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GRE Math Challenge #30-architect designs a room with
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18 Sep 2014, 12:13
18 Sep 2014, 12:13
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Question Stats:
81% (02:00) correct
18% (01:21) wrong based on 61 sessions
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An architect designs a room with a rectangular floor that meets the following requirements:
1. The length is twice the width
2. The area is at least 400 square meters
3. The perimeter is no more than100 meters
If w is the width of the floor, in meters, which of the following must be true for w?
A. \(15≤w≤20\)
B. \(15≤w≤√398\)
C. \(16.66667≤w≤20\)
D. \(√200/3≤w≤25/2\)
E. \(√200≤w≤50/3\)
1. The length is twice the width
2. The area is at least 400 square meters
3. The perimeter is no more than100 meters
If w is the width of the floor, in meters, which of the following must be true for w?
A. \(15≤w≤20\)
B. \(15≤w≤√398\)
C. \(16.66667≤w≤20\)
D. \(√200/3≤w≤25/2\)
E. \(√200≤w≤50/3\)
Re: GRE Math Challenge #30-architect designs a room with
[#permalink]
17 Feb 2020, 13:14
17 Feb 2020, 13:14
Expert Reply
Bump for further discussion
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Regards
Re: GRE Math Challenge #30-architect designs a room with
[#permalink]
16 Aug 2020, 18:47
16 Aug 2020, 18:47
Let: l=length, w=width
l=2w
area= l*w=2w*w=2w^2
area if greater than or equal to 400
therefore
2w^2>=400
w^2>=200
(1) w>=rad(200)
Perimeter=2(l+w)=2(2w+w)=2(3w)=6w
6w<=100
w<=100/6
(2) w<=50/3
from (1) and (2) we get
rad(200) <= w <= 50/3
Final Answer: E
l=2w
area= l*w=2w*w=2w^2
area if greater than or equal to 400
therefore
2w^2>=400
w^2>=200
(1) w>=rad(200)
Perimeter=2(l+w)=2(2w+w)=2(3w)=6w
6w<=100
w<=100/6
(2) w<=50/3
from (1) and (2) we get
rad(200) <= w <= 50/3
Final Answer: E
