## Lower Secondary Mathematics

PSLE marks the graduation of Primary school students and their entry into Secondary schools as teenagers. Discuss all issues about Secondary schooling here.
iwork
BrownBelt Posts: 622
Joined: Fri Nov 07,
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### Re: Lower Secondary Mathematics

EC72 wrote:Need help with the following question. Thanks in advance.

The length of a rectangle is 20 m, correct to the nearest metre. If the area of the field is 332m2, correct to the nearest square metre, find the least possible breadth of the field, giving your answer correct to 3 significant figures.

Breadth to be the least, then Area must be the least and Length the highest.

So Area = 331.5 m2 and Length = 20.4 m.

And you can work out the Breadth.

HAPPYH
KiasuGrandMaster Posts: 1532
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### Re: Lower Secondary Mathematics

Following

iwork
BrownBelt Posts: 622
Joined: Fri Nov 07,
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### Re: Lower Secondary Mathematics

tm1 wrote:wouldn't it be great if there were online 'tuition' services that help our kids with homework - like online tutors who can explain to our kids how to do math questions. homework these days are so difficult.
Well, something that I might want to explore to provide online coaching. Will have to get the right platform and experience how easy it is to provide group tuition and share information.

CheerioMama
OrangeBelt Posts: 31
Joined: Sun Feb 23,

### Re: Lower Secondary Mathematics

What is the market rate for 1 to 1 or group for Maths tuition at the tutor's home ?

iwork
BrownBelt Posts: 622
Joined: Fri Nov 07,
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### Re: Lower Secondary Mathematics

CheerioMama wrote:What is the market rate for 1 to 1 or group for Maths tuition at the tutor's home ?
For 1-1 at tutor's home, it could be cheaper by 10% or more since the tutor don't have to travel and hence save time and transport. But for group, as most likely it will be at tutor's home, the rate will remain the same. Well, unless the tutor provide group tuition at your house!

EC72
BlueBelt Posts: 220
Joined: Sun Oct 05,
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### Re: Lower Secondary Mathematics

iwork wrote:
EC72 wrote:Need help with the following question. Thanks in advance.

The length of a rectangle is 20 m, correct to the nearest metre. If the area of the field is 332m2, correct to the nearest square metre, find the least possible breadth of the field, giving your answer correct to 3 significant figures.

Breadth to be the least, then Area must be the least and Length the highest.

So Area = 331.5 m2 and Length = 20.4 m.

And you can work out the Breadth. PlayfulFairy
BlueBelt Posts: 223
Joined: Tue Oct 07,
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### Re: Lower Secondary Mathematics

The key to a code is given by 3 whole numbers. The key is transmitted from the HQ by 3 agents, who were given the product of two of the 3 numbers. Suppose that the 3 numbers transmitted are 432, 540 and 720. What's the key code?

How do I solve the code? Thanks!

justMommy
GreenBelt Posts: 134
Joined: Fri Nov 07,
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### Re: Lower Secondary Mathematics

However, online coaching fails to ''inspire'' which is one kids of this generation needs. Most of them are not naturally hardworking.

Pand
YellowBelt Posts: 26
Joined: Fri Jan 01,
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### Re: Lower Secondary Mathematics

Saw this posting and thought, I should Clarify it. I am a Maths teacher in a local secondary school. The mistake made here is quite common for those who have not undergone NIE training..anyway, thought would share the correct answer so that you can learn.

To obtain least breadth, Area should be minimum and Length should be maximum

Least area = 331.5m2
max length = 20.5 m.. ( not 20.4m..).. the correct answer should be actually 20.499999999
but 20.49999999...... is equal to 20.5m.

This is an important question that came out in the O levels which many students got it wrong.

If you want to know more...go and look for the proof of 0.999999....... = 1.

Hope this helps in some way.

Pand

iwork wrote:
EC72 wrote:Need help with the following question. Thanks in advance.

The length of a rectangle is 20 m, correct to the nearest metre. If the area of the field is 332m2, correct to the nearest square metre, find the least possible breadth of the field, giving your answer correct to 3 significant figures.

Breadth to be the least, then Area must be the least and Length the highest.

So Area = 331.5 m2 and Length = 20.4 m.

And you can work out the Breadth.

CoffeeCat
BrownBelt Posts: 512
Joined: Sat Feb 27,

### Re: Lower Secondary Mathematics

PlayfulFairy wrote:The key to a code is given by 3 whole numbers. The key is transmitted from the HQ by 3 agents, who were given the product of two of the 3 numbers. Suppose that the 3 numbers transmitted are 432, 540 and 720. What's the key code?

How do I solve the code? Thanks!
The key to solving this is to let the 3 numbers be represented by a, b and c where a < b < c.

Then there will be 3 possible products, a X b, a X c, b X c.
Note that a*b < a*c < b*c implying
a X b = 432
a X c = 540
b X c = 720

a X b X a X c X b X c = 432 X 540 X 720
a^2 X b^2 X c^2 = 432 X 540 X 720 = 167961600
Taking the square root of both sides of the above equation,
a X b X c = 12960
c = (a X b X c) / (a X b ) = 12960 / 432 = 30
a = (a X c) / c = 540 / 30 = 18
b = (b X c) / c = 720 / 30 = 24

Notice that this question ( & solution) is similar to the more common variant when you replaced the product by addition.