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

Long method, count backwards....

Shortcut, use Geometric Progression.
Nth term of a GP is, tn = a*r^n (where a is the first term of progression and r is the ratio of second term to first term)
We know, r = 2 (as speed doubles)
tn = 400 (speed at 90 sec)
n= 9 (since there will be 9 intervals of 10 sec in 90 sec)
so, substitute all these values in equation above.
400 = a * 2^9 or, 400 = a* 512
therefore, a = 400/512 = 25/32.

This method looks long but it only takes 10-15 sec to solve.

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

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.

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

Isn't geometric progression given by a*r^n-1?

Seems like this solution is based on compound interest
In that case Total = Principal(1+rate)^Time

so 400 = principal (1+2)^9

finally I'm getting wrong answer. Please tell me what's wrong with my approach

It is doubling, or it is increasing by 100%. So, rate will be 100% => 400=P(1+(100/100))^9=P(1+1)^9=P*2^9

Thanks a lot chetan2u
Have been scratching my head for a long time. Got it now

Solution:

A minute and a half is 90 seconds, which means the gyroscope has doubled its speed 9 times after the stopwatch started. Therefore, if we let x = the initial speed, we can create the following equation:

x * 2^9 = 400

x * 512 = 400

x = 400/512 = 25/32

Answer: E