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Molly's bank pays 10% compound interest on savings annually. Molly deposited $10,000 in her account at the start of last year, another$10,000 at the start of this year, and plans to deposit $10,000 more at the start of next year. If she makes no other deposits to or withdrawals from the account, what will be her balance at the end of next year? Incorrect. [[Snippet]] Since Molly's deposits alone amount to$30,000 in three years, this answer choice is too small. __POE__ and move on.
Incorrect. [[Snippet]] Since Molly's deposits alone amount to $30,000 in three years, this answer choice is too small. __POE__ and move on. Incorrect. [[Snippet]] This answer choice is actually Molly's opening balance next year. Since Molly's deposits alone amount to$30,000 in three years, this answer choice is too small. __POE__ and move on.
Correct. [[Snippet]] Last year, Molly deposited $10,000 so she earned >$$\text{Interest}_1 = \text{Principal} \times \text{Interest rate} \times \text{Time}$$ >$$\text{Interest}_1 = \10{,}000 \times 0.1 \times 1$$ >$$\text{Interest}_1 = \1{,}000$$. This year, her opening balance was $$\10{,}000 + \1{,}000 = \11{,}000$$, and she deposited$10,000. So interest earned this year will be applicable to $$\11{,}000 + \10{,}000 = 21{,}000$$. >$$\text{Interest}_2 = \21{,}000 \times 0.1 \times 1$$ >$$\text{Interest}_2 = \2{,}100$$ Next year, her opening balance will be $$\21{,}000 + \2{,}100 = 23{,}100$$, and she will deposit $10,000. So interest earned this year will be applicable to $$\23{,}100 + \10{,}000 = \33{,}100$$. >$$\text{Interest}_3 = \33{,}100 \times 0.1 \times 1$$ >$$\text{Interest}_3 = \3{,}310$$ Hence, her balance at the end of next year will be $$\33{,}100 + \3{,}310 = \36{,}410$$.$28,075
$29,500$33,100
$36,410$39,720