June Crash Course

June Crash Course

Chemistry, JC Chemistry, JC Mathematics, Mathematics

The team at The Culture SG has been really busy and we have a lot of things prepared to help you guys work for that A. First up! Crash course for June…

And we know it is a bit late to be announcing this on the site now, but we have really been caught up with preparing our students lately that we don’t have the time to properly update here. So here are the details for the Math Crash Course and the Chemistry Crash Course.

P.S. For SCIENCE students who wish to chiong in October, please take note that the H2 Chem/ Phy/ Bio Paper 4 (practical) is in October. So better start soon! Here are the details!

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For 3 hr lessons, they are priced at $105.

For 2 hr lessons, they are priced at $70.

Lessons will be held at:
Newton Apple Learning Hub
Blk 131, Jurong Gateway Road #03-263/265/267 Singapore 600131
Tel: +65 6567 3606

For math enquiries, you may contact Mr. Teng at +65 9815 6827.

For chem enquiries, you may contact Ms. Chan at +65 93494384.

For GP enquiries, you may contact Ms. Chen at +65 91899133.

Thinking Math@TheCulture #3

Thinking [email protected] #3

JC Mathematics, Mathematics

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by KS. More of him can be found here.

This is a question from 1976 A’levels Paper 2. I thought it is pretty interesting to discuss the question with a little extension.

(a) In how many ways can 5 copies of a book be distributed among 10 people, if no-one gets more than one copy?

(b) In how many ways can 5 different books be distributed among 10 people if each person can get any number of books?

So now, let us modify it a bit.

(c) In how many ways can 5 copies of a book be distributed among 10 people if each person can get any number of books?

Notice that the difference between (b) and (c) is that the book distributed is not identical. So for (c), we are pretty much distributing r identical balls to n distinct boxes. Whereas for (b) , we are pretty much distributing r distinct balls to n distinct boxes.

Thinking Math@TheCulture #2

Thinking [email protected] #2

JC Mathematics, Mathematics

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by KS. More of him can be found here.


(i) Find the two possible values of z such that z^2 = 1 + \sqrt{3}i, leaving your answer in exact form a + bi, where a and b are real numbers.

(ii) Hence or otherwise, find the exact roots of the equation

2w^2 + 2 \sqrt{6}w + 1 - 2 \sqrt{3} i = 0

Thinking [email protected] #1

JC Mathematics, Mathematics

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by KS. More of him can be found here.


Each card in a deck of cards bear a single number from 1 to 5 such that there are n cards bearing the number n, where n = 1, 2, 3, 4, 5. One card is randomly drawn from the deck. Let X be the number on the card drawn.

(i) Find the probability distribution of X.

(ii) Show that \mathbb{E}(X) = \frac{11}{3} and find \text{Var}(X).

Andrew draws one card from the deck, notes the number and replaces it. The deck is shuffled and Beth also draws on card from the deck and notes the number. Andrew’s score is k times the number on teh card he draws, while Beth’s score is the square of the number on the card she draws. Find the value of k so that the game is a fair one.

Probability Question #4

Probability Question #4

JC Mathematics

A gambler bets on one of the integers from 1 to 6. Three fair dice are then rolled. If the gambler’s number appears k times (k = 1, 2, 3), he wins $ k. If his number fails to appear, he loses $1. Calculate the gambler’s expected winnings

Thoughts on A’levels H2 Mathematics 2016 Paper 2

Thoughts on A’levels H2 Mathematics 2016 Paper 2

JC Mathematics

I’ll keep this short since we are all busy. One thing about paper 1 we saw, there were many unknowns.

So topics which I think will come out…

Differentiation – I think a min/max problem will come out, possibly with r and h both not given and asked to express r in terms of h. But students should revise a on the properties of curves with differentiation; given a curve equation with an unknown, for instance y=/frac{x^2+kx+1}{x-1}, find the range of k such that there is stationary points.

Complex Number – Loci will definitely come out. I’m saying they will combine with trigonometry.

Integration – Modulus integration hasn’t really been tested. Else a question on Area/ Volume could be tested, and I’ll say they need students to do some
Conics too.

For statistics, my students should have gotten the h1 stats this year. And if it’s an indicator, then it should not be a struggle.

