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C
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Difficulty:
95%
(hard)
Question Stats:
42% (03:21) correct
58%
(03:24)
wrong
based on 197
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A hall has space for t tables of equal length running along its center. Regulation requires a gap between any two tables of 1/3 length of each table. The also require an identical gap between a table and the wall at the end of the room. If the length of the hall is 296 feet and the length, in feet, of each table is an integer, how many tables can the hall accommodate ?
A. 3
B. 8
C. 9
D. 24
E. 37
A. 3
B. 8
C. 9
D. 24
E. 37
02 Mar 2023, 02:46
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Bunuel
Please look at the attached pic. below:
Attachment:
GMAT hall.jpg [ 6.8 KiB | Viewed 5864 times ]
Each of the rectangles \((1),(2),\) and \((3) \) represents a hall.
In hall # (1) we have two tables and 3 gaps
In hall # (2) we have three tables and 4 gaps
In hall # (3 ) we have four tables and 5 gaps
So the number of gaps is always ONE more than the number of tables
So if there are \(t\) tables then the number of gaps will be \(t+1\), we will be using this principle in the eqn. below.
If the length of each table is \(n\) then we can form an equation such as :
\(\frac{1}{3}n*(t+1) + n*t = 296\)
\(nt+n+3nt =888\)
\(4nt +n =888\)
\(n(4t+1) = 888\)
\(4t+1 = \frac{888}{n}\)
Now if both \(t\) and \(n\) are integers then n has to be a factor of \(888.\)
\(888 = 2*2*2*3*37\)
Using the options we can see only for \(t=9\) we have \(n\) as a factor of \(888\). \(( n = 24 )\)
Ans =C
Hope it helps.
09 Mar 2023, 05:23
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Bunuel
Solution:
- There are \(t\) tables and the hall has just enough space to accommodate these \(t\) tables
- Let the length of each table be \(l\) feet
- We will need \(t+1\) spaces of \(\frac{l}{3}\) feet each
- Thus, we can form the equation, \(tl+\frac{(t+1)l}{3}=296\)
\(⇒3tl+tl+l=888\)
\(⇒4tl+l=888\)
\(⇒4tl=888-l\)
\(⇒t=\frac{888-l}{4l}\)
\(⇒t=\frac{222}{l}-\frac{1}{4}\)
\(⇒t=\frac{222}{l}-0.25\) - The value of \(t\) will be integral when \(l=24\)
\(⇒t=\frac{222}{24}-0.25=9\)
Hence the right answer is Option C
