Q&A - PSLE Math

[tianzhu]

Hi, hope this helps.


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As for Q2, one of the solution is by using MD.

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As for Q3, you may wish to check your question again.Is there any missing information such as meeting at midway?

[lizawa]

_jas_:

Hi help needed for these questions! ASAP pls as my child has an exam coming up!!


1) A bus left town A for town B at a speed of 45km/h. At the same time, a van left Town B for Town A at the speed of 55km/h. Both vehicles travelled along the same road. how far apart were they 1 hour before they met (ans:100km)

TIA!

When the vehicles are 1 hour apart, means they each need to travel another 1 hour before they meet.

Dist = speed * time
Distance by bus in 1 hour = 45km
Distance by van in 1 hour = 55km
Distance apart (they are in opposite direction)= 45 + 55 = 100km

[Vanilla Cake]

_jas_:

2) At 09 00, a van left Town P for Town Q. After some time, a car left Town Q for Town P. The two vehicles met at 11 30. The ratio of the average speed of the van to the average speed of the car is 3:5.

a) What time did the car leave town Q (ans:10 00)
b) If the distance between town P and Town Q is 150 km, calculate the average speed of the van. (ans:30km/h)

As mentioned by tianzhu, there is missing information about the meeting point. Where did you get this question from? Based on the answers provided, let's assume that both the distances travelled by the van and the car till they met at 11 30 were the same.

(a)
Ratio of average speed of Van : Car -> 3 : 5
Units (distance) that the van had travelled till meeting at 11 30 -> 2.5 x 3 = 7.5
Time taken for the car to meet at 1130 -> 7.5 / 5 = 1.5 h
11 30............10 30................10 00
11 30............- 1 h .............- 30 min
Ans : The car left town Q at 10 00 h

(b)
15 units -> 150 km
7.5 units -> 75 km
75 / 2.5 = 30 km/h
Ans : Average speed of the van was 30 km/h

[tianzhu]

Thank you for your help.How do you go about interpreting this question?


Grace had 40 red and 48 green beads. She used all the beads to make into flower bookmarks. The number of red and green beads used for each bookmark was the same. What is the maximum number of bookmarks that she could make?

[lizawa]

tianzhu:

Thank you for your help.How do you go about interpreting this question?


Grace had 40 red and 48 green beads. She used all the beads to make into flower bookmarks. The number of red and green beads used for each bookmark was the same. What is the maximum number of bookmarks that she could make?

interesting, so you are suppose to make a good guess on how many beads are used for each flower bookmark.

use factors ? common factor for 40 and 48 = 1, 2, 4, 8

To make a flower bookmark, maybe 4 of each will be enough ? So 8 per flower bookmark.

Max no. of bookmarks will be 10.

[kiasiparent]

tianzhu:

Thank you for your help.How do you go about interpreting this question?


Grace had 40 red and 48 green beads. She used all the beads to make into flower bookmarks. The number of red and green beads used for each bookmark was the same. What is the maximum number of bookmarks that she could make?

The HCF of 40 and 48 is 8.

So the maximum number of bookmarks is 8

because 40 = 5+5+5+5+5+5+5+5 and 48 = 6+6+6+6+6+6+6+6

Each bookmark consists of 5 red and 6 green beads.

5 and 6 have no more common factors.

[lizawa]

kiasiparent:

tianzhu:

Thank you for your help.How do you go about interpreting this question?


Grace had 40 red and 48 green beads. She used all the beads to make into flower bookmarks. The number of red and green beads used for each bookmark was the same. What is the maximum number of bookmarks that she could make?

The HCF of 40 and 48 is 8.

So the maximum number of bookmarks is 8

because 40 = 5+5+5+5+5+5+5+5 and 48 = 6+6+6+6+6+6+6+6

Each bookmark consists of 5 red and 6 green beads.

5 and 6 have no more common factors.

But the question says the no. of red and green beads used for each bookmark was the same. so cannot be 5 and 6 respectively for each bookmark.

[corneyAmber]

mathsparks:

Thanks lizawa. I wonder why one of the neighbourhood school teacher marked a colleague's son wrong when he used algebra instead of model.

I think lizawa missed out this question buried amongst the mind-boggling math questions. She is right to say that PSLE accepts algebra method as long as the working and answer is correct.

However, in schools, there are more stringent rules because it depends on what the teacher is testing. Testing understanding of question and hence the solution or testing understanding of methods taught?

If the latter, then a child will be penalised for not using the \"taught\" method. This demonstrates the child is switching off in class when the teacher is teaching because there is alternative method that the child knows. So the child has failed the learning objectives.

[ChiefKiasu]

lizawa:

kiasiparent:

...So the maximum number of bookmarks is 8


because 40 = 5+5+5+5+5+5+5+5 and 48 = 6+6+6+6+6+6+6+6

Each bookmark consists of 5 red and 6 green beads.

5 and 6 have no more common factors.

But the question says the no. of red and green beads used for each bookmark was the same. so cannot be 5 and 6 respectively for each bookmark.

Hi lizawa, I think kiasiparent is correct. It's how we interprete the statement \"no. of red and green beads used for each bookmark was the same\".

If B=Total number of beads per bookmark=R+G,

It could mean, as you have interpreted, that each bookmark has exactly the same number of red beads as there are green beads, ie. R=G

Or, as kiasiparent has interpreted, that each bookmark comprises the same number of red beads and green beads, ie. R need not be = G, but the same number of R and the same number of G is used for each bookmark.

I tend to agree with the 2nd interpretation.

[lizawa]

ChiefKiasu:

lizawa:

[quote=\"kiasiparent\"]...So the maximum number of bookmarks is 8


because 40 = 5+5+5+5+5+5+5+5 and 48 = 6+6+6+6+6+6+6+6

Each bookmark consists of 5 red and 6 green beads.

5 and 6 have no more common factors.

But the question says the no. of red and green beads used for each bookmark was the same. so cannot be 5 and 6 respectively for each bookmark.

Hi lizawa, I think kiasiparent is correct. It's how we interprete the statement \"no. of red and green beads used for each bookmark was the same\".

If B=Total number of beads per bookmark=R+G,

It could mean, as you have interpreted, that each bookmark has exactly the same number of red beads as there are green beads, ie. R=G

Or, as kiasiparent has interpreted, that each bookmark comprises the same number of red beads and green beads, ie. R need not be = G, but the same number of R and the same number of G is used for each bookmark.

I tend to agree with the 2nd interpretation.[/quote]Hi tianzhu,

yeah, i think your interpretation makes sense .