Q&A - PSLE Math
-
Superbugs:
HiAitong CA1 2011 P6
Q5.
10/11 kg of sugar is packed into little bags. If each bag can contain only 1/7kg of sugar, how much sugar is left unpacked?
(Give your answer as a fraction in the simplest form)
10/11 divided by 1/7
I get 6 4/11
Why is the answer 4/77kg and not 4/11kg?
4/11 is the remainder, so the answer is 4/11 of 1/7 kg = 4/77 kg.
cheers. -
Superbugs:
Aitong CA1 2011 P6
Q5.
10/11 kg of sugar is packed into little bags. If each bag can contain only 1/7kg of sugar, how much sugar is left unpacked?
(Give your answer as a fraction in the simplest form)
10/11 divided by 1/7
I get 6 4/11
Why is the answer 4/77kg and not 4/11kg?
Hi Superbugs,
HTH..
6 bags of 1/7 kg ---6 x 1/7 = 6/7 or 66/77
10/11 - 66/77 --- 70/77 - 66/77 = 4/77 -
Superbugs:
I am rather confused by the solution. Why 24-23 becomes 1u? Whymathfreak:
Nan Hua CA1 P6 2011
Q15
A farm has some ducks and cows.
The ratio of the number of animals to the total number of legs is 8:23.
Express the number of ducks as a fraction of the cows.
Give your answer in the simplest form.
Pls assist with this question. Tks
Solution:
Pairing a cow and a duck, the two animals have 6 legs.
Hence in a situation where there are equal numbers of cows and ducks, the ratio is 2 : 6 = 8 : 24.
However, the ratio is now 8 : 23.
The difference in legs is 24-23=1 unit. Since the difference in number of legs between a cow and a duck is 2, we let 1 unit be 2 legs. By this assumption, the total number of animals is 16.
If we decrease 1 cow and increase 1 duck, the number of legs decreases by 2.
Hence, there are 7 cows and 9 ducks.
Ans: 9/7 = 1 2/7
1u=2legs? Thereafter I am lost! Pls help! Tks
Hi
here's another approach using area model...
http://www.flickr.com/photos/62167097@N02/6788698190/in/photostream
cheers. -
MathIzzzFun:
Tks a lot! Now I got it.
HiSuperbugs:
Aitong CA1 2011 P6
Q5.
10/11 kg of sugar is packed into little bags. If each bag can contain only 1/7kg of sugar, how much sugar is left unpacked?
(Give your answer as a fraction in the simplest form)
10/11 divided by 1/7
I get 6 4/11
Why is the answer 4/77kg and not 4/11kg?
4/11 is the remainder, so the answer is 4/11 of 1/7 kg = 4/77 kg.
cheers. -
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 -
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? -
![](\"http://i40.tinypic.com/fllsh0.jpg\")
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 -
![](\"http://i40.tinypic.com/15zvxh.jpg\")
Superbugs:
Hi,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?
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 -
ET-PH:
Hi,
Hi, can someone help on St Nic CA1 Q18b and Q18c. Thanks.
Ans : Q18b is Fig 62, Q18c is 750 sticks
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 -
Superbugs:
Hi,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
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
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