Q5 of 167 Page 19

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.

Given: total trees are 25 and equal distance between two adjacent trees are 5 meters


To find: the total distance the gardener will cover


As gardener is coming back to well after watering every tree


Distance covered by him to water 1st tree and return to the initial position is 10m + 10m = 20m


Now, distance between adjacent trees is 5m


Distance covered by him to water 2nd tree and return to the initial position is 15m + 15m = 30m


Distance covered by the gardener to water 3rd tree return to the initial postion is 20m + 20m = 40m


Hence distance covered by the gardener to water the trees are in A.P


A.P. is 20, 30, 40 ………upto 25 terms


Here first term,a = 20,common difference, d = 30 – 20 = 10


And n = 25


We need to find S25 which will be the total distance covered by gardener to water 25 trees


Formula used:



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


Therefore,






⇒ S25 = 25(140)


⇒ S25 = 3500


Hence, total distance covered by gardener to water trees is 3500 m


More from this chapter

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3

A man arranges to pay off a debt of ₹ 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the instalment.

 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

 6

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.

 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?