word problem please help

[spoon2562]

Here is the the problem:

I'm having troubling filling out the chart, and a possible equation.

The blue cooler I think should have .2 for the strength and 4 for pure concentrate.


Thanks

 

[mmm4444bot]

I'm having troubling filling out the chart

I won't know what kind of trouble you're having, unless you tell me.

Are there some words, in this exercise, that you do not understand? I mean, what is troubling you?

The blue cooler I think should have .2 for the strength and 4 for pure concentrate.

This is correct.

How did you determine 4 gallons? I mean, why can't you apply your same logic to the orange cooler?

The orange cooler contains a mixture of water and sports-drink contentrate. We're told that any amount withdrawn from this container will be 10% sports-drink concentrate and 90% water.

We need to withdraw x gallons.

Can you express 10% of x gallons? That expression represents the amount of pure sports-drink concentrate contained within the x gallons withdrawn from the orange cooler.

We're adding x gallons from the orange cooler to 20 gallons from the blue cooler. Can you express the sum of x gallons and 20 gallons? That expression represents the amount of 12% mixture we end up with.

Can you express 12% of that sum?

If you get this far, then you have enough information to write an equation because the volume of the two contributions of pure concentrate from the coolers must add up to the volume of pure concentrate in the final mixture.

Also, note that your solution for x will be some number of gallons, but the exercise requires you to report this amount in pints. (I'm assuming that all units are US liquid units.)

If you still need help, please ask specific questions. I'm not really sure where or why you're stuck.

Cheers

 

[spoon2562]

Re:

Thank you! , let me know if I am on the right path:
20*.2=4
so
I know the end result should be a 12% mixture and I know I already have 4% , 12-4=8 pure concentrate for oarnge

so: x= amount of oarnge , x*.10=8 x=80

The mixture forumla should be : x*.3=12 X= 40 gallons

so 40 gallons of mixture with .3 strength = 12% concentrate ?


If not could you please spell it out for me its been 4 hours on this one problem, I don't know why it has me mentally blocked.

 

[mmm4444bot]

No, that's not quite correct.

We remove x gallons from the orange cooler.

The number of gallons of pure concentrate in x gallons is 10% of x: 0.1x

There are 20 gallons in the blue cooler, and 20% of that is pure concentrate: 0.2(20) = 4

So, we are mixing 0.1x gallons of pure concentrate with 4 gallons of pure concentrate:

0.1x + 4 represents the total number of gallons of pure concentrate in the final mixture.

The volume of the final mix is x + 20 gallons. (NOTE: this is the volume of the final mixture, not the final concentrate. It contains both water and concentrate percentages contributed from each cooler.)

12% of this volume ALSO represents the number of gallons of concentrate in the final mixture.

Whenever we have two different expressions that we know both represent the same value, we can equate them.

0.1x + 0.2(20) = 0.12(x + 20)

0.1x + 4 = 0.12(x + 4)

Solve for x.

Convert to pints.

Cheers