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...

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