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primary 5 | Maths | Whole Numbers
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Choo
Choo

primary 5 chevron_right Maths chevron_right Whole Numbers chevron_right Singapore

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Date Posted: 3 years ago
Views: 604

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Tutor May Ho, Ex MOE Trained
Tutor May Ho, Ex Moe Trained's answer
63 answers (Tutor Details)
As 2 Red are between every 2 green. Start with 2 green. Study the pattern, decide a set or group. Then divide from 130. If 130÷4= 32 1/2, adjust to 32 2/4. Make sure denominator same as divisor 4, so it is 32 remainder 2. Hope you could understand.
Choo
Choo
3 years ago
The answer is 86
J
J
3 years ago
Each set should be like this :

GRR GRR GRR GRR GRR ..... GRR G


130 ÷ 3 = 43 R 1

(You can do long division and you will see the remainder of 1 at the end.

Alternatively, using the calculator,

130 ÷ 3 = 43⅓ or 43.33333....

There are 43 sets of 3 bulbs (GRR) and a remainder that would make up only ⅓ of a set

⅓ × 3 = 1

Or simply looking at ⅓ would already tell you its 1 bulb out of 3 in a set.

So remainder of 1


This reminder is the last bulb at the end. It's a green bulb.


Number of red bulbs

= 43 sets × 2 red bulbs per set
= 86
J
J
3 years ago
Tutor May might have misinterpreted the question.

In her grouping , there are no red bulbs between the green ones, which does not satisfy the condition that at least 2 red bulbs are in between every 2 green bulbs.
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annissa chan
Annissa Chan's answer
16 answers (Tutor Details)
I think the question wasn’t phrased ideally, but hope this clears up the confusion. Final answer should be 86. Key point is to note that in the pattern, the first set is not the same as the subsequent sets, as 2 red bulbs can be on the left and right of each green bulb.