Q&A - PSLE Math

[octopusmum]

Hi Vanilla cake

thks for telling me the solutions for the 2 maths questions. i simply did not have the time to search for the answers myself so have to trouble kind souls like you to help…

i need help again with the following question from Henry Park maths 2010 prelim paper. it is qtn 11 in paper 2:

The figure shows an empty tank. It is made from 2 containers. The containers are in the form of 2 cubes of sides 8 cm and 20 cm respectively. The small container is attached to the centre of one of the sides of the big container. 4 litres of water is poured into the big container such that water flows in to fill part of the small container. What is the height of the water level in the big container? Round off your answer to 2 decimal places.

it is easier to see the question with the diagram but i am not able to reproduce the diagram here…if you or anyone else happen to know the solution or know where to get the detailed solution, pls share.

thks v much

a very desperate and exhausted octopusmum

[kancheongmum]

octopusmum:

Hi Vanilla cake

thks for telling me the solutions for the 2 maths questions. i simply did not have the time to search for the answers myself so have to trouble kind souls like you to help...

i need help again with the following question from Henry Park maths 2010 prelim paper. it is qtn 11 in paper 2:

The figure shows an empty tank. It is made from 2 containers. The containers are in the form of 2 cubes of sides 8 cm and 20 cm respectively. The small container is attached to the centre of one of the sides of the big container. 4 litres of water is poured into the big container such that water flows in to fill part of the small container. What is the height of the water level in the big container? Round off your answer to 2 decimal places.

it is easier to see the question with the diagram but i am not able to reproduce the diagram here..if you or anyone else happen to know the solution or know where to get the detailed solution, pls share.

thks v much

a very desperate and exhausted octopusmum

Hi octopusmum

I take the liberty to help you with this question as DS also need help.
The basic concept is volume = L x B x H
what is L x B? Is the base area right. So keep this in mind
Total volume water 4L = 4000 cm3
volume of water below 8 cm cube = 20 x 20 x 6 ((20 - 8/2))=2400cm3
the remaining water above the 8cm cube will have different base area.
= 4000 - 2400 = 1600cm3
new base area is (20x20) + (8x8)= 464 cm2
vol of remaining water divide by this new base area you will get the height. 1600/464 = 3.45 cm
3.45 + 6 = 9.45 cm

Nanyang prelim 2010 Q11 is almost the same as above question. You may want to let your child try.

Let us know if you need help.

[Vanilla Cake]

octopusmum:

i need help again with the following question from Henry Park maths 2010 prelim paper. it is qtn 11 in paper 2:


The figure shows an empty tank. It is made from 2 containers. The containers are in the form of 2 cubes of sides 8 cm and 20 cm respectively. The small container is attached to the centre of one of the sides of the big container. 4 litres of water is poured into the big container such that water flows in to fill part of the small container. What is the height of the water level in the big container? Round off your answer to 2 decimal places.

Do take note that the diagram is not the original diagram as printed in the original question.

Refer to http://psle2010a.blogspot.com/2010/09/figure-shows-empty-tank.html

OR

from another helpful soul in OnSponge forum:

(20-8)cm/2 = 6cm\t\t\t\t
6cm x 20cm x 20cm = 2400cm³ (cubic centimetre)\t\t\t\t
\t\t\t\t
4 litres = 4000cm³\t\t\t\t
4000cm³ - 2400cm³ = 1600cm³\t\t\t\t
\t\t\t\t
1cm x 8cm x 8cm + 1cm x 20cm x 20cm = 464cm³\t\t\t\t
(1600cm³ / 464cm³/cm) + 6cm = 9.45cm (correct to 2 d.p.)\t

PS: I can understand your feelings as I took PSLE in 2008 and my younger sister will be taking PSLE in 2011.

[Dharma]

tianzhu:

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?

Hi

This question is quite similar to the balloon question in PSLE Maths 2009.

String of 2 big balloons is 90cm
String of 5 small balloons is is 1.2m
If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether?

Best wishes

2 big ballons at every 90cm of string.
1 big balloon => 45cm

5 small ballons at every 120cm of string
1 small balloon => 24cm

Ratio of length => Small : Big = 8 : 15
Ratio of qty => Small : Big = 15 : 8

15u – 8u = 105
7u = 105
1u = 15
Total no. of balloons = 23u = 23 x 15 = 345

[kancheongmum]

Hi all this question is from RGS 2010 prelim Q18

I believe it had been posted before in this forum and by Observer. Here is another suggested solution done by my DS which I find very simple and easy the understand. I have taught him to draw model like this for this type of question as he not good at doing the usual model. I have seen solutions provided by sponge forum with complicated models and workings which myself and some of you find difficult to understand. I hope this can help.

