Tutor MathsGuru: Ask me for your burning Maths questions!

[mathsguru]

Herbie:

Hi, Maths Guru and all


Can help to solve the qn below?

Amy and Tommy each have some money. If Amy spends $50 per day and Tommy spends $60 per day, Amy would still have $500 left when Tommy has spent all his money.
If Amy spends $60 per day and Tommy spends $50 per day, Amy would still have $280 left when Tommy spent al his money. How much money does Tommy have?

Tx

Hi Herbie,

Here's my solution.

Cheers,
MathsGuru

http://www.postimage.org/image.php?v=TsDKzgr

[mathsguru]

ck123:

Hi all


I need help with the model for this question.

The number of marbles in Box A, Box B and Box C was 195.
John added 60 marbles to those in Box A, doubled the number of marbles in Box B and halved the number of marbles in Box C.
The ratio of the number of marbles becomes 4:1:2.
What is the total number of marbles in the three boxes now?

Thanks

Hi Ck123,

Here's my solution. Hope it helps!


MathsGuru

http://www.postimage.org/image.php?v=TsDKQJr

[mathsguru]

Vanilla Cake:

Hi Mathsguru,

From http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=6113&postdays=0&postorder=asc&start=100 and his http://chillycrab.webs.com/, is it possible for you to compile all your questions and solutions from this thread into a blog/website for others to learn from you?

Hi Vanilla Cake,

Thanks for your suggestion! I've thought of this actually...just that it can be quite time-consuming to start and maintain it and I'm not exactly very IT-savvy in this area (I don't even have a blog! :)).

Might need more time to prepare my own website...hehe

Nevertheless, if it ever realises, I'll keep all of you informed!!

Cheers,
MathsGuru

[mathsguru]

SoWoW:

Hi,


Please help to clarify!

Mrs Wong wants to pack 100 buns in boxes for her son to take to his class party. Each box can hold up to a max. of 8 buns. What is the minimum no. of boxes she will need to pack all the buns?

Should the answer be 12 or 13?
If the answer is 12, can help to explain why 12?

Thanks alot

Hi SoWoW,

Like what Dharma mentioned, it should be 13. 100/8 = 12 R 4.

In this case, we need 12 + 1 more box to contain the remainder.

Hmm...maybe it's a coincidence. I remember another member, Chanpy, also had a similar question and the answer provided is also 1 less than what it should be.

Seriously hope it is a typo mistake, else it is very misleading to students who self-learn...

Regards,
MathsGuru

[mathsguru]

Vanilla Cake:

Hi Mathsguru,

Could you pls help to post your approach to solve this 5-mark question from http://www.orlesson.org/orp/09Ma/2009-Math-SA1-AiTong.pdf?
There are workings for the solution but both of us (my P5 younger sister and myself)cannot understand them :oops:.Pls refer to page 27 of the PDF for the workings.

Devi had 26 more $5 notes than $10 notes. After paying $480 for a camera with some of the $5 notes, she had 6 times as many $10 notes as $5 notes.
a) How many $5 notes did Devi have at first?
b) How much money did Devi have left?

Could the model solution be applicable for such problem sum?
Thanks for your help.

Hi Vanilla Cake,

Here's my solution.

Cheers,
MathsGuru

http://www.postimage.org/image.php?v=aVe9mL9

[Herbie]

thanks vanilla cake and maths guru for the providing the solutions.

[ck123]

Hi all,


Thanks for the model.

[Vanilla Cake]

Herbie:

thanks vanilla cake and maths guru for the providing the solutions.

Hi Herbie, you are most welcome.


Hi adhdadhd,
Thanks for your solutions and no worries about the small images.:D

Hi Mathsguru,
Thanks for your helpful and useful explanations.1 more question [5 marks] from http://www.orlesson.org/orp/09Ma/2009-Math-SA1-SCGS.pdf which needs your model solution. Today, I checked through your earlier solutions in the 85+ pages of this thread but could not find something similar to this one.

A coin box contained some twenty-cent and fifty-cent coins in the ratio 4 : 3. After 20 twenty-cent coins were taken out to exchange for fifty-cents coins of the same value and put back in the box, the ratio of the number of twenty-cents coins to the number of fifty-cents coins became 7 : 11. Find the sum of money in the box.

Thanks again for your kindness in helping us.
When you have time, it's good to have your own blog to group your Maths solutions by topics. This will help us to search and look out for similar questions with your worked solutions before contemplating to post our new questions. Just my 2 cents worth.

[Guan Hui]

20 x 20c>>> 8 x 50c


*note amount of money is always the same
for the 1st ratio 4:3
4x0.2+3x0.5=2.3(1 set of 1st ratio)
2nd ratio 7:11
7x 0.2 + 11x 0.5= 6.9(1 set of 2nd ratio)
since 1 set of the 1st ratio is 2.3 and 1 set of 2nd ratio is 6.9
1st ratio x 3
4:3=12:9
12:9 and 7:11 of 20c and 50c give the same amount of money.
But 12:9 must be able to give 20 20c and 8 50 c to become 7: 11

Using the 20c side
12-7=5(the 20c side of the ratio)
with the existing ratio, only able to give 5 20c coins.
we need 20… so,
20/5=4

Using the 50c side
11-9=2(the 50c side of the ratio)
with the existing ratio, only able to receive 2 50c coins.
we need 8… so,
8/2=4

(use either side is ok)
Using this we can know the sum of money = 6.9 x 4 = $27.60

Check
12:9=48:36(x4)
7:11=28:44(x4)
48-20=28
36+8=44
48x0.2+36x0.5=9.6+ 18 =$27.60
28 x 0.2+ 44 x 0.5= 5.6 + 22 = $27.60

[girlmum]

Hi Mathguru,


Pls help to draw the model for this P3 maths question:

1) Mrs Chan had 20 pencils. She gave each pupil 4 pencils.
a) How many pupils did she give the pencils to? Ans : 5
b) If each child broke 1 pencil, how many pencils were not
broken? Ans : 15

My DD2 knows how to do the number sentences but not the model. Pls assist. :?: Thanks.