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Secondary 1 | Maths
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I tried cutting each sausage into 3 parts to get 36 pieces that can be shared among 18 people for part i using 2 cuts per sausage and 24 cuts in total. I am curious why the answer key for part (i) gives a different answer and how to prove the answer in part (ii) please.
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II) use the same reasoning as above. Sausages ÷ number of persons = number of cuts
no. of cuts = 12 (sausages) ÷ 18 (persons)
= 2/3 cuts = unreasonable answer!
if m & n do not share any common factors, the no. of cuts is always (n-1).
if m & n have common factor, it becomes a bit more tricky.
firstly, we reduce the problem to the simplest fraction ...
for 12 sausages & 18 persons, it is equivalent to
6 groups of 2 sausages & 3 persons.
using idea above, for 2 sausages & 3 persons,
no. of cuts = (3-1). since there are 6 groups, total need 6 x (3-1) = 12 cuts.
to generalize the problem to m sausages & n persons ...
no. of cuts = HCF(m,n) x [n/HCF(m,n) - 1]
= n - HCF(m,n)
test with smaller numbers ...
sausage, persons --> no. of cuts
1, 5 --> 5-HCF(1,5) = 5 - 1 = 4
2,5 --> 5-HCF(2,5) = 5 - 1 = 4
5,5 --> 5-HCF(5,5) = 5 - 5 = 0
6,5 --> 5-HCF(6,5) = 5-1 = 4
5,10 --> 10-HCF(5,10) = 10-5 = 5
10,5 --> 5-HCF(10,5) = 5-5 = 0
4,10 ---> 10-HCF(4,10) = 10-2 = 8
10,4 ---> 4-HCF(10,4) = 4-2 = 2
for these smaller nos, it is relatively easy to verify that the no. of cuts calculates out correctly. the formula works correctly even when m>n and also when m & n have common factors.
applying to the original problem ...
for 12 sausages & 18 persons,
no. of cuts = 18-HCF(12,18) = 18-6 = 12
If it is actually not allowed, then the OP’s 24 cuts will be correct.
(Though in reality; I’m certain the ends of a sausage are somewhat spherical)
I once encountered a question where a figure, made of 1/4 of a square being removed (basically, a corner of the original square is removed), is supposed to be cut into four equal parts of the same shape or size. In that question, the solution did not allow for cutting a figure many times “to rearrange them later”