SOLUTION: ANGELA WANTS 20QUARTS OF 50% ANTIFREEZE. SHE HAS 40% SOLUTIO. SHE WANTS TO MAKE IT STRONGER BY ADDING PURE ANTIFREEZE. CONTRUCT TABLE....
TABLE FOUR COLUMS DOWN AND ACROSS
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-> SOLUTION: ANGELA WANTS 20QUARTS OF 50% ANTIFREEZE. SHE HAS 40% SOLUTIO. SHE WANTS TO MAKE IT STRONGER BY ADDING PURE ANTIFREEZE. CONTRUCT TABLE....
TABLE FOUR COLUMS DOWN AND ACROSS
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Question 154551: ANGELA WANTS 20QUARTS OF 50% ANTIFREEZE. SHE HAS 40% SOLUTIO. SHE WANTS TO MAKE IT STRONGER BY ADDING PURE ANTIFREEZE. CONTRUCT TABLE....
TABLE FOUR COLUMS DOWN AND ACROSS
-------CONCENTRATION ---AMOUNT----AMOUNT OF ANTIFREEZE
PURE
ANTIFREEZE
40%
TOTAL
WHAT NGELA
WANTS THE
GOAL
SOLVE WITH ONE VARIABLE SINCE THE TOTAL IS 100.
1QUART RELATES TO X AND THE OTHER TO 100-X Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of pure antifreeze that needs to be added
Then 20-x=amount of 40% solution that needs to be used
Now we know that the amount of pure antifreeze in the 40% solution(0.40(20-x)) plus the amount of pure antifreeze added (x) has to equal the amount of pure antifreeze in the final solution (20*0.50) So our equation to solve is:
0.40(20-x)+x=20*0.50 get rid of parens and simplify
8-0.40x+x=10 subtract 8 from each side
8-8-0.40x+x=10-8 collect like terms
0.60x=2 divide both sides by 0.60
x=3.33 qts -----------amount of pure antifreeze that needs to be added
20-x=20-3.33=16.67 qts------------amount of 40% antifreeze used
CK
16*67*0.40+3.33=10
6.668+3.333=10
10~~~10
