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ARIEN3228
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IanStewart
To fit in as many points as possible, we'll want them to be as close together as possible, so (if we can) we want to make the distance between points 1 meter. Along a diameter, we can place three points -- one at the center, and two on the circumference. If we have a point at the center, every other point will need to be on the circumference, to be 1 meter away from the center.

So now we want points around the circumference that are 1 meter apart, in a straight line. If you take two consecutive points around the circumference, if they're 1 meter apart, and the radius is 1 meter, you'd make an equilateral triangle by connecting the two circumference points and the center. So there will be a 60 degree angle at the center, and going around the circle, you'll be able to place 6 points around the circumference that are 1 meter apart in a straight line (since 6*60 = 360). Including the center, we have 7 points in total.

you can't put a point on the center of a disc as in the center there is simply a hole
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IanStewart
To fit in as many points as possible, we'll want them to be as close together as possible, so (if we can) we want to make the distance between points 1 meter. Along a diameter, we can place three points -- one at the center, and two on the circumference. If we have a point at the center, every other point will need to be on the circumference, to be 1 meter away from the center.

So now we want points around the circumference that are 1 meter apart, in a straight line. If you take two consecutive points around the circumference, if they're 1 meter apart, and the radius is 1 meter, you'd make an equilateral triangle by connecting the two circumference points and the center. So there will be a 60 degree angle at the center, and going around the circle, you'll be able to place 6 points around the circumference that are 1 meter apart in a straight line (since 6*60 = 360). Including the center, we have 7 points in total.

you can't put a point on the center of a disc as in the center there is simply a hole
I don't know how to react to this.
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gurmukh
Circumference of the circle =2×3.14×1
= 6.28
So 6 points on the circle and one at center total 7
Option E is the answer

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I don't think the argument (highlighted) is justified. The circumference is an arc, and you want them away 1 meter on a straight line. So in my opinion we cannot draw this conclusion. If we take IanStewart 's method above of drawing equilateral triangles, that makes much more mathematical sense. What do you think?
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Hello from the GMAT Club BumpBot!

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