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Chapter 0: Problem 21

You invested \(\$ 7000\) in two accounts paying \(6 \%\) and \(8 \%\) annual interest. If the total interest earned for the year was \(\$ 520,\) how much was invested at each rate?

Short Answer

Expert verified
\$2000 was invested at 6\% and \$5000 was invested at 8\%.

Step by step solution

01

Understand the Information and Formulate the Equations

The total investment of \$7000 is split between two accounts. Let's say that 'x' is the amount invested in the account with 6\% interest, and that the remaining amount is put in the account with 8\% interest. We know that the total amount invested (x plus the remaining amount), is \$7000. This gives us our first equation: \(x + (7000-x) = 7000\). We also know that the total interest earned is \$520 which comes from the interest received from both accounts. This gives us our second equation: \(0.06x + 0.08(7000 - x) = 520\).
02

Solve the Equations

From the first equation, we can see that no matter what the value of 'x' is, the total amount will always be \$7000, as \$7000 is the total investment amount. The challenge is to solve the second equation to find the value of 'x'. For that, we simplify the second equation: \(0.06x + 560 - 0.08x = 520\) which gives us \(0.02x = 40\)
03

Calculate the Amount Invested at Each Rate

Now we solve for 'x' in the equation \(0.02x = 40\) to find the amount invested at 6\%. Dividing both sides of the equation by 0.02, we get \(x = 2000\). Therefore, \$2000 was invested at 6\%. Now, we find the amount invested at 8\% by subtracting 'x' from the total investment amount: \(7000 - 2000 = 5000\). Therefore, \$5000 was invested at 8\%.

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