A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty if he has delayed the work by 30 days?

Solution:

The penalty for delay of completion by a day, Rs 200 and Rs 50 more each succeeding day.

The sum of the first n terms of an AP is given by Sₙ = n/2 [2a + (n - 1) d] or Sₙ = n/2 [a + l].

Here, a is the first term, d is a common difference and n is the number of terms.

Penalty for 1^st day = Rs. 200

Penalty for 2^nd day = Rs. 250

Penalty for 3^rd day = Rs. 300

By observation, penalties are in A.P. having the first term as 200 and common difference as 50 and number of terms as 30.

a = 200, d = 50, n = 30

Penalty that has to be paid if he has delayed the work by 30 days will be calculated as follows.

Sₙ = n/2 [2a + (n - 1) d]

S₃₀ = 30/2 [2 × 200 + (30 - 1) 50]

= 15 [400 + 1450]

= 15 × 1850

= 27750

Therefore, the contractor has to pay Rs. 27750 as a penalty.

☛ Check: NCERT Solutions for Class 10 Maths Chapter 5

Video Solution:

A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty if he has delayed the work by 30 days?

Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.3 Question 15 

Summary:

A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day, then the money that the contractor has to pay a penalty if he delays the work by 30 days is ₹ 27750.

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