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Mr. Lee has two fixed deposit accounts, one yielding 3% interest per year and the other 7% interest per year.If he has a total of $100,000 deposits and in a year his total interest is $4,500, how much money did he put into the 7% interest deposit?Note: Use the algebraic approach.Let the amount of money in the account with a yield of 7% = $p

Question

Mr. Lee has two fixed deposit accounts, one yielding 3% interest per year and the other 7% interest per year.If he has a total of 100,000depositsandinayearhistotalinterestis100,000 deposits and in a year his total interest is 4,500, how much money did he put into the 7% interest deposit?Note: Use the algebraic approach.Let the amount of money in the account with a yield of 7% = $p

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Solution

Let's denote the amount of money Mr. Lee put into the 3% interest deposit as xandtheamountheputintothe7x and the amount he put into the 7% interest deposit as p.

From the problem, we know two things:

  1. The total amount of money is 100,000.So,wehavetheequation:x+p=100,000. So, we have the equation: x + p = 100,000.

  2. The total interest in a year is 4,500.Theinterestfromthe34,500. The interest from the 3% account would be 0.03x and from the 7% account would be 0.07p. So, we have the equation: 0.03x + 0.07p = 4,500.

Now we have a system of two equations, and we can solve it step by step:

Step 1: Multiply the first equation by 0.03, we get: 0.03x + 0.03p = $3,000.

Step 2: Subtract the equation from Step 1 from the second equation, we get: 0.04p = $1,500.

Step 3: Solve for p, we get: p = 1,500/0.04=1,500 / 0.04 = 37,500.

So, Mr. Lee put $37,500 into the 7% interest deposit.

This problem has been solved

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