Q&A - PSLE Math

[Vanilla Cake]

Cheerfuldad:

Hi vlim,

Do u have the solution for paper 2 - Q5, Q12, Q15b, Q16,Q17b ?I tried to solve but could not, can you help!

Hi Cheerfuldad,
While waiting for vlim's response, pls refer to the suggested solutions as below:

Q5
Total area
A : B
100 : 120
5 : 6

Assume the common shaded area to be p
Unshaded area of A = 0.8u
Unshaded area of B = u

Solve by simultaneous equations,
0.8u + p = 5
u + p = 6

0.2u = 1
u = 5
p = 1

Percentage of rectangle B that is shaded = p/p+ux100%
= 1/(1+5)x100%
= 1/6 x 100% = 16 and 2/3%

Q12
Before
A : O
4u : 5u

Change (sold)
A : O
-170 : -1.25u (25/100x5u)

After
A : O
1 : 2

2(4u-170)=1(5u-1.25u)
8u-340=3.75u
4.25u = 340
u = 80

Number of apples left = 4u-170 = 4(80)-170 = 320-170 = 150
Number of oranges left = 5u-1.25u = 3.75u = 3.75x80 = 300
Number of fruits that Mr Tan had in the end = 150+300 = 450

Q15(b)
From part (a), you worked out that after 10 min, height of water level in tank A = 15 cm @ 8.10 am
Given that flow rate of water by Tap P = 1.2 litres/min (1200 cm3/min) and Tap B = 5 litres/min (5000 cm3/min). Calculate the base area for tank A and tank B = 40x20 and 50x40= 800 cm2 and 2000 cm2. From here, work out the increase in height for both tanks.
Increase in height per min for Tank A = 1200/800 = 1.5 cm/min
Increase in height per min for Tank B = 5000/2000 = 2.5 cm/min
At 8.10 am, height of Tank B = 0 and assume the time taken for both heights of the water level in both tanks to be equal as t.
1.5t+15=2.5t+0
t=15 min
Add 15 min to 8.10 am., will give 8.25 am.

(b) At 8.25 am, height of the water level in both tanks are equal.

Q16
Pls refer to solutions by http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&start=4210 on page 422.

Q17(b)
Pls refer to http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&start=4220 on page 423.

VC's mum

[Vanilla Cake]

pensiveowl:

Cindy had four times as many postcards as Annie. After Cindy gave 20% of her postcards to Jane and Annie gave 10% of her postcards to Jane, the number of Jane's postcards increased by 75%. Jane had 252 postcards in the end. How many postcards did Cindy have at first?

Working backwards.
In the end, Jane had 252 postcards (increased by 75%).
175% -> 252 postcards
100% -> 252/175x100 = 144 postcards for Jane at first.
Increase in the postcards for Jane = 252-144 = 108
10% from Annie + 20% from Cindy-> 108 postcards

Given that at first, Cindy had four times as many postcards as Annie ->
20% from Cindy = 4x20% = 80% from Annie.
10%+80% = 90%
90% from Annie -> 108 postcards
100% from Annie -> 108/90x100 = 120 postcards for Annie at first.

So, number of postcards that Cindy had at first -> 4x number of postcards that Annie had at first = 4x120 = 480 postcards.

VC's mum

[pensiveowl]

Vanilla Cake:

pensiveowl:

Cindy had four times as many postcards as Annie. After Cindy gave 20% of her postcards to Jane and Annie gave 10% of her postcards to Jane, the number of Jane's postcards increased by 75%. Jane had 252 postcards in the end. How many postcards did Cindy have at first?

Working backwards.
In the end, Jane had 252 postcards (increased by 75%).
175% -> 252 postcards
100% -> 252/175x100 = 144 postcards for Jane at first.
Increase in the postcards for Jane = 252-144 = 108
10% from Annie + 20% from Cindy-> 108 postcards

Given that at first, Cindy had four times as many postcards as Annie ->
20% from Cindy = 4x20% = 80% from Annie.
10%+80% = 90%
90% from Annie -> 108 postcards
100% from Annie -> 108/90x100 = 120 postcards for Annie at first.

So, number of postcards that Cindy had at first -> 4x number of postcards that Annie had at first = 4x120 = 480 postcards.

