Mixture Word Problems | MathHelp.com - Free Educational videos for Students in K-12 | Lumos Learning

Mixture Word Problems | MathHelp.com - Free Educational videos for Students in k-12


Mixture Word Problems | MathHelp.com - By MathHelp.com



Transcript
00:0-1 The mad scientist has one solution that is 30 acid
00:05 and another that's 18 acid . How much of each
00:10 should he use to get 300 liters of a solution
00:14 ? That's 21 acid . Well let's start things off
00:19 by setting up a chart across the top . We
00:23 have our mixture formula which in this case is amount
00:27 of solution times percent acid equals amount of acid .
00:33 Down the left side . We have are two different
00:36 types of solutions that add to our final mixture .
00:42 So for our amount of solution column We know that
00:45 we have 300 litres in our final mixture . So
00:50 we can split up the amount of solution for each
00:53 of our original solutions as X and 300 minus X
01:03 . For our percent acid column . We have 30
01:07 which is .30 or 0.3 18% , which is point
01:16 18 And we know that our mixture is 21 acid
01:22 or point 21 based on our formula , we now
01:27 multiply the first two columns together To fill out the
01:31 3rd column of the chart and we have 0.3 x
01:38 .18 times parentheses , 300 - X . and 300
01:46 times point to one which is 63 for our equation
01:55 we use the idea that the amount of acid in
01:59 each of our solutions will add to the amount of
02:03 acid in our final mixture . So that's .3 x
02:10 . Mhm . Yeah plus 0.18 Yeah , Times 300
02:18 -1 equals 63 solving from here . I would first
02:26 multiply both sides of the equation By 100 to get
02:31 rid of the decimals and eventually we find That x
02:35 equals 75 . Going back to our chart , X
02:42 represents the amount of 30 solution that we have .
02:47 So we know we have 75 leaders of 30 solution
02:57 . There are 18 solution . We have 300 -100
03:03 -75 which is 225 l Of 18 solution .
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