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Hi Bunuel - Can you check this question and clarify the question below the quoted text.

$3000 when invested for 2 years at the rate of 8% per year (compounded quarterly) will yield approx. $3500 at the end of second year. The principle value of the investment in simple interest is $5000.

Quote:
What is the approximate value of x, if both investments yield the same amount after 2 years

Does this mean the interest values are same at the end of two years? The amount (principle + interest) at the end of two years cannot be the same unless the investment that's invested in simple interest deprecates ?

Am I missing anything ?

The question says that both investments yield the same amount, meaning that they yield the same interest amount. Edited to avoid confusion. Thank you.
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Simple interest at the end of two years = \($5000 * 2 * \frac{x }{ 100}\)

Amount at the end of two years = \(3000(1+\frac{8}{400})^8\)

= \(3000(\frac{51}{50})^8\)

= \(3000(1.02)^8\)

At the end of two years

\(5000 * \frac{x}{100} * 2\) = \(3000(1.02)^8\) - 3000

\(5000 * \frac{x}{100} * 2\) = \(3000 (1.02^8 - 1)\)

\(\frac{x}{100}\) = \(\frac{3}{10} ((1+0.2)^8 - 1)\)

Approximately -

\(\frac{x}{100}\) = 0.3 * (1 + 0.16 - 1)

How ??
\((1+x)^n\) can be approximated to 1+nx; when x is very small
x = 0.48 * 100

x = 4.8

Approx 5

IMO C
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Bunuel why are we doing -3000?
gmatophobia
Simple interest at the end of two years = \($5000 * 2 * \frac{x }{ 100}\)

Amount at the end of two years = \(3000(1+\frac{8}{400})^8\)

= \(3000(\frac{51}{50})^8\)

= \(3000(1.02)^8\)

At the end of two years

\(5000 * \frac{x}{100} * 2\) = \(3000(1.02)^8\) - 3000

\(5000 * \frac{x}{100} * 2\) = \(3000 (1.02^8 - 1)\)

\(\frac{x}{100}\) = \(\frac{3}{10} ((1+0.2)^8 - 1)\)

Approximately -

\(\frac{x}{100}\) = 0.3 * (1 + 0.16 - 1)

How ??
\((1+x)^n\) can be approximated to 1+nx; when x is very small
x = 0.48 * 100

x = 4.8

Approx 5

IMO C
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Bunuel why are we doing -3000?
gmatophobia
Simple interest at the end of two years = \($5000 * 2 * \frac{x }{ 100}\)

Amount at the end of two years = \(3000(1+\frac{8}{400})^8\)

= \(3000(\frac{51}{50})^8\)

= \(3000(1.02)^8\)

At the end of two years

\(5000 * \frac{x}{100} * 2\) = \(3000(1.02)^8\) - 3000

\(5000 * \frac{x}{100} * 2\) = \(3000 (1.02^8 - 1)\)

\(\frac{x}{100}\) = \(\frac{3}{10} ((1+0.2)^8 - 1)\)

Approximately -

\(\frac{x}{100}\) = 0.3 * (1 + 0.16 - 1)

How ??
\((1+x)^n\) can be approximated to 1+nx; when x is very small
x = 0.48 * 100

x = 4.8

Approx 5

IMO C
Because \(3000(1.02)^8\) includes Principal + Interest earned, and we are told that the two Interests earned are equal, so we subtract the principal 3000 from \(3000(1.02)^8\)
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