Misc 18 - Chapter 8 Class 11 Sequences and Series
Last updated at April 16, 2024 by Teachoo
Miscellaneous
Misc 2
Misc 3
Misc 4 Important
Misc 5
Misc 6
Misc 7 Important
Misc 8
Misc 9
Misc 10 Important
Misc 11 (i) Important
Misc 11 (ii)
Misc 12 Important
Misc 13
Misc 14 Important
Misc 15 Important
Misc 16
Misc 17 Important
Misc 18 Important You are here
Question 1 Deleted for CBSE Board 2025 Exams
Question 2 Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams
Question 5 Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams
Question 7 Deleted for CBSE Board 2025 Exams
Question 8 Deleted for CBSE Board 2025 Exams
Question 9 Important Deleted for CBSE Board 2025 Exams
Question 10 Deleted for CBSE Board 2025 Exams
Question 11 Important Deleted for CBSE Board 2025 Exams
Question 12 Deleted for CBSE Board 2025 Exams
Question 13 Important Deleted for CBSE Board 2025 Exams
Question 14 Deleted for CBSE Board 2025 Exams
Miscellaneous
Last updated at April 16, 2024 by Teachoo
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