Q&A - PSLE Math

[MathIzzzFun]

yingxuan0806:

Tanks P and Q are each filled with some water. If water from Tank Q is poured into Tank P until the water in Tank P reaches the brim, there will be 5 litres of water left in Tank Q. If water from Tank P is poured into Tank Q until the water in Tank Q reaches the brim, there will be 17 litres of water left in Tank P. The ratio of volume of Tank P to the volume of Tank Q is 3 : 1. How many more litres of water are needed to fill both tanks to their brim?

can help me solve this question? if yes, can you please use ration n algebra?? preferably dont use model method. ty in advance

Hi

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cheers

[gnagna]

tianzhu:

gnagna:

Hi Tianzhu
Can explain how you get 38?

Hi

Good Morning.

This question appears friendishly easy at first glance.

But, it’s not so easy for PSLE kids.

Equal circles can be stacked in square or hexagonal packing such that each circle touches four or six others.

If you use square packing, you’ll get 36 circular shapes. With hexagonal packing, you’ll get 2 more circular shapes.

Best wishes

Thanks Tianzhu for pointing out.

[tianzhu]

gnagna:

Thanks Tianzhu for pointing out.

Hi

Good Morning.

You're welcome.

Best wishes

[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

[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

[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