Tutor MathsGuru: Ask me for your burning Maths questions!

[MOE Hater]

Vanilla Cake:

MOE Hater:

That confused me as well. I study in Henry Park but my teacher got her hands on the RGPS SA1 2010 paper. I completely did not understand when you said questions 13b, 14 and 15 were about circles when none were. As a matter of fact, there was no question 13b

Hi MOE Hater,
What I mean is that the marble question (answer=350) was from RGPS P6 SA1 2009 Paper 2 Q18. Do you have copies of the RGPS SA1 2010 paper? Perhaps you may wish to post Q13b, 14 and 15 on behalf of abc_parent and provide solutions for them since you mentioned that your teacher got her hands on these papers.Hope this will clear the confusion.

I don't think you got what i meant. Before mentioning the marbles question, she asked about question 13b, 14 and 15 which were said to be circles questions. In RGPS 2010, these questions were not circles question and there was no question 13b in the 2010 paper. I think the question 13b, 14 and 15 were from the 2009 paper.

[Vanilla Cake]

MOE Hater:

I don't think you got what i meant. Before mentioning the marbles question, she asked about question 13b, 14 and 15 which were said to be circles questions. In RGPS 2010, these questions were not circles question and there was no question 13b in the 2010 paper. I think the question 13b, 14 and 15 were from the 2009 paper.

OIC, maybe you could consider posting Q13, Q14 and Q15 from RGPS 2010 P6 SA1 Maths paper 2 and share your solutions if you have time for everyone including abc_parent to see.This will help to clear the confusion for everyone and show that these questions were not circles question and there was no question 13b in the 2010 paper. If you don't mind, pls also post some interesting questions from Henry Park 2010 P6 SA1 Maths paper 2.

[sachiko]

Hi MahtsGuru


I would appreciate if you could help me with the following qns:

3 tins, A, B and C, contained a total of 240 cookies. Some cookies from A were transferred to B and the number of cookies in B was doubled. Then some cookies from B were transferred to C and the number of cookies in C was doubled. As a result of this, there was an equal number of cookies in each tin. How many cookies were in each tin at first?

TIA.

Sachiko

[MOE Hater]

sachiko:

Hi MahtsGuru


I would appreciate if you could help me with the following qns:

3 tins, A, B and C, contained a total of 240 cookies. Some cookies from A were transferred to B and the number of cookies in B was doubled. Then some cookies from B were transferred to C and the number of cookies in C was doubled. As a result of this, there was an equal number of cookies in each tin. How many cookies were in each tin at first?

TIA.

Sachiko

As the exchange of cookies took place between the 3 tins, the total number of cookies would always be 240. As the no. of cookies in each tin in the end were the same, we can calulate the no. of cookies in each tin in the end.

The no. of cookies in each tin in the end = 240/3 = 80

As the no of cookies in Tin C were doubled before it became 80, we can calculate the no. of cookies in Tin C at first. From this, we can also tabulate the no. of cookies in tin B before the original no. was doubled.

Cookies added to Tin C = Cookies in tin C at first = 80/2 = 40

Cookies in Tin B after original no. doubled = 80 + 40 = 120

It is plain sailing from here. We can calculate the original no. of cookies in Tin B, and with the total no. of cookies and the original no of cookies in each Tin B and C, we can calculate the no. of cookies orginally in Tin A.

Cookies in Tin B at first = 120/2 = 60

Cookies in Tin A at first = 240 - 40 - 60 = 140

Alternatively, we can see that the no. of cookies in tin B at first is also the no. of cookies removed from Tin A. We also know the no. of cookies left in Tin A, so by adding both values we can derive the no. of cookies in Tin A at first

Cookies in Tin A at first = 80 + 60 = 140

Hope you can understand my rather wordy solution. Algebra can be used but to some children it is quite confusing.

[RedRedWine]

[quote="Dharma"][quote="RedRedWine"]Hi Maths Guru


I have 2 questions from Catholic High Prelim exan 2. Can you help tp solve?

1) At a swimmimg pool, the number of boys to the numbers of girls was 4:3. After 3/8 of the boys left and 12 more girls joined the group, there were 2/3 as many boys as girls. How many children were there at the swimming pool first?

TIA

Thank you MOE Hater, Dharma and Muffins.

[Muffins]

RedRedWine:

Hi Maths Guru


I have 2 questions from Catholic High Prelim exan 2. Can you help tp solve?

1) At a swimmimg pool, the number of boys to the numbers of girls was 4:3. After 3/8 of the boys left and 12 more girls joined the group, there were 2/3 as many boys as girls. How many children were there at the swimming pool first?

TIA

Thank you MOE Hater, Dharma and Muffins.

No prob!!!

[KSP]

Just wonder where Mathsguru is. Have not seen her answering questtion for a long time.

[starlight1968sg]

RedRedWine:

I have 2 questions from Catholic High Prelim exan 2. Can you help tp solve?


1) At a swimmimg pool, the number of boys to the numbers of girls was 4:3. After 3/8 of the boys left and 12 more girls joined the group, there were 2/3 as many boys as girls. How many children were there at the swimming pool first?

TIA

Thank you MOE Hater, Dharma and Muffins.

Is the ans: 64 boys and 48 girls ie 112 children?

[Brenda10]

RedRedWine:

Dharma:

[quote=\"RedRedWine\"]Hi Maths Guru


I have 2 questions from Catholic High Prelim exan 2. Can you help tp solve?

1) At a swimmimg pool, the number of boys to the numbers of girls was 4:3. After 3/8 of the boys left and 12 more girls joined the group, there were 2/3 as many boys as girls. How many children were there at the swimming pool first?

TIA

Thank you MOE Hater, Dharma and Muffins.

[/quote]Hi, my girl would like to give a try. Hopefully her steps and answer are correct.



\t\tB : G\t\t\t\t
\t\t4 : 3\t(At first)\t\t\t
\t\tx 2 : x 2\t(to get the denominator of 8 )
\t\t8 : 6\t\t\t\t
\t-3u : +12\t\t\t\t
\t\t5u : ?\t(This becomes 5u :7.5u aft 10u:15u)
\t\tx2 : x2\t\t\t\t
\t 10u: 15u\t(At the end 2:3)\t\t\t
\t\t\t\t\t\t\t

\t\t7.5u -\t6u\t= 1.5u\t
\t\t12 /\t1.5\t= 8\t

\tTherefore\t8u x\t8\t= 64\t(boy)
\t\t6u x\t8\t= 48\t(girl)

\t\t64 + 48 = 112\t


Answer: There were 112 children at first.

Thank you.

[Muffins]

Hi Brenda10, your daughter just managed to summarise my workings into a few steps! :shock: How old is your DD???