Q. 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 the 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.
Sol: Suppose the work is completed in n days. Since 4 workers went away on every day except the first day. ∴ Total number of worker who worked all the n days is the sum of n terms of A.P. with first term 150 and common difference – 4. Total number of worker who worked all the n days = n/2[2 x 150 + (n-1) x -4 ] = n (152 – 2n) If the workers would not have went away, then the work would have finished in (n – 8) days with 150 workers working on every day. ∴ Total number of workers who would have worked all n days = 150 (n – 8) ∴ n (152 – 2n) = 150 (n – 8) ⇒ 152n – 2n2 = 150n – 1200 ⇒ 2n2 – 2n – 1200 = 0 ⇒ n2 – n – 600 = 0 ⇒ n2 – 25n + 24n – 600 = 0 ⇒ n(n – 25) + 24 (n + 25) = 0 ⇒ (n – 25) (n + 24) = 0 ⇒ n – 25 = 0 or n + 24 = 0 ⇒ n = 25 or n = – 24 ⇒ n = 25 ( Number of days cannot be negative) Thus, the work is completed in 25 days. Q. Divide 32 into 4 parts which are in A.P. such that the product the product of means is 7:15. Let four numbers in A.P. be a – 3d, a – d, a + d and a + 3d. (a – 2d) + (a – d) + (a + d) + (a + 2d) = 32 ⇒ 4a = 32 ⇒ a = 8 Product of first and fourth term(extremes) = (a – 3d) (a + 3d) = (8 – 3d) (8 +32d) = 64 – 9d2 Product of second and third term (means) = (a – d) (a + d) = (8 – d) (8 + d) = 64 – d2 Product of extremes: product of means = 7:15 (Given) ( 64 – 9d2) / (64 – d2) = 7 : 15
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