Tutor MathsGuru: Ask me for your burning Maths questions!

[Dharma]

Ler:

Maha Bohi SA2 math 2010, Paper 2


Questions 10. Joanne and Tomomi bought the same bag from the same shop when they went shopping together . Johanne spents 75% of her money on the bag and Tomomi spent 2/3 of hers. What percentage of their total sum of money was the cost of a bag if the girls had $100 left altogether? (round off your answer to the nearest tenth).

Joanne => ¾ of her money
Tomomi => 2/3 of her money

¾ of Joanne = 2/3 of Tomomi
6/8 of Joanne = 6/9 of Tomomi

% of cost of 1 bag compared to total sum of money = 6/17 x 100% = 35.3% (rounded to nearest tenth)

[Vanilla Cake]

Ler:

Question 30, After 40% of the boys....

OK, last question from you. BTW, are you a parent and is your child taking PSLE 2010 exam?My DD2 will be taking PSLE exam next year (2011).
Using unit and part method,

Before
Boys : Girls
u : p

Change
40% of the boys had been promoted and there were as many boys as girls left.........

60% of the boys left and no change for the girls.

After
Boys: Girls
0.6u : p

Given that p=0.6u, so percentage of the pupils in the club that were girls originally-> p/(u+p)x100%
= 0.6u/(u+0.6u)x100%
= 0.6u/1.6ux100%
= 3/8x100% = 37½%

Ler:

I try to do the math, but i got it wrong. Need???

You may wish to refer to http://prischoolmaths.blogspot.com/ and http://psle2010a.blogspot.com/ for many worked Maths examples by Observer.
Thks.

VC's mum

[Vanilla Cake]

liketoeat:

Hi, I need help with this question. No answer given.


A group of people went hunting. On average, 6 people hunted 5 rabbits, 15 people hunted 2 deer and 10 people hunted 1 buffalo. The total number of animals hunted was 6 more than the total number of people who went hunting. How many people went hunting?

Thank you.

There is answer given for this question. Your question is from http://www.orlesson.org/orp/09Ma/2009-P6-Math-CA1-Rulang.pdf.
Answer is 90 people and Tianzhu and Dharma had solved and posted the solution for this question before in KSP forum.

Tianzhu/Dharma - Pls repost the solution or direct the link for your solution.
Thks

VC's mum

Suggested solution from DD2 (P5)

\"->\" means hunted
6p-> 5r
30p-> 25r
15p-> 2d
30p-> 4d
10p-> 1b
30p-> 3b

Total: 30p, 32 animals
Difference=2
6÷2=3
3x30=90

Answer : 90 people

[Dharma]

Vanilla Cake:

liketoeat:

Hi, I need help with this question. No answer given.


A group of people went hunting. On average, 6 people hunted 5 rabbits, 15 people hunted 2 deer and 10 people hunted 1 buffalo. The total number of animals hunted was 6 more than the total number of people who went hunting. How many people went hunting?

Thank you.

There is answer given for this question. Your question is from http://www.orlesson.org/orp/09Ma/2009-P6-Math-CA1-Rulang.pdf.
Answer is 90 people and Tianzhu and Dharma had solved and posted the solution for this question before in KSP forum.

Tianzhu/Dharma - Pls repost the solution or direct the link for your solution.
Thks

Suggested solution from DD2 (P5)

\"->\" means hunted
6p-> 5r
30p-> 25r
15p-> 2d
30p-> 4d
10p-> 1b
30p-> 3b

Total: 30p, 32 animals
Difference=2
6÷2=3
3x30=90

Answer : 90 people

VC's mum

6 people => 5 rabbits
30 people => 25 rabbits
15 people => 2 deers
30 people => 4 deers
10 people => 1 buffalo
30 people => 3 buffaloes

A group of 30 people can hunt (25 + 4 + 3) = 32 animals
Difference between the no. of animals and people = 32 – 30 = 2

If the difference between the no. of animals and people = 6
No. of people who went hunting = 30 x 3 = 90

[tianzhu]

Vanilla Cake:

liketoeat:

Hi, I need help with this question. No answer given.


