Tutor MathsGuru: Ask me for your burning Maths questions!

[kancheongmum]

ganchiong:

kancheongmum:

[quote=\"ksi\"]
container = 6x11x14 = 924
cuboid = 2x2x3 = 12
max. Number of cuboids = 924/12 = 77

Hi ksi and ganchiong
I don't think the answer is 77. You cannot divide the volume of the container by the volume of the cuboid. You have to approach this type of question this way. Minimize the remainder. Divide each dimension by each dimension. Try not to have remainder.

6/3 -- 2 cuboids
11/2 -- 5 cuboids remain 1 cm
14/2 -- 7 cuboids

Max no. 2 x 5 x 7 = 70 cuboids

Thank you all for your reply.

I am really confused... :? now.
Actually i tried to draw it out and the answer is 77. Any Maths Guru out there can clarify?[/quote]Hi ganchiong very very sorry, now I remember this question in one of the 2009 papers multiple choice. Yes the answer is 77. But still is not right to divide the volume. One of the tricky question. The answer depend on how you place the cuboids

6/2 --- 3 cuboids
11/3 --- 3 cuboids remainder 2cm*
14/2 --- 7 cuboids

3 x 3 x 7 = 63 cuboids

the remainder 2 cm part you still can get

2/2 ---1 cuboid
6/3 ---2 cuboids
14/2 ---7 cuboids

1 x 2 x 7 = 14 cuboids

Max no. 63 + 14 = 77 cuboids

sorry hope other gurus can give a better presentation of the answer



[/quote]

[Vanilla Cake]

Herbie:

Hi! Thanks Vc! I hv another qn. A fruitseller had some oranges, pears n durians. After selling some of them, there were 6 pears for every 5 oranges left and 7 oranges for every 3 durians left. He sold 168 oranges. He had 456 pears and oranges left. The no of pears sold was the Same as the no of durians sold. How many oranges had he at first? If the no of pears was twice the ö ofDurian at firSt, what the total no of pears and durians sold

Hi Herbie,
Could you pls advise the source of this question as I have problem solving using the given figure as \"456\"? Is this question from one of the GEP schools?Let's say this figure \"456\" is allowed to be changed to \"462\" and here's the solution:

Left
P : O
6 : 5
Multiply by 7
42 : 35

O : D
7 : 3
Multiply by 5
35 : 15

P : O : D
42 : 35 : 15

P+O = 462 instead of 456
(42+35)u = 77u
77u = 462
u = 6

Orange left: 35u = 35x6=210
Orange at first = 210+168=378
At first, the fruitseller had 378 oranges.

At first
P : 2n
O: 378
n

The difference between P and D must be the same since the number of pears sold = number of durians sold.

2n-n = 42u-15u
2n-n = 42x6-15x6
n = 162

At first
P: 2n = 2x162 = 324
n = 162

Left
P: 42u = 42x6 = 252
15u = 15x6 = 90

324-252 = 72
or
162-90 = 72

72x2 = 144
Total number of pears and durians sold = 144

The above explanation is long-winded in order to make sure that you are able to understand. In actual exam, you can skip some of these working steps. Let's hope Uncle Dharma and Uncle Tianzhu will come up with a better and simple explanation than mine.
Thks.

[corneyAmber]

kancheongmum:

Hi ganchiong very very sorry, now I remember this question in one of the 2009 papers multiple choice. Yes the answer is 77. But still is not right to divide the volume. One of the tricky question. The answer depend on how you place the cuboids


6/2 --- 3 cuboids
11/3 --- 3 cuboids remainder 2cm*
14/2 --- 7 cuboids

3 x 3 x 7 = 63 cuboids

the remainder 2 cm part you still can get

2/2 ---1 cuboid
6/3 ---2 cuboids
14/2 ---7 cuboids

1 x 2 x 7 = 14 cuboids

Max no. 63 + 14 = 77 cuboids

sorry hope other gurus can give a better presentation of the answer

It is very strange to set this question as MCQ as no working is required to be shown and most ppl may just choose 77. :? So what are they testing cos either way you calculate arrives at 77?

[Vanilla Cake]

ksi:

kancheongmum:

Hi ganchiong very very sorry, now I remember this question in one of the 2009 papers multiple choice. Yes the answer is 77. But still is not right to divide the volume. One of the tricky question. The answer depend on how you place the cuboids


6/2 --- 3 cuboids
11/3 --- 3 cuboids remainder 2cm*
14/2 --- 7 cuboids

3 x 3 x 7 = 63 cuboids

the remainder 2 cm part you still can get

2/2 ---1 cuboid
6/3 ---2 cuboids
14/2 ---7 cuboids

1 x 2 x 7 = 14 cuboids

Max no. 63 + 14 = 77 cuboids

sorry hope other gurus can give a better presentation of the answer

It is very strange to set this question as MCQ as no working is required to be shown and most ppl may just choose 77. :? So what are they testing cos either way you calculate arrives at 77?

