1. Class 11
2. Important Question for exams Class 11

Transcript

Misc 32 150 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 = (๐๐๐ก๐๐ ๐ค๐๐๐)/(๐๐ข๐๐๐๐ ๐๐ ๐๐๐ฆ๐  ๐ก๐ ๐๐๐๐๐๐๐ก๐ ๐ค๐๐๐) = 1/๐ This is the work done by 150 workers Work done by 1 worker in one day = 1/150๐ Given that In this manner it took 8 more days to finish the work i.e. work finished in (n + 8) days Therefore, 150/150๐ + 146/150๐ + 144/150๐ + โฆ + to (n + 8)term = 1 1/150๐ [150 + 146 + 142 + โฆ to (n + 8)term]= 1 150 + 146 + 142 + โฆ + to (n + 8)terms = 150n 150 + 146 + 142 + โฆ + to (n + 8) terms this is an AP, where first term (a) = 150 Common difference = 146 โ 150 = -4 We know that sum of n terms of AP Sn = ๐/2[2a + (n โ 1)d] Putting n = n + 8 , a = 150 , d = -4 Sn + 8 = (๐+8)/2[2(150) + (n + 8 โ 1)(-4)] = (๐+8)/2 [300 + (n + 7)(-4)] = (๐+8)/2 [300 โ 4 (n + 7)] = (2(๐+8))/2[150 โ 2(n + 7)] = (n + 8)[150 โ 2(n + 7)] = (n + 8)[150 โ 2n โ 14] = n(150 โ 2n โ 14) + 1200 โ 16n โ 112 = 150n โ 2n2 โ 14n -16n + 1200 โ 112 = โ 2n2 + 150n โ 14n โ 16n + 1200 โ 112 = โ 2n2 + 120n + 1088 Hence, 150 +146 +142 + โฆ to (n + 8)term = -2n2 + 120n + 1088 Also, we know that 150 +146 +142 + โฆ to (n + 8)term = 150n โ 2n2 + 120n + 1088 = 150n โ 2n2 + 120n โ 150n + 1088 = 0 โ 2n2 โ 30n + 1088 = 0 2n2 + 30n โ 1088 = 0 n2 + 15n โ 544 = 0 n2 + 32n โ 17n โ 544 = 0 n (n + 32) โ 17(n + 32) = 0 (n โ 17)(n + 32) = 0 Since n cannot be negative , n = โ 32 is not possible Hence n = 17 Work was completed in n + 8 days i.e. 17 + 8 = 25 days

Class 11
Important Question for exams Class 11