I recently wrote the second round of the South African Mathematics Olympiad Round 2. Only the top 100 students in the country progress to the final round.

Could you be ranked amongst the best? PROVE IT!

All answers are integer values between 000 and 999.

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TopNewestPROBLEM 20(hardest problem of the paper)There are 300 white boxes and

nred boxes in storage. Each box contains the same number of soccer balls. The total number of soccer balls in all of the boxes isn^2 + 290n - 2490. Determinen. – Mark Mottian · 3 years, 4 months agoLog in to reply

Problem 14

ABCD is a quadrilateral with parallel sides AB and CD. AB is longer than CD. If angle D is twice angle B, AD = 5 and CD = 3, then find the length of AB. – Mark Mottian · 3 years, 4 months ago

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– Xuming Liang · 3 years, 4 months ago

Extend \(CD\) to meet \(AD\) at \(E\), by the parallels, \(\angle CED=\angle D-\angle ECD=\angle D-\angle B=\angle B=\angle ECD\). Hence \(ED=DC=3,AB=AE=AD+DE=5+3=8\)Log in to reply

Problem 19

A and B ride at constant speeds in opposite directions around a circular track, starting at diametrically opposite points. If they start at the same time, meet one another for the first time after B has ridden 100 metres, and meet a second time 60 metres before A completes one lap, what is the circumference of the track in metres? – Mark Mottian · 3 years, 4 months ago

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\( R\omega_{2} t = 100 \dots (1)\)

\( \omega_{1} t + \omega_{2} t = \pi \dots (2)\)

Using \((1)\) and \((2)\), we can write

\( \displaystyle \frac{\pi R \omega_{2}}{\omega_{1} + \omega_{2}} = 100 \Rightarrow \frac{\pi R}{100} = 1 + \frac{\omega_{1}}{\omega_{2}} \dots (5) \)

Similarly, for second encounter, we can write,

\( R\omega_{1} t_{1} = 2\pi R - 60 \dots (3) \)

\( R\omega_{2} t_{1} = \pi R + 60 \dots(4) \)

Using \((3) \text{and} (4)\), we can write,

\(\displaystyle \frac{\omega_{1}}{\omega_{2}} = \frac{2\pi R - 60}{\pi R + 60} \dots (6)\)

Substituting eqn. \((6)\) in eqn. \( (5)\), we obtain,

\(\displaystyle \frac{\pi R}{100} = 1 + \frac{2\pi R - 60}{\pi R + 60} \Rightarrow \pi R = 240\)

Hence, the circumference of the circular path is \(480\). – Sudeep Salgia · 3 years, 4 months ago

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– Mark Mottian · 3 years, 4 months ago

A very unique approach!Log in to reply

Problem 11

If a die is rolled once, the probability that a 4 shows is (1/6). If three dice are rolled the probability that a 4 shows up exactly once is

x / 72. What is the value ofx? – Mark Mottian · 3 years, 4 months agoLog in to reply

The probability of a 4 doesn't show up is: \(\frac{5}{6}\)

Hence, the probability is

\(\displaystyle 3\big (\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6} \big )\)

We need to times three because there are three different case on which die should be showing 4. I'll try the other problems tomorrow, it's quite late already here.... – Christopher Boo · 3 years, 4 months ago

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– Mark Mottian · 3 years, 4 months ago

My exact reasoning! At the end of the day, x = 25.Log in to reply

Thanks for sharing these problems! Why don't you make it as a problem set instead of sharing a note? – Christopher Boo · 3 years, 4 months ago

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– Mark Mottian · 3 years, 4 months ago

Hi Christopher! I'll definitely share the other problems in a problem set. I've posted these questions here because I'm not sure if my answers are correct. I was hoping that I could compare my answers with your solutions so that I can achieve some peace of mind. Please help!Log in to reply

– Calvin Lin Staff · 3 years, 4 months ago

I think that a set of notes would likely make more sense (esp if you don't know what the answer is). This way, you can present several problems, and it would be easy for others to find / discuss.Log in to reply

– Christopher Boo · 3 years, 4 months ago

Oh ok!Log in to reply

@Christopher Boo – Mardokay Mosazghi · 3 years, 4 months ago

I agreeLog in to reply