A group of people went hunting. On average, 6 people hunted 5 rabbits, 15 people hunted 2 deer and 10 people hunted 1 buffalo. The total number of animals hunted was 6 more than the total number of people who went hunting. How many people went hunting?

Thank you.

There is answer given for this question. Your question is from http://www.orlesson.org/orp/09Ma/2009-P6-Math-CA1-Rulang.pdf.
Answer is 90 people and Tianzhu and Dharma had solved and posted the solution for this question before in KSP forum.

Tianzhu/Dharma - Pls repost the solution or direct the link for your solution.Thks

Hi

Please refer to pg 38 of this thread.
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=6373&postdays=0&postorder=asc&start=370

Best wishes

[liketoeat]

Thank you very much to VC’s mum, DD2, Dharma and Tianzhu for your very quick response. Thank you again.

[kancheongmum]

Vanilla Cake:

Ler:

Maha Bohi SA2 math 2010, Paper 2

Many thanks for working out most of the solutions. I hv problem understanding but still trying to work out.

Quick one as I need to go to school,my mum will continue for the rest of your problem sums.

Ler:

Question 7 - In the diagram show above, O is the centre of the circle and OAB is a right angle triange. Given that AB is 10cm and the shaded parts of the circle add up to 133 m square. Find the area of the circle.

Pls post the diagram otherwise how can others be expected to solve without reading the diagram? I notice that you have changed your question as the original question stated \"AB is 10m\" and the diagram showed length of AB labeled as 10m. The original question also stated 133 m² and this is where the confusion comes.So, I change all metres and metres square to cm and cm² to make life simple.

Square of length of AO+Square of length of OB=10cmx10cm (http://en.wikipedia.org/wiki/Pythagorean_theorem)
but length of AO=length of OB (radii of circle)
Assume length of AO be n, so
n²+n²=100
2n²=100
n²=50

Area of triangle AOB = 1/2xABxOB=1/2xn²=1/2x50=25 cm²
Since area of the shaded part of the circle was given as 133 cm², area of the circle = (25+133) cm² = 158 cm².

Need to go now and mum will continue the rest.

Hi Vanilla Cake and Ler

This Maha Bodhi prelim 2010 paper is very challenging and have a few very interesting questions. Vanilla Cake I don't think Q7 should be solved this way. Pythagoras' Theorem is not taught in P6 yet.

Area of the triangle = 1/2 x base x height
if the radius is given you can use that to work out the answer. But not given in this case. So look at the triangle in another way. You will note that it is 1/4 of the square inscribe in the circle.

base = 10m height = 10m/2 = 5 m

area of the circle = 133m3 + (1/2 x 10x5) =158m3

I always tell my ds, don't think too complicated when doing the questions. They will not give you anything out of the syllabus. Make use of the basic concept taught.

[Dharma]

kancheongmum:

Vanilla Cake:

[quote=\"Ler\"]Maha Bohi SA2 math 2010, Paper 2

Many thanks for working out most of the solutions. I hv problem understanding but still trying to work out.

Quick one as I need to go to school,my mum will continue for the rest of your problem sums.

Ler:

Question 7 - In the diagram show above, O is the centre of the circle and OAB is a right angle triange. Given that AB is 10cm and the shaded parts of the circle add up to 133 m square. Find the area of the circle.

Pls post the diagram otherwise how can others be expected to solve without reading the diagram? I notice that you have changed your question as the original question stated \"AB is 10m\" and the diagram showed length of AB labeled as 10m. The original question also stated 133 m² and this is where the confusion comes.So, I change all metres and metres square to cm and cm² to make life simple.

Square of length of AO+Square of length of OB=10cmx10cm (http://en.wikipedia.org/wiki/Pythagorean_theorem)
but length of AO=length of OB (radii of circle)
Assume length of AO be n, so
n²+n²=100
2n²=100
n²=50

Area of triangle AOB = 1/2xABxOB=1/2xn²=1/2x50=25 cm²
Since area of the shaded part of the circle was given as 133 cm², area of the circle = (25+133) cm² = 158 cm².

Need to go now and mum will continue the rest.

