SOLUTION: $5000.00 was invested in 2 different bonds. One bond yielded 5% in one year and the other yielded 6% in one year. The total yield was $280.00.
How much was invested in each bond
Algebra ->
Customizable Word Problem Solvers
-> Finance
-> SOLUTION: $5000.00 was invested in 2 different bonds. One bond yielded 5% in one year and the other yielded 6% in one year. The total yield was $280.00.
How much was invested in each bond
Log On
Question 68873: $5000.00 was invested in 2 different bonds. One bond yielded 5% in one year and the other yielded 6% in one year. The total yield was $280.00.
How much was invested in each bond?
I originally set up the problem as such:
.05A+.06B=280
I don't know what steps to take from there. I know what the answer is but do not understand how to figure on paper. .05A=280-.06B Other than plug in numbers and guess, I do not understand how to do.
Thank you Answer by rmromero(383) (Show Source):
Given :
$5000 was invested in 2 different bonds
one yields 5% in one year
the other yielded 6% in one year
The total yield was $280.00
Representation:
A = investment which yield 5% in dollars
B = investment which yield 6% in dollars
Write an Equation. Translate the given sentences to mathematical sentence.
A + B = $5000
0.05A + 0.06B = $280.00 <-- it is better if you multiply 100 each term so that you will not deal with decimal later on.
0.05A + 0.06B = $280.00 is the same as 5A + 6B = 28000
We can now use Substitution method to solve for the variable of the two equations.
Here are the Steps using Substition method
1. In either equation, solve for one variable in terms of the other.
A + B = $5000
A = -B + 5000
2. Substitute for that variable in the other equation. Solve.
5A + 6B = 28000
5(-B+5000) + 6B = 28000
-5B - 25000 + 6B = 28000
B = $ 3000 <<<<<<<< investment which yield 6% in dollars
3. Substitute the result from step 2 in either equation. Solve for the other variable.
A + B = $5000, B = $3000
A + 3000 = 5000
A = $ 2000 <<<<<< investment which yield 5% in dollars
4. Check the solution in both original equations.
A + B = $5000, A = $2000, B = $3000
2000 + 3000 = 5000
5000 = 5000 --------->> True