PSLE M2009 MATHEMATICS QUESTION

Mei and Lin were in a bicycle race. Mei was travelling at a constant speed of 20km/hr and they both did not change their speed. When Lin completed half the race, Mei was 3.5km ahead. Mei completed the race at 10.45am. What time did Lin complete the race?

Solution:

1st of all, I drew the model, marking out the midpoint.



Then I added in the details, when Lin travelled halfway, Mei was 3.5km ahead of her.



Now, with a distance and a speed, we can now find the time that Mei would have saved, if she was only travelling to the midpoint instead of travelling 3.5km more.

Mei would have saved 10.5 minutes.



So, I drew an imaginary scenerio. If Lin had reached the end point of the race, and Mei did not stop moving, this is where they would both be at.



So in this scenerio, Mei has travelled two times of 3.5km more than she should.



So she wasted a total of 10.5 minutes x 2 = 21 minutes.



21 minutes after 10.45am is 11.06am.



Method at a glance.

3.5km ÷ 20 km/h = 10.5 minutes

10.5 minutes x 2 = 21 minutes

10.45am + 21 minutes = 11.06am

PSLE 2009 MATHEMATICS QUESTION

Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim. Jim ate 12 sweets and Ken ate 18 chocolates.

The ratio of Jim's sweets to chocolates became 1 : 7 and the ratio of Ken's sweets to chocolates became 1:4. How many sweets did Ken buy?

Solution:

I started off with a model on Jim's ratio.



Since Jim ate 12 sweets, I added it in.



Then I drew Ken's ratio separately below. Take note that 1 unit of Ken is equals to 1 unit of Jim's sweet + 12 because Ken gave half of his sweet to Jim.



I drew the dotted red line to show the link.



Since Ken ate 18 chocolates, I added that in next.



So I cut every Ken's unit so that it is comparable to Jim's unit.



This is the tricky part. For each unit of Jim's chocolate, I cross out each white unit of Ken's chocolate.



So that leaves us with 3 units of Chocolate for Jim.



Those 3 units are actually made up of 12, 12, 12, 12 and 18.



So,



To find Ken's share of sweets,





Then I multiply by 2.



And there you have it. Ken had 68 sweets at first.

Isn't this way easier to explain to primary students, rather than using Simultaneous equations? Let me know what you think.

Method at a glance.

3 units --> 12 + 12 + 12 + 12 + 18
= 66

1 units --> 66 /3
= 22

22 + 12 = 34

34 x 2 = 68