SOLUTION: Your assistance is greatly appreciated. Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with an app

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Question 57310This question is from textbook Elementary and Intermediate algebra
: Your assistance is greatly appreciated.
Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate amount of 40% antifreeze solution. How many quarts of each should she use?This question is from textbook Elementary and Intermediate algebra

Found 3 solutions by aaaaaaaa, CrazyMan Jr., ptaylor:
Answer by aaaaaaaa(138) About Me  (Show Source):
You can put this solution on YOUR website!
p = quarts of pure antifreeze sol.
i = quarts of the impure one
The second equation needs some explanation, 50% of 20 is 10.
system%28p+%2B+i+=+20%2C+p+%2B+0.4i+=+10%29
Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
+system%28+%0D%0A++++1%5Cp+%2B+1%5Ci+=+20%2C%0D%0A++++1%5Cp+%2B+0.4%5Ci+=+10+%29%0D%0A++We'll use substitution. After moving 1*i to the right, we get:
1%2Ap+=+20+-+1%2Ai, or p+=+20%2F1+-+1%2Ai%2F1. Substitute that
into another equation:
1%2A%2820%2F1+-+1%2Ai%2F1%29+%2B+0.4%5Ci+=+10 and simplify:
EXPLAIN simplification of an expression
Your Result:
Simplify 1*(20/1-1*i/1)+0.4i=10
Your Answer:
  • Solutions: i=16.6666666666667.
  • Graphical form: Equation 1%2A%2820%2F1-1%2Ai%2F1%29%2B0.4i=10 was fully solved.
  • Text form: 1*(20/1-1*i/1)+0.4i=10 simplifies to 0=0
  • Cartoon (animation) form: simplify_cartoon%28+1%2A%2820%2F1-1%2Ai%2F1%29%2B0.4i=10+%29
    simplify_cartoon( 1*(20/1-1*i/1)+0.4i=10 )

Detailed explanation:

Look at highlight_red%28+1+%29%2A%2820%2F1-1%2Ai%2F1%29%2B0.4%2Ai=10.
Remove extraneous '1' from product highlight_red%28+1+%29
It becomes %2820%2F1-1%2Ai%2F1%29%2B0.4%2Ai=10.
Look at %2820%2Fhighlight_red%28+1+%29-1%2Ai%2F1%29%2B0.4%2Ai=10.
Remove extraneous '1' from product highlight_red%28+1%2F1+%29
It becomes %2820-1%2Ai%2F1%29%2B0.4%2Ai=10.
Look at %2820-highlight_red%28+1+%29%2Ai%2F1%29%2B0.4%2Ai=10.
Remove extraneous '1' from product highlight_red%28+1+%29
It becomes %2820-i%2F1%29%2B0.4%2Ai=10.
Look at %2820-i%2Fhighlight_red%28+1+%29%29%2B0.4%2Ai=10.
Remove extraneous '1' from product highlight_red%28+1%2F1+%29
It becomes %2820-i%29%2B0.4%2Ai=10.
Look at %28highlight_red%28+20+%29-i%29%2B0.4%2Ai=10.
Moved 20 to the right of expression
It becomes %28-i%2Bhighlight_green%28+20+%29%29%2B0.4%2Ai=10.
Look at highlight_red%28+%28-i%2B20%29+%29%2B0.4%2Ai=10.
Remove unneeded parentheses around terms highlight_red%28+-i+%29,highlight_red%28+20+%29
It becomes -highlight_green%28+i+%29%2Bhighlight_green%28+20+%29%2B0.4%2Ai=10.
Look at -i%2Bhighlight_red%28+20+%29%2B0.4%2Ai=10.
Moved 20 to the right of expression
It becomes -i%2B0.4%2Ai%2Bhighlight_green%28+20+%29=10.
Look at -highlight_red%28+i+%29%2Bhighlight_red%28+0.4%2Ai+%29%2B20=10.
Eliminated similar terms highlight_red%28+-i+%29,highlight_red%28+0.4%2Ai+%29 replacing them with highlight_green%28+%28-1%2B0.4%29%2Ai+%29
It becomes highlight_green%28+%28-1%2B0.4%29%2Ai+%29%2B20=10.
Look at %28-highlight_red%28+1+%29%2Bhighlight_red%28+0.4+%29%29%2Ai%2B20=10.
Added fractions or integers together
It becomes %28-highlight_green%28+6%2F10+%29%29%2Ai%2B20=10.
Look at highlight_red%28+%28-highlight_red%28+6%2F10+%29%29%2Ai+%29%2B20=10.
Remove unneeded parentheses around factor highlight_red%28+6+%29,highlight_red%28+1%2F10+%29
It becomes -highlight_green%28+6+%29%2Fhighlight_green%28+10+%29%2Ai%2B20=10.
Look at -highlight_red%28+6+%29%2Fhighlight_red%28+10+%29%2Ai%2B20=10.
Factors 6 and 10 have greatest common factor (GCF) of 2. Reducing fraction.
It becomes -highlight_green%28+3+%29%2Fhighlight_green%28+5+%29%2Ai%2B20=10.
Look at -3%2F5%2Ai%2B20=highlight_red%28+10+%29.
Moved these terms to the left highlight_green%28+-10+%29
It becomes -3%2F5%2Ai%2B20-highlight_green%28+10+%29=0.
Look at -3%2F5%2Ai%2Bhighlight_red%28+20+%29-highlight_red%28+10+%29=0.
Added fractions or integers together
It becomes -3%2F5%2Ai%2Bhighlight_green%28+10+%29=0.
Look at highlight_red%28+-3%2F5%2Ai%2B10+%29=0.
Solved linear equation highlight_red%28+-3%2F5%2Ai%2B10=0+%29 equivalent to -0.6*i+10 =0
It becomes highlight_green%28+0+%29=0.
Result: 0=0
Solutions: i=16.6666666666667.

Done!
So, we know that i=16.6666666666667. Since p+=+20%2F1+-+1%2Ai%2F1, p=3.3333333333333.

Answer: system%28+p=3.3333333333333%2C+i=16.6666666666667+%29.

Answer by CrazyMan Jr.(21) About Me  (Show Source):
You can put this solution on YOUR website!
0.4x+1x = 0.5
x+y = 20

0.4x+1y = 0.5 0.6y = -7.5 y = -12.5
0.4x+0.4y = 8 0.6y/0.6 = -7.5/0.6
Sub In
0.4x+1(-12.5) = 0.5
0.4x(+-)12.5 = 0.5
0.4x-12.5(+12.5) = 0.5(+12.5)
0.4x = 13
0.4x/0.4 = 13/0.4
x = 32.5
Proof - Sub In Again
0.4(32.5)+1(-12.5) = 0.5
13+-12.5 = 0.5
0.5 = 0.5
N.P.
J-Man

Answer by ptaylor(1661) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=amt of pure antifreeze that is mixed in
Then 20-x=amt of 40% antifreeze solution that is mixed in
We know that the amt of pure antifreeze that is mixed in (x)(1) plus the amt of pure antifreeze in the 40% solution that is mixed in (20-x)(.40) equals the amt of pure antifreeze in the final solution (20)(.50).
Thus, our equation to solve is:
(x)(1)+(20-x)(.40)=(20)(.50) Simplifying, we get:
x+8-.4x=10 or
.6x=2; 6x=20
x=3.3333 qts of pure antifreeze
20-x=20-3.3333=16.6667 qts of the 40% antifreeze solution
Hope this helps-----ptaylor