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M08-25

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M08-25  [#permalink]

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New post 16 Sep 2014, 00:38
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A
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  75% (hard)

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71% (01:44) correct 29% (02:12) wrong based on 148 sessions

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If the diameter of each of the car's wheels is 20 inches, what was the approximate average speed of the car in km/h if after the two-hour journey each wheel made 75,000 revolutions? (1 inch = 0.0254 meter)

A. 32
B. 41
C. 59
D. 74
E. 88

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Re M08-25  [#permalink]

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New post 16 Sep 2014, 00:38
1
Official Solution:

If the diameter of each of the car's wheels is 20 inches, what was the approximate average speed of the car in km/h if after the two-hour journey each wheel made 75,000 revolutions? (1 inch = 0.0254 meter)

A. 32
B. 41
C. 59
D. 74
E. 88


The wheel's radius in meters = \(\frac{20 \text{ inches } * 0.0254}{2} = 0.254 \approx 0.25\) meters.

The wheel's circumference = \(2\pi(0.25) = 0.5\pi \approx \frac{3}{2} = 1.5\) meters.

The distance in kilometers = \(75000*\frac{1.5}{1000} = 75*1.5 = 112.5 \approx 113\) km.

The average speed in km/h = \(\frac{113}{2} = 56.5\) km/h which is closest to choice C.


Answer: C
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Re: M08-25  [#permalink]

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New post 16 Jun 2015, 08:42
Hi Bunuel,

I am having trouble understanding this problem visually. Based on the solution, I can see we calculated the distance that the "wheel" travel. However, we want to calculate the distance that the car travel. Are they the same? I draw out how the wheel rotate on paper and having trouble to think of these two distances are the same. Could you help me clear this out?
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New post 16 Jun 2015, 10:44
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Tsuruga wrote:
Hi Bunuel,

I am having trouble understanding this problem visually. Based on the solution, I can see we calculated the distance that the "wheel" travel. However, we want to calculate the distance that the car travel. Are they the same? I draw out how the wheel rotate on paper and having trouble to think of these two distances are the same. Could you help me clear this out?


I thinks similar problems might help:
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
a-point-on-the-edge-of-a-fan-blade-that-is-rotating-in-a-plane-is-65505.html
m10-183873.html
at-the-same-time-sue-began-rolling-a-wheelbarrow-from-x-to-y-nancy-199350.html
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Re: M08-25  [#permalink]

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New post 21 Feb 2016, 18:13
The way I skipped some of the math:

piD = Circumference = 20pi

20 * 75,000 *pi = 150(plus a lot of zeroes) *pi

.025 = (1/4) minus a couple of zeroes. So take away a couple of zeroes from the big number and divide by four and you get close to 20. 20*pi = close to 60

Answer C
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New post 25 Feb 2018, 10:23
Bunuel wrote:
If the diameter of each of the car's wheels is 20 inches, what was the approximate average speed of the car in km/h if after the two-hour journey each wheel made 75,000 revolutions? (1 inch = 0.0254 meter)

A. 32
B. 41
C. 59
D. 74
E. 88


Beautiful question, and thank you for the explanation.

On the actual test, for questions like this in which there isn't an elegant way of dealing with pi and we are forced to compute its value, is 3 typically a safe estimation? Or should we go deeper with something like 3.14?
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New post 25 Feb 2018, 10:43
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jsheppa wrote:
Bunuel wrote:
If the diameter of each of the car's wheels is 20 inches, what was the approximate average speed of the car in km/h if after the two-hour journey each wheel made 75,000 revolutions? (1 inch = 0.0254 meter)

A. 32
B. 41
C. 59
D. 74
E. 88


Beautiful question, and thank you for the explanation.

On the actual test, for questions like this in which there isn't an elegant way of dealing with pi and we are forced to compute its value, is 3 typically a safe estimation? Or should we go deeper with something like 3.14?


It depends on the question and how widespread the options are. For this question it was a safe estimation because of that.
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Re: M08-25  [#permalink]

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New post 15 Mar 2018, 17:40
Bunuel, hello, the approximate method works everytime for this same type of question?
I also use the approximate method but only when it is necessary.
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New post 29 Apr 2018, 17:57
chetan2u Bunuel

Hi, could one of you guys please explain why "The distance in kilometers = 75000∗1.5/1000" we are dividing 1.5 by 1000? I was wondering where did we get 1000 from?
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New post 29 Apr 2018, 20:59
csaluja wrote:
chetan2u Bunuel

Hi, could one of you guys please explain why "The distance in kilometers = 75000∗1.5/1000" we are dividing 1.5 by 1000? I was wondering where did we get 1000 from?


We are converting meters to kilometers. 1.5 meters = 1.5/1000 kilometers.
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New post 08 May 2018, 03:10
Hi Bunuel,

Can't we estimate 1 inch=0.0254 as 1 inch=0.02=1/50. I've calculated this way and got 45km p/h avr.speed and selected B. Please have a look at my solution below:

1)Circumference of the wheel: 2πr= 20 inch * (≈3)≈60 inch. Thus, in 1 revolution car goes 60 inch;

2)Calculating 1 revolution in meters: Since 1 inch=0.0254≈0.02=1/50 meters, then 60 inch=60*1/50=6/5=1.2 meters. Thus, in 1 revolution car goes 1.2 meters;

3)Now, we have to calculate in 75000 revolution how long, in meters, car would go: 75000*6/5=90,000 meters which is equal to 90 km.

4) Calculating avr. speed: Since in 2 hours car goes 90 km, then in 1 hour it should go 45km per hour. Hence, the closest answer is B- 41 kmp/h.
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New post 23 Jun 2018, 01:43
Hi,

Why did we not consider 75000 rev in 2 hours?
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New post 23 Jun 2018, 06:36
Sumit1703@4 wrote:
Hi,

Why did we not consider 75000 rev in 2 hours?


1. We used 7,500 to calculate the distance covered in kilometers. You see (revolutions)*(circumference) = (distance covered)

2. We used 2 hours to calculate the rate: (rate) = (distance)/(time)

Hope it helps.
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Re: M08-25 &nbs [#permalink] 23 Jun 2018, 06:36
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