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.
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Did you calculate the amount deposited in one of the two accounts? Read the question carefully!

Incorrect.
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Incorrect.
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__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.
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Correct.
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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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