VC's mum

Thanks VC's mum

[MathIzzzFun]

pensiveowl:

Hi.. could someone pls help me with this question:


Cindy had four times as many postcards as Annie. After Cindy gave 20% of her postcards to Jane and Annie gave 10% of her postcards to Jane, the number of Jane's postcards increased by 75%. Jane had 252 postcards in the end. How many postcards did Cindy have at first?

Thanks,
PO

Hi

At first,
Cindy = 40u
Annie = 10u

Number of postcards given to Jane by
Cindy = 20% x 40u = 8u
Annie = 10% x 10u = 1u

So, 9u = 75/175 x 252 = 108, 1u = 12

At first, number of postcards Cindy had = 40 x 12 = 480

cheers.

[tianzhu]

Vanilla Cake:

Hi tianzhu,


Good morning,
2011 MGS P6 Prelim Maths is http://mwtuition.com.sg/free-resources online but without answers. I would like DD2 to work out these sums after her PSLE 2011 LC by tomorrow afternoon. Is it possible for you to go through her answers and advise if there are any mistakes? BTW, it seems that Q12 and Q14 from paper 2 cannot be solved due to incomplete information. Pls kindly look into them as well as the above request to seek help from you.

Best wishes
VC's mum

Hi VC's mum

Good Afternoon.

First, I am deeply humbled that you have faith in my “masak masak” way of doing maths.

I may not know the solution to every question and also time may be a consideration. I may not be able to give you the answers fast enough.

I thought the answer sheets are usually provided with past years papers. Perhaps a faster way is to cross check with current PSLE parents or kids who are doing the same paper.

Best wishes

[Vanilla Cake]

tianzhu:

I may not know the solution to every question and also time may be a consideration. I may not be able to give you the answers fast enough.

Hi tianzhu,
Thks for yr fast response. Have PM for you.
Yes, I understand that time is a big consideration.
Your efforts in helping out are deeply appreciated.

VC's mum

[smileplease]

Need help :


Chan cycled from the post office to the library at a constant speed. At the same time, Valerie cycled from the library to the post office at a constant speed. 20 minutes later, they passed each other. Chan took another 16 minutes to reach the library, while Valerie was still 1.53km away from the post office.
a) what was the speed of Valerie? Give your answer in m/min
b) what was the distance between the post office and the library?

[MathIzzzFun]

smileplease:

Need help :


Chan cycled from the post office to the library at a constant speed. At the same time, Valerie cycled from the library to the post office at a constant speed. 20 minutes later, they passed each other. Chan took another 16 minutes to reach the library, while Valerie was still 1.53km away from the post office.
a) what was the speed of Valerie? Give your answer in m/min
b) what was the distance between the post office and the library?

Hi

there are a couple of ways to solve this.... here's one using combined speed.

Chan & Valerie took 20 min to cycle the total distance from library to post office (when they passed each other).

after they passed each other, they cycle another 16 min and Valerie was still 1.53km away from post office..
so, 4 min = 1.53km
20 min = 5 x 1.53 km = 7.65 km so the distance between post office and the library is 7.65km.

Distance cycled in 36 min by Valerie = 7.65 km - 1.53km = 6.12km
Valerie's speed = 6120m/36 = 170 m/min

cheers.

[Alchemist]

Hi - need some help for this sum from SHHK maths prelim 2011 paper 2.


There were a total of 930 boys and girls at a carnival. After 4/5 of the boys and 3/4 of the girls left, there were 120 more girls than boys that remained. How many boys were at the carnival at first?

Thanks in advance.

[PiggyLalala]

vidhyalog:

Hi - need some help for this sum from SHHK maths prelim 2011 paper 2.


There were a total of 930 boys and girls at a carnival. After 4/5 of the boys and 3/4 of the girls left, there were 120 more girls than boys that remained. How many boys were at the carnival at first?

Thanks in advance.

This is what my ds will do for such question.
Let the number of boys be 5p and the number of girls be 4u.
therefore 5p + 4u = 930.-----(1)
4/5 boys left, that is, 1/5 boys stayed. Therefore p boys remained.
3/4 girls left, that is, 1/4 girls stayed. Therefore u girls remained.
There are 120 more girls than boys, that is, u = p + 120.------(1)
hence from (1) and (2): 5p + 4(p+120) = 930
5p + 4p + 480 = 930
9p = 450
p = 50.

Boy: 5p = 250.

this is a simplified simultaneous equation but I think most pupils can understand