Question

A computer solved several problems in succession. The time it took the computer to solve each successive problem was the same number of times smaller than the time it took to solve the preceding problem. How many problems were suggested to the computer if it spent \( 63.5 min \) to solve all the problems except for the first, \( 127 min \) to solve all the problems except for the last one, and \( 31.5 min \) to solve all the problems except for the first two?

Answer 1

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

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

So now according 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.