Royston Lee's answer to Maya's Primary 6 Maths Fractions Singapore question.
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What I did:
I rearranged the squares to find a pattern (on the top left of the picture)
I found the pattern to be
Figure 1 = 3 3 1
Figure 2 = 5 5 1 1
where Figure X = 2[3 + 2(x-1)] + x
Using this idea, I was able to solve for a & b.
Since 3 + 2(x-1) + x = no of squares
for (c), 142 = 2[3 + 2(x-1)] + x
I rearranged the squares to find a pattern (on the top left of the picture)
I found the pattern to be
Figure 1 = 3 3 1
Figure 2 = 5 5 1 1
where Figure X = 2[3 + 2(x-1)] + x
Using this idea, I was able to solve for a & b.
Since 3 + 2(x-1) + x = no of squares
for (c), 142 = 2[3 + 2(x-1)] + x
Date Posted:
2 years ago
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