good old plugging in numbers--ETS paper test

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wawatan: Master | Next Rank: 500 Posts; Posts: 171; Joined: Sat May 03, 2008 3:51 pm; Thanked: 8 times

by wawatan » Wed May 28, 2008 9:33 pm

one night a certain motel rented 3/4 of its rooms, including 2/3 of its air conditioned rooms. If 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned?

A) 20%
B) 33 1/3%
c) 35%
d) 40%
e) 80%

please explain why answer choice d is wrong.

OA is E

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Source: Beat The GMAT — Problem Solving | Wed May 28, 2008 9:33 pm

netigen: Legendary Member; Posts: 631; Joined: Mon Feb 18, 2008 11:57 pmFollowed by: 3 members; Thanked: 29 times

by netigen » Wed May 28, 2008 10:22 pm

Rooms not rented = 1/4
Total AC rooms = 3/5
AC rooms rented = 2/3 x 3/5 = 2/5
AC room not rented = 3/5 - 2/5 = 1/5

so out of 1/4 not rented rooms 1/5 are AC
so % = 1/5 x 4/1 x 100 = 80%

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mim3: Senior | Next Rank: 100 Posts; Posts: 58; Joined: Fri Apr 04, 2008 12:36 pm; Thanked: 4 times; GMAT Score:680

by mim3 » Thu May 29, 2008 6:09 am

netigen wrote:Rooms not rented = 1/4
Total AC rooms = 3/5
AC rooms rented = 2/3 x 3/5 = 2/5
AC room not rented = 3/5 - 2/5 = 1/5

so out of 1/4 not rented rooms 1/5 are AC
so % = 1/5 x 4/1 x 100 = 80%

Alternatively, the "smart" numbers method...

60= total number of rooms (common denominator of 3,4,5). Based on that:

3/4 of the rooms rented- 45
1/4 not rented- 15

3/5 have AC- 36
2/5 don't- 24

2/3 of AC rooms rented- 24
1/3 not rented- 12

ac rooms not rented/rooms not rented= 12/15= 4/5= 80%

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