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Re: In the figure above, a wheel rolls down a slope in a straight line. If [#permalink]
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Bunuel wrote:

In the figure above, a wheel rolls down a slope in a straight line. If point x touches the surface at 50-inch intervals, then the radius of the wheel is approximately

A. 16 inches
B. 12 inches
C. 10 inches
D. 8 inches
E. 7 inches

Attachment:
Untitled.png


50-inch intervals means that the circumference is 50 inch

So, 2πr = 50
Or, πr = 25
Or, \(\frac{22}{7}r\)= 25
Or, \(r\) =25*7/22
Or, \(r\) ~ 8

Hence answer is D.
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Re: In the figure above, a wheel rolls down a slope in a straight line. If [#permalink]
2 pi r= 50
r= 25/(22/7) = apprx 8

Ans : D
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Re: In the figure above, a wheel rolls down a slope in a straight line. If [#permalink]
CounterSniper wrote:
one rev =circumfrence of the circle=2pir
by the time x touches the surface , wheel would have travelled 2pir/2=pir
also pir=50 > r=50*7/22 =15.92 (A)


I understand the figure is not drawn to scale and also it is not mentioned in the question that X lies on half circumference.

Should we still take it as 2 pi R/2?
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Re: In the figure above, a wheel rolls down a slope in a straight line. If [#permalink]
My bad !!

Abhishek already explained 50 inches interval means 2pir =50

so even if we go by the diagram , it should have been pir =25
as x already covered half of the distance.

so r should be 8 and not 16 !!
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Re: In the figure above, a wheel rolls down a slope in a straight line. If [#permalink]
CounterSniper wrote:
one rev =circumfrence of the circle=2pir
by the time x touches the surface , wheel would have travelled 2pir/2=pir
also pir=50 > r=50*7/22 =15.92 (A)

I think the clue here is that the problem says "point x touches the ground in 50 inch intervals". So 2Πr=50 then r=8 ish. Hence the best answer is D.
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Re: In the figure above, a wheel rolls down a slope in a straight line. If [#permalink]
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Re: In the figure above, a wheel rolls down a slope in a straight line. If [#permalink]
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