One of the most important and exam oriented question from Chapter name- Arithmetic Progression

Class 10th

Chapter number- 9

Exercise :- 9.6

This type of question has been asked in previous years exams.

In this question we have been given that a man is employed to count Rs 10710.

He counts at the rate of Rs 180 per minute for half an hour.

After this, he counts at the rate of Rs 3 less every minute than the preceding minute.

Now we have to find the time taken by him to count the entire amount.

CBSE DHANPAT RAI publication

CBSE Mathematics Class 10th

Question 67

Total amount to be counted = Rs 10710.

Amount man would count at the rate of Rs 180 per minute for 1/2 hour = 180(30) = Rs 5400

Now amount left before the rate starts decreasing = 10710â€“5400 = Rs 5310

This amount of 5310 is counted at a rate of Rs 3 less every minute than the preceding minute.

So, first term(a) = 5310/30 = 177, common difference = â€“3 and S

_{nÂ }= 5310Now by using the formula of the sum of n terms of an A.P.

S

_{n}Â = n[2a +(n â€“ 1)d] / 2We get

=> n[2(177) + (n â€“ 1)(â€“3)] / 2 = 5310

=> n[357 â€“ 3n] = 10620

=> 3n

^{2Â }â€“ 357n + 10620 = 0=> n

^{2Â }â€“ 119n + 3540 = 0=> n

^{2Â }â€“ 60n â€“ 59n + 3540 = 0=> n(n â€“ 60) â€“ 59(n â€“ 60) = 0

=> n = 60 or n = 59

Ignoring n = 60 as n cannot be equal to or greater than 60. So, we get n = 59.

So, total time taken = 30+59 = 89 minutes.

Hence, time taken by him to count the entire amount is 89 minutes.