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

Question

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

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

BrentGMATPrepNow · Accepted Answer

This can be answered within a reasonable time frame by just charting each 10-second period. - After 90 seconds: 400 meters/second - After 80 seconds: 200 meters/second - After 70 seconds: 100 meters/second - After 60 seconds: 50 meters/second - After 50 seconds: 25 meters/second - After 40 seconds: 25/2 meters/second - After 30 seconds: 25/4 meters/second - After 20 seconds: 25/8 meters/second - After 10 seconds: 25/16 meters/second - START: 25/32 meters/second Answer: E For more on this straightforward technique, see our free video - https://www.gmatprepnow.com/module/gmat- ... /video/929 Cheers, Brent

chetan2u · Answer

Hi,
two ways you can do the Qs of this type..

1) working backwards.. 
here you will have to carry out HOW many steps- 90/10..
can be error prone, but the way would be..
after 
 90 secs - 400..
 80 secs - 200.., keep dividing by 2
 70 secs - 100 
 60 secs - 50 
 50 secs - 25 
40 secs - 25/2 
 30 secs - 25/4
 20 secs - 25/8 
10 secs - 25/16..
 initial - 25/32..
 
 2) straight ..

an easier and better, and less time consuming method

let the initial speed be x ..
it doubles in 10 secs, or it increases 2 times every 10 secs..
total 10 secs period= 90/10=9..
so final speed after 90 secs= x*2*2...9 times
therefore  x*2^9=400 ...
or  x*2^9=400=2^4*25 ...
 x=25*2^4/2^9=25/2^5=25/32.. 
ans E

utkarshthapak · Answer

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.

generis · Answer

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

pushpitkc · Answer

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.

ScottTargetTestPrep · Answer

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

sviral90 · Answer

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

Abir77 · Answer

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

chetan2u · Answer

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

Abir77 · Answer

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

ScottTargetTestPrep · Answer

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

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards