Tutor MathsGuru: Ask me for your burning Maths questions!

[zenztormz]

adhdadhd:

Having read recent posted solutions, which were mostly solved by using algebra, I tried solving some of them using models, but it seems impossible to, as they involved multiplications of units.


How do we then teach our kid, to consider giving up using model?

My DS is Pri 5 now, and I can't explain to him using algebra.

MOE has released a publication on MODELS ... it is available in popular bookshop, you can read on that before teaching your kid...but it is kinda expensive..

[CoffeeCat]

adhdadhd:

Having read recent posted solutions, which were mostly solved by using algebra, I tried solving some of them using models, but it seems impossible to, as they involved multiplications of units.


How do we then teach our kid, to consider giving up using model?

My DS is Pri 5 now, and I can't explain to him using algebra.

my opinion is that models used to be powerful in my generation but very limited now that there are so many variety of challenging problems sums involving fractions and ratio. A pure model method is only going to make things look untidy when you have to divide your model into 20 equal parts. It is possible to avoid algebra for most of the questions by using branching (fraction) or ratio method.
If I am not wrong that book is expensive but doesn't help much with the challenging ones, my opinion, might not be true.

[KSP]

CoffeeCat:

adhdadhd:

Having read recent posted solutions, which were mostly solved by using algebra, I tried solving some of them using models, but it seems impossible to, as they involved multiplications of units.


How do we then teach our kid, to consider giving up using model?

My DS is Pri 5 now, and I can't explain to him using algebra.

my opinion is that models used to be powerful in my generation but very limited now that there are so many variety of challenging problems sums involving fractions and ratio. A pure model method is only going to make things look untidy when you have to divide your model into 20 equal parts. It is possible to avoid algebra for most of the questions by using branching (fraction) or ratio method.
If I am not wrong that book is expensive but doesn't help much with the challenging ones, my opinion, might not be true.

Have you tried this book? http://www.kiasuparents.com/kiasu/files/file/maths_heuristics_books_utm.jpg\">

I think is very useful.

You can find out more from http://www.mathsheuristics.com/

[adhdadhd]

Hi Zenstorm, Coffecat and KSP.

Thank you for speedy response,
I will look at your suggestions.

[Almighty]

Hi,


Please help me solve this problem

Gin left Town P at 6.45 a.m. and travelled at 80Km/h along a highway. Two hours later, Mr Brown left Town P and travelled at 100Km/h along the same highway. At what time did he overtake Gin?

Ans : 4.45Pm

[Sun_2010]

Almighty:

Hi,


Please help me solve this problem

Gin left Town P at 6.45 a.m. and travelled at 80Km/h along a highway. Two hours later, Mr Brown left Town P and travelled at 100Km/h along the same highway. At what time did he overtake Gin?

Ans : 4.45Pm

When Mr. Brown starts, distance between Gin and him = 2 hr x 80kmph
= 160km
Difference in between Mr. Brown and Gin's speed = 100kmph - 80kmph
= 20kmph

So time taken for Mr.Brown to catch up with Gin = 160km / 20kmph
= 8hrs

which would 2hrs+8hrs after 6:45am = 4:45pm

[Dharma]

Almighty:

Hi,


Please help me solve this problem

Gin left Town P at 6.45 a.m. and travelled at 80Km/h along a highway. Two hours later, Mr Brown left Town P and travelled at 100Km/h along the same highway. At what time did he overtake Gin?

Ans : 4.45Pm

Both Gin and Brown travelled the same distance from
Town P when Brown overtook Gin. So, when distance is constant, speed is inversely proportional to the time taken.
Speed ratio => Gin : Brown = 4:5
Time ratio => Gin : Brown = 5:4

5u - 4u = 1u = 2 hrs
4u = 8hrs
Brown overtook Gin 8 hrs after he left Town B.

8 hrs after 8.45am => 4.45pm

[YLH88]

Hi Vanilla Cake, Dharma,


Thank you for the explanation!!

Have a good day

[Vanilla Cake]

Dharma:

abc_parent:

Q3) A set of dictionaries which fitted exactly 4 shelves, each 2.05m long, was replaced by a new set. Every 7 old dictionaries was replaced by 4 new ones, each 5 cm thick. If the new set of dictionaries also fitted the 4 shelves exactly, what was the difference in the number of books between the two sets?


