This is a rejoinder regarding the issue of 5 children weighing themselves two at a time to reduce the ..expenses.The weights revealed by the weighing machine in ascending order were 51.7 52 53.5 54 55 55.3 55.8 56.7 57 and 58.5.You were asked to determine the individual weights
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If the individual weights were a,b,c,d and e the combinations 2 at a time possible are a+b,a+c,a+d a+e,b+c,b+d,b+e,c+d,c+e and d+e Adding all 10 values will result in the sum 4(a+b+c+d+e).So a+b+c+d+e=(51.7+52+53.5+54+55+55.3+55.8+56.7+57+58.5)/4=549.5/4=137.375.Deducting a+b=51.7 and d+e=58.5 we get c=27.175.We can then find that a=24.825,b=26.875,d=28.675 and e=29.825
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If the individual weights were a,b,c,d and e the combinations 2 at a time possible are a+b,a+c,a+d a+e,b+c,b+d,b+e,c+d,c+e and d+e Adding all 10 values will result in the sum 4(a+b+c+d+e).So a+b+c+d+e=(51.7+52+53.5+54+55+55.3+55.8+56.7+57+58.5)/4=549.5/4=137.375.Deducting a+b=51.7 and d+e=58.5 we get c=27.175.We can then find that a=24.825,b=26.875,d=28.675 and e=29.825
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