There are more pupils in School A than School B. 30% of the pupils in School A is 45 more than 40% of the pupils in School B. If 10% of the pupils in School A leaves to join School B, there will be 200 more pupils in School A than School B.
(a) How many pupils are there in School B?
(b) How many percent less pupils are there in School B than School A?
Leave your answer as fraction in the simplest form.


30% ----3/10 40% ----2/5
equal fraction 3/10 — 6/20 (school A) 2/5 ---- 6/15 (school B)

from the equal fraction we compare A & B:
School A — total 20u + 150
for every 6u of A there is 45 more, 2u there is 15 more

School B ---- total 15u

with the above we can draw the model for better visual of the equation

At First:
School A
()()15
()()()()()()45
()()()()()()45
()()()()()()45
total: 20u + 150

School B
()()()()()()
()()()()()()
()()()
total: 15u

10% move from A to B, ()()15 shift from A to B
then there are 200 more pupils in School A

18u + 135 = 17u + 15 + 200
1u = 80
(a) pupils in School B 80 x 15u = 1200 pupils
(students please note that this is a "if " question so there is no actual transfer. I always highlight to my DS as he is very careless. Underline the word "if" to remind yourself)

(b) Pupils in School A = (80 x 20u) + 150 = 1750
1750 - 1200 = 550
(550/1750) x 100% = 31 3/7 %

Best wishes to all sitting for PSLE.

[Muffins]

Vanilla Cake:

PS: I can understand your feelings as I took PSLE in 2008 and my younger sister will be taking PSLE in 2011.

Good luck in advance to your DSis!!! for an all A* clean sweep :xedfingers:

[leesf]

atutor2001:

Vanilla Cake:

[quote=\"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 gaps between 2 lamp and 4 gaps between 2 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 = 5 + 11 x 4 = 49 No. of lamps = 4 + 11 x 3 = 37

(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) - x represents lamps
|--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.

Hi Vanilla Cake

Thank you for correcting my mistake. Got confused between trees and distances. Your method for finding the number of trees and lamps is so much better.

I have amended my working in red so as not to cause further confusion. Apology to anyone who got confused by me.[/quote]Thanks to Vanilla cake for clarification.

[tianzhu]

Hi


An interesting question on circle from onsponge.
You may wish to have a go at it.

http://www.onsponge.com/forum/35-thinkingmathonsponge/5482-p6-maths-question-.html

Best wishes

[octopusmum]

Hi Vanilla cake


thks once again for helping! i am sure your sis will do v well cause she has a v helpful and smart sister to guide her

Hi Kancheongmum
thks to you too...actually i have managed to figure out the answer just now but killed many of my brain cells on the way! i don't remember maths was this tough during my time!

actually, there is another question which also stumped my son and I, and surprise surprise, my son's maths teacher as well...it is from the Rosyth School 2010 prelim, question 15. i am not able to reproduce the question cause it is under geometry. there is a square with 2 lines cris crossing each other in it. suppose to find a shaded area in the square.

not sure if u can help again?! thks v much!

regards

[kancheongmum]

octopusmum:

Hi Vanilla cake


thks once again for helping! i am sure your sis will do v well cause she has a v helpful and smart sister to guide her

Hi Kancheongmum
thks to you too...actually i have managed to figure out the answer just now but killed many of my brain cells on the way! i don't remember maths was this tough during my time!

actually, there is another question which also stumped my son and I, and surprise surprise, my son's maths teacher as well...it is from the Rosyth School 2010 prelim, question 15. i am not able to reproduce the question cause it is under geometry. there is a square with 2 lines cris crossing each other in it. suppose to find a shaded area in the square.

not sure if u can help again?! thks v much!

regards

Hi Octopusmum,

You are most welcome. Yes this question Rosyth 2010 Q15, there are a few solutions posted here pg 194, 195 and Observer. Some of the solutions I personally feel that is not for pri level except for the one posted by Dharma. You may want to have a look. I took quite sometime to find a simple and logical way the explain to my DS.

get a midpoint of line AB and name it Z. Draw a line from point F to Z
AZ = ZB = AE = ED = 12/2= 6cm (because is a square)
tri AZF = tri ZBF (both triangles have equal base and height)

Diagonal AC is the line of symmetry

So tri AEF = tri AZF (AE=AZ, EF=FZ)
area of tri AEB = area of (tri AEF + tri AZF + tri ZBF) = 1/2x12x6=36cm2
tri AEF = 36/3 = 12 cm2
tri ADC = 1/2 x 12 x 12 = 72 cm2

shaded area = 72 - 12 = 60 cm2

Hope this help and let me know if you need further explanation.