Hi naeln,

I'm going to give you some 'hints' so that you can try this question again:

1) There's a 'visual' component to this question, so drawing a picture should help. Draw the 2 circles, draw a VERTICAL diameter in each circle, then draw a "belt" as described....You should have a picture with 2 identical circles and a rectangle that is formed by the two diameters and the space between the circles.
2) Since the belt "goes around" one half of each circle, you have to think about the circumference (and how long it is).
3) The length of the belt includes the two half-circles and the two lengths of the rectangle....
4) At some point, you're going to end up with an "ugly-looking" fraction. You'll have to do a little math to 'clean up' the fraction.

GMAT assassins aren't born, they're made,
Rich
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Circumference of a circle = 2rP
r=1/2d=1/4
The belt only goes around half of the circle, so the length of the belt to go around one wheel is 1/4P
There are two wheel, so it takes 1/4P*2=1/2P to go around both wheels
The rest of the belt cover the distance between the wheels from both sides, so the distance is (10-1/2P)/2=20-P/4

Situation is something like the picture depicts. So the belt which is 10 ft long will go around both the semi-circles (wheels) which are of the same dia.

So semicircular circumference *2 + distance between the centres*2 = 10

pi*1/4*2 + 2*d = 10

2d = 10-pi/2

d = (20-pi)/4

Answer is B

(10-1/2*⫪)/2=(20-⫪)/4
B

Given that \(d_1\)=\(d_2\)=\(d\)=1/2
Please refer the diagram as well.
Hence , \(d_1\)*⫪/2 =\(d_2\)*⫪/2 =\(d\)*⫪/2
Hence according to the question \(2x\)+\(2\)*\(d\)*⫪/2 =\(10\)
or \(2x\)+\(2\)*(1/2)*⫪/2 =\(10\)
or \(2x\)+⫪/2 =\(10\)
or \(4x\)+⫪=20
or \(4x\)=20 -⫪
or \(x\)=(20 -⫪)/4.................Hence option B.

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