Tutor MathsGuru: Ask me for your burning Maths questions!

[Maths Monster]

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[ttyh]

I’m sorry to post this but I’ve absolutely no clue on this P2 Math question.

It is a picture showing a mother cat watching 2 kitten playing a ball of wool.
Question : Write a multiplication story for this picture…

Pls help me!!!

[firebird]

Dear Vanilla cake


Good morning.

Very sorry for the incomplete question. Thank you detailed working.

With best regards
firebird

[Chrislee]

Good morning, please help me with the following sum.


I have 2 pairs of hamsters. Each pair with give birth to 2 pairs of hamsters in every 1 day. How many hamsters will there be 4 days later?
Answer: 324

Thanks a lot!

[CoffeeCat]

Chrislee:

Good morning, please help me with the following sum.


I have 2 pairs of hamsters. Each pair with give birth to 2 pairs of hamsters in every 1 day. How many hamsters will there be 4 days later?
Answer: 324

Thanks a lot!

Before Day 1: 2
After Day 1 : 4 (new born) + 2 = 6
After Day 2 : 12 (new born) + 6 = 18
After Day 3 : 36 (new born) + 18 = 54
After Day 4 : 108 (new born) + 54 = 162 pairs

162 pairs = 324 hamsters.
If you look at the pattern, 2*3 = 6, 6*3 = 18, 18*3= 54, 54*3=162

[Herbie]

I have one qn.

Kent had a collection of 160 toys. 11/20 of them were soft toys.
After he gave away some of the soft toys away, the no. of soft toys left made up of 2/5 of the remaining collection. How many soft toys did he give away?

Can help to provide the solution?
Tx

[CoffeeCat]

Herbie:

I have one qn.

Kent had a collection of 160 toys. 11/20 of them were soft toys.
After he gave away some of the soft toys away, the no. of soft toys left made up of 2/5 of the remaining collection. How many soft toys did he give away?

Can help to provide the solution?
Tx

At first, 88 soft toys, 72 non soft toys.
After giving some away, the 72 non soft toys make up 3/5 of remaining.
Therefore number of soft toys left is 48.
Number of soft toys given away = 40

[Sun_2010]

Hi,


DD is struck at this question . I tried it too- cant get it.
Can some one help?
Question 6 of the sample paper for APMOS
http://www.hci.sg/aphelion/apmops/sample_questions/Invitation%20Round%20Sample%20Questions.pdf

Thanks

[CoffeeCat]

Sun_2010:

Hi,


DD is struck at this question . I tried it too- cant get it.
Can some one help?
Question 6 of the sample paper for APMOS
http://www.hci.sg/aphelion/apmops/sample_questions/Invitation%20Round%20Sample%20Questions.pdf

Thanks

What a coincidence, I was explaining that question to a student at the time when you post this.
This happens to belongs to a class of geometry questions which solutions requires the construction of an equilateral triangle.
At first glance, the clue AB=DC look useless because the 2 lines are far away and usually we think of isoceles triangle when we heard of 2 equal lines in angles.
Imagine that there is a perpendicular line passing through the midpoint of BC, then reflect the triangle BDC about that line, so that there is now a new point D' in the interior of the triangle such that angle BD'C = angle CDB, and angle D'BC = angle DCB = 20 deg.
Notice CD = BD' = AB ( ABD' is isoceles) and since angle ABC=80 deg, angle ABD' = 80-20=60 deg, thus ABD' is an equilateral triangle.
We want to find angle BDC, which is equivalent to finding angle CD'B.
The next step will be hard to see if your equilateral triangle is not drawn properly, because the point D' is actually directly midway between A and B. angle BD'C = angle AD'C = (360-60)/2 = 150deg.

[Sun_2010]

CoffeeCat:

Sun_2010:

Hi,


DD is struck at this question . I tried it too- cant get it.
Can some one help?
Question 6 of the sample paper for APMOS
http://www.hci.sg/aphelion/apmops/sample_questions/Invitation%20Round%20Sample%20Questions.pdf

Thanks

What a coincidence, I was explaining that question to a student at the time when you post this.
This happens to belongs to a class of geometry questions which solutions requires the construction of an equilateral triangle.
At first glance, the clue AB=DC look useless because the 2 lines are far away and usually we think of isoceles triangle when we heard of 2 equal lines in angles.
Imagine that there is a perpendicular line passing through the midpoint of BC, then reflect the triangle BDC about that line, so that there is now a new point D' in the interior of the triangle such that angle BD'C = angle CDB, and angle D'BC = angle DCB = 20 deg.
Notice CD = BD' = AB ( ABD' is isoceles) and since angle ABC=80 deg, angle ABD' = 80-20=60 deg, thus ABD' is an equilateral triangle.
We want to find angle BDC, which is equivalent to finding angle CD'B.
The next step will be hard to see if your equilateral triangle is not drawn properly, because the point D' is actually directly midway between A and B. angle BD'C = angle AD'C = (360-60)/2 = 150deg.

Thanku Thank u , Coffee cat.
I did try shifting the trangle etc , but never thought of reflecting it. Frankly it took me abt 10 full min to understand a solution u have so neatly explained. So not only is a problem solved , I have learnt a new technique. So really :salute: to u.

Ah now i can sleep in peace