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AfilCa [17]
4 years ago
5

Adam drew two same size rectangles and divided them into the same number of eqal parts. He shaded 1/3 of one rectangle and 1/4 o

f the other rectangle. What is the least number of parts into which both rectangles could be divided?
Mathematics
1 answer:
Aleksandr-060686 [28]4 years ago
3 0
The problem needed to say that when he did the shading,
he only shaded WHOLE PARTS.

For the rectangle that he shaded 1/3 of, how many parts could it have ?
It could have 3, 6, 9, 12, 15, 18, 21, etc. parts ... the multiples of 3.

For the rectangle that he shaded 1/4 of, how many parts could it have ?
It could have 4, 8, 12, 16, 20, etc. parts ... the multiples of 4.

Both rectangles have the same number of parts.
What's the smallest number on both of those lists ?
It's called the "least common multiple of 3 and 4".
It's the smallest number that they both go into.
It's 12 .
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