\(Amount = Principal ( 1 + \frac{Rate}{100})^{No \ of \ Years}\)

\(Amount = x ( 1 + \frac{8}{100})^5\)

\(Amount = x ( 1.08)^5\)

Hence, correct answer will be (D) \(x ( 1.08)^5\)
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Hi OCDianaOC

This is the formula for compound interest when done yearly

\(A = P*(1 + \frac{r}{100})^{n}\)
where
n - number of years
r - rate of interest
A - Amount realized at the end of n years
P - Principal invested initially

Coming back to this problem

Pat invests x(Principal) at 8%(R) for 5(n) years

Therefore, Amount = x*\((1 + \frac{8}{100})^5\) = x*\((1.08)^5\)(Option D)

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You've got what it takes, but it will take everything you've got

Both are necessary, please go through this link to clear the concept

https://magoosh.com/gre/2012/compound-i ... n-the-gre/
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Abhishek....

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Done. Thank you.
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