Q&A - PSLE Math

[tianzhu]

Kiasu Friend:

The question was taken from a Maths Forum posted there by some member. So the speed on level ground could have been erroneously mentioned as 8 Km instead of 7.5 km. I am sorry for the error.

Hi

Could you help to provide a link to the Maths forum you mentioned so that members can have another avenue to broaden their knowledge?

Best wishes

[Kiasu Friend]

tianzhu:

Kiasu Friend:

The question was taken from a Maths Forum posted there by some member. So the speed on level ground could have been erroneously mentioned as 8 Km instead of 7.5 km. I am sorry for the error.

Hi

Could you help to provide a link to the Maths forum you mentioned so that members can have another avenue to broaden their knowledge?

Best wishes

I will contact my nephew to get the details of that Maths forum and post the details. Please bear with me till then.

Thank you.

[James Ang]

My P6 students told me their school’s prelims had a strange "hamster giving birth" problem sum, then I straightaway know it is the fibonacci sequence. This is a Maths olympiad style question. The sum is something like that;


A pair of hamsters can give birth to a pair of baby hamsters every month. Each pair of baby hamsters needs 1 month to mature and then this pair of hamsters will in turn give birth to a pair of baby hamsters every month. How many pairs of hamsters are there after 6 months? (4 marks)

1,2,3, 5, 8, 13, 21

the answer is 21 pairs of hamsters.

[James Ang]

From the look of the answer, the correct question should be to find the sum of number from 1 to 99. Meaning 1+2+3+...+ 97+98+99 =


Pair up 1 and 99, 2 and 98, 3 and 97, and so on, then there are 49 pairs of 100. But 50 does not have a pair, so answer is 100x49 + 50 = 4950


20 Years ago I learnt this kind of sequence in O levels Additional Maths under AP and GP, now it is a P4 problem.

tianzhu:

kiasuparents_user:

Hi gurus!

I am unsure of the solutions provided for the above question.

Q15c) It is a question to detect patterns.
Any gurus can help me visualise the patterns? Most most most appreciated! Don't have the diagram so please help if you can.

100 x 49 = 4900 -> how & why?
4900 + 50 - 4950 - how & why?

Hi

First thing first, if you’re looking for gurus, then you’re in the wrong thread.

Many of us here are parents who are sharing from their experience as one who had gone through the PSLE journey.

As for your question, it’ll be helpful if you could let us know the source (school/P5/6/year) so that we could search for more information.

Best wishes

[James Ang]

Correct, it doesn't matter how much uphill and downhill distance the cyclist travelled because the average speed of uphill/downhill is equal to the level ground speed, so the answer is 2 and 1/3 multiply by 6 = 14km

Dharma:

James Ang:

not sure if my speed problem has been solved, the correct answer is 14km.

Kiasu friend's question may not be correct, the uphill speed is 5km/h, the downhill speed is 15km/h, the possible way to solve is when the level ground speed is 7.5km/h, then the total distance travelled is 20km. What is the source of this question, perhaps I can find out if the question is correct or not. Cheers.
[quote=\"tianzhu\"]Hi

Your question is almost similar to JA's earlier question in this thread except for changes in the name and speeds of cyclist.The duration is also slightly different.

You may wish to check if figures(5,8 and 15) are in order.

Best wishes

Hi Kiasu Friend,

What JA is trying to say is that in order to solve this problem, there must be a certain relationship between the speeds that are given in the question for uphill, downhill and on level ground.

For JA’s Qn

(A/4 + B/6 + C/12) + (A/12 + B/6 + C/4) = 7/3
Solving
4(A+ B + C)/12 = 7/3
A + B + C = 7km

Total distance cycled by Jack = 7km x 2 = 14km

Take note :
1/6 = (1/4 + 1/12)/2

For Kiasu Friend’s question

If speed at level ground is X

1/X = (1/5 + 1/15)/2 =2/15
X = 7.5

If X = 7.5km/h

(A/5 + 2B/15 + C/15) + (A/15 + 2B/15 + C/5) = 8/3
4(A + B + C )/15 = 8/3

A + B + C = 10

Total distance travelled by Robert = 10km x 2 = 20km[/quote]

[tianzhu]

James Ang:

My P6 students told me their school's prelims had a strange \"hamster giving birth\" problem sum, then I straightaway know it is the fibonacci sequence. This is a Maths olympiad style question. The sum is something like that;


A pair of hamsters can give birth to a pair of baby hamsters every month. Each pair of baby hamsters needs 1 month to mature and then this pair of hamsters will in turn give birth to a pair of baby hamsters every month. How many pairs of hamsters are there after 6 months? (4 marks)

1,2,3, 5, 8, 13, 21

the answer is 21 pairs of hamsters.

Hi James

You’ve just introduced a big word(fibonacci sequence)which may be unfamiliar to many PSLE students.

For the benefits of members could you let us have the detailed breakdown of your solution.

Could this question be solved by systematic listing/tabulated list which PSLE students are more familiar with?

Best wishes

PS ---Hi moderator -----Please consider merging this thread with PSLE Maths to minimise scattering of information.

[tianzhu]

Hi moderator


Please help to delete.
Thank you

Best wishes

[James Ang]

I started a prelims maths topic to discuss the interesting questions that came out in recent P6 prelims which students say are tough, so that we can learn the new trend or pattern of questions. Fibonacci is a very basic number sequence common in Maths Olympiad books, so I guess it is worth knowing more, even finding out what the Fibonacci numbers are like.


To list;
After 6 months
1st pair hamsters --- 7 pairs
1st born hamsters --- 4 pairs
2nd born hamsters --- 3 pairs
3rd born hamsters --- 2 pairs
4th born hamsters --- 1 pair
1st baby pair of 1st born --- 2 pairs
2nd baby pair of 1st born --- 1 pair
1st baby pair of 2nd born --- 1 pair

Total 21 pairs

tianzhu:

James Ang:

My P6 students told me their school's prelims had a strange \"hamster giving birth\" problem sum, then I straightaway know it is the fibonacci sequence. This is a Maths olympiad style question. The sum is something like that;

A pair of hamsters can give birth to a pair of baby hamsters every month. Each pair of baby hamsters needs 1 month to mature and then this pair of hamsters will in turn give birth to a pair of baby hamsters every month. How many pairs of hamsters are there after 6 months? (4 marks)

1,2,3, 5, 8, 13, 21

the answer is 21 pairs of hamsters.

Hi James

You’ve just introduced a big word(fibonacci sequence)which may be unfamiliar to many PSLE students.

For the benefits of members could you let us have the detailed breakdown of your solution.

Could this question be solved by systematic listing/tabulated list which PSLE students are more familiar with?

Best wishes

PS ---Hi moderator -----Please consider merging this thread with PSLE Maths to minimise scattering of information.

[tianzhu]

[quote] pair of hamsters can give birth to a pair of baby hamsters every month. Each pair of baby hamsters needs 1 month to mature and then this pair of hamsters will in turn give birth to a pair of baby hamsters every month. How many pairs of hamsters are there after 6 months? (4 marks) [/quote]
Hi

I used a taulated list and got an answer of 13.

Dear fellow members, please help to check if my interpretation is correct.

Dharma,I'll appreciate if you could help to spot any mistakes in my table.

Best wishes

http://farm5.static.flickr.com/4102/4928674198_53b63b7647_b.jpg\">

[tianzhu]

Hi moderator


Please help to delete.

Thanks