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Difficulty:
55%
(hard)
Question Stats:
65% (02:57) correct
35%
(03:21)
wrong
based on 17
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A hotel has three rectangular meeting rooms. The first room is square. The second room is 3 feet narrower than the first room but has the same length, whereas the third room is 5 feet narrower and 2 feet longer than the first room.
If the area of the second room is 810 square feet, the area of the largest room is approximately what percent greater than the area of the smallest room?
A. 1.2%
B. 10.0%
C. 11.1%
D. 12.0%
E. 12.5%
If the area of the second room is 810 square feet, the area of the largest room is approximately what percent greater than the area of the smallest room?
A. 1.2%
B. 10.0%
C. 11.1%
D. 12.0%
E. 12.5%
11 Mar 2026, 00:33
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kevincan
First room is a square, as the length and width are both equal.
Second room has dimensions L , (W-3)
Third room has dimensions, (L+2) and (W-5)
The area do the second room : L * (W-3) = 810
The square above the value of 810 are 841 = 29^2 and 900 = 30^2.
Let’s factorize 810, we get 3*3*3*3*2*5 = 27 * 30
So, L has to be 30. And, W-3 = 27. Thus, W = 30.
First room : 30*30 = 900
Second room = 30 * 27 = 810
Third room = 32 * 25 = 800
900 > 810 > 800.
Largest room how much % greater than smallest room ?
= (Largest - Smallest)/ (Smallest room)
= (900-800)/800
=(1/8)
= 12.5%
Option E
