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The rate of spin of a certain gyroscope doubled every 10 seconds from

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The rate of spin of a certain gyroscope doubled every 10 seconds from the moment a particular stopwatch started. If after a minute and a half the gyroscope reached a speed of 400 meters per second, what was the speed, in meters per second, when the stopwatch was started?

A. 25/3
B. 25/4
C. 25/8
D. 25/16
E. 25/32
[Reveal] Spoiler: OA

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The rate of spin of a certain gyroscope doubled every 10 seconds from [#permalink]

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New post 17 Sep 2017, 03:39
Bunuel wrote:
The rate of spin of a certain gyroscope doubled every 10 seconds from the moment a particular stopwatch started. If after a minute and a half the gyroscope reached a speed of 400 meters per second, what was the speed, in meters per second, when the stopwatch was started?

A. 25/3
B. 25/4
C. 25/8
D. 25/16
E. 25/32


Let 'a' be the speed at the beginning. We have a series similar to a, 2*a, 4*a, 8*a ..... with the next term doubling after each unit (of 10seconds)
The rate of spin exhibits a geometric progression with a common ratio of (2*a/ a) = 2. [common ratio = second term / first term of the series]

General term of a Geometric Progression is given by:
term at the n^{th} unit = a*r^{n-1} -------for a=first term and r is common ratio >1

We are given n = 9 [time elapsed is 90 seconds i.e. 9*(unit of 10 seconds).]
value at 9th unit = 400 and r=2
Substituting,
400 = a(2)^{9-1}
400=a(2^8)
(2^4)(5^2) = a(2^8)
a= 25/16
"D"
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Bunuel wrote:
The rate of spin of a certain gyroscope doubled every 10 seconds from the moment a particular stopwatch started. If after a minute and a half the gyroscope reached a speed of 400 meters per second, what was the speed, in meters per second, when the stopwatch was started?

A. 25/3
B. 25/4
C. 25/8
D. 25/16
E. 25/32

I get E, 25/32. I'm posting this in an order different from that in which I solved. The first approach seems to me to be the simplest.

When the clock starts, the speed is x. After 10 seconds, the speed is 2x. After 20 seconds, the speed is 4x.
The whole sequence should be \(x * 2^9 = 512x\). Note there are 10 terms

00 seconds: x
10 seconds: 2x
20 seconds: 4x
30 seconds: 8x
40 seconds: 16x
50 seconds: 32x
60 seconds: 64x
70 seconds: 128x
80 seconds: 256x
90 seconds: 512x

400 = 512x

\(\frac{400}{512}\) = \(\frac{25}{32}\) = x

The doubling of the gyroscope's speed begins when the stopwatch starts. The prompt does not say that the speed of the gyroscope is zero at moment zero (0:00). The time at the beginning is zero. The speed is positive (from the answer choices).

I get Answer E if I start with 25/32 and double every 10 seconds. Note there are nine intervals

0 -10: 25/32 --> 25/16
10-20: 25/16 --> 25/8
20-30: 25/8 ---> 25/4
30-40: 25/4 ---> 25/2
40-50: 25/2 ---> 25
50-60: 25 ---> 50
60-70: 50 --> 100
70-80: 100 --> 200
80-90: 200 --> 400

I first solved using geometric progression.* But I struggled for a moment. Is 400 the 9th value? Or the 10th? I think it is the 10th. I count nine intervals. But I think there are 10 terms, that is, I think 400 is the value of the 10th term.

Number of terms, inclusive (if \(Time_0\) has a value, it should be subtracted from the value at \(Time_{90}\)), is \(\frac{(90-0)}{10} + 1\), which equals 10 terms.

\(A_{n} = A*r^{(n-1)}\)
\(n^{th}\) term = \(400\)
\(r = 2\)
\((n-1)\) = 9

Then

\(400 = A*2^{9}\)
\(\frac{400}{512} = A\)
\(A = \frac{25}{32}\)

From above, yet a third method (working from answer choice E), seems to indicate E.

Please correct me if I am wrong.

ANSWER E

*\(A_{n} = A*r^{(n-1)}\), where
\(A\) = beginning term
\(A_{n} = n^{th}\) term
\(r\) = common ratio

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If the rate doubled every 10 seconds, we have a total of 9 such ten seconds slots.

Let the initial rate of spin be x.

At the end of 10 seconds the rate is 2x
At the end of 20 seconds the rate is 4x
At the end of 30 seconds the rate is 8x
At the end of 40 seconds the rate is 16x
At the end of 50 seconds the rate is 32x
At the end of 60 seconds the rate is 64x
At the end of 70 seconds the rate is 128x
At the end of 80 seconds the rate is 256x
At the end of 90 seconds the rate is 512x

Therefore, 512x = 400.

x=\(\frac{400}{512} = \frac{25}{32}\)

genxer123, you are correct about the answer being Option E.
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The rate of spin of a certain gyroscope doubled every 10 seconds from [#permalink]

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New post 17 Sep 2017, 12:31
pushpitkc wrote:
genxer123, you are correct about the answer being Option E.

Thank you for responding. I appreciate it. (Many questions seem to go unanswered.) And you respond often. Kudos. :-)

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Re: The rate of spin of a certain gyroscope doubled every 10 seconds from [#permalink]

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Bunuel wrote:
The rate of spin of a certain gyroscope doubled every 10 seconds from the moment a particular stopwatch started. If after a minute and a half the gyroscope reached a speed of 400 meters per second, what was the speed, in meters per second, when the stopwatch was started?

A. 25/3
B. 25/4
C. 25/8
D. 25/16
E. 25/32


We can let the initial rate = x.

Thus, after 10 seconds the rate is 2x, after 20 seconds the rate is 4x (or 2^2 * x), and after 30 seconds the rate is 8x (or 2^3 * x). Thus, after 1.5 minutes, or 90 seconds, the rate is 2^9 * x = 512x. Since the rate after 1.5 minutes is 400 m/s:

512x = 400

x = 400/512 = 50/64 = 25/32.

Answer: E
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Re: The rate of spin of a certain gyroscope doubled every 10 seconds from   [#permalink] 22 Sep 2017, 11:46
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