I see we're working through this compound interest problem. It might look complex at first, but there's actually a clever shortcut built right into the question that makes this much simpler than it appears.
Let's break this down together:The problem gives you a golden rule: money doubles every \(\frac{70}{r}\) years when invested at \(r\)% interest. This is your key to solving this quickly!
Step 1: Find the doubling timeWith \(r = 8\)% interest, your money doubles every:
\(\frac{70}{8} = 8.75\) years
Step 2: Count how many times the money doublesIn 18 years, you'll have:
\(\frac{18}{8.75} \approx 2.06\) doubling periods
Notice how this is essentially 2 complete doublings? The 0.06 extra is so small we can ignore it for an approximation.
Step 3: Apply the doublingsStarting amount: $5,000
After first doubling (8.75 years): \(5,000 \times 2 = 10,000\)
After second doubling (17.5 years): \(10,000 \times 2 = 20,000\)
Since we're at approximately 18 years (just 0.5 years past the second doubling), and the question asks for an approximate value, our answer is
$20,000.
The answer is
(A).
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