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 A Sequential Reduction Problem (Posted on 2006-09-09)
A certain factory had decided to sequentially reduce its annual expenditure on a specified raw material over a period of three years, starting from 2003. The proposed sequential annual reductions were respectively P%, Q% and R%; where P< Q< R< 100, and the three numbers are positive integers in geometric progression.

For example, if the annual expenditure for 2002 was 1000 units and the sequential reductions for the years 2003 to 2005 were respectively 10%, 10% and 20%, then the amount spent on the raw material during 2005 would have been 1000*.9*.9*.8 = 648 units.

Thus, the amounts spent during 2003, 2004 and 2005 were respectively \$L, \$M, and \$N; the three numbers were positive integers in arithmetic progression.

Determine P, Q and R.

 Submitted by K Sengupta Rating: 2.0000 (1 votes) Solution: (Hide) (P,Q,R)=(16,20,25) EXPLANATION: Let the initial expenditure on the raw material be J dollars. Then, by conditions of the problem: L= J(1 - P/100) M = J(1 - P/100)(1 - Q/100) N = J(1 - P/100)(1 - Q/100)(1 - R/100) Also, by the problem: L+M = 2N or, J(1 - P/100) + J(1 - P/100)(1 - Q/100)(1 - R/100) = 2J(1 - P/100)(1 - Q/100) or,(1 - P/100) + (1 - P/100)(1 - Q/100)(1 - R/100) = 2(1 - P/100)(1 - Q/100) or, 1 + (1 - Q/100)(1 - R/100) = 2(1 - Q/100) or, - QR + 100(R - Q) = 0 or, (100 + R )(100 - Q)= 10,000 Now, 1<=Q,R < = 99 or, 101 < = 100 + R < = 199, and the only integer belonging to the interval [101,199] that divides 10,000 is 125. Consequently, 100 + R = 125, giving R = 25; so that: 100 - Q = 10000/125 = 80, giving Q = 20. Now, by the problem; P is less than Q which is less than Q and, P,Q and R corresponds to three consecutive terms of a Geometric Series. Hence, P = (20^2)/25 = 16 Consequently, (P,Q,R)=(16,20,25)

 Subject Author Date re: No Subject Kenny M 2006-09-09 22:57:12 No Subject Dej Mar 2006-09-09 21:37:19 re: solution Kenny M 2006-09-09 19:21:39 solution Stefan 2006-09-09 12:37:26

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