Interest problem

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Interest problem

by metallicafan » Tue Dec 18, 2012 5:34 pm
An investment receives 20% annual interest compounded semi-annually for 2 years. If its final value is $29,200, approximately what was the initial investment?

A $17000
B $18000
C $19000
D $20000
E $21000

A fast way to solve it? Thanks!

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by puneetkhurana2000 » Tue Dec 18, 2012 6:48 pm
29200 = x(1 + 10/100)^4.

Solving we get x = 19943.99.

Answer D.

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by GMATGuruNY » Tue Dec 18, 2012 7:28 pm
metallicafan wrote:An investment receives 20% annual interest compounded semi-annually for 2 years. If its final value is $29,200, approximately what was the initial investment?

A $17000
B $18000
C $19000
D $20000
E $21000

A fast way to solve it? Thanks!
20% interest compounded semi-annually means that 10% is earned every 6 months.
We can plug in the answers, which represent the initial investment.
Problems about compound interest tend to involve VERY ROUND numbers.
Thus, we should start with D:
1st 6 months: 20000 + 2000 = 22000.
Next 6 months: 22000 + 2200 = 24200.
Next 6 months: 24200 + 2420 ≈ 26600.
Next 6 months: 26600 + 2660 ≈ 29200.
Success!

The correct answer is D.
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by shreerajp99 » Wed Dec 19, 2012 10:12 am
Hi Mitch,

U use nos effectively while solving questions.What is ur approach when u see a quant question?How do u decide which option to be tried 1st/when to use numbers instead of variables given in the question?
I have seen that in inequalities/absolute values questions,inserting numbers can be a recipe for disaster.What's ur take on it?

Thanks,
Shreeraj

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by Scott@TargetTestPrep » Fri Nov 22, 2019 11:32 am
metallicafan wrote:An investment receives 20% annual interest compounded semi-annually for 2 years. If its final value is $29,200, approximately what was the initial investment?

A $17000
B $18000
C $19000
D $20000
E $21000

A fast way to solve it? Thanks!
We use the compound interest formula: A = P(1 + r/n)^nt, where A = the final value, P = the initial value, r = the interest rate (decimal), n = the number of compounding periods per year, and t = the number of years. We can create the equation:

29,200 = P(1 + 0.2/2)^[(2)(2)]

29,200 = P(1.1)^4

29,200 = 1.4641P

19,944 ≈ P

Answer: D

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