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primary 6 | Maths | Speed
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primary 6 chevron_right Maths chevron_right Speed chevron_right Singapore

Hi. Can someone please help to solve. Thank you!

Date Posted: 4 years ago
Views: 244
J
J
4 years ago
Travelling at 50m/min , ends up early by 1 min.
Travelling at 40/min, ends up late by 3 min.

Difference in time taken = 3 + 1 = 4 min.

So if she travels at 40m/min she will be 4 min slower/behind compared to 50m/min.
She has to walk more 4 more min to arrive at school.

4 min x 40m/min = 160m

This tells us she was 160m away from reaching school. She walked 160m less in the same time compared to 50m/min, where she reached already.


Difference in speed = 50m/min - 40min/min = 10min/min

She walks 10m less every min compared to 50min/min.

160m ÷ 10m/min = 16min

So in 16min, she ended up walking 160m less, was 160m behind/slower, and had 4 min more to walk.

This also tells us at 50m/min, that she reached school in 16 min .


Distance from the point she changed speed to school

= 16min x 50m/min
= 800m

Total distance
= 800m + distance of first 3 minutes she walked

= 800m + 3 x 40m/min
= 800m + 120m
= 920m

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AC Lim
Ac Lim's answer
12331 answers (A Helpful Person)
1st
Hope this helps
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AC Lim
Ac Lim's answer
12331 answers (A Helpful Person)
You may use time different for 2 different speed
Time different = 3+1=4mins
Sped different = 50-40= 10 m/min
Distance are same =D

Refer to red pen answer...
Hope you understand the working.