O-Level Additional Math

[tianzhu]

Herbie:

hi how to solve this qqn


a team of 8 men can build 3 swimming pools in 5 weeks. Find the no. Of men required to build 6 swimming pools in 4 weeks? Tq

Hi

One way is to use compound units which in this question is \"man-weeks\"

3 swimming pool ------ 8*5 ------ 40 man-weeks
6 swimming pools ------ 2*40 ------- 80 man-weeks

80/4 ------ 20 men

Best wishes

[Herbie]

hi tianzhu, many thanks!


I hv another qn.

15 men working 8 hours a day can build a bridge in 30 days.
How long will 9 men working 8 hours a day take to buiild the same ridge assuming that the rate at which they work remains unchanged? My ans is 50 days.

B. How many hours a day must 25 men work to build the same bridhe in 16 days!

Tq

[tianzhu]

Herbie:

15 men working 8 hours a day can build a bridge in 30 days.
How long will 9 men working 8 hours a day take to buiild the same ridge assuming that the rate at which they work remains unchanged? My ans is 50 days.

B. How many hours a day must 25 men work to build the same bridhe in 16 days!

Hi

Your answer of 50 days for part A is right.

For part B,
1 bridge ------ 15*30*8 ------- 3600 man-hours

Number of men*number of days*number of hours ------- 3600 man-hours

25*16*h ------- 3600 where h refers to the number of hours the men must work.

h ------- 9 hours.

Best wishes

[tutor4kids]

Hi, thank you for the solution.

Could you have me with the following question?

In the diagram below, ACB is a tangent to the circle at point C. This tangent is parallel to the line FE. Points F, G and H lie on the circle.
i) Prove that (EC)(CG) = (CF)^2.
ii) The line FE is extended to meet the circle at H. Prove (FE)(EH) = CF^2 - EC^2.

http://i52.tinypic.com/33zd7hv.jpg\">

Thank you.

[FrekiWang]

(i) (this is a very standard A Maths Question, Similarity is one of the standard approaches)

In FEC and GFC,
AngleEFC=AngleFCA=AngleFGC (hopefully you know why)
AngleFCE=AngeGCF
Thus FEC is similar to GFC
FE/GF=EC/FC=FC/GC
Cross-multiply 2nd and 3rd, we have EC x GC = FC^2(proven)

(ii) (usually (ii) is related to (i))
RHS=CF^2-EC^2
=EC x GC - EC^2 (from part (i))
=EC x (GC - EC)
=EC x GE
=FE x EH (intersecting chords thereom)
=LHS (proven)
Seriously, part(ii) is much easier. In the exam/test, if you cant prove part(i), you can always assume part(i) result and use it to prove part(ii)

[tutor4kids]

Hi,


Thank you very much for the prompt reply! I can understand the solution.
Thanks once again.

[FrekiWang]

tutor4kids:

Hi,


Thank you very much for the prompt reply! I can understand the solution.
Thanks once again.

Seriously, if you are a home tutor teaching A Maths, I would like to advise that you need more preparation for this topic. A lots of students are having problems in this topic so most likely they are going to ask a lot.

[tutor4kids]

Hi, thanks for the advice. I am not a home tutor for A Maths though, just seeking help on behalf of a friend. Anyway, thanks for your help.

[Herbie]

hi uncle tianzhu. Thanks very much.

[g131271]

Hi parents,


Does anyone of you can help me in this question? A million thanks to those who help to solve.

[b]A regular polygon has n sides. The size of each interior angle is 4 times the size of each exterior angle.
a. calculate the value of each exterior angle.
b. calculate the value of n.
c. hence find the sum of interior angles of polygon