Feeding Chickens - Arithmetical ProgressionDate: 7/6/96 at 20:53:51 From: Wong Cheong Siong Subject: Feeding Chickens Dear Dr. Math, 1) On a certain date a farmer has 3000 hens. One week later he sells 20 and he continues to do this each week.If each hen costs 8p per week to feed, calculate the total cost of feeding the hens up to the day when his stock reaches 500. 2) A farmer, starting with 2100 hens, sells ten at the end of the first week... after how many weeks will he have sold all his hens? Best Regards, Wong Cheong Siong Date: 7/8/96 at 11:24:12 From: Doctor Brian Subject: Re: Feeding Chickens 1) For week one, he has 3000 hens, so it costs 3000 * 8p For week two, he has 2980 hens, so it costs 2980 * 8p We need to sum up: 8p * (3000 + 2980 + 2960 + ... + 500) The sum in parentheses is arithmetic, first term = 3000, difference = -20. It's not too tough to show that there are 126 terms. One formula for the sum is (half the # of terms) * (sum of end terms) which would give us 63 * 3500 = 220500, which when multiplied by 8p, gives a total cost of 176400p. *I don't know how to convert foreign money, so you're on your own there* -Doctor Brian, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 7/8/96 at 5:53:55 From: Doctor Anthony Subject: Re: Feeding Chickens 1) The cost of feeding 3000 hens for a week is .08*3000 = 240 UKP (UKP = pounds sterling) The number of weeks is (3000-500)/20 = 125 Cost of feeding 500 hens for a week = .08*500 = 40 UKP Total cost = n/2(first + last) = (125/2)(240 + 40) = (125/2)*280 = 17500 UKP 2) This is an AP with a=10, d= 10 Sn = 2100, n is to be found Solve the equation (n/2)(2a + (n-1)d) = 2100 (n/2)(20 + (n-1)*10) = 2100 n(2 + n-1) = 420 2n + n^2 - n = 420 n^2 + n - 420 = 0 (n+21)(n-20) = 0 we can't have negative n, So we must have n = 20 weeks. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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