O-Level Elementary Math

[VisualTextEducation]

wei lun:

thks for your reply but need to use equation to solve this problem.

Hi wel lun,

Here you go:

Let the first digit be x,

Since the sum of the digits is 6, the second digit would be 6-x,

Therefore, the products of the digits would be = x(6-x)
= 6x-x^2

If the first digit is x, and it is a 2 digit number, x must be in the tens place, hence the NUMERICAL value of x would be = 10 x (x)
= 10x

Since the second digit is in the ones place, the NUMERICAL value of 6-x would be
= 1 x (6-x)
= 6-x

Hence, the NUMERICAL value of the 2 digit number would be = 10x + (6-x)
= 10x + 6 - x
= 9x +6

Since the products of the digits IS EQUAL to the 1/3 of the original number, we can construct the equation as follows,

6x-x^2 = (1/3) x (9x+6)

Rearranging the equation, we will have,

6x-x^2 = 3x+2
6x = x^2+3x+2
0 = x^2+3x-6x+2
0 = x^2-3x+2
x^2-3x+2 = 0
Factorising using the Cross technique you learnt in secondary 2,

You will get,

(x-2)(x-1)=0
x=2 or x=1 (first digit)
6-x=4 or x=5 (second digit)

Hence, the original number may be 24 or 15.

SOLVED! =D :imcool:

You may call me at 92220737 if you require any further assistance in these maths questions, I will be glad to help!

[savoury sweet]

Can someone pls enlighten me the following maths qn as I am really clueless when come to my gal’s Secondary maths qn. Thanks


Find the value of X
2 logx 4 - 3 log4 X = 1

[mum_sugoku]

savoury sweet:

Can someone pls enlighten me the following maths qn as I am really clueless when come to my gal's Secondary maths qn. Thanks


Find the value of X
2 logx 4 - 3 log4 X = 1

first change log base x to log base 4, ie
logx 4=(log4 4) /(log4 x) = 1/(log4 x)

let y= log4 x

hence eqn becomes 2[1/y] - 3y =1
or 2 - 3y^2 = y
or 3y^2 + y -2 =0
(3y-2)(y+1)=0

hence y=2/3, or y=-1

log4 x= 2/3 --> x=4^(2/3)

or log4 x =-1 --> x=4^(-1)

[savoury sweet]

Thank you so much mum_sugoku. Will show my gal today

[Rejoin]

Angle BOC = 64 x 2 = 128 degrees (because angle at the centre is 2x angle at the circumference)


Triangle BOC is isoceles. So that means angle OCB = angle OBC = (180 - 128)/2 = 26 degrees.

Angle AOC = 180 - 16 - 26 = 138 degrees.
Angle DOC = 180 - 138 = 42 degrees

Triangle OCD is isoceles. So that means angle OCD = angle DOC = (180-42)/2 = 69 degrees.

Angle ADB = 69-64 = 5 degrees.

Angle ABD = 180 - 5 - 16 = 159 degrees.

Angle DBC = 180 - 159 = 21 degrees.

[iamastudent]

Hi,

I need help for the following questions
1)
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part d.

My answers for part a,b,c:
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2)
http://i58.tinypic.com/mlqst2.jpg\">

3)
http://i62.tinypic.com/24bpxtf.jpg\">

4)
http://i60.tinypic.com/6px11l.jpg\">

5)
http://i60.tinypic.com/w2jayf.jpg\">

Part iv

My answers for i,ii,iii:
http://i58.tinypic.com/29b0k7c.jpg\">
http://i57.tinypic.com/2mn2b0p.jpg\">

6)
http://i61.tinypic.com/11hzamd.jpg\">

Thank you!!

[Goodluck889]

Can someone help me with this maths question?

Factorise 144z^2 + 121.

[Eddow]

Goodluck889:

Can someone help me with this maths question?

Factorise 144z^2 + 121.

It can't be factorise unless it is a minus(-) sign

[Rejoin]

Goodluck889:

Can someone help me with this maths question?

Factorise 144z^2 + 121.

This has a pair of complex roots. So, it is
(z + 11/12 i)(z - 11/12i), where i = sqrt(-1) is the imaginary number.

This is not in the O level syllabus, only in the A level syllabus.

[student101]

How to solve this compound interest problem!

A man deposited 2,000 into a bank that paid compound interest of 3.28% compound half yearly. He continued depositing $2,000 into the bank every six months.Calculate the amount of money he had after he made his fifth deposit of $2,000,giving your answer correct to the nearest cent.

I need help to solve this question in an easy way acceptable for O level. Also I have a doubt about whether the interest given(3.28%) is annual interest or half year interest as in the question,it is not mentioned clearly.
Thanks