Answer: 120 dictionaries

Total thickness of dictionaries = 4 x 205cm = 820cm

Each shelf will have 205cm/5cm = 41 new dictionaries

No. old dictionaries that were replaced by the new set per row = (41/4) x 7 = 287/4 = 71.75

Only 71 old dictionaries fit into each row of the shelf.
Total number of old dictionaries = 4 x 71 = 284
Total number of new dictionaries = 4 x 41 = 164

Difference in the number of dictionaries = 284 – 164 = 120

(You cannot tear off a portion of a dictionary to force it into a shelf !!!)

Dear Dharma,
Could you pls revisit this dictionaries problem sum again? Below is the comments made by a member in another forum:

The difference is in the understanding on the layout of the shelves. You have assumed the shelves are stacked vertical, perhaps in a book case. Where as, in http://www.postimage.org/image.php?v=Tsqkvjr interpretation the shelves would be placed horizontally side by side creating one 'super' shelf of 820cm in length.

Using your approach, vertical, there is no solution as we cannot take a fraction of a dictionary - ie 71.75 dictionaries. However, using the approach that the shelves are placed side by side we have ...
- 4 shelves x 205 cm = 820cm
- at end can fit 820/5 --> 164 dictionaries
- 164 is 164/4 = 41 sets
- 41 sets x 7 = 287 old dictionaries
- difference is 287 - 164 = 123

The difference in the number of dictionaries between the two sets is 123.

This 5-mark question was from Singapore Chinese Girls' School 2009 P6 SA1 Maths paper 2 Q17 and the worksheet's answer was 120.Hope to have comments/interpretations from other members in this forum about this dictionaries problem sum on whether the shelves are stacked vertical or placed horizontally side by side etc.

Submitted by VC's mum

[Dharma]

Vanilla Cake:

Dharma:

[quote=\"abc_parent\"]Q3) A set of dictionaries which fitted exactly 4 shelves, each 2.05m long, was replaced by a new set. Every 7 old dictionaries was replaced by 4 new ones, each 5 cm thick. If the new set of dictionaries also fitted the 4 shelves exactly, what was the difference in the number of books between the two sets?


Answer: 120 dictionaries

Total thickness of dictionaries = 4 x 205cm = 820cm

Each shelf will have 205cm/5cm = 41 new dictionaries

No. old dictionaries that were replaced by the new set per row = (41/4) x 7 = 287/4 = 71.75

Only 71 old dictionaries fit into each row of the shelf.
Total number of old dictionaries = 4 x 71 = 284
Total number of new dictionaries = 4 x 41 = 164

Difference in the number of dictionaries = 284 – 164 = 120

(You cannot tear off a portion of a dictionary to force it into a shelf !!!)

Dear Dharma,
Could you pls revisit this dictionaries problem sum again? Below is the comments made by a member in another forum:

The difference is in the understanding on the layout of the shelves. You have assumed the shelves are stacked vertical, perhaps in a book case. Where as, in http://www.postimage.org/image.php?v=Tsqkvjr interpretation the shelves would be placed horizontally side by side creating one 'super' shelf of 820cm in length.

Using your approach, vertical, there is no solution as we cannot take a fraction of a dictionary - ie 71.75 dictionaries. However, using the approach that the shelves are placed side by side we have ...
- 4 shelves x 205 cm = 820cm
- at end can fit 820/5 --> 164 dictionaries
- 164 is 164/4 = 41 sets
- 41 sets x 7 = 287 old dictionaries
- difference is 287 - 164 = 123

The difference in the number of dictionaries between the two sets is 123.

This 5-mark question was from Singapore Chinese Girls' School 2009 P6 SA1 Maths paper 2 Q17 and the worksheet's answer was 120.Hope to have comments/interpretations from other members in this forum about this dictionaries problem sum on whether the shelves are stacked vertical or placed horizontally side by side etc.

Submitted by VC's mum[/quote]
Hi VC’s mum,

Think the difference in the answers is due to the fact that the assumption made on the type of shelf is not the same. I’ve assumed the shelf to be closed end (conventional type) whereas some others like Mathsguru had assumed the sides are open.

If the sides are closed, then we have 205cm length per shelf. To re-arrange the new dictionaries that are 5mm thick and we are only able to fit in 71 dictionaries per row of shelf. (even if the shelves are placed side by side)

Yes, you are also correct to say that the 4 shelves that are open ended can be joined horizontally to get a long shelf of 820cm length. Here, the 0.75 of dictionary that does not fit into the closed end shelf will fit into this open ended shelf giving space for (0.75 x 4 = 3 additional dictionaries)

Therefore, if shelf is closed ended, we get 120 dictionaries and if it is open ended then 3 additional dictionaries can be squeezed in to get 123 dictionaries.

I guess the question is quite ambiguous and both answers are correct.