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Wednesday, July 1, 2015

UPSC Maths Problems Challenge : A Problem A Day! (Day 19)

Today's problem is from Arithmetic Progressions,which as I read is one of the medium priority topics for IAS prelims.Academically,I have been introduced to A.P while I was in Class X but we do have it here in Class XI and I am still a mere beginner who often acts as if she has lost navigation in mid-sea in case of solving A.P/G.P series problems.Though I do not think I have ever seen questions that hard in prelims,like the series that involve summation of squares and even harder sums with numbers like 1/2*3 + 1/3*4,to admit it,I have not yet mastered them at all.But the word problems and shorter sums can be done quite easily.

Today's Problem(01/07/2015)

Question Credit :

(2011) A contract on construction job specifies a penalty for delay in completion of the work beyond a certain date IS as follows : Rs. 200 for the first day,  Rs. 250 for the second day, Rs. 300 for the third day etc., the penalty for each succeeding day being 50 more than that of the preceding day. How much penalty should the contractor pay if he delays the work by 10 days?

We find from the problem that the remuneration of each worker increases by 
(300 - 250) = 
(250 - 200) = 
50 every day.This clearly specifies that the sum has to be solved with the help of Arithmetic Progression,since there is a common difference between the terms.
Now,S10 = 10/2 { 2(200) + (10 - 1)(50)} = 5 (400 + 450) = 5 (850) = 4250.
[Using formula S
n = n/2 { 2a +  (n - 1)(d) } ]
So,the penalty that the contractor has to pay if he delays work by 10 days is ₹ 4250.

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