Question

Help .


..

Answer 1

These kind of problems sometimes confuse me, but here is an attempt. The 50 gallon final solution will consist of x gal at 60% + (50-x) at 10% which will = 50 gal at 50% solution. Don't know if that is the correct approach or not. But will blindly continue.

Answer 2

.6x + 5-.1x =50
.5x=45
x=22.5 gal of 60% solution and 50-22.5 =27.5 of the 10% solution

Answer 3

the answer remains 22.5 gallons of the 60% mixture
and 27.5 gal of the 10% mixture.

Answer 4

There may be more eloquent procedure, but this worked.

Answer 5

i just entered and both were wrong :/ thanks for trying though

Answer 6

I see the error
.6x + 5 - .1x=.5*50=25
.6x - .1x =25-5=20
.5x=20
x=40 or 40 gal of 60%
50-40 = 10 gal of 10%
that should work

Answer 7

correct! Can i show you one more?

Answer 8

New verification lol. 40 gal at 60% consist of 24 gal solution and 16 gal water
10 gal at 10% consist of 1 gal solution and 9 gal water
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50 25

Answer 9

Surely.

Answer 10

That will be a snap, now that I am on track.
Let x = number of gallons of the 20 % solution.
30 -x = number of gallons of the 80 % solution

We can now state the following .2x +.8(30-x)=30*.6 (percentages converted to decimal)
.2x +24-.8x=18
-.6x=-6
x=10 10 gallons of 20% solution
30-10=20 gallons of the 80% solution

Answer 11

got it correct. Thank you very much. I've been struggling with those problems a lot lately.

Answer 12

I guess we should verify:

10 gal @ 20% is 2 gal fertilizer 8 gal fill
20 gal @80% is 16 gal fertilizer 4 gal filler
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18 12

Answer 13

18 is 60% of 30 so it is a good bingo.

Answer 14

Yes they can be challengeing, especially if you get on other subjects, and then come a few months later lol. good luck with em.

Answer 15

Thank you very much :*

Answer 16

You're welcome