.13x + .05(64000-x)=6000
solve for x.
I have been racking my head with this one and don't know how to approach it.
The question is:
A bank loaned out $64,000, part of it at the rate of 13% per year and the rest at a rate of 5% per year. If the interest received was $6000, how much was loaned at 13%.
I know the formula is I=PrT but i don't seem to be able to apply it to this problem.
Thanks so much for the help!!
4 answers
Wrong formula for this.
Let x = amount loaned at 13%.
Then 64,000-x = amt loaned at 5%.
0.05x + 0.13(64,000-x) = 6000
solve for x. I get 35,000 @ 13% (and 29,000 @ 5%).
Check my thinking. Check my work. At the end of the problem, multiply the amount at 13% and the amount at 5%, add the two interests to see that it totals $6,000.
Let x = amount loaned at 13%.
Then 64,000-x = amt loaned at 5%.
0.05x + 0.13(64,000-x) = 6000
solve for x. I get 35,000 @ 13% (and 29,000 @ 5%).
Check my thinking. Check my work. At the end of the problem, multiply the amount at 13% and the amount at 5%, add the two interests to see that it totals $6,000.
Thank you so much, I feel confident that I can tackle this type of problem now.
0.05x + 0.13(64,000-x) = 6000
this is incorrect..
since you said that x is the amount of money loaned at 13% the equation should be;
0.13x + .05(64000-x)=6000
your answer though is correct
35K @ 13% and 29K @ 5%
this is incorrect..
since you said that x is the amount of money loaned at 13% the equation should be;
0.13x + .05(64000-x)=6000
your answer though is correct
35K @ 13% and 29K @ 5%
1 answer