Posted by White Group Mathematics on April 12, 2018 at 4:45 AM | comments (10) |
Hi Mr Koh,
I came across your website while seeking someone to coach my daughter in H2 Maths. She is currently studying at TJC and will be taking her A Levels at the end of this year. At the present moment she has someone helping her with the subject but thus far she has failed in most of her tests and exams (since JC1). Quite frankly we are at a loss, because my daughter has been working very hard all this while, yet she isn't witnesing any significant improvement in her grades, which is rather discouraging. Will it be too late for her to catch up? Please do also advise with regards to your tuition fee rates, thanks.
Parent X
Hi,
Thank you for your inquiry. Your daughter is studying in a top tier junior college, understandably it will be a rollercoaster ride for her. Admittedly it may be demoralising to consistently experience setbacks in attempting to do well for the subject, however she mustn't lose sight of the ultimate goal which is to score decently in the upcoming A Levels. Her inability to achieve better grades at the present moment could be attributed to a variety of factors - poor math foundation, chronic carelessness, nerves during actual test sittings among others.
That being said, there is certainly time left to salvage the currrent situation, then again I do need to personally review her work process to better discern the underlying troubles. To a certain extent, one might take comfort (not too much though) in the fact that each year numerous JC students have to endure this disheartening phase of their learning journey no thanks to schools who somehow feel inclined to set far higher standards internally as compared to that of the actual A Levels.
I presently charge $ X /hr, do have a proper discussion with your daughter and keep me updated accordingly. Thanks and god bless.
Best Regards,
Mr Koh
Posted by White Group Mathematics on April 4, 2018 at 2:05 AM | comments (0) |
The problem is presented as follows:
A student entering a junior college has to offer 4 subjects, of which at least one must be chosen from the basket of science-related subjects ( Maths, Physics, Chemistry And Biology) and at least one being chosen from the basket of arts-related subjects ( Economics, Literature, History And Geography). How many subject combinations in all are available?
Here are my workings.
Case 1
1 art and 3 science subjects : 4 × (4 ×3 ×2) = 96
(My reasoning: there are 4 possible art subjects to choose from; also each time the student picks one science subject, the number of science subjects still left available for choosing decreases by one, hence the 4 ×3 ×2 portion of the working)
Case 2
2 science and 2 art subjects : (4 ×3 ) ×( 4×3) = 144
Case 3
1 science and 3 art subjects: 4 × (4 ×3 ×2) = 96
Total possible number of combinations = 96 + 144 +96 = 336
However, the actual answer given is merely 68. How is this right? I have scrutinized my workings repeatedly, and can't for the love of god sniff out any mistake. Perhaps the answer supplied is wrong?
Please advise, thanks.
Student X
Hi,
Your manner of computation unfortunately is erroneous, because it has actually included permutation, ie the sequence in which the subjects are chosen matters based on your workings, when it shouldn't. Please find the amended version below.
Case 1
1 art and 3 science subjects : 4C1 × 4C3 =16
(If I were to permutate accordingly, then I would arrive at 16 × 3! =96, which is equivalent to your answer previously obtained for this particular case)
Case 2
2 science and 2 art subjects : (4C2 ) ×(4C2) = 36
(If I were to permutate accordingly, then I would arrive at 36 × 2!× 2! =144, which is equivalent to your answer previously obtained for this particular case)
Case 3
1 science and 3 art subjects: 4C1 × 4C3 =16
(If I were to permutate accordingly, then I would arrive at 16 × 3! =96, which is equivalent to your answer previously obtained for this particular case)
∴ Total possible number of combinations = 16 + 36 + 16 = 68 (shown)
Hope this clarifies. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on January 12, 2018 at 12:20 AM | comments (1) |
Hello Mr Koh,
I recently encountered a problem in my holiday revision package which asked us to discover the equations of the tangents fo the curve x^2- axy +y^2 =a which are parallel to the y-axis. I managed to compute an expression via implicit differentiation for the gradient function, ie dy/dx = (ay-2x)/ (2y-ax) which I am not sure if it is correct; I also do not really know how to proceed from here. Could you please advise?
Student X
Hi,
The expression for dy/dx obtained is absolutely accurate. With regards to finding the necessary tangent equations, you shall have to set dx/dy=0 , thereafter inject the result arising from this back into the original curve equation. Kindly find my set of workings in full below, hope it helps. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on November 28, 2016 at 4:40 AM | comments (0) |
Hi Mr Koh it's XXX. I have a question for you:
Why did the teacher use cross product instead of dot product? Thank you.
The way the respective dot and cross product results arise are attributed to the orientation of the known direction vector. In the context you cited, the known direction vector, ie the normal to the plane is vertically placed, hence things are phrased as such.
