Posts made by 0xBB40E64E

0xBB40E64E

8228:
chanbi:
[quote=\"8228\"]
I don't mind my ds uses it as e-dictionary. For online research or watching video, PC at home is a preferred choice as I don't want his eye-sight to get much worse. I can make him to agree to some rules but it will be difficult to enforce them if he chooses not to follow.
There is apps out there that can be use to control the usage. I am using this app called \"screen time\" (for android). It can set each day total usage time, what apps that can't be access during sch hrs, what apps that can't be access after bed time.. And you can set it differently on weekend.

Most likely I will get him an android phone too. Will find out more about the app. Thanks.[/quote]Just a tip, from personal experience, apps like screen time are hilariously easy to disable. I feel that a parent should trust their child can exercise self-control and know if their own usage is affecting their studies, rather than using such an app. IMHO, using parental control apps show a total lack of confidence a parent has in their child.
Of course, if his phone gets confiscated in school, or he shows poor self control, then a parental control app might be a solution, but if this is your son's first phone, why not give him a chance?

0xBB40E64E

MummyMoo:
May I know the URL of EMB pls? Ds doesn't know how to log onto EMB.

May I also ask where to submit the form to get the car decal pls?

Many thanks in advance

EMB: http://messages.hci.edu.sg/smb/hs_student/

0xBB40E64E

tinghao_sg:
My son is also Sec 1 student and will take bus 67 to HCI by himself started from next week. Bus 67 (CCK bus interchange) can directly go to HCI. But I heard the traffic is always jammed in that area, the bus takes longer time to HCI. Can any senior HCI student from CCK area suggest what time should be in CCK bus interchange in the morning in order to arrive HCI at 7.15am?

About 1h, so 6.15 at CCK should be fine. I heard the main problem is along CCK road especially at the junction towards Upp Bt Timah Rd, and the rather unpredictable frequency of 67.

If possible, waiting along the bus stop on Upp Bt Timah Rd near Bt Panjang should be much faster, and will also provide 3 additional bus services, but of course you have to take the LRT first, and with all the construction works at Bt Panjang, it might not be convenient to do so.

0xBB40E64E

koguma:
Need advise here.

Anyone who has kid who is involved in a sport but the sport is not a CCA in the school ?

example : you represent your school in Golf competition, but you pick up golf on your own.

Does the school recognise your kid competition result and award the cca points accordingly?

Thanks.

If you talk to the head of CCA, he should be able to put it in.

Q Pan:
lexis:
Hi,

My son will be taking public transport to school - MRT to Botanics then take a bus. May I know which is the correct bus stop to drop at and if the walkway from the bus stop to school is sheltered?

Thanks!
Your son should alight at the last Hwa Chong bus stop (Gate 3) just before Namly Avenue. The entire school is sheltered and interlinked, starting from the bus stop, so don't worry.

Just to add on, it will be the 5th stop after the MRT station, and the bus stop name is opp Natl JCFrom the MRT, you can take [list]

0xBB40E64E

ljk:
Hi HCI parents,

What is the usual dismissal time in Sec 1? My son isn't taking any 3rd language. I saw the 2013 timetable and there was AEP/MSP/PRP/etc. from 2:20 - 4:50pm. What do those acronyms stand for? Thanks.

AEP: Art Elective Program
MSP: Malay Special Program
PRP is some reading thing iirc

Basically it's the special programs

0xBB40E64E

mummytofour:
0xBB40E64E:
[quote=\"mummytofour\"]Hi, my daughter has some problems with her sec one maths questions so could anyone help her? thx!

1)Given that 106480=2^4 × 5 × 11^3 , find the smallest value for x where 106480x is a perfect square and x is an integer.
2)Find two numbers if their HCF (highest common factor) is 5 and their LCM (lowest common multiple) is 175.
3)The lowest common multiple of 6,9, and x is 126. Find the two possible values of x which are odd numbers.
4)The HCF (highest common factor) of two numbers is 1 and the LCM of these numbers is 51. What are the two numbers?
5)The product of the ages of a group of people from 20 years to 25 years is 11088. Find
(i) the number of people in the group,
(ii) the sum of their ages.
1)
For a number to be a perfect square, the power of the prime factors must be a multiple of 2, aka 2^2 x 3^4 is fine, but 2^2 x 3^3 is not.

So, for 106480, you make the powers all even,
2^4 × 5 × 11^3 => 2^4 x 5^2 x 11^4
what you added was 5 x 11 = 55.
To check, 106480 x 55 = 5856400 which is a perfect square

2)
HCF = 5
LCM = 5^2 x 7

As the LCM is 5, both numbers must be multiples of 5

No-1: 5
No-2: 5

As LCM has a 5^2, one of the numbers must have a multiple of 5^2. Both numbers cannot be multiples of 5^2 as if they are, the LCM would be 5^2 instead of 5.

No-1: 5^2
No-2: 5

As LCM has a 7, one of the numbers must be a multiple of 7. As explained above, both numbers cannot be multiples of 7.

No-1: 5^2 = 25
No-2: 5 x 7 = 35

25, 35

3)
No-1 : 2 x 3
No-2 : 3^3
No-X : ?
LCM : 2 x 3^2 x 7

As none of the numbers there are multiple of 7, but the LCM is a multiple of 7, x must be a multiple of 7.

