Q&A - PSLE Math

[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 .

[tianzhu]

Thank you for your help.

http://farm4.static.flickr.com/3301/3505808671_2c3cf52507_o.jpg\">

[corneyAmber]

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[suiyuan]

Thank you for your help.


Simon cycled at an average speed of 10km/h from his house to the park. On reaching the park, he cycled back home along the same route at an average speed of 8km/h. He took 1h 12 mins for the entire journey. How long did he take to cycle from the park back to his home?

[Vanilla Cake]

suiyuan:

Simon cycled at an average speed of 10km/h from his house to the park. On reaching the park, he cycled back home along the same route at an average speed of 8km/h. He took 1h 12 mins for the entire journey. How long did he take to cycle from the park back to his home?

Distance between Simon's house and the park was constant.

( Distance = Speed x Time )

Let T1 be the time taken to travel from Simon's house to the park and
let T2 be the time taken to travel from the park back to Simon's house.

10 x T1 = 8 x T2
T2 / T1 = 10 / 8
-> T2 : T1 = 5 : 4 (reduced to simplest form)

Total time taken for the entire journey (T1+T2) = 1 h 12 min = 72 min
9 parts -> 72 min
T2 = 5/9 x 72 = 40 min

Time taken for Simon to cycle from the park back to his home was 40 min

PS : The above explanation is detailed whereas in actual exam, a few working steps are actually required.

[tweet]

Thank you for your help.


1/2 of the number of pupils going to the Airport is equal to 2/3 of the number of pupils going to the Bird Park.The number of pupils going to the Bird Park is 1/4 the number of pupils going to the zoo.

(a) What is the ratio of the number of pupils going to the Airport to the number of pupils going to the Bird Park to the number of pupils going to the zoo.

(b) 4 more pupils went to the Airport than the Bird Park.
How many pupils went to the zoo?

TIA