Q&A - PSLE Math

[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.

[tianzhu]

cycjca:

Thanks for you help. Really appreciate it.

Hi

You're welcome.

Best wishes

[leesf]

Thanks to atutor2001 and teachingmum for your solutions.

[tianzhu]

Hi


Three interesting Speed questions.

Best wishes

Q1
Last Monday, Danny walked from home to market. For the first 5 minutes, he was walking at an average speed of 35m/min. When he realized that he was going` to be late by 4 minutes, he quickly increased his speed by 10m/min. As a result, he was early by 2 minutes. Find the distance between market and home.

Q2
Jacki’s mum leaves home at 12.35pm every day and arrives 25 minutes later at his school to pick him up. She always cycles at the same speed along the same route. One day Jacki was dismissed early at 12.30pm, so he started walking home at 6km/h, hoping to meet her along the way. He met her at x and got onto her bicycle. They reached home 15 minutes earlier than before. How far did Jacki walked before his mum picked him up?

Q3
Tony and Charles took part in a car race. Tony drove at a speed of 90km/h. Both of them did not change their speed throughout the race. When Charles had covered 1/3 the distance, Tony was 15km in front of him. Tony reached the finishing line at 9.35 a.m. At what time did Charles reach the finishing line?

[Vanilla Cake]

leesf:

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?

atutor2001:

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]

Lowest common multiple of 6 and 8 is 24.
24÷6=4
4+1=5
So, there are 5 angsana trees for the first 24m.

24÷8=3
3+1=4
So, there are 4 street-lamps for the first 24m.

From the above workings, you can see that there is 1 additional angsana tree for every 24m.
Since there are 12 more angsana trees than street-lamps (given in the question itself), length of the street -> 24÷1x12=288m
(a) If there is one street-lamp and one angsana tree on both ends of the street, length of street = 288m

288÷8=36
36+1=37 street-lamps.
Number of angsana trees -> 37+12=49.
(b) Numbers of street-lamps and angsana trees are 37 and 49 respectively.

[Vanilla Cake]

atutor2001:

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.

teachingmum:

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

Just to add more explanations:

Mrs Lee's case
Distance travelled in 4 min -> 4 minx 40 m/min = 160m (Heading to meet Cynthia)
Distance travelled in 20 min -> 20 min x 40m/min = 800m (Heading to nowhere?)

Cynthia's case
Distance travelled in 4 min to meet Mrs Lee -> To be calculated.
So, Cynthia had to travel another 16 min (20min-4min) to catch up with Mrs Lee and the distance covered = 160m+800m = 960 m.

16 min -> 960m
4 min -> 960÷4=240m
240m was the distance that Cynthia travelled in 4 min to meet Mrs Lee.

Finally, distance between school and library = (240+160)m = 400m

[Rockstar]

tianzhu:

Hi


Three interesting Speed questions.

Best wishes

Q1
Last Monday, Danny walked from home to market. For the first 5 minutes, he was walking at an average speed of 35m/min. When he realized that he was going` to be late by 4 minutes, he quickly increased his speed by 10m/min. As a result, he was early by 2 minutes. Find the distance between market and home.

Q2
Jacki’s mum leaves home at 12.35pm every day and arrives 25 minutes later at his school to pick him up. She always cycles at the same speed along the same route. One day Jacki was dismissed early at 12.30pm, so he started walking home at 6km/h, hoping to meet her along the way. He met her at x and got onto her bicycle. They reached home 15 minutes earlier than before. How far did Jacki walked before his mum picked him up?

Q3
Tony and Charles took part in a car race. Tony drove at a speed of 90km/h. Both of them did not change their speed throughout the race. When Charles had covered 1/3 the distance, Tony was 15km in front of him. Tony reached the finishing line at 9.35 a.m. At what time did Charles reach the finishing line?

Solution for question 1

New speed after 5 minutes------35 m/min+10 m/min=45 m/min

Slower Speed:Faster Speed
35 : 45
7 : 9


Slower speed time : Faster speed time
9 : 7

35 m/min (4 minutes late)
<---------------------------------------->
[----------------------------------]
^<-------------------->
45 m/min (2 minutes early)
5 min
(35 m/min)


Difference in time ------- 4 + 2 =6

2u--------6
1u--------3
9u--------27

27min + 5 min(The first five minutes he walked in 35 m/min) = 32 min

32 x 35 = 1120
The distance is 1120 m