SOLUTION: I've been working on profit equations and am starting to understand, but this one added something new and I don't know how to deal with it.
Question - Rose invested some of her $1
Question 182696This question is from textbook Teaching Textbooks
: I've been working on profit equations and am starting to understand, but this one added something new and I don't know how to deal with it.
Question - Rose invested some of her $15,000 in bonds that made a 6% profit and the rest in bonds that made a 10% profit. If the profit on the 10% bonds was $1,000 more than the profit on the 6% bonds, how much did she invest in the 6% bonds?
I think that if x= amount invested on the 6% bonds then
.06x = .10(15,000 -x), but I don't know where to put the extra $1000 profit? Does it even belong in the equation?
I tried .06x = .10+1,000(15,000-x), but that is a mess. Thanks for your help! This question is from textbook Teaching Textbooks
You can put this solution on YOUR website! ok you have the right idea...many times when they throw something new into a problem it throws us off just a little. We know that we have two unknowns. The amounts invested at 6 and 10%. So we know we have to come up with 2 equations to solve.....lets, as you say, call the amount invested at 6%, x and the amount invested at 10% lets call y. Here is what we know without much thinking
:
x+y=15000...............eq 1
:
so we have one equation under our belt.
:
now typically in this type of problem we would have another equation of the form
:
.06x+.1y=(profit). But they through a twist in this one. We know the profit of one in terms of the other...... we know the profit for the 6% is .06x and we know the profit for 10% is .1y. But we also know that the profit from the 10% is 1000 more than the profit from the 6%, so lets write it out
:
.06x=.1y+1000..........eq 2
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now we have 2 equations
:
x+y=15000...............eq 1
.1y=.06x+1000..........eq 2
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rewrite eq 1 to and plug that value into eq 2
:
.1y=.06(15000-y)+1000
:
.1y=900-.06y+1000
:
.16y=1900
:
$amount invested at 10%
:
$amount invested at 6%
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