Problem Solving Question #2

JC Mathematics

These series of questions target on students’ abilities to solve it. They are common questions used in job interviews and interviewers are more interested in learning how the interviewee derive the answer, instead of the answer itself. So do feel free to share your method of solving the problem. Remember, we are interested in how you solve it. 🙂

A frog is in the foot of a 100m well. Each time the frog jumps, it moves up by 2m and then down by 1m. How many jumps are required for the frog to get out of the well?

Ans: 99

Probability Question #3

JC Mathematics

A man write five letters, one each to A, B, C, D, and E. Each letter is placed in a separate envelope and sealed. He then addresses the envelopes, at random, on each to A, B, C, D, and E.

  1. Find the probability that the letter to A is in the correct envelope and the letter to B is in an incorrect envelope.
  2. Find the probability that the letter to A is in the correct envelope, given that the letter to B is in an incorrect envelope.
  3. Find the probability that both of the letter to A and B are in incorrect envelopes.

Ans: \frac{3}{20}; \frac{3}{16}; \frac{13}{20}

Summation Question #2

JC Mathematics

This is a question from AJC/C2/MY/P1/10. Quite challenging. But once you get the hang of it, you should be capable of solving other variations.

Using the formula for sin(A+B), prove the sin(3A)=3sinA-4sin^{3}A.

(i) Hence show that \sum_{r=1}^{n} \frac{4}{3^r} sin^{3}(3^{r}\theta)=sin(3\theta)-\frac{1}{3^n} sin(3^{n+1}\theta)

(ii) Deduce the sum of the series \sum_{r=1}^{n} sin^{2}(3^{r}\theta)cos(3^{r}\theta)

Ans: \frac{1}{4} [cos(3\theta)-cos(3^{n+1}\theta)]

APGP Question #1

JC Mathematics

This is a question from A-level 1999 Paper 1. A very simple question that tests you if you know your stuffs. 🙂

The r^{th} term of a series is 3^{r-1} + 2r. Find the sum of the first n terms.

Ans: \frac{1}{2} (3^{n}-1)+n(n+1)

Summation Question #1

JC Mathematics

This is a modified summation question taken from HCI. I think it forces the students to think out of the box, and we all know how scary trigonometry and summation is together.

Given \sum_{r=1}^{n}cos[(2r-1){\theta}] = \frac{sin(2n\theta)}{2sin\theta}, show that

sin{\theta}+3sin{3\theta}+5sin{5\theta}+...+99sin{99\theta} = \frac{cos{\theta}sin{100\theta}-100sin{\theta}cos{100\theta}}{2sin^{2}\theta}

Hint: try writing out what you’re given and contrast it with what you are attempting to show. 🙂

Probability Question #2

JC Mathematics

This is question in relation to the birthday paradox we discussed earlier.

A room contains n randomly chosen people.

  1. Assume that a randomly chosen people is equally likely to have been born on any day of the week. The probability that the people in the room were all born on different days of the week is denoted by P.
    1. Find P when n=3.
    2. Show that P=\frac{120}{343} when n=4
  2. Assume now that a randomly chosen person is equally likely to have been born in any month of the year. Find the smallest value of n such that the probability that the people in the room were born in different months of a year is less than \frac {1}{2}.

Ans: \frac{30}{49}; 5

Statistics Question #1

JC Mathematics

This is a question taken from the 2008 H1 paper. I think it is a simple question that test students on their understanding of what standard deviation mean.

An examination is marked out of 100. It is taken by a large number of candidates. The mean mark, for all candidates is 72.1 and the standard deviation is 15.2. Give a reason why a normal distribution, with this mean and standard deviation, would not give a good approximation to the distribution of the marks.

N.B. We can effectively do Central Limit Theorem here, since n is sufficiently large. So why is it still not a good approximation?

Hint:

Credits: www.hhcpa.com
Credits: www.hhcpa.com