Word problems-algebra?
I am struggling with a couple word problems. The text book is very vague as to how to do these problems.
1) A lab technician mixed a 660mL solution of water and alcohol. If 3% of the solution is alcohol, how many mL of alcohol were used? And how many mL of water?
2)9500 is invested, part at 12% and part at 9%. For a certain year, the total yield is 1032.00. How much is invested at each rate?
Step by step instructions would be fantastic, I usually can do these ok, it's just getting them set up properly. Thanks so much!
Comments
A lab technician mixed a 660mL solution of water and alcohol
Total solution: 660ml
3% of the solution is alcohol:
3% of 660 = 0.03 * 660 = 19.80ml
how many mL of water
total solution - amount of alkohol = 660-19.80 = 640.20ml <= ans1
2. the total yield is 1032.00:
say they made amount F1 at the rate of r1 interest and
amount F2 at the rate of r2 interest
(F1+F2) - P = 1032.00 where P = 9500 = (P1+P2)
F1+F2 = 1032 +9500 = 10532
F1 = P1(1+r1)^1 : assuming invested for a year at r1% interest rate
=P1(1+0.12)
F2 = P2(1+r2)^1 : assuming invested for a year at r2% interest rate
= P2(1+0.09)
F1+F2 = P1(1.12) + (9500-P1)1.09 = 10532
P1 = (10532-10355)/0.03 = 157/0.03 = 5233
P2 = 9500-5233 = 4267
How much is invested at each rate?
ans: 5233 at 12% and 4267 at 9%
1) the total is 660 mL.
3% of 660 = 19.8 mL is alcohol.
97% = 660 - 19.8 = 640.2 mL of water
2)
A + B = 9500
.12 A + .09 B = 1032
or
12 A + 9 B = 103200
mult. eqn 1 by 9
9 A + 9 B = 85500
subtract this from eqn 2
3 A = 103200-85500 =17700
A = 5900
B = 9500 - 5900 = 3600
For percentages you divide by 100 (cos that'll give you 1%)
Then multiply your answer by the % you want to find out...
3600