Problem A:
Jack has $800 invested in two accounts. One pays 5% interest per year and other pays 10%
interest per year. The amount of yearly interest is the same as he would get if the entire $800 was
invested at 8.75%. How much does he have invested at each rate?
Problem B:
Jill has 800 L of acid solution. She obtained it by mixing some 5% acid with some 10% acid. Her final
mixture of 800 L is 8.75% acid. How much of each of the 5% and 10% solutions did she use to get her final
mixture?
a.) Solve Problem A
b.) Solve Problem B
c.) Explain the similarities between the processes used in solving Problems A and B
In problem (a), let x represent the amount invested at 5% interest, and in problem (b), let y represent
the amount of 5% acid used.
a.) Step 1: Read the problem, we are required to determine the amount invested at each account.
Step 2 : Assign the variable. Then organize the information in the table.
Let x= amount invested at 5% interest.
Then, 800−x= amount invested at 10% interest.
Principal⋅Interest rate=Interest5%x⋅0.05=0.05x10%800−x⋅0.10=0.10(800−x)
Step 3: Write an equation from the last column of the table
0.05x+0.10(800−x)=0.0875(800)
Step 4: Solve
0.05x+80−0.10x=70−0.05x=70−80−0.05x=−10x=200
Then, by substitution
800−x=800−200=600
Step 5: State the answer
In other words, the amount invested at 5% and 10% interest rates is $250 and $600 respectively.
b.) Step 1: Read the problem, we are required to determine quantity of each solutions used.
Step 2 : Assign the variable. Then organize the information in the table.
Let x= amount of 5% acid used.
Then, 800−y= amount of 10% acid used.
Liters of solution⋅Percent Concentration=Quantity5%y⋅0.05=0.05y10%800−y⋅0.10=0.10(800−y)
Step 3: Write an equation from the last column of the table
0.05y+0.10(800−y)=0.0875(800)
Step 4: Solve
0.05y+80−0.10y=70−0.05y=70−80−0.05y=−10y=200
Then, by substitution
800−y=800−200=600
Step 5: State the answer
In other words, the final mixture she must use is 200 L of 5% acid solution and 600 L
of 10% acid solution.
c.) In general, solving part A and B is similar in a way that the total amount or quantity
is equated with the sum of the individual amounts or quantities of each condition.
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