Q&A - PSLE Math

[Snowball99]

Snowball99:

carebear:

[quote=\"stary\"]Hi all, need help for the following questions


1) Mrs Yeo had 20% more books than Mdm Lim. Miss Tang had 40% fewer books than Mrs Yeo. Mrs Yeo and Mdm Lim gave Miss Tang some books in the ratio of 1:2. As a result, Miss Tang had 50% more books than before. Given that Miss Tang had 96 more books than Mdm Lim in the end, how many books did Miss Yeo give to Miss Tang?

2)Last Monday morning, Edward walked from home to school. For the first 2 minutes, he was walking at an average speed og 40m/min. When he realised that he was going to be late by 3 minutes, he quickly increased his speed by 10m/min. As a result, he was early be 1 minute. Find the distance between his home to his school.[/quote

I think you should use the ratio of speed and time. (inversely related) for the 2nd question.

Speed:
4:5
Time:
5:4
The difference between the times is 4 minutes.
Thus, 1u time = 4 min
So,
5u= 20 min
Distance= 40m/min for 20 min
= 800m

Cheers,
Snowball99




http://www.flickr.com/photos/63342911@N03/6093555012/

[/quote]

[stary]

Snowball99:

carebear:

[quote=\"stary\"]Hi all, need help for the following questions


1) Mrs Yeo had 20% more books than Mdm Lim. Miss Tang had 40% fewer books than Mrs Yeo. Mrs Yeo and Mdm Lim gave Miss Tang some books in the ratio of 1:2. As a result, Miss Tang had 50% more books than before. Given that Miss Tang had 96 more books than Mdm Lim in the end, how many books did Miss Yeo give to Miss Tang?

2)Last Monday morning, Edward walked from home to school. For the first 2 minutes, he was walking at an average speed og 40m/min. When he realised that he was going to be late by 3 minutes, he quickly increased his speed by 10m/min. As a result, he was early be 1 minute. Find the distance between his home to his school.

I think you should use the ratio of speed and time. (inversely related)


Speed:
4:5
Time:
5:4
The difference between the times is 4 minutes.
Thus, 1u time = 4 min
So,
5u= 20 min
Distance= 40m/min for 20 min
= 800m

Managed to do 1st question only.


http://www.flickr.com/photos/63342911@N03/6093555012/

[/quote]Thanks Snowball99... (you've got the arragement wrong)

[tianzhu]

SaPan:

thank you so much tianzhu. :salute:

Hi

Good Morning.

You're welcome.

Best wishes

[carebear]

stary:

Thanks Snowball99... (you've got the arragement wrong)

I think what Snowball99 actually wanted to post is this :

Snowball99:

stary:

Hi all, need help for the following questions

1) Mrs Yeo had 20% more books than Mdm Lim. Miss Tang had 40% fewer books than Mrs Yeo. Mrs Yeo and Mdm Lim gave Miss Tang some books in the ratio of 1:2. As a result, Miss Tang had 50% more books than before. Given that Miss Tang had 96 more books than Mdm Lim in the end, how many books did Miss Yeo give to Miss Tang?

2)Last Monday morning, Edward walked from home to school. For the first 2 minutes, he was walking at an average speed og 40m/min. When he realised that he was going to be late by 3 minutes, he quickly increased his speed by 10m/min. As a result, he was early be 1 minute. Find the distance between his home to his school.

carebear:

2nd question solved.
Sorry for the poor quality cos I am too tired .

http://www.flickr.com/photos/63342911@N03/6093193929/

I think you should use the ratio of speed and time. (inversely related)


Speed:
4:5
Time:
5:4
The difference between the times is 4 minutes.
Thus, 1u time = 4 min
So,
5u= 20 min
Distance= 40m/min for 20 min
= 800m

[gijg]

Hi ,


I need some help with this speed question:
Mike and Steve started jogging at the same time from pt A to pt B , respectively along a running track in opp direction . Mike met Steve when he is 8km from A . Both of them turn back after reaching A and B . They met again at 4km from B. Find the dist from pt A to pt B .

Thank you .

[carebear]

gijg:

Hi ,


I need some help with this speed question:
Mike and Steve started jogging at the same time from pt A to pt B , respectively along a running track in opp direction . Mike met Steve when he is 8km from A . Both of them turn back after reaching A and B . They met again at 4km from B. Find the dist from pt A to pt B .

Thank you .

Not enough info given.
Is the speed of Mike and Steve specified in the question?
Do we assume they are jogging at the same speed?

[MathIzzzFun]

gijg:

Hi ,


I need some help with this speed question:
Mike and Steve started jogging at the same time from pt A to pt B , respectively along a running track in opp direction . Mike met Steve when he is 8km from A . Both of them turn back after reaching A and B . They met again at 4km from B. Find the dist from pt A to pt B .

Thank you .

Hi

though the question is not very clear, but it is same as this one which was discussed in an earlier post - http://www.flickr.com/photos/62167097@N02/6085706330/in/photostream

cheers.

[tianzhu]

Hi moderator.


Please delete this post.Question has been revised.

Thanks.

Best wishes

[gijg]

Hi


I got my ds to recheck the question . Here is it , there was some typo error in the 1st q posted .

Mike and Steve started jogging at the same time from pt A and pt B , respectively along a running track in opp direction . Mike met Steve when he is 8km from A . Both of them turn back after reaching A and B . They met again at 4km from B. Find the dist from pt A to pt B . The answer is 20 and none of his frens were able to solve it .

[tianzhu]

gijg:

Hi


I got my ds to recheck the question . Here is it , there was some typo error in the 1st q posted .

Mike and Steve started jogging at the same time from pt A and pt B , respectively along a running track in opp direction . Mike met Steve when he is 8km from A . Both of them turn back after reaching A and B . They met again at 4km from B. Find the dist from pt A to pt B . The answer is 20 and none of his frens were able to solve it .

Hi

Yes, now with the revised question, it is solvable.

Whether we are on a circular track or jogging on a straight route, we can still get the same answer.

To make life easier, consider a straight run with two joggers starting from opposite ends A and B.

As in any speed question,it's useful to draw a diagram and put in the information bit by bit.

With logical reasoning, we can get the answer of 20km.

At the point when two joggers meet, we get the distance between A and B.

At the first interval of time, Mike covered 8km, while Steve covered 12km.

At the second period, Mike covered another 8km; Steve did a U turn and is now 4km from A.

At the third period, Mike did a U turn and is now 4km away from B, Steve covered another 12km to meet Mike at B.

So total distance ------- 8+8+4 ------20

Best wishes