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Reverse Plugging In: What to Do with an Ugly Number in the Middle Answer?

Julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest. After two years, she earned $120 and $126 from the simple interest account and the compound interest account respectively. If the interest rates for both accounts were the same, what was the **total** amount of Julie's initial savings?
Incorrect. [[Snippet]] Did you calculate the amount deposited in one of the two accounts? Read the question carefully!
Incorrect. [[Snippet]]
Incorrect. [[Snippet]]
__Alternative Explanation__: Numbers in the answer choices and the specific question call for __Reverse PI__. Start with answer choice D since the other answer choices do not hold comfortable numbers. If the total amount deposited in both accounts is $1,200, then Julie deposited $600 in each account. Since interest earned on the simple interest account is $120, she earned $60 for each of the two years, which is 10% of 600. Based on this, the compound interest earned on $600 over two years should be >$$I = \$60 + \$60 + 10\% \text{ of } \$60$$ >$$I = \$60 + \$60 + \$6$$ >$$I = \$126$$. Hence, this is the correct answer.
Incorrect. [[Snippet]]
Correct. [[Snippet]] The $6 interest earned on the interest was calculated out of the total interest earned on the first year, which is $60. Divide 6 by 60 to find the interest rate: >$$\frac{6}{60} \times 100 = 10%$$. Given that interest rate is 10%, Julie's investment in the simple interest account is >$$\text{Interest} = \text{Principal} \times \text{Rate} \times \text{Time}$$ >$$\$120 = P \times 10\% \times 2$$ >$$\$120 = P \times 20\%$$ >$$\$120 = P \times \frac{20}{100}$$ >$$P = \$120 \times \frac{100}{20}$$ >$$P = \$600$$. Hence, the initial investment in both accounts was $1,200 ($$\$600 + \$600$$).
$600
$720
$1,080
$1,200
$1,440
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