SOLUTION: I need help with a couple of word problems. The first one is - Someone invests $50,000 for one year in two different banks with one account yeilding 5% and the other 11% intrest pe

Question 61343: I need help with a couple of word problems. The first one is - Someone invests $50,000 for one year in two different banks with one account yeilding 5% and the other 11% intrest per year. A total of $4300 intrest was earned. Find the ammount deposited in each account. Also There is a problem about solutions of alcohol. Soulution A is 30% alcohol solution B is 75% alcohol. How much of each solution should be used to make 100L. of a solution that is 50% alcohol?
Found 2 solutions by checkley71, ptaylor:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
4300=.05X+.11(50,000-X)
4300=.05X+5500-.11X
.06X=5500-4300
.06X=1200
X=1200/.06
X=20,000 IS THE INVESTMENT FOR 5%THEN 50,000-20,000=30,000 INVESTED @ 11%
PROOF
4300=.05*20,000+.11*30,000
4300=1000+3300
4300=4300
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
FIRST PROBLEM:
Let x=amount invested in the first account that yields 5%
Then 50,000-x=amt invested in the second account that yields 11%.
Now we know that the interest earned on the first account (x)(.05) plus the interest earned on the second account (50,000-x)(.11) must equal the total interest earned ($4300). Thus, our equation to solve is:
(x)(.05)+(50,000-x)(.11)=4300 Simplifying, we get:
.05x-.11x=-5500+4300 Further simplifying, we have:
-.06x=-1200 and
x=$20,000 in first account that draws 5%
50,000-x=50,000-20,000=30,000 in second account that draws 11%
Note: We used the old familiar equation I=PRT to find the amount of interest that each account earned where P=x and (50,000-x) respectively and R=5% and 11% respectively and T=1.
SECOND PROBLEM:
Let x=amount of Solution A (30% alcohol)
Then 100-x=amount of Solution B (75% alcohol)
Now we know that the amount of pure alcohol in Solution A (x)(.30) plus the amount of pure alcohol in Solution B (100-x)(.75) has to equal the amount of pure alcohol in the final mixture (100)(.50). Thus, our equation to solve is:
(x)(.30)+(100-x)(.75)=(100)(.50) Simplifying, we get:
.30x+75-.75x=50 further simplifying, we have
-.45x=-25
x=55.5555 liters of Solution A (30% alcohol)
100-x=100-55.5555=44.4444 liters of Solution B (75% alcohol)
Hope this helps----ptaylor

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