Q&A - PSLE Math

[tianzhu]

Maths Questions from http://www.channelnewsasia.com/discussion/forum.htm


1. Two sisters, Melissa and Nicole, started brisk-walking at 6.45a.m. from their house to a nearby park. Melissa walked at an average speed of 90m/min while Nicole's average speed was 60m/min. After arriving at the park, Melissa rested for 10 minutes and then walked back home. If Melissa met Nicole at 7.40 a.m. on her return journey, what was the distance between their house and the park?

2. A group of people met at a party. Each person shook hands with everyone else. Mr Li shook hands with 3 times as many men as women. Mrs Li shook hands with 4 times as many men as women. How many men and how many women were there at the party?

[corneyAmber]

http://i3.photobucket.com/albums/y74/miamia2004/P6Mathwalkingdistance-1.jpg\">

[ChiefKiasu]

Wow… thanks for the great effort, ks2me!

[corneyAmber]

ChiefKiasu:

Wow... thanks for the great effort, ks2me!

Errr :oops: ....Chief...most importantly, is the answer correct?

[csc]

ks2me:

http://i3.photobucket.com/albums/y74/miamia2004/P6Mathwalkingdistance-1.jpg\">

Wow! Nice presentation but I came up with a different answer.

Time taken by Melissa shd be 55-10 = 45 min(after deducting 10 min rest)
Distance travelled by Melissa shd be 90 x 45 min = 4050 m

Hence, 4050 + 3300 =7350
7350/2 = 3675 m (distance between house and park)

What do you all think?

[tianzhu]

Q1)3675m is the correct answer.


Best wishes

[lizawa]

tianzhu:

Maths Questions from http://www.channelnewsasia.com/discussion/forum.htm


3. A group of people met at a party. Each person shook hands with everyone else. Mr Li shook hands with 3 times as many men as women. Mrs Li shook hands with 4 times as many men as women. How many men and how many women were there at the party?

Answer : no. of men = 16; no. of women = 5

See working :

http://www.postimage.org/image.php?v=gx2w0utS

[suiyuan]

Q47

Please help to establish the number pattern for Q47 in this link.
http://www.wendykoh.com/08/primary6-aitongca1-maths.pdf

[lizawa]

suiyuan:

Q47

Please help to establish the number pattern for Q47 in this link.
http://www.wendykoh.com/08/primary6-aitongca1-maths.pdf

Stage Squares
1 ===> 1
2 ===> square (2+1) - (1*4) = 5
3 ===> square (3+2) - ((2+1) *4) = 13
4 ===> square (4+3) - ((3+2+1)*4) = 25
5 ====> square (5+4) - ((4+3+2+1)*4) = 41
10 ===> square (10+9) - ((9+8+7+6+5+4+3+2+1) *4) = 181
100 ==> square (100+99) - ((99+98+97+...+1)*4) = 19801

nth stage == > square (n+(n-1)) - ((sum of (n-1, n-2, n-3....1)*4)

So you take the area of the whole figure at each stage- (no of squares cut of each edge * 4 edges).

to solve 99+98+...1 at primary level, no need to teach AP/GP,
just re-arrange 1+2+.. 99 and place them under 99+98... +1
So you get 1 term as (99+1), (98+2) etc. ie. each term is 100. Total 99 terms. So sum = (100*99) / 2

[ApronMama]

suiyuan:

Q47

Please help to establish the number pattern for Q47 in this link.
http://www.wendykoh.com/08/primary6-aitongca1-maths.pdf

How about this series ? Answer same as Lizawa's

Stage Squares
1 ===> 1
2 ===> 1+4(2-1) = 5
3 ===> 1+4(2-1)+4(3-1) = 13
4 ===> 1+4(2-1)+4(3-1)+4(4-1)=25
5 ===> 1+4(2-1)+4(3-1)+4(4-1)+4(5-1)=41
.
.
10 ===>1+4(2-1)+4(3-1)+4(4-1)+4(5-1)+...+4(9-1)=181
.
.
nth ===>1+4(2-1)+....+4(n-1)

if n=100, we get
100 ===>1+4(2-1)+....+4(100-1)
=1+4*1+4*2+...+4*99
=1+4(1+2+....+99)
=1+4(4950)
=19801
{ to do 1+2+3+...+99 same as Lizawa's method to pair up (1+99), (2+98),... you get 49*pairs of 100 +50 = 4950}