QuestionAugust 7, 2025

9. Word Problem: An investment grows by 10% annually. If you start with 1,000 what's the balance after 2 years? a) 1,100 b) 1,210 C) 1,320 d) 1,500

9. Word Problem: An investment grows by 10% annually. If you start with 1,000 what's the balance after 2 years? a) 1,100 b) 1,210 C) 1,320 d) 1,500
9. Word Problem: An investment grows by 10% annually. If you start with 1,000 what's the balance after 2 years? a) 1,100 b) 1,210 C) 1,320 d) 1,500

Solution
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Answer

\ 1,210 Explanation 1. Calculate the growth factor The annual growth rate is 10\%, so the growth factor is 1 + 0.10 = 1.10. 2. Apply the growth factor for 2 years Use the formula for compound interest: A = P \times (1 + r)^n. Here, P = 1000, r = 0.10, and n = 2. So, A = 1000 \times (1.10)^2. 3. Compute the final balance Calculate 1000 \times 1.10^2 = 1000 \times 1.21 = 1210.

Explanation

1. Calculate the growth factor<br /> The annual growth rate is $10\%$, so the growth factor is $1 + 0.10 = 1.10$.<br />2. Apply the growth factor for 2 years<br /> Use the formula for compound interest: $A = P \times (1 + r)^n$. Here, $P = 1000$, $r = 0.10$, and $n = 2$. So, $A = 1000 \times (1.10)^2$.<br />3. Compute the final balance<br /> Calculate $1000 \times 1.10^2 = 1000 \times 1.21 = 1210$.
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