Since no start time was given, do we assume the distance of marathon to be 42km? 42.2km? Or can it be solved with some methods?

I'm not sure. The question seems to be short of something

because they ran at the same speed, if benny was 3km ahead of coco when coco was at 1/3, it means benny would have been 9km "ahead" of coco when coco crossed the finish line. that means benny crossed the finish line 54mins before coco (because he ran at 10km/h) therefore, coco crossed the finish line at 9.09am

Thanks a lot Marcus! Now I get the picture. Many thanks

It helps to draw models to have a better understanding of the relationship between the fractions. Here's how you can start. I have shown Eunice's blue pens and Iris' green pens. After this, complete the models to show the complete 'whole' for both of them.

Thanks Vei Li Soo

Just to add on, Your child knows that the LCM is required but did not use the correct fraction of green pens for Eunice.

Is the working and answer correct ?

Yes, Pauline Vong Nyit Li 👍

Thank you. My daughter did, so I'm just checking if she did it correctly. Tqvm

Thanks to all.

Pauline Vong Nyit Li your daughter did an excellent job.

This question deals with a tricky concept, that in questions with constant but unknown speeds and fixed distance, you can play around with the distance. Step 1: the distance is the same, so convert the fractions to common denominator. So Eunike had travelled 12/15 when Arthur had travelled 10/15. Step 2: this is tricky, because you aren't given any other indication of time. While primary school students do not learn algebra (in problem sums), that doesn't mean they can't throw an unknown number in. So let's assume the distance to be 60km (to make calculations with 2, 3, 4, 5, 12, 10, 15 friendly). Step 3: So when Eunike had travelled 48km, Arthur had travelled 40km. If Eunike travelled 12km/h faster, this means they had travelled for 40mins. Step 4: Therefore, Arthur speed is 60km/h. I understand this is a very tricky concept, I would expect this to be from an elite's school paper. PM me if you require more explanation on this concept.

That's nice explanation Marcus Ng.... Thanks...

Thanks Marcus Ng

Basically, the fractions of the distance travelled by both Eunike and Arthur provide the ratio of their average speeds. Since Eunike and Arthur travelled 12/15 and 10/15 of the distance respectively, the ratio of their respective average speeds is 12:10 or simply 6:5. With respect to 6:5, one can easily see that 1 part of the ratio us equal to 12km/h. Since Arthur's speed is 5 parts, his speed is 60km/h.

This is a favourite question for exams, I've seen it in almost all papers. This is slightly tricky, as it deals with both money amounts and quantity amounts at the same time. Step 1: Find out how much he spent in total on pears and apples. As the grand total is $95, you should get $45 for pears, $50 for apples. Step 2: Assume that pears and apples are the same cost. So if they were bought in the ratio 3:5, the cost ratio should also be 3:5. This means if I paid $45 for pears, I should pay $75 for apples. Step 3: Now we come back to reality. I only paid $50 for apples actually. The missing $25 accounts for the fact that apples are $0.50 cheaper. Hence, I can derive that I bought 50 apples. ($25/$0.50) Step 4: 50 apples for $50, 1 apple=$1. Hence, 1pear=$1.50. Double-check: 50 apples means I bought 30 pears. If 1apple=$1 and 1pear=$1.50, I would have spent $50 on apples and $45 on pears.

Marcus Ng simple and concise. Excellent.