CBSE Class 11-science Answered

Let there be n problems as that was fed to the computer and let t be the time it takes to solve the first question.

To find n  

So now according to the question the time taken to solve the successive questions will given by the series  

t- (t / x), t- (2t / x),....., t- ( (n-1) t )/x.

now , according to the question 

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

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

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

Now , let us subtract eq(1) by eq (3), we get 

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

 Let us subtract eq(2) by eq (1), we get

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

Now lets us evauate  eq( 2)

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

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

 => 63.5 x * [1 - (( n - 2 ) /  ( 2x ) ) ] = 127 ( putting the value                                                                                   obtained in eq (5) )

=> (n-1) t = 63.5x

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

=> [2x-n-2] = 4

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

Using eq(4) , we get 

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

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

So we can say that

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

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

therefore

=> 63.5n + 254 = 64n - 64

=> 0.5n = 318.

=> n = 636 questions would be fed to the computer in total.