Moved - word problem - Mixture/Solution

[robert mclaughlin]

( one over time) =4over 3hrs .ok========== solution a is 20% alcohol and solution b is 70% alcohol. How much of each is needed to make 50 liters of a 60% solution? I have tried this 10 different ways, if it was a mix of 70% alc. And water , the problem would be easier, but the two levels have me thrown. Can someone give me the first step on this and i can most likely take it from there.

 

[HallsofIvy]

Generally, it is better to start a new thread to ask a new question. Many times, people, seeing that a question has been answered, will not continue to read additional posts.

( one over time) =4over 3hrs .ok========== solution a is 20% alcohol and solution b is 70% alcohol. How much of each is needed to make 50 liters of a 60% solution? I have tried this 10 different ways, if it was a mix of 70% alc. And water , the problem would be easier, but the two levels have me thrown. Can someone give me the first step on this and i can most likely take it from there.

Let x be the amount, in liters, of the 20% solution and let y be the amount, in liters, of the 20% solution.

There are x+ y total liters and that must be equal to 50.

There are .2x+ .7y liters of alcohol in that solution and that must be equal to 60% of 50.

 

[soroban]

Hello, robert mclaughlin!

Solution A is 20% alcohol and solution B is 70% alcohol.
How much of each is needed to make 50 liters of a 60% solution?

Let \(\displaystyle x\) = liters of solution A which is 20% alcohol.
. . This contains \(\displaystyle 0.20x\) liters of alcohol.

Then \(\displaystyle 50-x\) = liters of solution B which is 70% alcohol.
. . This contains \(\displaystyle 0.70(50-x)\) liters of alcohol.

The mixture will contain: \(\displaystyle \color{blue}{0.2x + 0.7(50-x)}\) liters of alcohol.


But we know the mixture will be 50 liters which is 60% alcohol.
. . It will contain: \(\displaystyle 0.6(50) = \color{blue}{30}\) liters of alcohol.


We just described the final amount of alcohol in two ways.

There is our equation! . \(\displaystyle \text{ . . . }\;\;0.2x + 0.7(50-x) \;=\;30\)

 

[robert mclaughlin]

mixture

ok, lets see if i can adapt it...

.7x+.2(50-x)= .60 (50)

.7+10-?x = 30

.3 * 50 = 15

30+15 = 45

45 liters is the first solution

50 - 45 = 5 liters (2nd sol)

i think im close ,but ?????? (missing a step) or 2

 

[Deleted member 4993]

( one over time) =4over 3hrs .ok==========


solution a is 20% alcohol and solution b is 70% alcohol. How much of each is needed to make 50 liters of a 60% solution?

Always define WHAT do you need to find as an unknown variable

Volume of 20% alcohol solution in final mixture = T liters

Volume of 70% alcohol solution in final mixture = S liters

Then

S + T = 50

T = 50 - S................................................................................(1)

Then

Total alcohol present - individually in 20% solution (of T liters = 0.2*T) and 70% solution (of S liters = 0.7*S) - must be equal to the alcohol present in the final 60% solution.

0.2 * T + 0.7 * S = 0.6 * 50....................................................(2)

Using (1) in (2)

0.2 * (50-S) + 0.7 * S = 0.6 * 50

10 - 0.2*S + 0.7 * S = 30

0.5 * S = 20

S = 40................................................................................(3)

using (3) in (1)

T = 50 - 40 = 10

So

We will need 10 liters of 20% solution and 40 liters of 70% solution to make 50 liters of 60% solution ............................ANSWER


I have tried this 10 different ways, if it was a mix of 70% alc. And water , the problem would be easier, but the two levels have me thrown. Can someone give me the first step on this and i can most likely take it from there.

.

 

[robert mclaughlin]

mixture

thanks, let me work this formula with some other numbers, and i think i'll have it ( until i get to quadratics)

 

[HallsofIvy]

ok, lets see if i can adapt it...

.7x+.2(50-x)= .60 (50)

Yes, this is correct.

.7+10-?x = 30

I don't know what you mean by this. Did you forget the "x" with .7? And .2( 50- x)= 10- .2x
You should have .7x+ 10- .2x= 30. .7x- .2x= .5x so this is .5x+ 10= 30. Subtracting 10 from both sides, .5x= (1/2)x= 20 so that x= 40. The other amount is 50- 40= 10.

If you have 40 liters of 70% alcohol, then you have .7(40)= 28 liters of alchohol. If you have 10 liters of 20% alcohol, you have .2(10)= 2 liters of alcohol. Combining them, you have 28+ 2= 30 liters of alcohol in 20+ 30= 50 liters so you have 30/50= 3/5= .6 so this new solution is, in fact, 60% alcohol.



.3 * 50 = 15

30+15 = 45

45 liters is the first solution

50 - 45 = 5 liters (2nd sol)

i think im close ,but ?????? (missing a step) or 2[/QUOTE]