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Chapter 1: Problem 22

You invested \(\$ 11,000\) in two accounts paying \(5 \%\) and \(8 \%\) annual interest. If the total interest earned for the year was \(\$ 730,\) how much was invested at each rate?

Short Answer

Expert verified
So, $6000 was invested at 5% and $5000 was invested at 8% interest rate.

Step by step solution

01

Setting up the equations

Let's call the money placed in the bank at 5% 'x' and at 8% 'y'. From the problem, we know that x + y = 11000. We also know that the total interest earned from both accounts is $730 or 0.05x + 0.08y = 730.
02

Solving the equations

Now, we have two equations with two variables, which can be solved by substitution or elimination method. But before we do it, we could simplify second equation by multiplying each side by 100 to remove the decimal points, which gives us 5x + 8y = 73000.
03

Solve for y

Choose one equation to express one variable in terms of the another. Let's use the first equation and express y in terms of x as y = 11000 - x. Replace y in the decimal-free equation and get the equation: 5x + 8(11000 - x) = 73000.
04

Calculate x

Solving the above equation by first distributing the 8, then simplifying on both sides and finally isolating x, gives x = 6000.
05

Calculate y

Now, substitute x = 6000 in y = 11000 - x, it provides y = 5000.

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