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 terms00 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 intervals0 -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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Formerly genxer123