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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink]
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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has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested.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(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.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]
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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]
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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]
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]
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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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink]
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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]
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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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An investment has been growing at a fixed annual rate of 20% since it [#permalink]
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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.


(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]
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


Hi ZoltanBP! Why are you dividing "x" per "1.2" in the second statement? Why not just x? Tks! :thumbsup:
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink]
Will2020 wrote:
Why are you dividing "x" per "1.2" in the second statement? Why not just x?


Because growth factor = new value / old value, it must be true that old value = new value / growth factor. In my solution, x is the new value, and 1.2 is the annual growth factor, so the old value, or the value of the invetsment one year ago, must be x/1.2.
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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 investment is compounded. We need the amount now.
Annual rate of interest, r = 20%

(1) The value of the investment has increased by 44% since it was first made.

This means that the investment has completed 2 yrs (20% compounded over 2 yrs will give 44%). But we don't know the principal. We don't know how much was invested. So we don't know what it has become now. Not sufficient alone.

(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.

If one year ago, 600 had been withdrawn, today the investment would be less by 600 and by the interest earned in the year on this 600 (which is 20% of 600 i.e. 120). So the total investment amount would have been less by $720 today. This 720 is 12% of the amount i.e.
12% of Total Amount = 720
Total Amount = $6000
Sufficient alone. (Note that we don't really need to do these calculations in the actual exam but we should know that we can calculate the total amount)

Answer (B)
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink]
KarishmaB 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 investment is compounded. We need the amount now.
Annual rate of interest, r = 20%


(1) The value of the investment has increased by 44% since it was first made.

This means that the investment has completed 2 yrs (20% compounded over 2 yrs will give 44%). But we don't know the principal. We don't know how much was invested. So we don't know what it has become now. Not sufficient alone.

(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.

If one year ago, 600 had been withdrawn, today the investment would be less by 600 and by the interest earned in the year on this 600 (which is 20% of 600 i.e. 120). So the total investment amount would have been less by $720 today. This 720 is 12% of the amount i.e.
12% of Total Amount = 720
Total Amount = $6000
Sufficient alone. (Note that we don't really need to do these calculations in the actual exam but we should know that we can calculate the total amount)

Answer (B)

Will2020


Hi KarishmaB! Regarding statement (1), I did some quick math to understand your explanation that 20% compounded in 2 years will give 44%. In fact, I used a smart number - $100 compounded by 20% annually - generating an amount of $144 (44% more than $100); but how can I realize that faster? How did you rationalise this? Thank you! :please:
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink]
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Will2020 wrote:
KarishmaB 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 investment is compounded. We need the amount now.
Annual rate of interest, r = 20%


(1) The value of the investment has increased by 44% since it was first made.

This means that the investment has completed 2 yrs (20% compounded over 2 yrs will give 44%). But we don't know the principal. We don't know how much was invested. So we don't know what it has become now. Not sufficient alone.

(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.

If one year ago, 600 had been withdrawn, today the investment would be less by 600 and by the interest earned in the year on this 600 (which is 20% of 600 i.e. 120). So the total investment amount would have been less by $720 today. This 720 is 12% of the amount i.e.
12% of Total Amount = 720
Total Amount = $6000
Sufficient alone. (Note that we don't really need to do these calculations in the actual exam but we should know that we can calculate the total amount)

Answer (B)

Will2020


Hi KarishmaB! Regarding statement (1), I did some quick math to understand your explanation that 20% compounded in 2 years will give 44%. In fact, I used a smart number - $100 compounded by 20% annually - generating an amount of $144 (44% more than $100); but how can I realize that faster? How did you rationalise this? Thank you! :please:


It it were simple interest, we would have seen an increase of 20% every year and hence in 2 years, the amount would have increased by 40%, in 3 years by 60% and so on.
Since it is compound interest, the amount will increase by 20% at the end of first year but at the end of second year, it will be somewhat more than 40%. Had 3 years passed, then the amount would have increased by more than 60%. Since we know that the amount increased by 44%, it stands to reason that 2 years must have passed. Also, (6/5)*(6/5) = 36/25 = 144/100 (a successive increase of 20%s gives an effective increase of 44%) or we can use the formula (Total increases is (a + b + ab/100)% where a = 20 and b = 20)

Once you practice enough questions, you are able to quickly relate simple numbers.
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An investment has been growing at a fixed annual rate of 20% since it [#permalink]
Answer OPTION B

Official explanation -


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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink]
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


Can anyone please explain where this 720 come from?

Thanks in advance!
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Re: An investment has been growing at a fixed annual rate of 20% since it [#permalink]
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Rebaz wrote:
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


Can anyone please explain where this 720 come from?

Thanks in advance!


$600 was withdrawn a year ago. This amount, in one year at a 20% interest rate, would amount to $600*1.2 = $720.
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