A man is employed to count Rs.10710. He counts at there 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. Find the time taken by him to count the entire amount.
Total amount for counting = Rs. 10710
In 1 min he counts =Rs. 180
In half an hour (30 min) he count = Rs.180×30 =Rs.5400
Amount left count after half an hour, Sn= Rs.10710 –Rs. 5400 = 5310
Now, in 31st min he count 3 less than preceding minute = Rs.180 – Rs.3 = Rs.177
In 32nd min he count 3 less than preceding minute = Rs. 177 –Rs. 3 =Rs. 174
Then, Arithmetic progression formed is 177, 174,…
Here, difference between minutes = 174 – 177 = -3
We know, sum of n terms, Sn= (n/2) [2a + (n – 1) d]
5310 = (n/2) [2(177) + (n – 1) -3]
5310 × 2 = n [354 – 3n + 3]
10620 = 354n – 3n2 + 3n
10620 = 357n – 3n2
⇒ 3n2 – 357n + 10620 = 0
On taking 3 common from the complete equation, we get,
⇒ 3 (n2 – 119n + 3540) = 0
⇒ n2 – 119n + 3540 = 0
Now, from factoring by splitting the middle term method, we get,
⇒ n2 – 60n – 59n + 3540 = 0
⇒ n (n – 60) – 59 (n -60) = 0
⇒ (n – 59) (n – 60) = 0
Therefore, either n = 59 or n = 60