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Secondary 1 | Maths
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Haylee Denize Singh
Haylee Denize Singh

Secondary 1 chevron_right Maths chevron_right Singapore

could someone help me with this?

Date Posted: 3 years ago
Views: 182

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You'll need to find the LCM of 60 and 80 first.
60 = 2 × 30 = 2 × 2 × 15 = 2 × 2 × 5 × 3 = 2² × 5 × 3
(Or even more quickly, 60 = 4 × 15 = 2² × 3 × 5)
80 = 2 × 40 = 2 × 2 × 20 = 2 × 2 × 2 × 10 = 2 × 2 × 2 × 2 × 5 = 2⁴ × 5
(Or even more quickly, 80 = 16 × 5 = 2⁴ × 5)
LCM = 2⁴ × 5 × 3 = 16 × 5 × 3 = 80 × 3 = 240
This means that after 240s, they will both be at the starting point again. (This is the answer for i)
ii)
Number of laps the faster car does in 240s = 240s ÷ 60s per lap
= 4
Number of laps the slower car does in 240s = 240s ÷ 80s per lap
= 3
So, for every 240s, the faster car does 1 lap more than the slower car, ans is 1 lap ahead of it. (4 - 1 = 3)
In order to be 5 laps ahead,
5 ÷ 1 = 5
They need to travel for 5 times/sets of 240s.
Time needed = 5 × 240s
= 1200s
= (1200 ÷ 60) min
= 20 min
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