D5W refers to dextrose 5% in water
Source:
https://www.drugs.com/dextrose-5-in-water.html
https://www.rnpedia.com/nursing-notes/fundamentals-in-nursing-notes/d5w-dextrose-5-water-iv-fluid/
It would be ideal if your teacher provided this information in the problem itself. It's probably implied that this knowledge is learned beforehand earlier in the class.
To do this via Alligation (not to be confused with the word "allegation"), we'll write the percentages 50% and 5% along the left-most column.
For the sake of simplicity, I'll omit the percent signs and imply they are percentages since each value is a percentage from now on.
In the second column will be the value 10 to represent the 10% we're aiming for.
The third column of values represents the difference or delta from one percentage value in the first column to the 10% in the second column.
The gap from 5% to 10% is 5%
The gap from 50% to 10% is 40%
So we'll have 5 and 40 in the third column
I'm following similar steps shown on slide 6 from this link here
https://www.slideserve.com/lixue/alligation
These values form an X shape like shown here
Take note of the color coding to see how the values link up.
The values 5 and 40 in the far right column represent the ratio of how we'll mix the 50% dextrose solution and D5W solution.
We'll have 5 parts of the 50% dextrose solution, and 40 parts of the D5W
The ratio 5:40 reduces to 1:8 when dividing both parts by the GCF 5.
So it's equivalent to saying we have 1 part 50% dextrose solution, and 8 parts D5W
We have 1+8 = 9 parts total.
1/9 of that is the 50% dextrose solution
8/9 of that is the D5W
1/9 of 250 = (1/9)*250 = 27.778 mL is the approximate amount of the 50% dextrose solution needed.
8/9 of 250 = (8/9)*250 = 222.222 mL is the approximate amount of D5W needed
Unfortunately 9 isn't a factor of 250, so we won't get whole number results.
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Another approach:
You need 250 mL bags containing 10% dextrose.
That means you require the final product to have 0.10*250 = 25 mL of pure dextrose.
x = amount of solution from the D5W
y = amount of solution from the 50% dextrose batch
The two amounts must add up to 250 mL
x+y = 250
y = -x+250
0.05x = amount of pure dextrose from the D5W batch
0.50y = amount of pure dextrose from the 50% dextrose batch
0.05x+0.50y = total amount of pure dextrose
That total amount is the 25 mL discussed earlier.
0.05x+0.50y = 25
0.05x+0.50(-x+250) = 25 .... replace y with -x+250
0.05x-0.50x+125 = 25
-0.45x = 25-125
-0.45x = -100
x = -100/(-0.45)
x = 222.222
We get the same value mentioned in the previous section.
You should find that (8/9)*250 = -100/(-0.45); both produce the same approximate value of 222.222
Then use this to find y
y = -x+250
y = -222.222+250
y = 27.778
which was mentioned earlier as well.
While this algebraic way is helpful to see another angle to this problem, it's likely best to stick to Alligation when it comes to determining the correct mixing ratio (so that the process is quick/easy). I recommend understanding how both methods work.