Q&A - PSLE Math

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Superbugs:

Nanyang CA1 2011

Q28
A group of 40 pupils took part in the Maths Olympiad competition. Their average score was 85 marks. The average score for the boys was 90 marks. The average score for the girls was 70 marks. How many girls participated in the competition?

Total score 40x85=3400
Difference in average score between boys and girls =90-70=20
Assuming all boys, total score 40x90=3600
Difference in score 3600-3400=200
No of girls = 200/20 =10
Is the understanding and answer correct? Any other method?

Hi,

You may use the Ratios Method, which is found in Chapter 2 of the book \"Raffles Maths Tricks - Volume 3\", please see the example attached.

Cheers

Speed Maths Centre
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=43&t=25433&start=10

[speedmaths.012624com]

ET-PH:

http://i40.tinypic.com/fllsh0.jpg\">


Hi, can someone help on St Nic CA1 Q18b and Q18c. Thanks.

Ans : Q18b is Fig 62, Q18c is 750 sticks

Hi,

To solve such questions, you should use \"Look for a Formula\":

Q18b
Figure Number = (Number of Triangles + 1) / 2
(123 + 1) / 2 = 124 / 2 = 62
Answer: Figure 62

Q18c
For (circles and) sticks, you should start from Figure 2.
Figure 2 = 10 sticks
Figure 3 = 15 sticks
Figure 4 = 20 sticks
and so on.

Since the difference between the number of sticks is \"Plus 5\", there is a \"Times 5\" in the formula.

Number of sticks= (Figure Number - Figure 2) x 5 + 10
[10 being the number of sticks for Figure 2]

So for Figure 150, the number of sticks will be:
(150 - 2) x 5 + 10
= 148 x 5 + 10
= 740 + 10
= 750 (sticks)

Hope this helps.

Cheers


Speed Maths Centre
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=43&t=25433&start=10

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Superbugs:

Aitong CA1 2011

Q17.
A coin box contained only twenty-cent and fifty-cent coins in the ratio of 4:5. When 16 fifty-cent coins were taken out and replaced by some twenty-cent coins, the number of fifty-cent coins left in the box was 7/8 of the twenty-cent coins. The total value of all the coins remained the same. Find the sum of money in the coin box.

20cent: 50cent
4:5
16 fifty cent coins taken out = $8. Therefore, 40 twenty cent coins put in, also $8 value
Then ratio of no of coins
20cent: 50cent
8:7
I am stuck here on forming the equations as the first ratio deals with value while the final ratio is on qty

Hi,

Both ratios deal with quantity/number.
You need to know the number of coins in order to calculate the total value.

You may use the Bags and Boxes Method to solve ALL the Double-Ratios problems for PSLE Maths:
(The Bags and Boxes Method can be found in the book: Raffles Maths Tricks - Volume 1; first printed in 2006.)

(twenty-cent and fifty-cent coins in the ratio of 4:5)
At first, let there be 4 Bags of 20-cent coins, and let there be 5 Bags of 50-cent coins.
(The number of coins in each of the 9 Bags is the same)

(The total/combined value of all the [20-cent and 50-cent] coins remained the same.)
16 fifty-cent coins were taken out and 40 twenty-cent coins were added.

(The number of fifty-cent coins left in the box was 7/8 of the twenty-cent coins)
Finally, let there be 8 Boxes of 20-cent coins, and let there be 7 Boxes of 50-cent coins.
(The number of coins in each of the 15 Boxes is the same)

20-cents coins:
4 Bags + 40 coins = 8 Boxes
(x7)
28 Bags + 280 coins = 56 Boxes

50-cents coins:
5 Bags - 16 coins = 7 Boxes
(x8)
40 Bags - 128 coins = 56 Boxes

40 Bags - 128 coins = 56 Boxes = 28 Bags + 280 coins
40 Bags - 128 coins = 28 Bags + 280 coins
12 Bags - 128 coins = 0 Bag + 280 coins
12 Bags = 280 + 128 = 408 coins
1 Bag = 408 / 12 = 34 coins

At first, there were 4 Bags = 4 x 34 = 136 twenty-cent coins.
At first, there were 5 Bags = 5 x 34 = 170 fifty-cent coins.

Can you work from here?

Cheers


Speed Maths Centre
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=43&t=25433&start=10

[MathIzzzFun]

ET-PH:

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Hi, can someone help on St Nic CA1 Q18b and Q18c. Thanks.

Ans : Q18b is Fig 62, Q18c is 750 sticks

Hi
(b)
the number of triangles in actually the odd number series 1,3,5,7,9...which can be represented by the formula
2N - 1, where N = 1, 2,3,4.. ie 1st odd number, 2nd odd number etc.

