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Tom saved $10,000 at a constant compound interest rate of r percent an

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New post 06 Dec 2018, 00:31
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[Math Revolution GMAT math practice question]

Tom saved $10,000 at a constant compound interest rate of r percent annually. After 10 years, the balance is double the principal. What will be the balance 30 years after the deposit?

A. $30,000
B. $40,000
C. $50,000
D. $60,000
E. $80,000

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Re: Tom saved $10,000 at a constant compound interest rate of r percent an  [#permalink]

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New post 06 Dec 2018, 04:24
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The question says,the balance got doubled in 10 years of time.This means Balance after 10 years is 10,000*2= 20,000$ .Going by the same logic, and since in 20 years, the balance will be double of the balance in 10 years, which means 20,000*2= 40,000$. In another 10 years, that means, at the end of 30 years, the balance will again double up. The final balance will thus become 40,000*2= 80,000$.

Hence Answer is E
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New post 09 Dec 2018, 19:04
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=>

After \(10\) years, we have \(10,000(1+r)^{10} = 20,000\) or \((1+r)^{10} = 2.\)
So, after \(30\) years the balance is \(10,000(1+r)^{30} = 10,000((1+r)^{10})^3= 10,000*2^3 =10,000*8 = 80,000.\)

Therefore, the answer is E.
Answer: E

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Re: Tom saved $10,000 at a constant compound interest rate of r percent an  [#permalink]

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New post 20 Jan 2019, 17:20
MathRevolution wrote:
=>

After \(10\) years, we have \(10,000(1+r)^{10} = 20,000\) or \((1+r)^{10} = 2.\)
So, after \(30\) years the balance is \(10,000(1+r)^{30} = 10,000((1+r)^{10})^3= 10,000*2^3 =10,000*8 = 80,000.\)

Therefore, the answer is E.
Answer: E


Hello MathRevolution !

Where does the next formula come from?

\(10,000(1+r)^{10} = 20,000\)

Thank you in advance.

Kind regards!
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Re: Tom saved $10,000 at a constant compound interest rate of r percent an   [#permalink] 20 Jan 2019, 17:20
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