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primary 6 | Maths
| Algebra
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Date Posted: 3 years ago
Views: 537
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Hope this helps! When you're stuck, don't give up. Trial and error may work :-)
Date Posted:
3 years ago
Is there another method?
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The important thing to consider is the number of red beads left after subtracting the black beads. Using the number of red beads left, we check if they are multiples of the given options. The last line that I've written is just to show you that there can only be a maximum of 4 black boxes because each black box contains 70 beads and there are only 300 beads altogether.
Date Posted:
3 years ago
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