Question 1193423: Hi
2 lengths of fencing 8m and 10m long were used along a path of length 228m. How many of each length of fencing were used for the pathway.
Thanks
Answer by greenestamps(13209)   (Show Source):
You can put this solution on YOUR website!
The least common multiple of the two lengths is 40m.
To find all the possible solutions, do the following:
(1) Find one solution by inspection; and
(2) Starting with that first solution, find others using the least common multiple of 40 by adding 4 lengths of 10m each and subtracting 5 lengths of 8m each, or by subtracting 4 lengths of 10m each and adding 5 lengths of 8m each.
I'll get you started....
With a total length of 228m, one obvious solution is 1 length of 8m and 22 lengths of 10m each: 22(10)+1(8)=220+8=228.
To find the "next" solution, replace 4 of the 10m lengths with 5 of the 8m lengths: 18(10)+6(8)=180+48=228.
Then continue with that strategy to find all the possible solutions.
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The preceding is a general elementary method for solving problems like this where you need to make a sum of A using multiples of two numbers B and C.
For this particular problem, there is a method that is perhaps easier and faster due to the fact that one of the lengths is 10m.
Any number of lengths of 10m each will have a total length in meters that is a number with units digit 0. Since the total length of the fence is 228m, the sum of the lengths of the 8m pieces must have units digit 8. And that means the possible numbers of pieces of length 8m must be one of the numbers 1, 6, 11, 16, ....
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