1st ans : 7

As x must be multiple of 7, the other number must be 3 x 7, as we can't use 2 x 7 (must be odd)

2nd ans : 21

7 , 21

4)
3,17

5i) 3

5ii)67

Sorry, I'll add the explanations soon. These questions are actually quite interesting! Never seen some of them before and also quite challanging

Hi 0xBB40E64E, thank you so much for your answers and explanation!
:thankyou:
My daughter understood it after seeing your solution

By the way, the questions are from a sec one math assessment book...and there are answers at the back and all of your answers are right, except for question 3 . Its supposed to be 21 and 63, but you got 21 right! :rahrah: Anyway thanks so much because i didnt know how to explain it to her
:thankyou:[/quote]
You're welcome

It's good to know my answers are correct, builds self-confidence haha

Also, 63 is acceptable, but 7 is also right!

http://i.imgur.com/xDU7iiI.png\">
There can actually be multiple answers for this question =P

0xBB40E64E

mummytofour:
Hi, my daughter has some problems with her sec one maths questions so could anyone help her? thx!

1)Given that 106480=2^4 × 5 × 11^3 , find the smallest value for x where 106480x is a perfect square and x is an integer.
2)Find two numbers if their HCF (highest common factor) is 5 and their LCM (lowest common multiple) is 175.
3)The lowest common multiple of 6,9, and x is 126. Find the two possible values of x which are odd numbers.
4)The HCF (highest common factor) of two numbers is 1 and the LCM of these numbers is 51. What are the two numbers?
5)The product of the ages of a group of people from 20 years to 25 years is 11088. Find
(i) the number of people in the group,
(ii) the sum of their ages.

1)
For a number to be a perfect square, the power of the prime factors must be a multiple of 2, aka 2^2 x 3^4 is fine, but 2^2 x 3^3 is not.

So, for 106480, you make the powers all even,
2^4 × 5 × 11^3 => 2^4 x 5^2 x 11^4
what you added was 5 x 11 = 55.
To check, 106480 x 55 = 5856400 which is a perfect square

2)
HCF = 5
LCM = 5^2 x 7

As the LCM is 5, both numbers must be multiples of 5

No-1: 5
No-2: 5

As LCM has a 5^2, one of the numbers must have a multiple of 5^2. Both numbers cannot be multiples of 5^2 as if they are, the LCM would be 5^2 instead of 5.

No-1: 5^2
No-2: 5

As LCM has a 7, one of the numbers must be a multiple of 7. As explained above, both numbers cannot be multiples of 7.

No-1: 5^2 = 25
No-2: 5 x 7 = 35

25, 35

3)
No-1 : 2 x 3
No-2 : 3^3
No-X : ?
LCM : 2 x 3^2 x 7

As none of the numbers there are multiple of 7, but the LCM is a multiple of 7, x must be a multiple of 7.

1st ans : 7

As x must be multiple of 7, the other number must be 3 x 7, as we can't use 2 x 7 (must be odd)

2nd ans : 21

7 , 21

4)
LCM : 3 x 17

HCF = 1 means that the two numbers do not share any common factors (other then 1)

Ans must be 3 and 17 then.

3,17

5i)
11088 = 2^4 x 3^2 x 7 x 11

As the age of the people MUST be between 20-25

The age of the first person must be 11 x 2 = 22 as 11 is less then 20, but multiplying 11 by any other number then 2 will lead to it being over 25.

1st person: 22

Remaining ages : 2^3 x 3^2 x 7

Age of the second person has to be 7 x 3 as 7 x 2 is less then 20 and
7 x 2^2 or 7^3 are over 25.

2nd person : 21

Remaining ages : 2^3 x 3

2^3 x 3 is the last person age. Amy other combination will lead to a number below 20.

3rd person: 24

Ans: 3 people

5ii)
22+21+24=
67

I'll admit these are pretty interesting and challenging questions. Had much fun working on them

0xBB40E64E

wonderm:
ljk:
Thanks wonderm and 0xBB40E64 for your clarification.

Just to clarify, would the marks from physics, chemistry and biology be divided equally to get an average and the average score is then added to calculate the msg?
For Sec 1 and Sec 2, Science is considered as one subject. Each term there will only be one science test. Physics, chemistry and biology are taught during science lessons. For my two boys, the ways school covered the 3 sciences were not the same. So they may twist it again next year. Don't worry about the details for now.

To further elaborate, there will only be one exam paper (LSS), and as for the mark allocation for the 3 modules, IIRC it's split evenly, but even if it's not, it's beyond anyone's control, so no point worrying about that.

If you are really worried, you can from time to time ask to view their ePortfolio.

0xBB40E64E

ljk:
When is Sec 1 registration at HCI? Is it 23 Dec?

To maintain in IP, one has to get MSG < 3? Or is it 3 and below? How is msg score calculated? So worrying...have to closely monitor my child's progress to ensure he stays in the IP program. Stress!

MSG= Mean Subject Grade aka Average grade of all subjects

Eg:
EL:A2
Math:A1
CL:B3
Chem:A1
Physics: B3
History: B4

MSG= (Sum of all subject grade) / (Number Of Subjects)
= (2+1+3+1+3+4) / (6)
= (14) / 6
= 2.33

A MSG below 3 aka Ave grade less then a B3 should be fine.