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30 May 2026, 10:48
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Today, Laura invested a total of $12,500 at two banks. She invested part of the money at Bank A, which pays 9% annual interest compounded annually, and the remainder at Bank B, which pays 8% annual interest compounded semiannually. How much did Laura invest in Bank A?
(1) After 1 year, the total value of Laura’s two investments at the banks will be $13,541.
(2) After 2 years, the amount in the Bank B investment will be $8,728.34 greater than the amount in the Bank A investment.
(1) After 1 year, the total value of Laura’s two investments at the banks will be $13,541.
(2) After 2 years, the amount in the Bank B investment will be $8,728.34 greater than the amount in the Bank A investment.
30 May 2026, 10:48
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Official Solution:
Today, Laura invested a total of $12,500 at two banks. She invested part of the money at Bank A, which pays 9% annual interest compounded annually, and the remainder at Bank B, which pays 8% annual interest compounded semiannually. How much did Laura invest in Bank A?
Use the compound interest formula:
\(\text{Final balance} = \text{Principal} * (1 + \frac{r}{100C})^{Ct}\)
where \(r\) = annual interest rate as a percent, \(C\) = number of compounding periods per year, and \(t\) = number of years.
Here:
Bank A: \(C = 1\), \(r = 9\)
Bank B: \(C = 2\), \(r = 8\)
Let \(x\) be the amount Laura invested in Bank A. Then \(12500 - x\) is the amount invested in Bank B.
(1) After 1 year, the total value of Laura’s two investments at the banks will be $13,541.
After 1 year:
Bank A value = \(x * 1.09\)
Bank B value = \((12500 - x) * (1.04)^2\)
So (1) gives:
\(x * 1.09 + (12500 - x) * (1.04)^2 = 13541\)
This is one linear equation in one variable \(x\), so \(x\) can be determined uniquely. Sufficient.
(2) After 2 years, the amount in the Bank B investment will be $8,728.34 greater than the amount in the Bank A investment.
After 2 years:
Bank A value = \(x * (1.09)^2\)
Bank B value = \((12500 - x) * (1.04)^4\)
Statement (2) says that the Bank B amount is 8728.34 greater than the Bank A amount, so:
\((12500 - x) * (1.04)^4 - x * (1.09)^2 = 8728.34\)
This is again one linear equation in one variable \(x\), so \(x\) can be determined uniquely. Sufficient.
Answer: D
Today, Laura invested a total of $12,500 at two banks. She invested part of the money at Bank A, which pays 9% annual interest compounded annually, and the remainder at Bank B, which pays 8% annual interest compounded semiannually. How much did Laura invest in Bank A?
Use the compound interest formula:
\(\text{Final balance} = \text{Principal} * (1 + \frac{r}{100C})^{Ct}\)
where \(r\) = annual interest rate as a percent, \(C\) = number of compounding periods per year, and \(t\) = number of years.
Here:
Bank A: \(C = 1\), \(r = 9\)
Bank B: \(C = 2\), \(r = 8\)
Let \(x\) be the amount Laura invested in Bank A. Then \(12500 - x\) is the amount invested in Bank B.
(1) After 1 year, the total value of Laura’s two investments at the banks will be $13,541.
After 1 year:
Bank A value = \(x * 1.09\)
Bank B value = \((12500 - x) * (1.04)^2\)
So (1) gives:
\(x * 1.09 + (12500 - x) * (1.04)^2 = 13541\)
This is one linear equation in one variable \(x\), so \(x\) can be determined uniquely. Sufficient.
(2) After 2 years, the amount in the Bank B investment will be $8,728.34 greater than the amount in the Bank A investment.
After 2 years:
Bank A value = \(x * (1.09)^2\)
Bank B value = \((12500 - x) * (1.04)^4\)
Statement (2) says that the Bank B amount is 8728.34 greater than the Bank A amount, so:
\((12500 - x) * (1.04)^4 - x * (1.09)^2 = 8728.34\)
This is again one linear equation in one variable \(x\), so \(x\) can be determined uniquely. Sufficient.
Answer: D
