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shvm_sin7
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An investment has been growing at a fixed annual rate of 20% since it 21 Nov 2024, 07:43
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I assumed that the number of years of investment to be t. I believe this should be done in the equation based on quetion stem info. Please help Bunuel
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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We're told that if $600 is withdrawn one year ago, the investment would be worth 12% less than what it's worth today.

We can let (x−600) represent the $600 withdrawn.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$600 was withdrawn a year ago.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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An investment has been growing at a fixed annual rate of 20% since it 21 Nov 2024, 08:51
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shvm_sin7
I assumed that the number of years of investment to be t. I believe this should be done in the equation based on quetion stem info. Please help BunuelAn 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has been growing at a fixed annual rate of 20% since it was first made; no portion of the investment has been withdrawn, and all interest has been reinvested.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(2) If one year ago $600 had been withdrawn, today the investment would be worth 12% less than it is actually now worth.
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You are overcomplicating the problem. The second statement says that if $600 had been withdrawn 1 year ago, today the investment would be worth 12% less than it is actually worth now. These $600 would have grown to $600 * 1.2 = $720 today.

So, the investment would be worth $720 less today than its actual value. Since $720 represents 12% of the investment's current worth, the actual value of the investment today is $720 / 0.12 = $6,000.
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An investment has been growing at a fixed annual rate of 20% since it 23 Nov 2025, 06:45
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