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# In the figure above, two security lights, L1 and L2, are located 100

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Math Expert
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In the figure above, two security lights, L1 and L2, are located 100  [#permalink]

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11 Nov 2019, 01:31
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45% (medium)

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72% (02:14) correct 28% (02:36) wrong based on 31 sessions

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In the figure above, two security lights, L1 and L2, are located 100 feet apart. Each illuminates an area of radius 100 feet, and both are located 60 feet from a chain-link fence. What is the total length s of fence, in feet, illuminated by the two lights?

(A) 260
(B) 240
(C) 220
(D) 200
(E) 180

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Re: In the figure above, two security lights, L1 and L2, are located 100  [#permalink]

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11 Nov 2019, 08:50
Bunuel wrote:
In the figure above, two security lights, L1 and L2, are located 100 feet apart. Each illuminates an area of radius 100 feet, and both are located 60 feet from a chain-link fence. What is the total length s of fence, in feet, illuminated by the two lights?

(A) 260
(B) 240
(C) 220
(D) 200
(E) 180

Join $$L_1$$ and $$L_2$$

Step I - Take $$L_1L_2EC$$. It is a parallelogram so $$L_1L_2=EC=100$$
Step II - Drop a perpendicular at D, so $$\triangle {L_2DE}$$ is right angled triangle with two sides in ratio 60:100, so third side will be 80 as it is a 3:4:5 triangle, and BE=80+80=160.
Step III - Find overlap or BC.
$$BE=BC+CE.....160=BC+100...BC=60$$

WE have to find AE.
$$AE=AB+BC+CE=100+100+60=260$$ OR
AE=2*160-overlap=2*160-60=260

A
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In the figure above, two security lights, L1 and L2, are located 100  [#permalink]

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Updated on: 11 Nov 2019, 10:11
1
Archit3110 wrote:
TestPrepUnlimited wrote:
The radius is 100 since their distance apart = radius = 100 ft. The distance of 60 ft represents the perpendicular length from the fence to any of the centers. So we can confirm each circle illuminates 80 * 2 = 160 ft of the fence using the 60 - 80 - 100 triangle. Finally, we need to subtract the overlap from the two fences. Note however the overlap cannot be more than 80, so the total fence length must be at least 160 + 160 - 80 = 240. We can only choose A.

Attachment:
2cir.png

TestPrepUnlimited
see highlighted part ; option B is 240

Archit3110
Only possibility exists is A, since B would mean one circle passing through point of intersection of perpendicular from fence(chord) to center of other circle(L1 or L2). This would gravely deform the diagram.

On the other side, if you are good at drawing circles to scale, you can observe that in the 60 - 80 - 100 triangle for L1 the perimeter of circle with center L2 bisects the side 80 in roughly a ratio of 1:3.

Thus the common part of fence ~ $$\frac{3}{4} * 80 = 60$$.
Trying to quote 'Ron'

Hence A.
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Originally posted by lnm87 on 11 Nov 2019, 09:20.
Last edited by lnm87 on 11 Nov 2019, 10:11, edited 1 time in total.
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Re: In the figure above, two security lights, L1 and L2, are located 100  [#permalink]

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11 Nov 2019, 22:15
1
The length of the fence covered by the light is: S = 80+100+80=260 feet.

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In the figure above, two security lights, L1 and L2, are located 100  [#permalink]

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Updated on: 11 Nov 2019, 09:55
The radius is 100 since their distance apart = radius = 100 ft. The distance of 60 ft represents the perpendicular length from the fence to any of the centers. So we can confirm each circle illuminates 80 * 2 = 160 ft of the fence using the 60 - 80 - 100 triangle. Finally, we need to subtract the overlap from the two fences. Note however the overlap cannot be more than 80, so the total fence length must be more than 160 + 160 - 80 = 240. We can only choose A.

Attachment:

double circles.png [ 8.27 KiB | Viewed 541 times ]

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Originally posted by TestPrepUnlimited on 11 Nov 2019, 05:42.
Last edited by TestPrepUnlimited on 11 Nov 2019, 09:55, edited 1 time in total.
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Re: In the figure above, two security lights, L1 and L2, are located 100  [#permalink]

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11 Nov 2019, 07:03
TestPrepUnlimited wrote:
The radius is 100 since their distance apart = radius = 100 ft. The distance of 60 ft represents the perpendicular length from the fence to any of the centers. So we can confirm each circle illuminates 80 * 2 = 160 ft of the fence using the 60 - 80 - 100 triangle. Finally, we need to subtract the overlap from the two fences. Note however the overlap cannot be more than 80, so the total fence length must be at least 160 + 160 - 80 = 240. We can only choose A.

Attachment:
2cir.png

TestPrepUnlimited
see highlighted part ; option B is 240
Math Expert
Joined: 02 Aug 2009
Posts: 8296
Re: In the figure above, two security lights, L1 and L2, are located 100  [#permalink]

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11 Nov 2019, 08:51
TestPrepUnlimited wrote:
The radius is 100 since their distance apart = radius = 100 ft. The distance of 60 ft represents the perpendicular length from the fence to any of the centers. So we can confirm each circle illuminates 80 * 2 = 160 ft of the fence using the 60 - 80 - 100 triangle. Finally, we need to subtract the overlap from the two fences. Note however the overlap cannot be more than 80, so the total fence length must be at least 160 + 160 - 80 = 240. We can only choose A.

Attachment:
2cir.png

At least 240 means answer can be 240 or 260?? So A or B possible
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In the figure above, two security lights, L1 and L2, are located 100  [#permalink]

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11 Nov 2019, 09:58
Thank you for pointing that out, I have fixed my wording. The intent was to avoid doing any other calculations by seeing the overlap is certainly less than 80, so the fence length > 240.

Archit3110 wrote:
TestPrepUnlimited wrote:
The radius is 100 since their distance apart = radius = 100 ft. The distance of 60 ft represents the perpendicular length from the fence to any of the centers. So we can confirm each circle illuminates 80 * 2 = 160 ft of the fence using the 60 - 80 - 100 triangle. Finally, we need to subtract the overlap from the two fences. Note however the overlap cannot be more than 80, so the total fence length must be at least 160 + 160 - 80 = 240. We can only choose A.

Attachment:
2cir.png

TestPrepUnlimited
see highlighted part ; option B is 240

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Source: We are an NYC based, in-person and online GMAT tutoring and prep company. We are the only GMAT provider in the world to guarantee specific GMAT scores with our flat-fee tutoring packages, or to publish student score increase rates. Our typical new-to-GMAT student score increase rate is 3-9 points per tutoring hour, the fastest in the world. Feel free to reach out!
In the figure above, two security lights, L1 and L2, are located 100   [#permalink] 11 Nov 2019, 09:58
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