2N - 1 = 123, N = 124 / 2 = 62 --> Fig 62 has 123 triangles.

side track a bit....note that the sum of 1st N odd numbers is simply N x N
Sum of 1st odd number (N=1) --> 1
Sum of 1st two odd numbers (N=2) 1+3 --> 4 = 2 x2
Sum of 1st three odd numbers (N=3) 1+3+5=9 = 3 x 3

(c)...

fig 1 -> 4 sticks
fig 2 --> 10 sticks
fig 3 --> 15 sticks

From Figure 2 onwards, number of sticks = Figure number x 5

For Figure 150, number of sticks = 150 x 5 = 750

cheers.

[thinkgmama]

Hi,


I would like to ask

36+2 ------ 38 ($2 is already added)
6+4 ------80 ( $4 is already added)

How come at the end,

Shoes ----- 38+2 ------ 40 (why do we need to add $2 again?)
Handbag --------80+4 ------84 (why do we need to add $4 again?)

Please kindly help to explain. Thank you very much.

tianzhu:

homeworkmummy:

I need help for another 1 Qs from Nan Hua paper.

Mrs Samy had a sum of money to spend. She spent 1/2 of her money plus $4 on a handbag. She then spent 1/2 of the remaining money plus $2 on a pair of shoes. Finally she spent 2/3 of what was left plus $1 on a skirt. She was then left with $11.
(a) How much did each item cost?
(b) How much money did she have at first?

Hi

You may use “Working backwards” and show it using MD.

1/3 ------- 11+1 ------ 12
3/3 -------12*3 ------ 36

36+2 ------ 38
38*2 ------ 76

76+4 ------80
80*2 ------ 160

Skirt ----- 24+1 ------ 25
Shoes ----- 38+2 ------ 40
Handbag --------80+4 ------84

Amount of money at first ------160

Best wishes

[MathIzzzFun]

thinkgmama:

Hi,


I would like to ask

36+2 ------ 38 ($2 is already added)
6+4 ------80 ( $4 is already added)

How come at the end,

Shoes ----- 38+2 ------ 40 (why do we need to add $2 again?)
Handbag --------80+4 ------84 (why do we need to add $4 again?)

Please kindly help to explain. Thank you very much.

tianzhu:

[quote=\"homeworkmummy\"]I need help for another 1 Qs from Nan Hua paper.

Mrs Samy had a sum of money to spend. She spent 1/2 of her money plus $4 on a handbag. She then spent 1/2 of the remaining money plus $2 on a pair of shoes. Finally she spent 2/3 of what was left plus $1 on a skirt. She was then left with $11.
(a) How much did each item cost?
(b) How much money did she have at first?

Hi

You may use “Working backwards” and show it using MD.

1/3 ------- 11+1 ------ 12
3/3 -------12*3 ------ 36

36+2 ------ 38
38*2 ------ 76

76+4 ------80
80*2 ------ 160

Skirt ----- 24+1 ------ 25
Shoes ----- 38+2 ------ 40
Handbag --------80+4 ------84

Amount of money at first ------160

Best wishes

[/quote]Hi

hope this model will help ...

http://i41.tinypic.com/zwto7.jpg\">

cheers.

[YumYum]

Hi, pls help with the following :

Sandy and Nicole had $60 at first.Sandy and Nicole spent some money. The amount of money Sandy spent was twice the amount of money Nicole spent. Nicole had $10 more then Sandy in the end. What Nicole had was also twice of what Sandy had in the end.
a) Find out how much did Nicole spend.
b) Find out how much Sandy had at first.
Thnks!

[MathIzzzFun]

YumYum:

Hi, pls help with the following :

Sandy and Nicole had $60 at first.Sandy and Nicole spent some money. The amount of money Sandy spent was twice the amount of money Nicole spent. Nicole had $10 more then Sandy in the end. What Nicole had was also twice of what Sandy had in the end.
a) Find out how much did Nicole spend.
b) Find out how much Sandy had at first.
Thnks!

Hi

In the end, Nicole had twice as much as Sandy and Nicole had $10 more than Sandy.

So, Nicole --> $20, Sandy --> $10, total spent --> $30
Total amount spent --> $60 - $30 = $30
Sandy spent twice as much as Nicole, Nicole's spending --> 1/3 x $30 = $10
At first, Sandy --> $10 + $20= $30

cheers.

[ET-PH]

Thanks MathIzzzFun and Speedmaths.com

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kamom:

Help :please:


http://img.photobucket.com/albums/v606/shirleywoon/SH-Sep2011/MathOlympiadTest1e.jpg\">

THANKS A MILLION

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Hi kamom,

The answer is 55 degrees.

Please see diagram.

First, draw a square, say 1 cm by 1 cm.
Next, draw an isosceles triangle on the right, and use trigonometry to find the value of lengths a, b and c.
Then draw another isosceles triangle on the left, and use trigonometry to find the value of lengths d, e and f.

g = c - e

w + x = 70 degrees

tan (w + 70 degrees) = (f + 1 + b) / g

x = (70 degrees - w) = 55 degrees.

The solution I used involves trigonometry, which is normally beyond P6 pupils.

May I know the source of this question?

Cheers


Speedmaths.com
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=43&t=25433&start=10


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