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Bunuel
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AryamaDuttaSaikia
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AryamaDuttaSaikia
Answer : D

Is C<200?

Area of each conference room = C

Area of each rectangular office (considering "Equal") = R

C>R

2C + 6R = 1200

S1 -

C>2R

Even if we take C= 2R, R= 120 , Since we know C>2R, then C>240. Sufficient.

S2-

2 (l+b) < 46

l+b<23

Maximum value could be l=12 , b=11

so, lb= 132 (maximum area of the rectangle)

2C+6*132=1200

C=204 (>200)

Sufficient.

It does not mention there are 6 equal offices. Please assist on the same.
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I agree. The wording should have made 2 things clear:

1) All offices have the same area
2) the perimeter of each room (instead of using the word length) is less than 46 or worded the prompt as the insides of the room measured or something to that tune.

Request Bunuel to kindly look into this.

Best,
SS18


shekyonline
AryamaDuttaSaikia
Answer : D

Is C<200?

Area of each conference room = C

Area of each rectangular office (considering "Equal") = R

C>R

2C + 6R = 1200

S1 -

C>2R

Even if we take C= 2R, R= 120 , Since we know C>2R, then C>240. Sufficient.

S2-

2 (l+b) < 46

l+b<23

Maximum value could be l=12 , b=11

so, lb= 132 (maximum area of the rectangle)

2C+6*132=1200

C=204 (>200)

Sufficient.

It does not mention there are 6 equal offices. Please assist on the same.
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AryamaDuttaSaikia
Answer : D

Is C<200?

Area of each conference room = C

Area of each rectangular office (considering "Equal") = R

C>R

2C + 6R = 1200

S1 -

C>2R

Even if we take C= 2R, R= 120 , Since we know C>2R, then C>240. Sufficient.

S2-

2 (l+b) < 46

l+b<23

Maximum value could be l=12 , b=11

so, lb= 132 (maximum area of the rectangle)

2C+6*132=1200

C=204 (>200)

Sufficient.
statement 2: fixes the perimeter= 46= 2(L + B) which means L + B= 23, however it doesn't mention that the sides have to be integers. as you have taken values 12,11 , to prove that C > 200, there can also be values like 12.9,10.1 which gives area of office 140.61=R and area of conference room=178.17=C<200 ,thus making statement 2 insufficient
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Bunuel
A certain floor of an office building has 1,200 square feet of floor area allocated to 2 equally sized conference rooms and 6 rectangular offices. If no office can be larger than a conference room, does each conference room have a floor area of less than 200 square feet?

(1) Each conference room must have a smaller area than the combined areas of any three offices and a larger area than the combined areas of any two offices.
(2) The four walls of each office have a combined length of less than 46 feet.

Kudos for a correct solution.

First, I don't think the offices must be equal in area, since it mentions equally sized conference rooms only. But I don't think it matters for this problem anyway.

1) Area any 2 Offices < C < Any 3 offices
If 2C = 400, then remaining area is 800/6 ≈ 133 average. Any two rooms will be at least 266, so there's no way to fulfill our constraint with 200 or lower area for conference rooms. Definite NO to prompt, sufficient.

To double check the idea, if you reduce the area of some offices when 2C=400, say 50 for 2 offices, the 4 others must make up that 100 extra area (now 175 each) so we can't fulfill our constraint either.

2) P Office < 46. The largest area for a quadrilateral is a square, 46/4=11.5 --> 11.5*11.5 = 132.25 max Area. (stopped calculating here since it's the same idea as above).
132.5*6 = 793.5 so the remaining area is 406.5 = 2C, which means C must be more than 200. Sufficient.
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