# A LEVEL MATHEMATICS

### The Mailbox

#### Wiped out over tiny P&C problem 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?

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

#### Finding equations of tangents parallel to y-axis, not sure how to proceed 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

#### Length of projection-confused with whether to use cross or dot product 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

#### Gradient of line and vectors 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

#### Conics and applications of differentiation 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

#### Integrating cosec ^4 x wrt x 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

#### Poisson distribution approximated to normal distribution: Supply and demand problem 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

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

#### Locus of points equidistant from x-y and x-z plane Posted by White Group Mathematics on November 4, 2015 at 10:10 AM comments (0)

Hi Mr Koh, Student X

I have drafted a detailed set of explanations-please see as attached below. Hope it helps. Peace. Best Regards,

Mr Koh

#### Discovering vertices of a rectangle in an Argand diagram Posted by White Group Mathematics on August 4, 2015 at 9:40 PM comments (1)

Hi Mr Koh,

How do I do this question- In an Argand diagram, the points A, B, C and D represent the complex numbers a,  -2+5i,  c   and  3/2 - 1/2 i   respectively. Given that ABCD is a rectangle described in the clockwise sense with BC=2AB, find a and c.

I tried drawing B and D on the same diagram but they don't even have a right angle between them, so how can it form a rectangle? And I can't place points A and C on the diagram because they said ABCD is read clockwise but it is also mentioned BC=2AB....so I am lost. Haha. Thanks Mr Koh!

Student X

Based on the construct described, the rectangle's length is twice its breadth. I would recommend drawing a generic upright rectangle instead of trying to place the points specifically in Argand space. Do note that vectors are involved in the solving process. I have drafted a solution template for you: Hope this helps. Peace.

Best Regards,

Mr Koh

#### Discerning winning probabilities in national lottery Posted by White Group Mathematics on July 19, 2015 at 10:40 PM comments (0)

Hi,

Here in my country, when you buy lottery tickets, you try to guess the six (6) combination of numbers out of 42 numbers. Unlike in Sweepstakes, the order of numbers does not really matter.

http://www.philippinepcsolotto.com/6-42-lotto-result-summary

Some say that the probability of winning the jackpot is simply 6/42 = 14.29%. That is high!

Since I was in high school, that formula seemed to make sense for me. After all, what is the probability of guessing one of the six combinations? 1/42. That is the probability of guessing each number in the combination of six. That is the result when you add the probability for each number.

(1/42) X 6 = 6/42

1/42 + 1/42 + 1/42 + 1/42 + 1/42 + 1/42 = 6/42

But I've come across another possibility today. Without repetitions of numbers, you can create 5,245,786 combinations of six digits from 1 to 42. What is the probability that you will guess the right combination out of more than 5 million combinations? 1 out of 5,245,786. That is 1.906 X 10 raised to negative 7.

I'm confused now because both solutions seem to make sense, but this second idea makes more sense. Each one is a computation based on a different point of view.

I hope you can enlighten me on this one.

Thanks,

Student X

The first manner of calculating the required probability which you cited is flawed. If say, each of these 6 combinations can be recycled, the probability would be computed as (1/42)^6 = 1/5489031744.

However, I am assuming they cannot be recycled, as such there will be 42 possible combinations for the first number, 41 for the second, 40 for the third, so on and so forth. That said, the probability would be computed as (1/42)*(1/41)*(1/40)*(1/39)*(1/38)*(1/37)= 1/3776965920.

You may therefore question: why isn't this equivalent to 1/5245786, which is considerably larger in fact? The reason is (1/42)*(1/41)*(1/40)*(1/39)*(1/38)*(1/37) actually examines the instance when sequencing of numbers matters, as opposed to 1/( 42 C 6) = 1/5245786 which doesn't pay any regard to the ordering of numbers. Clearly, the former would involve much greater stakes and thus even lower likelihood of winning.

Hope this clarifies. Peace.

Best Regards,

Mr Koh