Question

13. A construction company will be penalised each day for delay in construction of a
bridge. The penalty will be `4000 for the first day and will increase by `1000 for each
following day. Based on its budget, the company can afford to pay a maximum of
`1,65,000 towards penalty. Find the maximum number of days by which the completion
of work can be delayed

Answer 1

Hi ,

According to the problem given ,

penalty of first day = 4000

increase penalty each day = 1000

maximum penalty = 165000

Now ,

each day penalties are ,

4000 , 5000 , 6000, .... this sequence is

in A. P

first term = a = 4000

common difference = a2 - a1

d = 5000 - 4000 = 1000

Let number of days company paid the

penalty = n

Sum of penalties( Sn ) = 165000

n/2 [ 2a + ( n - 1 )d ] = 165000

n/2 [ 2 × 4000 + ( n -1 ) × 1000 ] = 165000

( 1000n/2 ) [ 8 + n - 1 ] = 165000

n( n + 7 ) = ( 165000 × 2 )/( 1000 )

n( n + 7 ) = 330

n² + 7n - 330 = 0

n² + 22n - 15n - 330 = 0

n( n + 22 ) - 15( n + 22 ) = 0

( n + 22 )( n - 15 ) = 0

n + 22 = 0 or n - 15 = 0

n = -22 or n = 15

n should not be negative .

n = 15

Therefore ,

Number of days company delayed the

work = n = 15

I hope this helps you.

: )