Adam drew two same size rectangles and divided them into the same number of equal parts. He shaded one third of one rectangle and one fourth of other rectangle. what is the least number of parts into which both rectangles could be divided?

Question

Virginia Salinas

Mathematics

23 December, 13:33

Adam drew two same size rectangles and divided them into the same number of equal parts. He shaded one third of one rectangle and one fourth of other rectangle. what is the least number of parts into which both rectangles could be divided?

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Aden Osborn · Answer 1

Twelve. You must find the lowest common multiple of 3 and 4 here, and that is twelve because the prime factorization of them is 3 and 2x2 so 2x2x3 is 12.