Q6 of 167 Page 19

A man is employed to count ₹ 10710. He counts at the rate of ₹ 180 per minute for half an hour. After this he counts at the rate of ₹ 3 less every minute then the preceding minute. Find the time taken by him to count the entire amount.

Given: Amount to be counted is Rs. 10710


To find: Time taken by man to count the entire amount


He counts at the rate of Rs. 180 per minute for half an hour or 30 minutes


Amount to be counted in an hour = 180 * 30 = Rs. 5400


Amount to be left = 10710 – 5400 = 5310


⇒ Sn = 5310


After an hour, rate of counting is decreasing at Rs. 3 per minute. This rate will form an A.P.


A.P. will be 177, 174, 171,……………………


Here a = 177 and d = 174 – 177 = –3


Formula used:



where a is first term, d is common difference and n is number of terms in an A.P.


Therefore,






⇒ 10620 = 3n(119 – n)




⇒ n2 – 59n – 60n + 3540 = 0


⇒ n(n – 59) – 60(n – 59) = 0


⇒ (n – 59)(n – 60) = 0


⇒ n = 59 or 60


We will take value of n = 59 as at 60th min he will count Rs.0


Therefore, total time taken by him to count the entire amount = 30 + 59 = 89 minutes


More from this chapter

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4

A manufacturer of the radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find

(і) the production in the first year


(іі) the total product in the 7 years and


(ііі) the product in the 10th year.


Solution |||


(і) the production in the first year


Answer:


Given: 600 and 700 radio sets units are produced in third and seventh year respectively


To find: the production in the first year i.e. a


⇒ a3 = 600 and a7 = 700


Formula used:


For an A.P., an is nth term which is given by,


an = a + (n – 1)d


where a is first term, d is common difference and n is number of terms in an A.P.


Therefore,


a3 = a + (3 – 1)d


⇒ 600 = a + 2d……………………(1)


a7 = a + (7 – 1)d


⇒ 700 = a + 6d


⇒ a = 700 – 6d………………………(2)


Now put this value of a in equation (1):


⇒ 600 = 700 – 6d + 2d


⇒ 600 – 700 = – 6d + 2d


⇒ –100 = –4d



⇒ d = 25


Put d = 25 in equation (2):


⇒ a = 700 – 6(25)


⇒ a = 700 – 150


⇒ a = 550


Production in the first year = a = 550


(іі) the total product in the 7 years

 5

There are 25 trees at equal distances of 5 meters in a line with a well, the distance of well from the nearest tree being 10 meters. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.

 7

A piece of equipment cost a certain factory ₹ 600, 000. If it depreciates in value 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?

 8

A farmer buys a used tractor for ₹ 12000. He pays ₹ 6000 cash and agrees to pay the balance in annual instalments of ₹ 500 plus 12% interest on the unpaid amount. How much the tractor cost him?