Primary School Maths question

[Leepee]

Also there is the possibility at the very last days...whereby the grass growth rate may not reach maturity to feed the hungry sheeps.

 

Bearing in mind that the grwoing grass is not a 'On/Off' kind of product, unlike a whole piece of animal feed like grains or meat that can be produced from a factory as a single edible item for the sheep to comsume!

 

 

 

 

 

[Parkingidiot]

alamak..... you all ai pee ai qi you ai dua liap nee...

 

dun like that leh... nowadays primary school maths not so simple... need to do problem sums, but cannot use algebra.

[Santav]

This is a good solution. But this is using algebra, rite?

If I remember correctly, in Pri school they do not want us to use algebra to solve the problems.

Is there another way to solve this kind of problem without using algebra? ie. i remember my kid used to do sums drawing little bars and things to represent the sheep and the number of days to feed. Something similar in concept to the simultaneous equations above, but yet manage to solve without bring X, Y and Z to represent the unknown numbers as in algebra.

yes, it's called modelling....my P5 boy is using this method....which i'm totally at a lost.....

[Koren]

sheep hit sexual maturity @ 6 mths of age ~150 day.. by then.. there'll be more sheeps.. heehee..

[Josho]

dun like that leh... nowadays primary school maths not so simple... need to do problem sums, but cannot use algebra.

 

Talk abt problem sums... it always take 3 steps to get the answer, while me.... I take big number minus or divide the small one, thats my answer during primary sch. LOL Thats how 'good' my math is last time.

 

Look at the current chinese assessment book... OMG... Sweat.. now parent need to be clever in order to catch up the current education system

[Amuze]

Thanks Gods this is nots an Englishs questions.

[Parkingidiot]

yes, it's called modelling....my P5 boy is using this method....which i'm totally at a lost.....

 

yep...me, too... lost...

i found a website that discuss about this "model" type of math problems solving.

 

For those interested:

http://www.teach-kids-math-by-model-method.com/

[Sturtles]

Using constant gradient equation:

Initial + (-number of sheeps + Growth rate) * number of days = 0

 

Plug in the two given conditions and you will get them.

 

 

Bro, you are good

 

I was thinking along the lines of a HDB loan installment, where interest rate(growth rate) is the constant, but the actual monthly compound interest(physical grass growth by weight/blade/whatnot) actually decreases as monthly repayment is made(eaten).

 

Quite difficult to form an equation in this case

 

But for exams, I guess they only look at the answer, and not the soundness of the question.

[Kusje]

Damm....I am in NUS and yet I still cannot derive the formula to calculate this 'easy' math!!!

 

Surely you jest. Clearly the standards in NUS are too low. Curiously, are you in arts? I hear the people there aren't too smart.

[Mivec9]

Hi Guys, my primary 6 cousin showed me a math question of his over the weekend, and because I'd flunk it everytime, I wasn't able to help him

 

It goes like this.

 

A grass field can feed 200 sheeps for 100 days. 150 sheeps for 150 days. Assuming the grass grows at a constant amount daily, how many days can it feed 100 sheeps.

 

 

Let A = no. of Initial Grass (if it can be counted!)

 

Let B = no. of Grass it grows per day

---------------------------------------------------------------------------------

 

Using general form:

 

Total Grass = feed Sheeps for no. of days

 

(no. of Initial Grass) + (New Grass grown for no. of Days) = (Sheeps X no. of days)

 

Hence: (A) + (B X no. of days) = (Sheeps X no. of days)

 

Given, A + 100B = 200 X 100

A + 100B = 20,000

 

Also, A + 150B = 150 X 150

A + 150B = 22500

 

Hence, subtracting them, we get 50B = 2500

B = 50 (no. grass growing each day)

 

Putting this into A + 100B = 20,000

A + 100(50) = 20,000

A = 15,000 (initial no. grass)

 

Let C = no. of days 100 sheeps can feed

 

A + C(B) = 100 X C

15,000 + C(50) = 100 X C

15,000 = 50C

Hence C = 300 days (ans)

 

 

P.S. This should not be a P6 Question, more like a Sec 1 Qn as it involves a variable which is changing daily. This should not be under P6 syllabus. Common Qn will be what Leepee mentioned (no. workers to build a houses in no. of days)

Edited March 5, 2009 by Mivec9

[Beavc]

paisay is the ans 200 instead of 300?

 

they tell u grass keep growing so constant and no need to consider...

 

simply

 

200 sheep = eat for 100 days

150 sheep = eat for 150 days

100 sheep = eat for 200 days

 

correct anot? surely not so cheem right?

[Wind30]

it is 300 lah.

 

It is actually one of the more interesting questions I have seen.

 

Actually it is not that difficult to get it if u look at it simpler.

 

initial grass + 100days of growth will give u 200 x 100 sheep days of grass.

initial grass + 150days of growth will give u 150 x 150 sheep days of grass.

 

so 50 days growth is 150 x 150 - 200 x100 = 2500 sheep days.

 

1 day growth is 50 sheep days meaning that u can minus 50 sheep from the count as those 50 sheep are just eating from the growth.

 

therefore initial grass = (200-50) x 100 = 15000 sheep days.

 

so how long will it last 100 sheep ? is 150000/50=300 as the other fifty sheep is eating growth.

 

if u look at it from another angle, all the variables are not changing day by day. and it invovles simple + - * /

[Allangenex]

Another question:

 

Fanny bought 3 storyboooks and 2 boxes of crayons.She gave $100 to the cashier and received $34 change. If a storybook cost 3 times as much as a box of crayons, what was the total cost of a storybook and 5 boxes of crayons?

[Wind30]

this question is trivial.

 

anyway, I find that most kids have problems when there is no point of reference. A Classic example is a question involving 2 moving parties. Kids can calculate speed and distance very well provided there is a point of reference. Once u change the question to 2 moving parties and ask them when they cross each other or something like that, they fail to see how to do it.

 

 

The most impt thing is to understand everything is relative. u just have to make one party stationary and transfer the movement to the other party.

 

Same for the sheep question. U have to get the relationship between sheep and growing grass. Once u understand that the fact that the grass grows just mean that u minus of 50 sheep, the problem becomes trivial.

 

PS: I am top student for my "O" levels many many many years ago :)

[Babyckh]

just calculate all items as crayons and u'll get the ans.

 

1 storybook = 3 box of crayons.

 

ans should be storybook = $18

 

crayon = $6 x 5 = $30

 

 

[Mivec9]

this question is trivial.

 

anyway, I find that most kids have problems when there is no point of reference. A Classic example is a question involving 2 moving parties. Kids can calculate speed and distance very well provided there is a point of reference. Once u change the question to 2 moving parties and ask them when they cross each other or something like that, they fail to see how to do it.

 

 

The most impt thing is to understand everything is relative. u just have to make one party stationary and transfer the movement to the other party.

 

Same for the sheep question. U have to get the relationship between sheep and growing grass. Once u understand that the fact that the grass grows just mean that u minus of 50 sheep, the problem becomes trivial.

 

PS: I am top student for my "O" levels many many many years ago :)

 

Hmmm, I was trying to say u r really good too. So far, I spotted a few good math folks here. Porker is good too. Ha, I am rite, u r top student!