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M10-28

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M10-28  [#permalink]

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New post 16 Sep 2014, 00:42
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The radius of the front wheels of a cart is half that of the rear wheels. If the circumference of the front wheels is 1 meter and the cart traveled 1 kilometer, how many revolutions did the rear wheels make?

A. \(\frac{250}{\pi}\)
B. \(\frac{500}{\pi}\)
C. 250
D. 500
E. 750

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Re M10-28  [#permalink]

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New post 16 Sep 2014, 00:43
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Official Solution:

The radius of the front wheels of a cart is half that of the rear wheels. If the circumference of the front wheels is 1 meter and the cart traveled 1 kilometer, how many revolutions did the rear wheels make?

A. \(\frac{250}{\pi}\)
B. \(\frac{500}{\pi}\)
C. 250
D. 500
E. 750

The circumference of the rear wheels is twice that of the front wheels, i.e. 2 meters. Thus, the rear wheels made \(\frac{1000}{2} = 500\) revolutions.

Answer: D
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Re: M10-28  [#permalink]

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New post 25 Sep 2014, 07:04
Bunuel wrote:
Official Solution:

The radius of the front wheels of a cart is half that of the rear wheels. If the circumference of the front wheels is 1 meter and the cart traveled 1 kilometer, how many revolutions did the rear wheels make?

A. \(\frac{250}{\pi}\)
B. \(\frac{500}{\pi}\)
C. 250
D. 500
E. 750

The circumference of the rear wheels is twice that of the front wheels, i.e. 2 meters. Thus, the rear wheels made \(\frac{1000}{2} = 500\) revolutions.

Answer: D


I thought we have to divided the distance i.e. 1000 meters with circumference i.e. 4\({\pi}\) and hence the answer should be \(\frac{250}{{\pi}}\)
but I guess in the solution you have divided the the distance with radius. Is it so?
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Re: M10-28  [#permalink]

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New post 25 Sep 2014, 07:25
2
b2bt wrote:
Bunuel wrote:
Official Solution:

The radius of the front wheels of a cart is half that of the rear wheels. If the circumference of the front wheels is 1 meter and the cart traveled 1 kilometer, how many revolutions did the rear wheels make?

A. \(\frac{250}{\pi}\)
B. \(\frac{500}{\pi}\)
C. 250
D. 500
E. 750

The circumference of the rear wheels is twice that of the front wheels, i.e. 2 meters. Thus, the rear wheels made \(\frac{1000}{2} = 500\) revolutions.

Answer: D


I thought we have to divided the distance i.e. 1000 meters with circumference i.e. 4\({\pi}\) and hence the answer should be \(\frac{250}{{\pi}}\)
but I guess in the solution you have divided the the distance with radius. Is it so?


No. We are given that the circumference of the front wheels is 1 meter and the radius of the front wheels is half that of the rear wheels.

Thus, the circumference of the front wheels is \(2\pi{r}=1\) and the circumference of the rear wheels is \(2\pi{(2r)}=2*2\pi{r}=2\).

So, we are dividing the distance with the circumference.

Hope it's clear.
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Re: M10-28  [#permalink]

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New post 19 Jan 2015, 11:25
are basically tryimg to find how many circumferences in the distance the cart travelled? So is resolution a full 360 degree circum?

thanks
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Re: M10-28  [#permalink]

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New post 21 Jan 2015, 00:27
Bunuel wrote:
Emiliya43 wrote:
are basically tryimg to find how many circumferences in the distance the cart travelled? So is resolution a full 360 degree circum?

thanks


One 360-degree revolution = the distance of one circumference.



What is the basic relation between Distance, Revolution, Circumference?

Is it Distance = No of Revolutions * Circumference ? Is this correct?
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Re: M10-28  [#permalink]

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New post 21 Jan 2015, 01:03
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buddyisraelgmat wrote:
Bunuel wrote:
Emiliya43 wrote:
are basically tryimg to find how many circumferences in the distance the cart travelled? So is resolution a full 360 degree circum?

thanks


One 360-degree revolution = the distance of one circumference.



What is the basic relation between Distance, Revolution, Circumference?

Is it Distance = No of Revolutions * Circumference ? Is this correct?


Yes, that;s correct.

Check similar questions to practice:
a-circular-rim-28-inches-in-diameter-rotates-the-same-number-65106.html
two-wheels-are-connected-via-a-conveyor-belt-the-larger-133697.html
how-long-in-minutes-did-it-take-a-bicycle-wheel-to-roll-130026.html
what-is-the-number-of-360-degree-rotations-that-a-bicycle-wh-166668.html
the-circumference-of-the-front-wheel-of-a-cart-is-30-ft-long-90588.html

Hope it helps.
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M10-28  [#permalink]

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New post 21 Jan 2015, 02:40
I guess it needs less math thinking than rational thinking. Hope I will explain well enough. Let's put the information we have together:

Radius of front wheel: r
Radius of rear wheel: 2r

This means that when the front wheel makes 1 whole revolution, the rear one makes 2. That because its diameter (2 radii) is twice as large, so it covers twice the distance of the front wheel at the same time (if the weels were not attached, when the front wheel covers 5 meters, the back would cover 10). Put differently, since the wheels are actually attached to the same vehicle, the distance the rear wheel has covered in the same amount of time with the front is half the distance of the front (if the front wheel covered 10 meters, the read wheel covered 5).

Now, we know that the circumference is 1 meter (so, in 1 meter the cart makes 1 revolution) and that the cart covered 1
km.

In others words:
1 revolution in 1 meter
x revolutions in 1000 meters?
x = 1000 for the front wheel, so x = 500 for the rear wheel.
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Re: M10-28  [#permalink]

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New post 25 May 2016, 06:42
Distance = Revolutions*Circumference ---> Revolutions = Distance/Circumference = 1km/Circumference

We need to find the circumference of the back tire using a relation to the circumference of the front tire.

Radius of front wheel (Rf): 1 meter = 2*Pi*Rf ---> Rf = 1/(2*Pi) meters
Radius of back wheel (Rb): 2*Rf = 1/Pi meters since it is stated that the radius of the back tire is twice that of the front
Circumference of back wheel (Cb) = 2*Pi*Rb = 2 meters

Revolutions = 1km/2 meters = 1*10^3 meters/2 meters = 500 meters
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Re: M10-28  [#permalink]

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New post 20 Aug 2016, 10:42
Nice question, you don't even have to lift the pen to answer it. :-D
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Re: M10-28  [#permalink]

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New post 28 Jul 2017, 09:40
Very cool - its like being tricked with word changes. The question talks about radius first and then mentions (*casually*) the circumferences. I was busy trying to calculate the 2 pi R; whereas its already there! :D
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New post 20 Sep 2018, 12:33
I don't agree with the explanation. Question says radius is half- not circumference
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New post 20 Sep 2018, 20:41
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Re: M10-28  [#permalink]

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New post 08 Oct 2018, 13:20
Bunuel wrote:
Emiliya43 wrote:
are basically tryimg to find how many circumferences in the distance the cart travelled? So is resolution a full 360 degree circum?

thanks


One 360-degree revolution = the distance of one circumference.


I'm confused as to how the pie gets cancelled out. That confusion seems to be baked into the answer choices, hence me picking " 500/pie" rather than 500
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Re: M10-28 &nbs [#permalink] 08 Oct 2018, 13:20
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