Hi Vanilla Cake and Ler

This Maha Bodhi prelim 2010 paper is very challenging and have a few very interesting questions. Vanilla Cake I don't think Q7 should be solved this way. Pythagoras' Theorem is not taught in P6 yet.

Area of the triangle = 1/2 x base x height
if the radius is given you can use that to work out the answer. But not given in this case. So look at the triangle in another way. You will note that it is 1/4 of the square inscribe in the circle.

base = 10m height = 10m/2 = 5 m

area of the circle = 133m3 + (1/2 x 10x5) =158m3

I always tell my ds, don't think too complicated when doing the questions. They will not give you anything out of the syllabus. Make use of the basic concept taught.[/quote]kancheong mum is correct. Nothing wrong to use Pythagoras but it is not necessary to use it.
Always try to keep things simple and that will make life easy for us.

[Vanilla Cake]

kancheongmum:

Hi Vanilla Cake and Ler


This Maha Bodhi prelim 2010 paper is very challenging and have a few very interesting questions. Vanilla Cake I don't think Q7 should be solved this way. Pythagoras' Theorem is not taught in P6 yet.

Area of the triangle = 1/2 x base x height
if the radius is given you can use that to work out the answer. But not given in this case. So look at the triangle in another way. You will note that it is 1/4 of the square inscribe in the circle.

base = 10m height = 10m/2 = 5 m

area of the circle = 133m3 + (1/2 x 10x5) =158m3

Hi kancheongmum,

Thks for your highlight. I think that Vanilla Cake (Sec 2) must have forgotten what are taught at P6 level and apply the right method to solve this question - Maha Bodhi School P6 Prelim Paper P2 Q7. :oops:

This type of question is common and has appeared quite a number of times in past years' school exams.If necessary, extend the triangle AOB to form a square within the circle for better visualisation.
AB is 10 m which is the side of the square ie the base of the triangle AOB.To find the height which is 1/2 of the side of the square, just divide 10 by 2 = 5 m
Area of triangle = 1/2x10x5=25 m²

BTW, I think that there is a typo error in this question. All units should be all in cm instead of m.

Hi Dharma,
Thks for your understanding and will let Vanilla Cake knows about this.

VC's mum

[Ler]

Vanilla Cake:

kancheongmum:

Hi Vanilla Cake and Ler


This Maha Bodhi prelim 2010 paper is very challenging and have a few very interesting questions. Vanilla Cake I don't think Q7 should be solved this way. Pythagoras' Theorem is not taught in P6 yet.

Area of the triangle = 1/2 x base x height
if the radius is given you can use that to work out the answer. But not given in this case. So look at the triangle in another way. You will note that it is 1/4 of the square inscribe in the circle.

base = 10m height = 10m/2 = 5 m

area of the circle = 133m3 + (1/2 x 10x5) =158m3

Thks for your highlight. I think that Vanilla Cake (Sec 2) must have forgotten what are taught at P6 level and apply the right method to solve this question - Maha Bodhi School P6 Prelim Paper P2 Q7. :oops:

This type of question is common and has appeared quite a number of times in past years' school exams.If necessary, extend the triangle AOB to form a square within the circle for better visualisation.
AB is 10 m which is the side of the square ie the base of the triangle AOB.To find the height which is 1/2 of the side of the square, just divide 10 by 2 = 5 m
Area of triangle = 1/2x10x5=25 m²

BTW, I think that there is a typo error in this question. All units should be all in cm instead of m.

Hi Dharma,
Thks for your understanding and will let Vanilla Cake knows about this.

VC's mum

Hi VC, VC mum, Kancheongmum, and Dharma

My kid is taking PSLE this year. She is so weak in math, so am I. When I have given her to do the paper, she almost done every sum wrong. Very sad , cant help her. I personally feel that Mohd Bohi sat the paper very challenging, i cant do at all. Very worried that she will fail the math paper. I happen came across this website yesterday, i was so happy. Sincerely, I like to thank you all for helping me to solve these questions. I am still studying how to explain to her. I might have more questions to ask for help. ....Thanks. (Very sorry , a lot of typing error, next time, i will type out the questions.).

Ler