Of 120 hotel rooms rented one night, some were suits rented for $115

Let x be suites and y be double rooms

X+y=120
Y=120-x
115x+85y=10890
115x+85(120-x)=10890
30x=690
X=23

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Let the number of suits rented = x
Then number of double rooms rented = 120-x

Revenue from suits = 115*x
Revenue from double rooms = 85*(120-x)
Total revenue = 115x + 85(120-x) = 10890
=> 115x + 10200 - 85x = 10890
=> 30x = 690
=> x = 23

Thus, 23 suits were rented.
A) is the answer.


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I used the similar approach, but divided everything by 5 at the beginning:

1) S+D=120

2) 23*S+17*D=2178

Substitute D for S:
23*S+2040-17*S=2178

6*S=138

S=23

We let s = the number of suites rented and d = the number of double rooms rented. We can create the equations:

s + d = 120

and

115s + 85d = 10,890

Multiplying the first equation by 115, we have:

115s + 115d = 13,800

Subtracting the second equation from the new equation, we have:

30d = 2910

d = 97

Since s + d = 120 and we know that d = 97, then we have:

s + 97 = 120

s = 23

Answer: A
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For those who want the answer without multiple calculations, here is the approach:

there is 115 (S type) on one side, and 85 (D type) on the other. The number of S type rooms and D type rooms gotta be in balance so that their total is 120 and sum of products is 10 890. Let's find out average room price and see deviation of each room type from the average:

10 890 total revenue / 120 total number of rooms =~ 91. Obviously, within 120 total, the number of D type is bigger, since 91 is much closer to 85 than to 115. Deviation between 85 and 91 is 6, and deviation between 115 and 91 is 24 => 24 vs 6 is same as 4 vs 1. This means that the number of D type rooms is 4 times bigger.

S + 4S=120 => S=~24

It is a little bit indirect, and I would say risky if you do not practice that a lot before actual exam, but it might be helpful to develop statistical thinking

--
Have a nice day

Assuming the suits that were rented = X.
The double rooms that were rented = 120 - X.
The total revenue generated = $ 115 ( X ) + $ 85( 120 - X) = $ 10890.
115X + 10200 - 85X = 10890.
30X = 690.
X = 23.

Let x be one type of room and y be another type of room

x+y = 120 (gives us total number of rooms), y= 120-x

115x + 85y = 10890

115x + 85(120-x) = 10890

x= 23

create 2 equations

1. S = suits
d= double rooms

hence S+d= 120 ---- (1)

then ,

115s + 85d = 10890
divide everything by 5

23s+17d=2178 ---- (2)

multiply (1) by 17 and then subtract it from 2

you will get


(23s+17d)- (17s+17d) = 2178-2040
6s=138
s=23

This is relatively a straightforward problem. The only issue I see here is trying to calculate the actual values. Lets see if we can help this time consuming solving!

We can easily reach the equations,
X+Y=120
115X+86Y=10890

If you see the options they are quite far apart even though the numbers given are not going to help the calculation. We can however, substitute arbitrary values to get approximates.
eg: substitute X=30
then equation is =115*30+ 86*90
=3450+7740
=11190
This value is clearly greater than given value of 10890$.
To reduce the value obtained we will have to reduce the value of X in the equation as X is multiplied by a higher number as compared with Y.(Increase proportion of Y in X+Y if you will)
Only 1 option fits the bill!
i.e. X<30
Option A) X=23
Therefore Ans-A.