I expect PnC and probability to be combined. Conditional Probability in a poisson question should be tested too, so do revise it well. For hypothesis testing, students should be careful of their formula and read really carefully about the alternative hypothesis. Also, :9 know that the formulas for poison PDF and binompdf are both given in mf15. Lastly, know when to use CLT.

All the best!

Random Questions from 2016 Prelims #13

Random Questions from 2016 Prelims #13

JC Mathematics

NYJC/2/11

On a typical weekday morning, customers arrive at the post office independently and at a rate of 3 per 10 minute period.

(i) State, in context, a condition needed for the number of customers who arrived at the post office during a randomly chosen period of 30 minutes to be well modelled by a Poisson distribution.

(ii) Find the probability that no more than 4 customers arrive between 11.00 a.m. and 11.30 a.m.

(iii) The period from 11.00 a.m. to 11.30 a.m. on a Tuesday morning is divided into 6 periods of 5 minutes each. Find the probability that no customers arrive in at most one of these periods.

The post office opens for 3.5 hours each in the morning and afternoon and it is noted that on a typical weekday afternoon, customers arrive at the post office independently and at a rate of 1 per 10 minute period. Arrivals of customers take place independently at random times.

(iv) Show that the probability that the number of customers who arrived in the afternoon is within one standard deviation from the mean is 0.675, correct to 3 decimal places.

(v) Find the probability that more than 38 customers arrived in a morning given that a total of 40 customers arrived in a day.

(vi) Using a suitable approximation, estimate the probability that more than 100 customers arrive at the post office in a day.

2016 A-level H1 Mathematics (8864) Paper 1 Suggested Solutions

2016 A-level H1 Mathematics (8864) Paper 1 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. KS will hold no liability for any errors. Comments are entirely personal opinions.

As these workings and answers are rushed out asap, please pardon me for my mistakes and let me know if there is any typo. Many thanks.
I’ll try my best to attend to the questions as there is H2 Math Paper 2 coming up.

Question 1
(i)
\frac{d}{dx} [2 \text{ln}(3x^2 +4)]

= \frac{12x}{3x^2+4}

(ii)
\frac{d}{dx} [\frac{1}{2(1-3x)^2}]

= \frac{6}{2(1-3x)^3}

= \frac{3}{(1-3x)^3}


Question 2

2e^{2x} \ge 9 - 3e^x

2u^2 + 3u - 9 \ge 0

(2u-3)(u+3) \ge 0

\Rightarrow u \le -3 \text{~or~} u \ge \frac{3}{2}

e^x \le -3 (rejected since e^x > 0) or e^x \ge \frac{3}{2}

\therefore x \ge \text{ln} \frac{3}{2}


Question 3
(i)

Graph for 3i
Graph for 3i

(ii)
Using GC, required answer = -1.606531 \approx -1.61 (3SF)

(iii)
When x = 0.5, y = 0.35653066

y -  0.35653066 = \frac{-1}{-1.606531}(x-0.5)

y = 0.622459 x +0.04530106

y = 0.622 x + 0.0453 (3SF)

(iv)

\int_0^k e^{-x} -x^2 dx

= -e^{-x} - \frac{x^3}{3} \bigl|_0^k

= -e^{-k} - \frac{k^3}{3} + 1


Question 4
(i)
y = 1 + 6x - 3x^2 -4x^3

\frac{dy}{dx} = 6 - 6x - 12x^2

Let \frac{dy}{dx} = 0

6 - 6x - 12x^2 = 0

1 - x - 2x^2 = 0

2x^2 + x - 1 = 0

x = -1 or \frac{1}{2}

When x =-1, y = -4

When x = \frac{1}{2}, y = 2.75

Coordinates = (-1, -4) \text{~or~} (0.5, 2.75)

(ii)
\frac{dy}{dx} = 6 - 6x - 12x^2

\frac{d^2y}{dx^2} = - 6 - 24x

When x = -1, \frac{d^2y}{dx^2} = 18 > 0. So (-1, -4) is a minimum point.

When x = 0.5, \frac{d^2y}{dx^2} = -18 \textless 0. So (0.5, 2.75) is a maximum point.

