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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.

 See The Solution Submitted by K Sengupta Rating: 2.0000 (1 votes)

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 solution | Comment 1 of 4

Let a be the annual expenditure for 2002. Then L=a(1-P/100) for 2003, M=L(1-Q/100)=a(1-P/100)(1-Q/100) for 2004 and N=M(1-R/100)=a(1-P/100)(1-Q/100)(1-R/100) for 2005.

L, M, N, being in arithmetic progression we have 2M=L+N or

2a(1-P/100)(1-Q/100)=a(1-P/100)+a(1-P/100)(1-Q/100)(1-R/100). It results 2(1-Q/100)=1+(1-Q/100)(1-R/100). After the calculus we have 100Q-100R+QR=0.

Let r be the ratio of the geometric progression. Then R=Qr and the last equality becomes 100Q-100rQ+rQ^2=0. It results that Q=100-100/r. Because Q is integer and 0<Q<100, r must be between 1 and 2. A solution is r=1.25. So Q=20, P=16, R=25.

 Posted by Stefan on 2006-09-09 12:37:26

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