Q&A - PSLE Math

[atutor2001]

leesf:

Hi, appreciate if you could help on the following questions.


http://postimage.org/image/j1m6ikhw/

Q2) Cynthia walked to the library from school and at the same time, Mrs Lee walked to school from the library. If they walked towards each other, they would meet 4 min later. If they walked in the same direction, it would take Cynthia 20min to catch up with Mrs Lee whose walking speed was 40m/min. How far was the library from the school? Assume that Mrs Lee's walking speed was constant at all times.

Q3) A street-lamp is erected at an interval of 8m on one side of the street. On the other side of the street, an angsana tree is planted at every 6m. There are 12 more angsana trees than street-lamps.
a) Ho long is the street if there is one street-lamp and one angsana tree on both ends of the street.
b) How many street-lamps and angsana trees are there?

Thanks

Q2) I have to assume that if they walk in the same direction, than Mrs Lee is walking away from the library to don't know where.

Let C be Cynthia's speed and L be Mrs Lee's speed and D is the distance from school to library.

D/(C+L) = 4
D = 4C + 4L
D = 4C + 4x40
D = 4C + 160

Time needed to catch up (T) = D/(C-L)
20 = D/(C-40)
20(C-40) = D
20C - 800 = D
D = 20C - 800

4C + 160 = 20C - 800
960 = 16C
C = 60 (m/min)

D = 4x60 + 160 = 400m (distance between school & library)

Above is the most traditional method for speed :lol: Just for fun sake, please don't use it, I am sure there is more creative way to solve this question.



Q3 The 1st common multiple of 6 & 8 is 24 (i.e. 3 lamp and 4 trees)
4 - 3 = 1
Therefore, for there to be 12 more trees than lamp we need to repeat this interval 12 times i.e. the 12 common multiples of 6 & 8

(a) Street length = 12 x 24 = 288m

(b) No. of trees = 4 + 11 x 3 = 37 No. of lamps = 3 + 11 x 2 = 25

(after the first group, the no. of trees & lamps is reduced by 1 because there is already a \"reference tree/lamp\" from the end of the last group\"

e.g.

x x x x)x x x)x x x)
|--24-|--24-|--24-|[/u]

[teachingmum]

I have a diff ans to the speed question…



16 mins -> 160 + 800
= 960
4 mins -> 960 / 4
= 240
Library to School:
240m + 160m = 400m

[atutor2001]

teachingmum:

I have a diff ans to the speed question...



16 mins -> 160 + 800
= 960
4 mins -> 960 / 4
= 240
Library to School:
240m + 160m = 400m

I calculated wrongly and have edited it but our answer is still different. I will check again. Thanks

PS

Just checked. You are right, make so many calculation errors. Must be becoming senile. Your approach to the solution is much better and should be the preferred way. :celebrate:

[Vanilla Cake]

leesf:

http://postimage.org/image/j1m6ikhw/

Area of square : Area of rectangle
1 : 3
5 : 15

Area of unshaded square : Area of unshaded rectangle
2 : 7
4 : 14

Shaded area -> (5-4)u= u or (15-14)u = u
u=20 cm²
Area of square = 5u = 5x20cm² = 100 cm²
10 cm x 10 cm = 100 cm²
Length of the square = 10 cm

[kancheongmum]

Vanilla Cake:

firebird:

4) The Quesion 15 Paper 2 has a diagram. Need to find the shaded figure. Very sorry, I do not have the scanner to upload the question. Please help if you can.

Dear firebird,
Pls refer to solution by http://psle2010a.blogspot.com/2010/09/area-p6-2010-sa2-rosyth-p2-q15.html and pls refer to http://1.bp.blogspot.com/_EG8oMs7TIYA/TIrJ3Pk8mkI/AAAAAAAAASM/eLmp3ZHjcek/s1600/MathsP610AreaRosythA15.jpg for bigger view.

OR

Solution by http://4.bp.blogspot.com/_hEGsaDYqgAQ/TIuqCEjKWlI/AAAAAAAABY0/qIYx5Bn65ns/s1600/qn.jpg:

Let AG be x cm.
Then GB = 12-x
Triangle AGF is similar to Triangle ABC, hence AG = GF = x
Tiangle BGF is similar to Triangle BAE, hence
BG/GF = BA/AE
=> (12-x)/x = 12/6 (since AB = 12, AE = 6)
=> 12-x = 2x
=> x=4
Area of Triangle DEF = 1/2 x ED x AG = 1/2 x 6 x 4 = 12
Area of Triangle DFC = 1/2 x DC x (BC - GF) = 1/2 x 12 x (12-4) = 48
Hence, Area of shaded region = 12 + 48 = 60 cm2.

Hi Vanilla Cake

I hope you don't mind about my comments. The solution you posted is too complicated and I think similar triangle is not in the syllabus. Your method is more for secondary level. I am looking at the question because DS cannot solve. Using Observer's solution to explain to him, he still cannot understand. Finally after looking at the diagram for sometime I used line of symmetry AC (diagonal of the square) and the rest of the solution fall into place.

[tianzhu]

kancheongmum:

Finally after looking at the diagram for sometime I used line of symmetry AC (diagonal of the square) and the rest of the solution fall into place.

Hi kancheongmum

Good Morning

I can’t resist logging in just to say Thank you for a brilliant solution.

You've given us a key to unlock this question.A simple move equivalent to a well measured pass floated by a top footballer in a soccer match.

Best wishes

[tianzhu]

firebird:

Dear Tianzhu sir


Thank you very much for your answer.

With best regards
Firebird

Hi firebird

Good Morning.

You're welcome.

Best wishes

[Dharma]

http://postimage.org/image/2iacuden8/

[Vanilla Cake]

firebird:

4) The Quesion 15 Paper 2 has a diagram. Need to find the shaded figure. Very sorry, I do not have the scanner to upload the question. Please help if you can.

Hi firebird,

http://www.onsponge.com/forum/35-thinkingmathonsponge/5406-rosyth-school-2010-p6-prelim-paper-2-q15-4-marks.html.

VC's mum

[cycjca]

tianzhu:

cycjca:

1) Mrs Chua has just enough money to buy 4 kg of crabs & 3 kg of prawns. However she bought 2 kg of crabs & 4 kg of prawns & had $4.50 left. If she exchanged the 3 kg of crabs with 3 kg of prawns, how much money would she have left?

If 2kg of crabs is changed to 3kg of crabs, here's the solution.

Hope this helps.

Best wishes

http://farm5.static.flickr.com/4133/5019157330_02c85ba4c0_b.jpg\">

Hi Tianzhu

sorry my tardy response. It should be 3 kg. Thanks for you help. Really appreciate it.