield $900 (3 times $300) in interest EACH YEAR
- a 4d dollar investment would yield $1200 (4 times $300) in interest EACH YEAR
- etc.

From here there are two approaches.

APPROACH #1
We want the ANNUAL interest to be $800.
This means we must invest an amount that is BETWEEN 2d dollars and 3d dollars [since $800 is BETWEEN $600 and $900].
When we check the answer choices, only E, which can be written as (8/3)d, is BETWEEN 2d dollars and 3d
So, the correct answer must be E


APPROACH #2
To increase the ANNUAL interest from $300 to $800, we must invest 800/300 TIMES as much money.
800/300 = 8/3, so we must invest (8/3)d dollars [aka 8d/3 dollars]
Answer = E

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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 15 Mar 2018, 18:06
Bunuel wrote:
An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period. In terms of d, what dollar amount invested at the same rate will yield $2,4,00 over a 3 year period?
A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3

Simple interest = principal * interest rate * time, where "principal" is the starting amount and "rate" is the interest rate at which the money grows per a given period of time (note: express the rate as a decimal in the formula). Time must be expressed in the same units used for time in the Rate.

An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period --> \(d*\frac{k}{100}*2=600\) --> \(k=\frac{30,000}{d}\)

What (x) dollar amount invested at the same rate will yield $2,400 over a 3 year period --> \(x*\frac{k}{100}*3=2,400\) --> \(x*\frac{30,000}{100d}*3=2,400\) --> \(x=\frac{8d}{3}\).

Answer: E.

Or another way: In order interest to be 4 times more in 1.5 times longer time period then investment must be 4d/1.5=8d/3.

Answer: E.

Bunuel.

I'm more interested in the "Another way" to solve it. I get the "4 times more" but I can't get to the "1.5 times longer". How do you calculate this?

Thanks.

Best.
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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 15 Mar 2018, 19:52
MarkHanna wrote:
Bunuel wrote:
An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period. In terms of d, what dollar amount invested at the same rate will yield $2,4,00 over a 3 year period?
A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3

Simple interest = principal * interest rate * time, where "principal" is the starting amount and "rate" is the interest rate at which the money grows per a given period of time (note: express the rate as a decimal in the formula). Time must be expressed in the same units used for time in the Rate.

An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period --> \(d*\frac{k}{100}*2=600\) --> \(k=\frac{30,000}{d}\)

What (x) dollar amount invested at the same rate will yield $2,400 over a 3 year period --> \(x*\frac{k}{100}*3=2,400\) --> \(x*\frac{30,000}{100d}*3=2,400\) --> \(x=\frac{8d}{3}\).

Answer: E.

Or another way: In order interest to be 4 times more in 1.5 times longer time period then investment must be 4d/1.5=8d/3.

Answer: E.

Bunuel.

I'm more interested in the "Another way" to solve it. I get the "4 times more" but I can't get to the "1.5 times longer". How do you calculate this?

Thanks.

Best.


(3 year period)/(2 year period) = 1.5.
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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 16 Mar 2018, 02:30
Thank you Bunuel! :-)
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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 19 Mar 2018, 15:14
144144 wrote:
An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 over a 3 year period?

A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3


Recall that the amount of simple interest, I, is equal to the principal, P, times the annual interest rate, r, times the number of years, t. That is, I = Prt.

We are given that I = 600, P = d, r = k/100 and t = 2, so we have:

600 = d(k/100)(2)

300 = dk/100

dk = 30,000

However, we are being asked for the dollar amount invested, in terms of d, at the annual interest rate of k percent that will yield $2,400 interest over a 3-year period. If we let this dollar amount be x, then

2,400 = x(k/100)(3)

800 = xk/100

xk = 80,000

Since dk = 30,000 and xk = 80,000, k = 30,000/d and k = 80,000/x. Thus, we can equate 30,000/d and 80,000/x and isolate x.

30,000/d = 80,000/x

30,000x = 80,000d

3x = 8d

x = 8d/3

Answer: E
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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 03 Apr 2018, 11:23
why aren't we using the general formula of: (1+x)*d+(1+x)^2*d??
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Re: An investment of d dollars at k percent simple annual  [#permalink]

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New post 30 May 2018, 11:35
Bunuel wrote:
An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period. In terms of d, what dollar amount invested at the same rate will yield $2,4,00 over a 3 year period?
A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3

Simple interest = principal * interest rate * time, where "principal" is the starting amount and "rate" is the interest rate at which the money grows per a given period of time (note: express the rate as a decimal in the formula). Time must be expressed in the same units used for time in the Rate.

An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period --> \(d*\frac{k}{100}*2=600\) --> \(k=\frac{30,000}{d}\)

What (x) dollar amount invested at the same rate will yield $2,400 over a 3 year period --> \(x*\frac{k}{100}*3=2,400\) --> \(x*\frac{30,000}{100d}*3=2,400\) --> \(x=\frac{8d}{3}\).

Answer: E.

Or another way: In order interest to be 4 times more in 1.5 times longer time period then investment must be 4d/1.5=8d/3.

Answer: E.


hi pushpitkc -

one question, why do we multiply by 2 like this \(d*\frac{k}{100}*2=600\) shoudnt we put 2 as exponent as per formula \(600 = d(1+\frac{k}{100})^2\)

can you explain why am i thining so ? :-)

thankns!
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An investment of d dollars at k percent simple annual  [#permalink]

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New post 30 May 2018, 11:45
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dave13 wrote:
Bunuel wrote:
An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period. In terms of d, what dollar amount invested at the same rate will yield $2,4,00 over a 3 year period?
A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3

Simple interest = principal * interest rate * time, where "principal" is the starting amount and "rate" is the interest rate at which the money grows per a given period of time (note: express the rate as a decimal in the formula). Time must be expressed in the same units used for time in the Rate.

An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period --> \(d*\frac{k}{100}*2=600\) --> \(k=\frac{30,000}{d}\)

What (x) dollar amount invested at the same rate will yield $2,400 over a 3 year period --> \(x*\frac{k}{100}*3=2,400\) --> \(x*\frac{30,000}{100d}*3=2,400\) --> \(x=\frac{8d}{3}\).

Answer: E.

Or another way: In order interest to be 4 times more in 1.5 times longer time period then investment must be 4d/1.5=8d/3.

Answer: E.


hi pushpitkc -

one question, why do we multiply by 2 like this \(d*\frac{k}{100}*2=600\) shoudnt we put 2 as exponent as per formula \(600 = d(1+\frac{k}{100})^2\)

can you explain why am i thining so ? :-)

thankns!
D


Hey dave13

This problem deals with simple interest.
The formula for simple interest - Principal * Interest * \(\frac{R}{100}\) where R is Rate of interest

Hope this helps you!
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