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05 Dec 2018, 23:31
00:00

Difficulty:

35% (medium)

Question Stats:

70% (01:41) correct 30% (02:08) wrong based on 67 sessions

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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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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 18 Oct 2018 Posts: 60 Location: India Concentration: Finance, International Business GPA: 4 WE: Business Development (Retail Banking) Re: Tom saved$10,000 at a constant compound interest rate of r percent an  [#permalink]

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06 Dec 2018, 03:24
1
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 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6967 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Tom saved$10,000 at a constant compound interest rate of r percent an  [#permalink]

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09 Dec 2018, 18:04
1
=>

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

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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 12 Sep 2017 Posts: 137 Re: Tom saved$10,000 at a constant compound interest rate of r percent an  [#permalink]

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20 Jan 2019, 16:20
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 $$10,000(1+r)^{30} = 10,000((1+r)^{10})^3= 10,000*2^3 =10,000*8 = 80,000.$$

Hello MathRevolution !

Where does the next formula come from?

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

Kind regards!