Brandon decided to paint some of the rooms at his 2424-room inn, Brandon's Place. He discovered he needed \frac{3}{5} 5 3 of a can of paint per room. If Brandon had 33 cans of paint, how many rooms could he paint?
+980 I can only make some sense of your question. I'm assuming he can paint a room with 3 fiths of a can of paint.
P=paint cans, r=rooms
$$\frac{3}{5}p:1r$$ Now we need to make p=1. We can do this by dividing both sides by 3/5.
(Note that $$\frac{3}{5}p:1r$$ can also be written as $$3p:5r$$)
$$1p:1\frac{2}{3}r$$ So if we have 33 paint cans (multiply both sides by 33), he can paint. . .
$$33p:55r$$
55 rooms.
+980 I can only make some sense of your question. I'm assuming he can paint a room with 3 fiths of a can of paint.
P=paint cans, r=rooms
$$\frac{3}{5}p:1r$$ Now we need to make p=1. We can do this by dividing both sides by 3/5.
(Note that $$\frac{3}{5}p:1r$$ can also be written as $$3p:5r$$)
$$1p:1\frac{2}{3}r$$ So if we have 33 paint cans (multiply both sides by 33), he can paint. . .
$$33p:55r$$
55 rooms.
