Q&A - PSLE Math

[tianzhu]

Hi

Hope this helps.

http://farm4.static.flickr.com/3425/3372425316_efdc87dfd3_o.jpg\">

Hope my answers are right.

[small]

Thanks Tianzhu, I had no answer but I think yours are correct.

[ChiefKiasu]

small:

Thanks Tianzhu, I had no answer but I think yours are correct.

Tianzhu's answers are correct. The sequence is an arithmetic progression of odd numbers, and the sum of which is n * n, where n is the \"row\". To visualize why it is so, check out this http://merganser.math.gvsu.edu/calculus/summation/odds.html.

General rule of thumb formulae:
Sum of 1st n even numbers = n * (n+1)
Sum of 1st n odd numbers = n * n

[Emelyn]

Priskhipo:

Emelyn:

I mean in the model, you drew Ali and Billy having the same number of cards.


I don't understand why.

Maybe I am really slow..... :?

Hi Emelyn,
You can draw a smaller unit excluding 28 too. See if you can understand my MD better.

http://www.postimage.org/image.php?v=gx22GUy0

Hi All,

sorry for missing in action. Was away for March school holidays. I got the following solution from the enrichment class teacher. See if you can understand.


http://www.postimage.org/image.php?v=Pq5o2kS

HOpe you can understand.

[small]

Thanks for your time.


Question 1:
Theodore, Gareth and Linus ytook part in a 100-meter race. When Theodore completed the race, Gareth and Linus were 15m and 30m away from the finishing line respectively. How far Linus from the finishing line when Gareth finished the race? (note: all the boys were travelling at a comstant speed throughout the race)

Question2:

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

:?:

[tianzhu]

Hi, hope this helps.


Q1. Linus was 15m away. Difference between Gareth and Linus was (30-15).

Q2. Length of one side of equilateral triangle = 12
Area of triangle CYW = ½6.83 = 10.2 (XW=3, midpoint)
Area of triangle CXW = 25.8 – 10.2 = 15.6 cm2

[chengsmummy]

Hi,


Q1, the answer should not be 15m, because Linus and Gereth are travelling on the different speed, so the difference should not remain the same.
assume they took 1 unit of time to complete the first 85 and 70m respectively, then Gereth will take 15/85 unit time to finish the remaining 15m.
then Linus will travel 15/85 * 70 = 12.35m at the same time. so when Gereth at the finish line, Linus was 30-12.35 = 17.65m away.

tianzhu:

Hi, hope this helps.

Q1. Linus was 15m away. Difference between Gareth and Linus was (30-15).

Q2. Length of one side of equilateral triangle = 12
Area of triangle CYW = ½*6.8*3 = 10.2 (XW=3, midpoint)
Area of triangle CXW = 25.8 – 10.2 = 15.6 cm2

[tianzhu]

chengsmummy:

Hi,


Q1, the answer should not be 15m, because Linus and Gereth are travelling on the different speed, so the difference should not remain the same.
assume they took 1 unit of time to complete the first 85 and 70m respectively, then Gereth will take 15/85 unit time to finish the remaining 15m.
then Linus will travel 15/85 * 70 = 12.35m at the same time. so when Gereth at the finish line, Linus was 30-12.35 = 17.65m away.

Hi, Thank you for the correction.

[kiasiparent]

tianzhu:

Hi, hope this helps.


Q1. Linus was 15m away. Difference between Gareth and Linus was (30-15).

Q2. Length of one side of equilateral triangle = 12
Area of triangle CYW = ½*6.8*3 = 10.2 (XW=3, midpoint)
Area of triangle CXW = 25.8 – 10.2 = 15.6 cm2

Hi,

For Q2, the question is wrong.

By pythagoras theorem, we will know that CX =sqrt root(144 - 36) = sqrt 108 = 10.39 cm

By similar triangle, CX = 6.8 x 2 = 13.6 cm

[AnnT2009]

Both questions are from CHIJ 2008 SA1.


Q1 Cabrini had 2 squares, M and N. Square M and N have perimeter in the ration 7:5. The lenghts and bradths of both squares are whole numbers.
She cut square N into 5 equal strips and put 4 strips side by side with square M to form a new shape. Find the perimeter of the new shape.

Q2 There are some rubber balls and plastic balls in a bag. If 1 rubber ball were to be removed from the bag, then 1/7 of the remaining balls in the bag would be rubber balls. If 2 plastic balls were to be removed from the bag instead of 1 rubber ball. then 1/5 of the balls left behind would be rubber balls. If another 5 rubber balls were to be put into the bag, what fraction of all the balls would be runner balls?