Following the recent discussion on what constitutes a ‘premium’ question (i.e. questions that reward you with lovely marshmallows during joss sticks sessions at The Temple), here’s one for the sake of an example:

In the diagram, ABCD is a square of 5cm. Find the area of the shaded region.

Such a beauty. Simplicity at its finest. Strictly speaking, even an (smart) E-Maths student should be able to do this.

However, just like the tendancy of a Zen Master to speak in profound riddles, questions of such minimalistic nature are often hidden death traps meant to snare unsuspecting students.

Unfortunately in exams where time is at a premium (no pun intended), many students still choose to be overwhelmed by long questions (i.e. those with parts (a) to (z)) and, like sailors drawn to sirens, couldn’t resist the allure of ‘short and simple’ questions – only to get stuck reflecting on this one question till the clock runs out.

Remember: Long questions are meant to help you. (Too) Short questions might kill you.

To illustrate further, would you rather see this in your exam?

Why did the chicken cross the road? [12 marks]

Or this?

A chicken is being chased by a cat (top speed 3.5km/h). With the cat just 10m behind, the chick stops in front of a road and sees a Ferrari Enzo (top speed 350km/h) speeding towards its direction from 1km away.

Calculate the time the cat takes to reach the chick, assuming chick is still deliberating at the roadside. [5 marks]

Calculate the time the Ferrari Enzo takes to reach the crossing point at top speed. [5 marks]

Hence, explain why did the chicken cross the road. [2 marks]

WARNING: Kids, premium questions are dangerous! If you encounter one in your exam, skip and do the rest first*, and come back only when you have the time. Time (and not your ego) is of the essence!

Best attempted at home or at Miss Loi’s Temple in anticipation of marshmallows. Not in your exam!

*applicable only if you’re not a Math Genius or cyborg

distance traveled by cat = 10m at a speed of 3.5kph (or you can say at 35/36 meter per second) Therefore, time taken by the cat to reach the chick is 10/(35/36)=10.28s

Similarly, distance covered by the ferrari enzo car at 350kph(3500/36 meter per second)=1km=1000m Implies, time taken by the car to reach the crossing point=10.28s again !

That implies that the chick crossed the road so that when the car is at the crossing point the cat would be forced to stay on the other side of the road be'cuz at that very instant the car would be crossing right in front of the cat.

This problem was fairly easy, except the one thing that I couldn't understand... HOW DID THE CHICK DO IT... Must have got some superpowers to judge the distances ! and do all the calculations while its life is jus on the tip off his head ready to jump-off anytime as the cat is chasing her ... Hha, Thats what got me interested in answering this one.

Good going Miss Loi (I wonder why you like to converse as a third person about yourself ! ) Keep educating the students, make a better world tomorrow.

Let the point of intersection of the two circular parts be X.

XC=XD=radius=5 cm

Therefore, triangle XCD is an equilateral triangle.

Yes this is the crucial part! Students need to know how to 'partition' the area first. In this case it's the triangle plus the two sectors. Next students need to know that this triangle is equilateral, either by appreciating that XC and XD are radii of the quadrants (like you mentioned) or by simply appreciating that the angle DXC subtended to the intersection point of the two quadrants is 60^{o}.

area of triangle XCD=0.5*5*5*sin 60=10.83

Since the sectors of a circle are identical, total area of two sectors=

Area of sector = , reproduced for the benefit of readers here

Shaded area=

Minor correction: unit of area should be cm^{2} instead of units^{3} as you've written. Any reason why you wrote units^{3}?

This question can also be solved via integration, using the equation of the circle as x^{2}+y^{2}=5^{2} Integrate y with respect to x, with lower limit 0 and upper limit 2.5, then multiply by two.

Substract this from 25 cm^{2} to get the final answer.

Equation of circle is not in the O-Level syllabus this year. But then again you might be in Sec Three. But if you are indeed, you shouldn't have learnt integration yet. So Miss Loi concludes that you're overaged!!!

* Please refer to the relevant Ten-Year Series for the questions of these suggested solutions.

About Miss Loi

Miss Loi is a full-time private tutor in Singapore specializing in O-Level Maths tuition. Her life’s calling is to eradicate the terrifying LMBFH Syndrome off the face of this planet. For over years she has been a savior to countless students … [read more]

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We were fortunate to have Miss Loi take my son under her wings in June this year and in a matter of four short months, his grade jumped to an A2 … [read more]

## 9 Comments

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distance traveled by cat = 10m at a speed of 3.5kph (or you can say at 35/36 meter per second)

Therefore, time taken by the cat to reach the chick is 10/(35/36)=10.28s

Similarly, distance covered by the ferrari enzo car at 350kph(3500/36 meter per second)=1km=1000m

Implies, time taken by the car to reach the crossing point=10.28s again !

That implies that the chick crossed the road so that when the car is at the crossing point the cat would be forced to stay on the other side of the road be'cuz at that very instant the car would be crossing right in front of the cat.

This problem was fairly easy, except the one thing that I couldn't understand... HOW DID THE CHICK DO IT... Must have got some superpowers to judge the distances ! and do all the calculations while its life is jus on the tip off his head ready to jump-off anytime as the cat is chasing her ... Hha, Thats what got me interested in answering this one.

Good going Miss Loi (I wonder why you like to converse as a third person about yourself ! )

Keep educating the students, make a better world tomorrow.

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Abhinav, you're supposed to be doing the question on the shaded area - not join the chick on the roadside admiring Ferrari cars! lol

Having said that, your reasons (for the chick's decision) alone are enough to get you a PhD!

P.S. Chicks in Singapore chicks are all blessed with superpowers! hahaha

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1.08 units^3 [3 s.f.]

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Impressive young jedi.

But show working you must, or get a marshmallow you won't. 🙂

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Let the point of intersection of the two circular parts be X.

XC=XD=radius=5 cm

Therefore, triangle XCD is an equilateral triangle.

Yes this is the crucial part! Students need to know how to 'partition' the area first. In this case it's the triangle plus the two sectors. Next students need to know that this triangle is equilateral, either by appreciating that XC and XD are radii of the quadrants (like you mentioned) or by simply appreciating that the angle DXC subtended to the intersection point of the two quadrants is 60

^{o}.area of triangle XCD=0.5*5*5*sin 60=10.83

Since the sectors of a circle are identical, total area of two sectors=

Area of sector = , reproduced for the benefit of readers here

Shaded area=

Minor correction: unit of area should be cm

^{2}instead of units^{3}as you've written. Any reason why you wrote units^{3}?This question can also be solved via integration, using the equation of the circle as

x^{2}+y^{2}=5^{2}Integrate y with respect to x, with lower limit 0 and upper limit 2.5, then multiply by two.

Substract this from 25 cm

^{2}to get the final answer.Equation of circle is not in the O-Level syllabus this year. But then again you might be in Sec Three. But if you are indeed, you shouldn't have learnt integration yet. So Miss Loi concludes that you're overaged!!!

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Jye Yee, tidied up your workings for readability and added some comments in red.

Great work but Miss Loi thinks you're overaged! :S

Now open your mouth and standby for marshmallows ...

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I'm a J2 student... Shame on me. Haha.

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Why did the chicken cross the road [12]

Answer: Because the chicken wanted to get to the other side of the road (12 marks)

anyway, nice short question...and a nice website

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