SOLUTION: Can someone help me with this problem?
Please help with this problem
6300 is invested, part of it at 10% and part of it at 8%. For a cerain year, the total is 566. How much w
Algebra.Com
Question 446191: Can someone help me with this problem?
Please help with this problem
6300 is invested, part of it at 10% and part of it at 8%. For a cerain year, the total is 566. How much was invest at each rate
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
The equation is
0.10x + 0.08*(6300 - x) = 566.
Just solve for x (that would be the portion invested at 10%).
RELATED QUESTIONS
Please help me with this problem. I do not know how to set it up.
How can $5,400 be... (answered by alicealc,aka042)
Part of $8,000 was invested at 10% interest and the rest at 12%. If the annual income... (answered by richwmiller)
Can someone help me with this?
Carla invested $20,000, part at 6% and part at 15%. If (answered by scianci)
I need help with this problem please:
Lane invested $14,000, part at 15% and part at... (answered by checkley75)
Extra help needed for this problem.
A manufacturer employs 24 part-tome workers. This... (answered by stanbon)
Mike invested $7000 for one year. He invested part of it at 8% and the rest at 12%. At... (answered by jojo14344)
$6300 is invested, part of it at 10% and part of it at 6%. For a certain year the total... (answered by Maths68)
Help me with this in a hurry please? I need it explained all problems algebraically... (answered by venugopalramana)
I am in need of help with this word problem.
Angus invested $22,000, part at 3% and... (answered by stanbon)