(iii)

Graph for 4iii
Graph for 4iii

x-intercept = (-1.59, 0) \text{~and~} (1, 0) \text{~and~} (-0.157, 0)

(iv)
Using GC, \int_0.5^1 y dx = 0.9375


Question 5
(i)
Area of ABEDFCA = \frac{1}{2}(2x)(2x)\text{sin}60^{\circ} -  \frac{1}{2}(y)(y)\text{sin}60^{\circ}

2\sqrt{3} = \sqrt{3}x^2 - \frac{\sqrt{3}}{4}y^2

2 = x^2 - \frac{y^2}{4}

4x^2 - y^2 =8

(ii)

Perimeter = 10

4x+2y + (2x-y) = 10

6x + y = 10

y = 10 - 6x

4x^2 - (10-6x)^2 = 8

4x^2 - 100 +120x -36x^2 = 8

32x^2 -120x+108=0

x=2.25 \text{~or~} 1.5

When x=2.25, y = -3.5 (rejected since y >0)

When x=1.5, y = 1


Question 6
(i)
The store manager has to survey \frac{1260}{2400} \times 80 = 42 male students and \frac{1140}{2400} \times 80 = 38 female students in the college. He will do random sampling to obtain the required sample.

(ii)
Stratified sampling will give a more representative results of the students expenditure on music annually, compared to simple random sampling.

(iii)
Unbiased estimate of population mean, \bar{x} = \frac{\sum x}{n} = \frac{312}{80} = 3.9

Unbiased estimate of population variance, s^2 = \frac{1}{79}[1328 - \frac{312^2}{80}] = 1.40759 \approx 1.41


Question 7
(i)
[Venn diagram to be inserted]

(ii)
(a)
\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B) = 0.8
(b)
\text{P}(A \cup B) -  \text{P}(A \cap B) = 0.75

(iii)
\text{P}(A | B')

= \frac{\text{P}(A \cap B')}{\text{P}(B')}

= \frac{\text{P}(A) - \text{P}(A \cap B)}{1 - \text{P}(B)}

= \frac{0.6 - 0.05}{1-0.25}

= \frac{11}{15}


Question 8
(i)
Required Probability = \frac{1}{2} \times \frac{3}{10} \times \frac{2}{9} = \frac{1}{30}

(ii)
Find the probability that we get same color. then consider the complement.

Required Probability = 1 - \frac{1}{30} - \frac{1}{2} \times \frac{5}{10} \times \frac{4}{9} - \frac{1}{2} \times \frac{2}{10} \times \frac{1}{9} - \frac{1}{2} \times \frac{4}{6} \times \frac{3}{5} - \frac{1}{2} \times \frac{2}{6} \times \frac{1}{5} = \frac{11}{18}

(iii)
\text{P}(\text{Both~balls~are~red} | \text{same~color})

= \frac{\text{P}(\text{Both~balls~are~red~and~same~color})}{\text{P}(\text{same~color})}

= \frac{\text{P}(\text{Both~balls~are~red})}{\text{P}(\text{same~color})}

= \frac{1/30}{7/18}

= \frac{3}{35}


Question 9
(i)
Let X denote the number of batteries in a pack of 8 that has a life time of less than two years.

X \sim B(8, 0.6)
(a)
Required Probability =\text{P}(X=8) = 0.01679616 \approx 0.0168

(b)
Required Probability =\text{P}(X \ge 4) = 1 - \text{P}(X \le 3) = 0.8263296 \approx 0.826

(ii)
Let Y denote the number of packs of batteries, out of 4 packs, that has at least half of the batteries having a lifetime of less than two years.

Y \sim B(4, 0.8263296)

Required Probability =\text{P}(Y \le 2) = 0.1417924285 \approx 0.142

(iii)
Let W denote the number of batteries out of 80 that has a life time of less than two years.

W \sim B(80, 0.6)

Since n is large, np = 48 > 5, n(1-p)=32 >5

W \sim N(48, 19.2) approximately

Required Probability

= \text{P}(w \ge 40)

= \text{P}(w > 39.5) by continuity correction

= 0.9738011524
\approx 0.974


Question 10
(i)
Let X be the top of speed of cheetahs.
Let \mu be the population mean top speed of cheetahs.

