On a Saturday night, each of the rooms at a certain motel was rented

Although this is a longer way but this is what had hit me when doing the question. Hope this helps some of you
So here it goes,

Let $40 rooms be 'a' in count
Let $60 rooms be 'b' in count
Let total rent be 'c'

Hence from first line of the prompt we can create this equation -
40a + 60 b = c - (1)

Now the second says, we removed 10 rooms from $60 lot and had put it out for $40 slots and rent was reduced by 25%, hence our (1) becomes as follows -
\(40 (a + 10) + 60 (b - 10) = \frac{3}{4}c\)
\(40a + 400 + 60b - 600 = \frac{3}{4}c\)
\(40a + 60b - 200 = \frac{3}{4}c\) - (2)

Using (1) and (2),

\(c - 200 = \frac{3}{4}c\)
\(\frac{1}{4}c = 200\)
c = 800

Hence B.

Even though it is a little time consuming but can be used when nothing strikes.

Solution:

If we let x = the number of $40 rooms and y = the number of $60 rooms, then the total rent for the motel is 40x + 60y. If 10 of the rooms that were rented for $60 had instead been rented for $40, then the total rent the motel charged for that night would have been reduced by 25 percent:

40(x + 10) + 60(y - 10) = (40x + 60y) * 3/4

160(x + 10) + 240(y - 10) = 120x + 180y

160x + 1600 + 240y - 2400 = 120x + 180y

40x + 60y = 800

Alternate Solution:

If 10 rooms were rented for $40 instead of $60, then the total rent would have been 10 * (60 - 40) = 10 * 20 = $200 less. We are given that this reduction corresponds to 25% of the total rent; thus, the total rent is 200/0.25 = $800.

Answer: B

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