I am glad that I discovered this channel because it gives me an opportunity to get back to basics in mathematics. I have worked as a math tutor off and on over a four- decade time interval starting at Rutgers State University at New Brunswick, NJ and ending where I now am, at San Diego City College. Because of the pandemic I am on open- ended unpaid leave of absence until this pandemic ends. I plan to return to that job starting this Autumn. Your video will give me a leg up to being ready to return to my part time job which means so much to me, so I thank you kindly for being there when I need you.
Mixture Formula: (Concentration 1 x volume 1) + (concentration 2 x volume 2) = (concentration 3 x volume 3) Note: you don't need to use decimals. You can use integers. You'll get the same answer. Volume 1 + volume 2 = volume 3 Let x = volume 1 x + volume 2 = 10 Therefore: volume 2 = 10 - x. This is the key. From here it's simple to set up the equation. 5x + 20(10 - x) = 10 x 12.5 Basic algebra from here. x = 5 ounces
Because the average of 5% and 20% is (5 + 20)/2 = 12.5% and 12.5% is what you want doesn't that just mean the same amount of both ? 10oz = 5oz of both.
Here is a similar word problem I got when I attended Seneca college: How many litres of a 20.0% (w/w) solution of an acid should be added to 40.0L of an 80.0% (w/w) solution of the same acid to make a 60.0% (w/w) solution. It is assumed that volumes are "additive" when mixed.
Just recognize that 1oz @ .2 plus 1 oz @ .05 together has .25oz hcl per 2 oz of volume, meaning .125 per oz, 1:1 ratio of volume of the two concentrations is correct. Then, in order to get 10 oz volume @ the correct mix we need 5 *(1:1), so 5oz of each concentration = ((5*.05 hcl)+(5*.2hcl))/10oz volume = .25ozh + 1.0ozh in 10oz vol, or 1.25/10 = .125 or 12.5%
Trying to get a grasp on the system, checked what would happen if you would mix 'em 1 to 1. The average of 5/100 and 20/100 is 12.5/100. So equal parts. Done :-)
I did this by adding the 0.05 and 0.2 then x by 5. This gives 1.25 but I'm stuck on why I need to divide by 10 to get the .125 ounces. Obviously because of the 10 ounces struggling to understand how that bit works. Even on the video it jumps over this when x by 100 to get rid of decimals.
note that : ( 20 + 5 ) / 2 = 12.5 so that means that x and 10 - x must be the same value. THererfore, x = 10 - x 2x = 10 x = 5 As a result, 5 ounces of the 5% solution is needed.
Don't know about all the steps so I suppose it doesn't count...but I did that in my head in about 15 seconds. Tried 6 and 4 first, that didn't work. 5 x 20 is 100 and 5 x 5 is 25. 125/10 is 12.5%
I am glad that I discovered this channel because it gives me an opportunity to get back to basics in mathematics. I have worked as a math tutor off and on over a four- decade time interval starting at Rutgers State University at New Brunswick, NJ and ending where I now am, at San Diego City College. Because of the pandemic I am on open- ended unpaid leave of absence until this pandemic ends. I plan to return to that job starting this Autumn. Your video will give me a leg up to being ready to return to my part time job which means so much to me, so I thank you kindly for being there when I need you.
Mixture Formula: (Concentration 1 x volume 1) + (concentration 2 x volume 2) = (concentration 3 x volume 3)
Note: you don't need to use decimals. You can use integers. You'll get the same answer.
Volume 1 + volume 2 = volume 3
Let x = volume 1
x + volume 2 = 10
Therefore: volume 2 = 10 - x. This is the key. From here it's simple to set up the equation.
5x + 20(10 - x) = 10 x 12.5
Basic algebra from here.
x = 5 ounces
i love algebra word problems
Because the average of 5% and 20% is (5 + 20)/2 = 12.5%
and 12.5% is what you want
doesn't that just mean the same amount of both ?
10oz = 5oz of both.
Here is a similar word problem I got when I attended Seneca college:
How many litres of a 20.0% (w/w) solution of an acid should be added to 40.0L of an 80.0% (w/w) solution of the same acid to make a 60.0% (w/w) solution. It is assumed that volumes are "additive" when mixed.
i get 20 L of the 20% solution.
You don't have to use a different variable of "y". He could have used 10-x instead....
Just recognize that 1oz @ .2 plus 1 oz @ .05 together has .25oz hcl per 2 oz of volume, meaning .125 per oz, 1:1 ratio of volume of the two concentrations is correct.
Then, in order to get 10 oz volume @ the correct mix we need 5 *(1:1), so 5oz of each concentration = ((5*.05 hcl)+(5*.2hcl))/10oz volume = .25ozh + 1.0ozh in 10oz vol, or 1.25/10 = .125 or 12.5%
0.5
Trying to get a grasp on the system, checked what would happen if you would mix 'em 1 to 1. The average of 5/100 and 20/100 is 12.5/100. So equal parts. Done :-)
THe alligation method
A_1 / A_2 = ( B_1 - B ) / ( B - B_2 )
A_1 / A_2 = ( 20 - 12. 5 ) / ( 12.5 - 5 )
A_1 / A_2 = 7.5 / 7.5 = 1 / 1
so the two parts of 10 must obey this : therefore, x / ( 10 - x ) = 1 / 1
10 - x = x
2x = 10
x = 5
Seems like 5oz of each gets you there. 5oz(.05h)+ 5oz(.25h) = 1.25h/10oz =.125 h/1 oz * 100% 12.5%h/oz
{0.5*20}= 0.100 {12.5÷100}= 25
I did this by adding the 0.05 and 0.2 then x by 5. This gives 1.25 but I'm stuck on why I need to divide by 10 to get the .125 ounces. Obviously because of the 10 ounces struggling to understand how that bit works.
Even on the video it jumps over this when x by 100 to get rid of decimals.
0.1.17
0.5
12.5
note that : ( 20 + 5 ) / 2 = 12.5
so that means that x and 10 - x must be the same value. THererfore, x = 10 - x
2x = 10
x = 5
As a result, 5 ounces of the 5% solution is needed.
How did he get the 1.25? He went from .125(10) to 1.25 in the equation.
I got it right by my setup is wrong: .05x + .20X = 10(.125) simplified to 25x = 1.25 - help!!!
Don't know about all the steps so I suppose it doesn't count...but I did that in my head in about 15 seconds. Tried 6 and 4 first, that didn't work.
5 x 20 is 100 and 5 x 5 is 25. 125/10 is 12.5%
I am so confused on these mixture word problem.
Talked waaay to much before anything helpful. Over 5 minutes at the beginning.
Wow:D
Very long winded.
Get to the damned point
Suggest that your presentation should go directly to the point i.e. with lesser ad libs. Train of thought would seem toget distracted