Arnold K H Tan's answer to Caleb's Secondary 1 Maths Singapore question.
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I) use logical reason by using dummy numbers. Say each sausage is 1 m long, 12 ÷ 18 = 2/3 m per person. Hence only 12 cuts needed (portions of 1/3 and 2/3.
II) use the same reasoning as above. Sausages ÷ number of persons = number of cuts
II) use the same reasoning as above. Sausages ÷ number of persons = number of cuts
Date Posted:
8 months ago
if apply (II) to the original question, you get ...
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
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
The “2 sausage and 3 people” idea with 2 cuts in it (each sausage is cut in the ratio 2 : 1) assumes that two 1/3 sausage pieces can be put together to form a 2/3 sausage, but I am not sure if for this question such joining of pieces is allowed or not.
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”
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”
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