On the other hand, if you consider the typical projection of a given vector onto a line, the known direction vector d is seen running horizontally instead of vertically. Therefore, the dot and cross product results are switched.
I have crafted two separate diagrams (attached below) for your reference, hopefully that will clarify matters.
Best Regards,
Mr Koh
Posted by White Group Mathematics on September 28, 2016 at 4:50 AM | comments (0) |
Hi Mr Koh,
I chanced upon your website online and I'd really like to engage you as my private tutor. I'm currently a JC2 student studying in SAJC. I know this is very last minute, but I really do need the help; in summary, I've tried really hard to work on my Math, however I really don't see any improvements in my grades and I feel like I need someone to tutor me over this last month.
I'm currently living at MacPherson, near Taiseng MRT station and I'm willing to meet outside my home if it is more convenient for you. I will understand if you are completely packed and cannot accommodate any more students. Apologies for this late message. Please get back to me soon-thank you and have a nice day!
Student X
Hi,
Thank you for your message. Firstly, if you work with me, bear in mind I cannot produce short term miracles and more than a fair bit of effort has to be invested on your part to secure a reasonably positive outcome at the A Levels. Secondly, while commuting to your residence isn't an issue, I am pretty swarmed right now, so please undersand it is only fair I accord priority to my existing students schedule-wise.
Thirdly, kindly note that a fee premium will be charged for extremely last minute engagements. If you still sincerely wish to set things in motion, give me a holler and we will take it from there. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on August 26, 2016 at 2:25 AM | comments (0) |
Hi Mr Koh,
I can't seem to get around this problem, could you please help?
Student X
Gradient of the line = -4/3 by rewriting its equation in the form y=mx+c. As such, it is as good as saying that for every 1 unit one moves in the x direction, it also has to move -4/3 units in the y direction. This general movement is hence represented by i - 4/3j.
Since OA is parallel to the line, then we say vector OA= k (i - 4/3j) , where k is a real valued constant. Given the magnitude of vector OA is 20 units, we can further surmise that k √ [ 1² + (-4/3)² ] = 20, solving this gives k=12 or -12. Thus vector OA= 12 (i - 4/3j) =12i -16j or -12 (i - 4/3j) = -12i+16j (shown)
Hope this helps. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on February 13, 2016 at 9:05 PM | comments (0) |
I am having considerable difficulty grappling with this OCR problem. Could you kindly assist?
Student X
Worked it out for you, hope it helps. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on February 2, 2016 at 6:00 AM | comments (0) |
I was given some basic calculus revision questions to do for an undergrad module, but since there's nothing new in them, so I decided to give these questions an A-level tag. I've been working on them for a while now, but I'm stuck at this. Any help would be appreciated; I'm not particularly bothered if you want to post the answers as it's not an assessed homework, and I'm starting to get tired of these.
Student X
Please find the integral worked out as follows. Hope it helps. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on January 24, 2016 at 11:50 AM | comments (0) |
Sales of cooking oil bought in a shop during a week follow a Poisson distribution with mean 100. How many units should be kept in stock to be at least 99% certain that supply will be able to meet demand? Use normal approximation to Poisson distribution, if appropriate.
I have already approximated the distribution to Y ~ N (100, 100). Then, I have tried to use P(Y>y) = 0.99 . That didn't work so I'm trying to use P( Y = y) = 0.99. However, with this, I'm stuck with P [ (y-100.5)/10 < Z < (y - 99.5)/10] = 0.99
The correct answer is 123. Please help.
Student X
You kinda got your understanding of stuff mixed up. Firstly, recognize that it is the demand that varies, not the supply. It is the amount demanded that follows a Poisson distribution, which is subsequently approximated to that of a normal distribution.
The part about defining the original underlying distribution modelling demand as Y ~ N (100, 100) is correct.
If you let k be the number of units you wish to store as stock, then the required inequality satisfying the requirements of the question should be P(Y ≤ k) ≥ 0.99, which after adjusting for continuity correct would yield P(Y ≤ k+0.5) ≥ 0.99
As such, k+0.5 ≥ InvNorm of a normal distribution curve with mean and variance both =100 *
ie after standardization, solve (k+0.5-100)/ 10 ≥ InvNorm ( 0.99, 0, 1)
Hope this clarifies. Peace.
* The graphic calculator should help you compute the inequality in k here; unless you are using basic tables, then standardization is needed.
Best Regards,
Mr Koh
Posted by White Group Mathematics on November 4, 2015 at 10:10 AM | comments (0) |
Hi Mr Koh,
I am unsure how about how the solution for part (a) is formulated. Could you please help?
Student X
I have drafted a detailed set of explanations-please see as attached below. Hope it helps. Peace.
Best Regards,
Mr Koh