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primary 6 | Maths | Measurement
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J
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primary 6 chevron_right Maths chevron_right Measurement chevron_right Singapore

Pls help with both questions

Date Posted: 1 year ago
Views: 219
sstrike
Sstrike
1 year ago
(a)
19 ÷ 7 = 2 R5
Need 3 boxes of short
Since 5 is more than 3,
Need 1 box of long
Least amount
= 3 × $2.50 + 1 × $3.20
= $7.50 + $3.20
= $10.70 (Ans)

Method #1
(b)
LCM【5 & 7】→ 5×7 = 35
total # of candels【NOT # of boxes】
Long→ 7u×5 = 35u
Short→ 7u×7-21 = 35u-21
35u+35u-21
= 70u -21 < 50 → u = 1
Long : 7×$3.2 = $22.4
Short : (5-3)×$2.5 = $5
$22.4+$5 = $27.40 (Ans)


Method #2
(b)
(49 - 21) ÷ 2
= 28 ÷ 2
= 14
Short candles is either 7 or 14. (Multiples of 7)
If short = 7, then long = 7+21 = 28 (rejected because not multiple of 5)
If short = 14, long = 14+21 = 35 is a multiple of 5
Total spent
= 14/7 × $2.50 + 35/5 × $3.20
= 2 × $2.50 + 7 × $3.20
= $5 + $22.40
= $27.40 (Ans)
sstrike
Sstrike
1 year ago
PSLE Math - 2018

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Cheryl Chia
Cheryl Chia's answer
7 answers (Tutor Details)
1st
This is using trial and error method