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Misc 18 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed. Let total work = 1 and let total work be completed in = n days Work done in 1 day = (𝑇𝑜𝑡𝑎𝑙 𝑤𝑜𝑟𝑘)/(𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑎𝑦𝑠 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑤𝑜𝑟𝑘) = 1/𝑛 This is the work done by 150 workers Work done by 1 worker in one day = 1/150𝑛 Given that In this manner it took 8 more days to finish the work i.e. work finished in (n + 8) days Therefore, 150/150𝑛 + 146/150𝑛 + 144/150𝑛 + … + to (n + 8)term = 1 1/150𝑛 [150 + 146 + 142 + … to (n + 8)term]= 1 150 + 146 + 142 + … + to (n + 8)terms = 150n 150 + 146 + 142 + … + to (n + 8) terms this is an AP, where first term (a) = 150 Common difference = 146 – 150 = -4 We know that sum of n terms of AP Sn = 𝑛/2[2a + (n – 1)d] Putting n = n + 8 , a = 150 , d = -4 Sn + 8 = (𝑛+8)/2[2(150) + (n + 8 – 1)(-4)] = (𝑛+8)/2 [300 + (n + 7)(-4)] = (𝑛+8)/2 [300 – 4 (n + 7)] = (2(𝑛+8))/2[150 – 2(n + 7)] = (n + 8)[150 – 2(n + 7)] = (n + 8)[150 – 2n – 14] = n(150 – 2n – 14) + 1200 – 16n – 112 = 150n – 2n2 – 14n -16n + 1200 – 112 = – 2n2 + 150n – 14n – 16n + 1200 – 112 = – 2n2 + 120n + 1088 Hence, 150 +146 +142 + … to (n + 8)term = -2n2 + 120n + 1088 Also, we know that 150 +146 +142 + … to (n + 8)term = 150n – 2n2 + 120n + 1088 = 150n – 2n2 + 120n – 150n + 1088 = 0 – 2n2 – 30n + 1088 = 0 2n2 + 30n – 1088 = 0 n2 + 15n – 544 = 0 n2 + 32n – 17n – 544 = 0 n (n + 32) – 17(n + 32) = 0 (n – 17)(n + 32) = 0 Since n cannot be negative , n = – 32 is not possible Hence n = 17 Work was completed in n + 8 days i.e. 17 + 8 = 25 days

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.