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A computer solved several problems in succession . the time it took the computer to solve the solve each successive problem was the same no of times smaller than the time it took to solve the preceding problem.how many promblems were suggested to the computer if it spend 63.5 min to solve all the problems except for the first, 127min to solve all the problems except for the last one,and 31.5 min to solve all the problems except for the first two.

utkarsha sonje , 13 Years ago
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anser 1 Answers
Ashwin Muralidharan IIT Madras

Last Activity: 13 Years ago

Hi Utkarsha,

 

Let there be n problems as that was fed to the comuter.

Let t be the time it had taken to solve the first question.

 

So now aaccording to the problem the time taken to solve the successive questions will be t-t/x, t-2t/x,....., t-(n-1)t/x.

 

There are three variables in the problem - n,t,x.

You need to solve for n.

There are three conditions in the problem. So now use the three conditions and form equations.

Then solve for n.

 

t-t/x + t-2t/x + t-3t/x + ....... t-(n-1)t/x = 63.5 ----------- (1)

t + t-t/x + t-2t/x +...... + t-(n-2)t/x = 127 ---------- (2)

t-2t/x + t-3t/x + ........ + t-(n-1)t/x = 31.5 --------- (3)

 

From the three equations, solve for n.

 

(1)-(3):

t-t/x = 32 ------------- (4)

 

(2)-(1):

(n-1)t/x = 63.5 ------------(5)

 

From (2):

(n-1)t - (t/x)(n-2)(n-1)/2 = 127

or (n-1)t*[1 - (n-2)/(2x)] = 127

63.5x*[1 - (n-2)/(2x)] = 127 -------- from (5) (n-1)t = 63.5x

or x[1 - (n-2)/(2x)] = 2

or [2x-n-2] = 4

or x-1 = (n+4)/2 ----------- (6)

 

Using (4):

(1-1/x)[63.5x/(n-1)] = 32

(x-1)[63.5/(n-1)] = 32 -------- (7)

So 63.5/(n-1) [n+4] = 64

or 63.5(n+4) = 64(n-1)

So 63.5n + 254 = 64n - 64

or 0.5n = 318.

or n = 636.

 

So 636 Questions would be fed to the computer.

That solves this question.

 

Hope it helps.

 

Regards,

Ashwin (IIT Madras),

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