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An investment has been growing at a fixed annual rate of 20% since it
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gmatt1476
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An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  21 Sep 2019, 14:52
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An investment has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested. How much is the investment now worth?

(1) The value of the investment has increased by 44% since it was first made.
(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.



DS24931.01
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ZoltanBP
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  28 Sep 2019, 16:06
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gmatt1476 wrote:
An investment has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested. How much is the investment now worth?

(1) The value of the investment has increased by 44% since it was first made.
(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.

DS24931.01


The annual growth factor for the investment is 1.2. Let x be the current value of the investment. The original question: x=?

1) We know that the combined growth factor for the investment is 1.44, but no information is given about actual $ values. Thus, we can't get a unique value to answer the original question. \(\implies\) Insufficient

2) We know that the value of the investment was x/1.2 one year ago, and we can set up an equation about the effect of the hypothetical withdrawal.

(x/1.2-600)(1.2)=0.88x

Thus, we could get a unique value to answer the original question. \(\implies\) Sufficient

Answer: B
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  28 Sep 2019, 22:45
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ZoltanBP wrote:
gmatt1476 wrote:
An investment has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested. How much is the investment now worth?

(1) The value of the investment has increased by 44% since it was first made.
(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.

DS24931.01


The annual growth factor for the investment is 1.2. Let x be the current value of the investment. The original question: x=?

1) We know that the combined growth factor for the investment is 1.44, but no information is given about actual $ values. Thus, we can't get a unique value to answer the original question. \(\implies\) Insufficient

2) We know that the value of the investment was x/1.2 one year ago, and we can set up an equation about the effect of the hypothetical withdrawal.

(x/1.2-600)(1.2)=0.88x

Thus, we could get a unique value to answer the original question. \(\implies\) Sufficient

Answer: B


Although your solution is neat and clean, but I have a question over here. Whether we are not assuming here that interest is getting compounded annually, though the question simply says that rate of interest is annual ?
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  29 Sep 2019, 03:55
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abhishekmayank wrote:
Whether we are not assuming here that interest is getting compounded annually, though the question simply says that rate of interest is annual ?


We don't need to assume anything about the compounding frequency of the interest because the 20% rate in the question is a fixed annual growth rate and statement 2) refers to a situation exactly one year ago. A 20% fixed annual growth rate (or a 20% effective annual interest rate) can be achieved with different nominal interest rates with different compounding frequencies.

However, if statement 2) referred to a situation not exactly an integer year ago, then the compounding frequency would have an effect on the answer, and statement 2) would be Insufficient.
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  29 Sep 2019, 09:12
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gmatt1476 wrote:
An investment has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested. How much is the investment now worth?

(1) The value of the investment has increased by 44% since it was first made.
(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.



DS24931.01


If you haven't already, you should know the interest formula \(A = P (1 + r)^t\) (you can discard n because it's only compounded annually, unless stated otherwise)

R= 20%
\(A = P (1.20)^t\)
We are trying to find the Value of A, in order to find A, we need to know P and t.

1.) Value increased by 44%. By using logic, we know that the value increased by 44%, which means we can determine how much time has passed if an investment is earning 20% a year, but we don't know the value because we don't know the starting (P) or current value. INSUFFICENT

2.) Convert Statement 2 into math terms:
(P-600)(1.20)^(t-1) = 0.88[P*(1.20)^t]

Divide the whole equation by 0.88, and then divide (1.20)^(t-1).
(P-600)/0.88 = P(1.20)^t / (1.20)^(t-1)
The t cancels out, and now you can solve for P. SUFFICIENT!

Answer is B
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  30 Sep 2019, 05:05
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Annual rate: 20%
Interest earned gets reinvested every year.
Final amount= ?

St1: Let Principal be x.
@20%
Final after 1 year= 1.2x
Final after 2 years= 1.2*1.2x= 1.44x, a return of 44%.
So, time of investment is 2 years.
We need Principal amount to calculate 1.44x
Insufficient.

St2: Let Principal be x and time of investment be t years.
At 20%, Final will be x*(1.2)^t
Given, in (t-1)th years, $600 was withdrawn. So, amount left should be x*(1.2)^t-1 - 600.
Now, Final after t years will be:

{x*(1.2)^t-1 - 600}1.2 = 0.88*x*(1.2)^t

Calculating Final amount = x*1.2^t = $6000
Sufficient.

Ans B
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  01 Oct 2019, 21:58
ZoltanBP wrote:
abhishekmayank wrote:
Whether we are not assuming here that interest is getting compounded annually, though the question simply says that rate of interest is annual ?


We don't need to assume anything about the compounding frequency of the interest because the 20% rate in the question is a fixed annual growth rate and statement 2) refers to a situation exactly one year ago. A 20% fixed annual growth rate (or a 20% effective annual interest rate) can be achieved with different nominal interest rates with different compounding frequencies.

However, if statement 2) referred to a situation not exactly an integer year ago, then the compounding frequency would have an effect on the answer, and statement 2) would be Insufficient.


Thanks for the answer ! Even for 1 year there is an effect on compounding. Suppose that the interest is getting compounded half yearly then growth factor Each Year will not be 1.2 as mentioned, but would be 1.21 and so the answer will vary.
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  03 Oct 2019, 15:04
abhishekmayank wrote:
Thanks for the answer ! Even for 1 year there is an effect on compounding. Suppose that the interest is getting compounded half yearly then growth factor Each Year will not be 1.2 as mentioned, but would be 1.21 and so the answer will vary.


The annual growth factor cannot be 1.21 because it's fixed at 1.2 by the 20% fixed annual growth rate.

Whether the 1.2 annual growth factor is the result of a 20% nominal annual interest rate compounded yearly or it is the result of a certain, less than 20% nominal annual interest rate compounded half-yearly doesn't matter because statement 2) refers to a situation exactly one year ago for which we can apply the originally given growth factor.
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An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  15 Oct 2019, 09:09
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Let P is the principal amount, and investment niw worth is A

Statement 1-
P(1.2)^n=1.44P
n= 2 years
We can't say anything about P or A

Statement 2-
600+(20/100)*600= 0.12* A
We can find A



gmatt1476 wrote:
An investment has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested. How much is the investment now worth?

(1) The value of the investment has increased by 44% since it was first made.
(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.



DS24931.01
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  03 Jan 2020, 00:26
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gmatt1476 wrote:
An investment has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested. How much is the investment now worth?

(1) The value of the investment has increased by 44% since it was first made.
(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.

DS24931.01



Let P be the inital amount, n be the no. of yrs and A be the current worth
Given r = 20%
We need to find the value of "A"

(1) The value of the investment has increased by 44% since it was first made.
A = (1.44)P --(p)
We know that, A = P*(1.2)^n --(q)
based eq (p) and (q) we can deduce the value of 'n'.
But we still need value of P to find "A"
Hence, Insufficient.

(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.
Let K be the value of the investment one year ago.
K - 600 = (0.88)A --(m)
also, K(1.2) = A --(n)
based on eq (m) and (n) both value of K and A can be computed
Hence, Sufficient

OA <-- B
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  09 Feb 2020, 17:43
reynaldreni wrote:
gmatt1476 wrote:
An investment has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested. How much is the investment now worth?

(1) The value of the investment has increased by 44% since it was first made.
(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.

DS24931.01



Let P be the inital amount, n be the no. of yrs and A be the current worth
Given r = 20%
We need to find the value of "A"

(1) The value of the investment has increased by 44% since it was first made.
A = (1.44)P --(p)
We know that, A = P*(1.2)^n --(q)
based eq (p) and (q) we can deduce the value of 'n'.
But we still need value of P to find "A"
Hence, Insufficient.

(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.
Let K be the value of the investment one year ago.
K - 600 = (0.88)A --(m)
also, K(1.2) = A --(n)
based on eq (m) and (n) both value of K and A can be computed
Hence, Sufficient

OA <-- B




I have a question about your solution. Should the equation (m) be (K-600)*1.2 = 0.88A since (K-600) is last year and in order to compare with this year, we need to multiply with the interest rate.
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  27 Jun 2020, 01:49
reynaldreni wrote:
gmatt1476 wrote:
An investment has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested. How much is the investment now worth?

(1) The value of the investment has increased by 44% since it was first made.
(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.

DS24931.01



Let P be the inital amount, n be the no. of yrs and A be the current worth
Given r = 20%
We need to find the value of "A"

(1) The value of the investment has increased by 44% since it was first made.
A = (1.44)P --(p)
We know that, A = P*(1.2)^n --(q)
based eq (p) and (q) we can deduce the value of 'n'.
But we still need value of P to find "A"
Hence, Insufficient.

(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.
Let K be the value of the investment one year ago.
K - 600 = (0.88)A --(m)
also, K(1.2) = A --(n)
based on eq (m) and (n) both value of K and A can be computed
Hence, Sufficient

OA <-- B




Shouldn't you have eqn (m) --> (K - 600)x1.2 = (0.88)A
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  25 Aug 2020, 03:50
jsound996 wrote:

Divide the whole equation by 0.88, and then divide (1.20)^(t-1).
(P-600)/0.88 = P(1.20)^t / (1.20)^(t-1)
The t cancels out, and now you can solve for P. SUFFICIENT!


Could you please show this last step of your operation?
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  09 Sep 2020, 14:38
I see a point here, maybe, that none of the solution takes into account in statement 2, that we will need to add the amount of interest of 1 year, after the withdrawal of 600$. The statement tells that the now amount after withdrawal of 600 last would be .88 of the amount if no withdrawal of 600 was done.

How can we simply make this eqn: (P-600)(1.20)^(t-1) = 0.88[P*(1.20)^t]???

Someone please explain, i am going nuts over this question.
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  02 Nov 2020, 20:55
Statement 2 is sufficient. One needs to write down the equations and realize that we get two equations in two variables for statement 2.
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  12 Nov 2020, 02:52
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(1) Clearly insufficient

(2)

Let \(x\) = how much investment was worth one year ago
\(1.2x\) = how much the investment is worth now

We're told that if $600 is withdrawn one year ago, the investment would be worth 12% less than what it's worth today.

We can let \((x-600)\) represent the $600 withdrawn.

\(1.2(x-600) = 1.2x - 720\)
\(1.2x - 720\) is \(12%\) less than \(1.2x\).

\(1.2x - 720 = (0.88)(1.2x)\)
We can then solve for \(x\) and determine the value of \(1.2x\).

SUFFICIENT.

Answer is B.
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  06 Jun 2021, 01:23
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If you assume it was X 1 year ago, to be 1.2X in the older scenarios, then
Alternate equation can also be

(X-600) * 1.2 = 1.2X * 0.88
Solving this, X = 5000

But bear in mind you need 1.2X (the value of the investment today), so the answer becomes 1.2X = 6000

But yeah, this is DS, so you dont really need to solve this till the end.
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink] New post  13 Jun 2021, 12:01
wishmasterdj wrote:
If you assume it was X 1 year ago, to be 1.2X in the older scenarios, then
Alternate equation can also be

(X-600) * 1.2 = 1.2X * 0.88
Solving this, X = 5000

But bear in mind you need 1.2X (the value of the investment today), so the answer becomes 1.2X = 6000

But yeah, this is DS, so you dont really need to solve this till the end.


I used this approach. The actual rate will equal out and isnt even necessary.

Equation becomes: 0,12x = 600
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink]
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