Question

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows. RS 200 for the first day, RS 250 for the second day, Rs 300 for third day and so on. If the contractor pays RS 27750 as penalty find the no of days for which the construction work is delayed

Answer 1

Answer:

The number of days is 552.

Step-by-step explanation:

The penalty of first day = 200 Rs

Penalty for second day = 250 Rs

Penalty for third day = 300 Rs

Given that, the contractor pays 27750 Rs for the penalty.

The number of days (n) =?

We can solve this problem by Arithmetic Progression.

Here First term of A.P (a) = 200 Rs

The second term of A.P. = 250 Rs

Common difference = Second term - first term

d = 250 - 200

d = 50

Last term = 27750

We know that-

The last term of A.P-

$T_{n} = a + (n - 1)d$

27750 = 200 + (n - 1)×50

27750 - 200 = (n - 1)×50

27550 = (n - 1)×50

$n-1=\dfrac{27550}{50}$

n - 1 = 551

n = 551 + 1

n = 552

Hence, the number of days is 552.

Answer 2

Answer:

The penalty of first day = 200 Rs

Penalty for second day = 250 Rs

Penalty for third day = 300 Rs

Given that, the contractor pays 27750 Rs for the penalty.

The number of days (n) =?

We can solve this problem by Arithmetic Progression.

Here First term of A.P (a) = 200 Rs

The second term of A.P. = 250 Rs

Common difference = Second term - first term

d = 250 - 200

d = 50

Last term = 27750

We know that-

The last term of A.P-

27750 = 200 + (n - 1)×50

27750 - 200 = (n - 1)×50

27550 = (n - 1)×50

n - 1 = 551

n = 551 + 1

n = 552

Hence, the number of days is 552

Step-by-step explanation: