Mathematics 4024/01 May June 2009

[Benjamin]

Hello,
I solved the question paper. I came to some problems to which I was unable to answer them. You may like to download the past paper straight away or can individually solve them by the following links,
Past paper:
http://www.xtremepapers.net/CIE/Cambrid ... 9_qp_2.pdf
The questions are 11,15, and 16.
The questions are individually posted as well:
1)http://img411.imageshack.us/img411/3176/maths2.jpg
2)http://img217.imageshack.us/img217/5695/maths1.jpg
3)http://img198.imageshack.us/img198/9893/maths3.jpg

I shall be highly glad to you.
Regards,
Benjamin.

 

[SMRJ]

What is your main issue in these questions? Please clarify each problem in every question so that i could accordingly assist you. :shock:

 

[Zishi]

Hello,
I solved the question paper. I came to some problems to which I was unable to answer them. You may like to download the past paper straight away or can individually solve them by the following links,
Past paper:
http://www.xtremepapers.net/CIE/Cambrid ... 9_qp_2.pdf
The questions are 11,15, and 16.
The questions are individually posted as well:
1)http://img411.imageshack.us/img411/3176/maths2.jpg
2)http://img217.imageshack.us/img217/5695/maths1.jpg
3)http://img198.imageshack.us/img198/9893/maths3.jpg

I shall be highly glad to you.
Regards,
Benjamin.

In the heading you've asked answers for question paper 1 and have given the link to question paper 2. Please always give the correct link first. But, here you go:

Q-11) For part (a) to find height of smaller bucket, take the ratios of height and diameters. That is 8/16=x/30 The 'x' will be the height of larger bucket. Secondly, for part (b) take the ratios of cube of diameters to their volumes, that is (8/16)^3=Volume of small bucket/volume of larger bucket

Q-15) First convert all the times in minutes, then add all of them and then convert the TOTAL time in minutes and seconds to get answer of part (a). For part (b) divide the sum of the time in minutes that you got in part (a) by '4'. Then convert the result into minutes and seconds. Lastly, for (c) part subtract the smallest time from the largest time, and then convert it into minutes and seconds.

Q-16) For part (a), put x=4 in f(x)=12-5x. For part (b) put f(x) = 17, which would be 12-5x=17. And as for part (c) make x the subject of formula and then substitute 'x' for 'y' in that formula to get the inverse.

I'm kinda bad at explanation at forums, but I hope that you'll get it.