H_0: \mu = 95

H_1: \mu \neq 95

Under H_0, \bar{X} \sim N(95, \frac{4.1^2}{40})

Test Statistic, Z = \frac{\bar{X}-\mu}{\frac{4.1}{\sqrt{40}}} \sim N(0,1)

Using GC, p=0.0449258443 \textless 0.05 \Rightarrow H_0 is rejected.

(ii)
H_0: \mu = 95

H_1: \mu > 95

For H_0 to be not rejected,

\frac{\bar{x}-95}{\frac{4.1}{\sqrt{40}}} \textless 1.644853626

\bar{x} \textless 96.06 \approx 96.0 (round down to satisfy the inequality)

\therefore \{ \bar{x} \in \mathbb{R}^+: \bar{x} \textless 96.0 \}


Question 11
(i)
[Sketch to be inserted]

(ii)
Using GC, r = 0.9030227 \approx 0.903 (3SF)

(iii)
Using GC, y = 0.2936681223 x - 1.88739083

y = 0.294 x - 1.89 (3SF)

(iv)
When x = 16.9, y = 3.0756 \approx 3.08(3SF)

Time taken = 3.08 minutes

Estimate is reliable since x = 16.9 is within the given data range and |r|=0.903 is close to 1.

(v)
Using GC, r = 0.5682278 \approx 0.568 (3SF)

(vi)
The answers in (ii) is more likely to represent since |r|=0.903 is close to 1. This shows a strong positive linear correlation between x and y.


Question 12
Let X, Y denotes the mass of the individual biscuits and its empty boxes respectively.

X \sim N(20, 1.1^2)

Y \sim N(5, 0.8^2)

(i)
\text{P}(X \textless 19) = 0.181651 \approx 0.182

(ii)
X_1 + ... + X_12 + Y \sim N(245, 15.16)

\text{P}(X_1 + ... + X_12 + Y > 248) = 0.2205021 \approx 0.221

(iii)
0.6X \sim N(12, 0.66^2)

0.2Y \sim N(1, 0.16^2)

Let A=0.6X and B = 0.2Y

A_1+...+A_12+B \sim N(145, 5.2528)

\text{P}(142 \textless A_1+...+A_12+B \textless 149) = 0.864257 \approx 0.864


2016 A-level H2 Mathematics (9740) Paper 2 Suggested Solutions

2016 A-level H2 Mathematics (9740) Paper 2 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. KS will hold no liability for any errors. Comments are entirely personal opinions.

Numerical Answers (click the questions for workings/explanation)

Question 1: 0.0251 \text{~m/min}
Question 2: \frac{x^2}{2} \text{sin}nx + \frac{2x}{n^2} \text{cos}nx - \frac{2}{n^3} \text{sin}nx + C;~ a = 2 \text{~or~} 6;~ \pi (\frac{2}{5} - \text{ln} \frac{3}{\sqrt{5}})
Question 3: a + \frac{3}{4} - \text{sin}a - \text{cos}a + \frac{1}{4} \text{cos}2a;~ \frac{1}{4}(\pi + 1)^2
Question 4: z = 2.63 + 1.93i, ~ 3.37 + 0.0715i ;~8^{\frac{1}{6}}e^{i(-\frac{\pi}{12})}, ~8^{\frac{1}{6}}e^{i(-\frac{3 \pi}{4})}, ~8^{\frac{1}{6}}e^{i(\frac{7\pi}{12})};~ n=7
Question 5: \frac{11}{42};~ \frac{4}{11};~ \frac{4}{1029}
Question 6: 60;~ 10;~ \{ \bar{x} \in \mathbb{R}, 0 \textless \bar{x} \le 34.8 \};~ \{ \alpha \in \mathbb{R}, 0 \textless \alpha \le 8.68 \}
Question 7: 24;~ 576;~ \frac{1}{12};~ \frac{5}{12}
Question 8: r = -0.980;~ c = -17.5;~ d = 91.8;~ y = 85.9
Question 9: a = 7.41;~ p = 0.599 \text{~or~} 0.166;~ 0.792
Question 10: 0.442;~ 0.151 ;~ 0.800;~ \lambda = 1.85

 

Relevant materials

MF15

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