In CHS, the lower primary years (P1 to P3) are not demanding. There was no need at all for DS to stay back after school. On some school days, the lower primary students are not called back at all e.g. Sports Day, Swimming Carnival, etc.


That being said, any CHS parent can tell you that it is precisely due to this "lack of need to stay back after school in CHS" that we have to undertake to ensure that their kids spent quality time at tuition centres/tuition/enrichment programmes. Sigh…How else are our kids going to do those tough papers if we as parents don’t invest in the extras? (I am sure you have seen them or bought them.)

Kudos to all the hardworking parents at CHS!

ruohoo97:

Yes, I find it strange too. It's from ACSJ P6 2

SA1, 2011. I don't have standard answer as it was given by my DD's teacher for this June holiday homework.

The answer is 18. The paper is from ACS (Primary).

Dear Applelyn1,


Age difference remains the same.

In this case, 39 - 9 = 30

When the father is twice as old, the father is 2 units and the son is 1 unit.

The difference -> 2u - 1u = 1u

As difference in age always remains the same,
1u -> 30

Therefore, the son is 30 yrs old now.

To find the current age of the father, find 2u.

That is, 2u x 30 -> 60

The father is 60 years old.

Hope this helps.

Mrs Alan

http://i43.tinypic.com/6gy2l3.jpg\">

YumYum:

Hi, can anyone help with this Qn:


Ashley, Betsy and Carlie shared some sweets.
Ashley received 11/13 of the total number of sweets Betsy and Carlie received.
Betsy received 1/2 of the total number of sweets Ashley and Carlie received.
Carlie had 12 fewer sweets than Ashley.
How many sweets did the 3 girls share altogether?

Thanks

Dear Yum Yum,

Hope this helps:-)

Mrs Alan

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MathAche:

Can anybody provide the model for this problem?


John has 3 times as many stamps as Peter. When John loses 420 stamps and Peter loses 40 stamps, Peter has twice as many stamps as John. How many stamps does each of them have at first?

Dear MathAche,

Hope this helps.

YumYum:

Hi all,


Need help with this Qn:

Four different whole numbers are arranged in ascending orders (from the smallest to the largest). The average of these four whole numbers is 14. The average of the last 3 numbers is 15 and the average of the first 3 numbers is 12 2/3 (twelve two third). If the third number is odd, find the third number.

thanks!

Dear YumYum,

Sum of the 4 numbers --> 14 x 4 = 56
Sum of the last 3 numbers --> 15 x 3 = 45
Sum of the first 3 numbers --> 12 2/3 x 3 = 38

Therefore, the 1st number --> 56 - 45 = 11
Therefore, the 4th number --> 56 - 38 = 18
Sum of middle 2 numbers --> 56 - 11 - 18 = 27

Thinking process
(a) The third number must be odd.
(b) The middle 2 numbers must add up to 27
(c) The numbers are arrange in ascending order. That means the 2nd number must be greater than 11 and the 3rd number must be less than 18.

Hence: 12 + 15 = 27

The third number must be 15.

Hope this helps.

http://i42.tinypic.com/xm8n08.jpg\">

Chan09:

Q7. I have some red & blue ribbons in a bottle. If I add in 20 red ribbons, 60% of my ribbons are blue. If I add in another 60 blue ribbons, 75% of my ribbons are blue. How many ribbons have I in the bottle ??


Ans: 180 but I don't understand why deduct 20 red if I use the UTM method, can this be done by model? As I got 160 if I use model?

Dear Chan09,

Usually I get my DS to think about what remains the same or what has changed when dealing with problems sums. Also bear in mind, get your DD/DS to simplify the numbers when dealing with percentages.

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cmm:

Sue has more than 50 hairpins.

If she puts them in packets of 7, she has 5 hairpins left.
If she puts them in packets of 9, she has 6 hairpins left.
How many hairpins does Sue have?

Dear cmm,

Hope this helps.

Dear Big Devil,


My son in P5 also mentioned this. But there has been no official statement from the school. If this is the case, shouldn’t the p6 parents be informed by now? Puzzled:-(

Dear Evergreen,


For the second qn, let your child know that there is an unknown whole, a part of 127 is removed and the remaining part left is 343. To find the whole, combine the 2 parts to give the whole.

For the third qn, the same concept applies. 674 is the part removed from the whole. If the remaining part is 279, adding the two parts form the answer to the whole.

Part + Part = Whole
Whole - Part = Part

Best regards.