Plugging In: Invisible Plugging In - Fractions

Company _X_ sold one-third of its total units on day one. Of the remaining units, one-fourth were sold on day two and two-fifths on day three. If the remaining units were sold on day four, what fraction of the total units did the company sell on day four?
Incorrect. [[snippet]] Did you simply subtract $$\frac{1}{4}$$ and $$\frac{2}{5}$$ from $$\frac{2}{3}$$? Pay close attention to which total number of remaining units these fractions are taken out of.
Incorrect. [[snippet]]
Correct. [[snippet]] The number of units the company has is the invisible variable. Therefore, __Plug In__ a good number, such as 60 (to find it, just multiply the bottoms of the fractions in the problem: $$3\times 4\times 5=60$$). On day one, $$\frac{1}{3}\cdot 60=20$$ units were sold. Days two and three are next. Note the tricky phrasing of the question: the fraction sold on day three ($$\frac{2}{5}$$) is also taken "out of the remaining units" on day 2 and not out of the remaining units on the successive day three. Thus, out of the 40 units that remain, $$\frac{1}{4}$$ (i.e., 10) were sold on day two, and $$\frac{2}{5}$$ (i.e., 16) were sold on day three. And so, $$40-10-16=14$$ units were sold on day four. Hence, the fraction of units sold on day four is $$\frac{14}{60}$$, or $$\frac{7}{30}$$.
Incorrect. [[snippet]]
Incorrect. [[snippet]] Did you subtract the units sold on day two and three directly from the total units sold? Notice that the units sold on day two and three are given as fractions of the remainder left after subtracting the units sold on day one.

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