Hi ksi,
This question is from http://www.orlesson.org/orp/09Ma/2009-P6-Math-SA2-PLMGS.pdf- P6 Prelim 2009 Paper 1 Q14 set on 25 August 2009.

ganchiong:

Need help in the following question:

Jeffrey wants to pack some identical cuboids into a container. What is the maximum number of cuboids Jeffrey can pack into the container?
Dimension for cuboid:2cmx2cmx3cm and container: 6cmx11cmx14cm
TIA

Hi ganchiong,
No offence intended. It's a good idea if you can post the complete question as the original question has a diagram with it. If you have difficulties in posting the diagram, pls ask for help or quote the source of your question. This will help potential problem solver/s in one way or another and avoid unnecessary confusion.

Excellent solution by http://psle2010a.blogspot.com/2010/07/volume-p6-2009-sa2-mgs-q-14.html posted on 10 July 2010.

[firebird]

Dear Maths guru


Good morning.

Please help me on Henry Park 2010 Preliminary paper 2. Howe Q18a) 224.5 cm b) answer given 30 pie cm squad:

A piece of wire is used to make the figure shown below. Inside the big semi circle are 2 circles and 3 small semi circles, all of which have the same radius.
a) Find length of wire used. Correct your answer to 2 decimal places.
b) Find the total saded area. Express your answer in terms of pie.

For part a) my child is able to answer but not part b) and I am lost too.

Please help and very sorry asI am unable to load up the diagram.

Thank you
Firebird

[Vanilla Cake]

firebird:

Dear Maths guru


Good morning.

Please help me on Henry Park 2010 Preliminary paper 2. Howe Q18a) 224.5 cm b) answer given 30 pie cm squad:

A piece of wire is used to make the figure shown below. Inside the big semi circle are 2 circles and 3 small semi circles, all of which have the same radius.
a) Find length of wire used. Correct your answer to 2 decimal places.
b) Find the total saded area. Express your answer in terms of pie.

For part a) my child is able to answer but not part b) and I am lost too.

Please help and very sorry asI am unable to load up the diagram.

Thank you
Firebird

Hi Firebird,
As I am in rush and need to go to church, I am not able to load up the diagram. Sorry but here's my suggested solution :

Circumference of small semi-circle = pi x 12 x 1/2 = 6 pi
Circumference of big semi-circle = pi x 36 x 1/2 = 18 pi
Total length = 7x6pi + 36 +18 pi = 60pi +36 = 224.50 cm ( 2 dec pls)
(a)Length of the wire used = 224.50 cm ( 2 decimal places)

Area of big semi-circle = 1/2 x pi x 18 x 18 = 162 pi
Area of the 2 circles and 3 semi-circles = 1/2x pix 6x6 x7 = 126 pi
Area of A and B = (162pi -126 pi) divided by 3 = 12pi
Area of small semi-circle = 1/2x pi x 6 x 6 = 18 pi
(b)Total shaded area = 18pi + 12pi = 30 pi cm2

Note:
1. Straight line for the 2 circles not counted.
2. Area A - 1 big unshaded area on the left side of the diagram and Area B - 1 small unshaded area on the left side of the diagram.

Will try to link up to Uncle Observer's blog for a better explanation.
Sorry for this messy solution while waiting for Mathsguru's solution.
:oops:

[Daddy]

Thank Herbie & ksi

[Ngserene617]

Hi Mathsguru,


Pls help this question:
1. Miss Siti wants to give some sweets to some pupils.
If she gives 9 sweets to each pupil, she will have 285 sweets left.
If she gives 17 sweets to each pupil, she will have 133 sweets left.
a) How many pupils are there?
b) How many sweets does Miss Siti have?

Thanks alot !

mathsguru:

Original Title: Let MathsGuru Answer Your Burning Maths Questions!

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[Herbie]

Ans: (285 - 133)/ 17-9 = 19 pupils.

Ans: 19 pupils x 9 sweets each + 285 sweets left =456 sweets

[firebird]

Dear Vanilla cake


Good afternoon.

Thank you very much for your solution for Henry Park Q18 paper 2, SA 2010 question.

With best regards
Firebird