A construction company will be penalised each day for delay in constru

1.	A construction company will be penalised each day for delay in construction of a bridge. The penalty will be Rs 4000 for the first day and will increase by Rs 1000 for each following day. Based on its budget, the company can afford to pay a maximum of Rs 1,65,000 towards penalty. Find the maximum number of days by which the completion of work can be delayed.
Answer» Penalty for the first day (a) = Rs 4000 Increased rate for every day (d) = Rs 1000 Maximum amount of penalty (S_n) = Rs 1,65,000 S_n = (n/2) [2a + (n - 1) 1000] 165000 = (n/2) [2(4000) + (n – 1) 1000] = (n/2) [8000 + 1000 n – 1000] 165000 = (n/2) (7000 + 1000n) 330000 = 7000 n + 1000 n² 0 = 1000 n² + 7000 n – 330000 ÷ 1000 on both sides, n² + 7n – 330 = 0 (n + 22) (n – 15) = 0 n = -22 or n = 15 n = -22 (not possible) ∴ Maximum number of days for which the work can be delayed is 15 days.

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

Answer»

Penalty for the first day (a) = Rs 4000

Increased rate for every day (d) = Rs 1000

Maximum amount of penalty (S_n)

= Rs 1,65,000

S_n = (n/2) [2a + (n - 1) 1000]

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

= (n/2) [8000 + 1000 n – 1000]

165000 = (n/2) (7000 + 1000n)

330000 = 7000 n + 1000 n²

0 = 1000 n² + 7000 n – 330000 ÷ 1000 on both sides,

n² + 7n – 330 = 0

(n + 22) (n – 15) = 0

n = -22 or n = 15

n = -22 (not possible)

∴ Maximum number of days for which the work can be delayed is 15 days.