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creativ13 [48]
3 years ago
11

Pensils are sold in packages of 10 and erasers are sold in packages of 6. What is the least number of pensils and erasers you ca

n buy so that there is one pencil for each eraser with none left over.
Mathematics
1 answer:
asambeis [7]3 years ago
8 0
To understand the problem consider the following cases.

i)

If we buy 1 packet of pencils, and 2 packet of erasers, 

we have 1*10=10 pencils and 2*6=12 erasers.

ii)

If we buy 3 packet of pencils, and 4 packet of erasers, 

we have 3*10=30 pencils and 4*6=24 erasers.


So let a and b be the correct number of packets of pencils and erasers respectively.

That is the least numbers a and b, such that 10a=6b.

10a=6b

divide both sides by 10:

a= \frac{6b}{10}= \frac{3b}{5}

divide both sides by b:

\frac{a}{b} =  \frac{3}{5}

the ratio a:b cannot be simplified any further. This means that the smallest (natural) numbers a and b such that \frac{a}{b} = \frac{3}{5} are a=3 and b=5


Answer:


3 packages of pencils, 5 packages of erasers.

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