Question 1025670: pLEASE HELP ME I AM SO CONFUSED BEEN OUT OF SCHOOL FOR OVER 30YRS AND NEVER HAD PROBLEMS LIKE THIS
1.Maricopa's Success scholarship fund receives a gift of $ 120000. The money is invested in stocks, bonds, and CDs. CDs pay 4 % interest, bonds pay 2 % interest, and stocks pay 7.7 % interest. Maricopa Success invests $ 35000 more in bonds than in CDs. If the annual income from the investments is $ 5835 , how much was invested in each account?
Maricopa Success invested $ in stocks.
Maricopa Success invested $ in bonds.
Maricopa Success invested $ in CDs.
2.A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 35% and the third contains 55%. He wants to use all three solutions to obtain a mixture of 75 liters containing 45%acid, using 3 times as much of the 55% solution as the 35% solution. How many liters of each solution should be used?
The chemist should use
liters of 25% solution,
liters of 35% solution, and
liters of 55% solution.
Found 2 solutions by josgarithmetic, MathTherapy: Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! The two questions are almost the same kind. I will do just your #2 mixture percents question.
x,y,z for the low, medium, and high(er) concentration acids. The quantities of them IN LITERS to use.
...using 3 times as much of the 55% solution as the 35% solution.
This means , and from this, .
Make the material sum equation.
, which you may use later.
Substituting for z, the material sum is also ,
.
Make the percentage relationship equation for the mixture to prepare.
, which you might need effort to understand.
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Again substitute for z in this percents equation.

and you should reduce this, since it can be reduced.
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Summarize your equations as a system, keeping in mind, you temporarily eliminated z.
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This is your system to solve for x and y. This IS a linear system; just multiply the members of the percents equation by 75.
If you not remember how to use Elimination Method, then just use Substitution Method.
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or
----------the 35% acid, liters to use
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FIND x.

---------liters to use of the 25% acid
Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website! pLEASE HELP ME I AM SO CONFUSED BEEN OUT OF SCHOOL FOR OVER 30YRS AND NEVER HAD PROBLEMS LIKE THIS
1.Maricopa's Success scholarship fund receives a gift of $ 120000. The money is invested in stocks, bonds, and CDs. CDs pay 4 % interest, bonds pay 2 % interest, and stocks pay 7.7 % interest. Maricopa Success invests $ 35000 more in bonds than in CDs. If the annual income from the investments is $ 5835 , how much was invested in each account?
Maricopa Success invested $ in stocks.
Maricopa Success invested $ in bonds.
Maricopa Success invested $ in CDs.
2.A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 35% and the third contains 55%. He wants to use all three solutions to obtain a mixture of 75 liters containing 45%acid, using 3 times as much of the 55% solution as the 35% solution. How many liters of each solution should be used?
The chemist should use
liters of 25% solution,
liters of 35% solution, and
liters of 55% solution.
Let amount invested in CDs be C
Then amount invested in bonds = C + 35,000
Thus, amount invested in stocks = 120,000 - (C + C + 35,000), or 120,000 - (2C + 35,000), or 120,000 - 2C - 35,000, or 85,000 - 2C
We then get: .04(C) + .02(C + 35,000) + .077(85,000 - 2C) = 5,835
.04C + .02C + 700 + 6,545 - .154C = 5,835
.04C + .02C - .154C = 5,835 - 7,245
- .094C = - 1,410
C, or amount invested in CDs = , or 
Amount invested in bonds: $15,000 + 35,000, or 
Amount invested in stocks: $120,000 - (15,000 + 50,000) = $120,000 - 65,000, or 
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2.
Let amount of 35% acid to be mixed, be T
Then amount of 55% acid to be mixed = 3T
Amount of 25% acid to be mixed = 75 - (T + 3T), or 75 - 4T
We then get: .35(T) + .55(3T) + .25(75 - 4T) = .45(75)
.35T + 1.65T + 18.75 - T = 33.75
.35T + 1.65T - T = 33.75 - 18.75
T, or amount of 35% acid to be mixed = 
Amount of 55% acid to be mixed: 3(15), or 
Amount of 25% acid to be mixed: 75 - (15 + 45) = 75 - 60, or
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