SOLUTION: A chemist has a barrel with 15% salt solution and another barrel with a 24% salt solution. If he wants to create 25 gallons of an 18% salt solution, how many gallons of each solu

Question 1102693: A chemist has a barrel with 15% salt solution and another barrel with a 24% salt solution. If he wants to create 25 gallons of an 18% salt solution, how many gallons of each solution will he need?
Found 3 solutions by josgarithmetic, ikleyn, richwmiller:
Answer by josgarithmetic(39627)   (Show Source): You can put this solution on YOUR website!

-

---------gallons of the 24% salt
-
--------gallons of the 15% salt
Answer by ikleyn(52864)   (Show Source): You can put this solution on YOUR website!
.

x + y = 25 (1) (gallons of the total mixture) 0.15x + 0.24x = 0.18*25 (2) (mass of salt, calculated based on mass of each mixture) Equivalently x + y = 25 (3) 15x + 24y = 18*25 (4) Solve by Elimination method. For it, multiply eq(3) by 15 (both sides). 15x + 15y = 15*25 (5) 15x + 24y = 18*25 (6) Next subtract (5) from (6). You will get 9y = 18*25 - 15*25 = 3*25 ====> y = = = . Answer. gallons of the 24% solution and 25 - = gallons of the 15% solution. Check. 15* + 24* = 450. 18*25 = 450. ! Correct !

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    - More Mixture problems
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    - Word problems on mixtures for alloys
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Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
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Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
We want to know how many gal. at 15% to mix with 24%.
We need a total of 25 gal. at 18%
Using the method of alligation.

24 3 1/3*25=8 1/3 gal of 24% solution 18 8 1/3+16 2/3= 25.0 gal of 18% solution 15 6 2/3*25=16 2/3 gal of 15% solution 9

check
0.15 * 16 2/3 + 0.24 * 8 1/3 = 0.18*25
2.5 + 2.0 = 4.5
4.5 = 4.5
ok

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