Primary School Maths question

[Sturtles]

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.

[Newbie_driver]

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.

 

is it 200 days

 

assuming that the no. of days increased proportionally to the amount of reduction of sheeps.

[Triple2onroad]

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.

 

Can feed for 300 days.

 

Initial grass field can feed 15000 sheep-days

Growth rate is 50 per day

[Sturtles]

That's what I thought, but the answer is 300.

 

The key point is that the grass grows

[Sturtles]

Can feed for 300 days.

 

Initial grass field can feed 15000 sheep-days

Growth rate is 50 per day

 

Wow. how did you get the growth rate and sheep days?

[Newbie_driver]

Can feed for 300 days.

 

Initial grass field can feed 15000 sheep-days

Growth rate is 50 per day

 

hmm... can explain it mathematically.... sound so chim...

[Triple2onroad]

Wow. how did you get the growth rate and sheep days?

 

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.

[Kusje]

Do a simultaneous equation.

[Grx80]

Let X=initial amount of grass, Y=daily grass growth rate.

 

X + 100Y = 200 * 100 -> Equation 1

X + 150Y = 150 * 150 -> Equation 2

 

Solving above equations give X = 15000, Y = 50

 

If there's 100 sheeps which can be fed in Z days, => X + Y*Z = 100 * Z (X = 15000, Y = 50)

=> Z=300 days

 

[Fkadir]

This is a primary 6 question? WOW! I think I used to do this kinda math in Sec 4. Goes to show how old I am!

[Newbie_driver]

Let X=initial amount of grass, Y=daily grass growth rate.

 

X + 100Y = 200 * 100 -> Equation 1

X + 150Y = 150 * 150 -> Equation 2

 

Solving above equations give X = 15000, Y = 50

 

If there's 100 sheeps which can be fed in Z days, => X + Y*Z = 100 * Z (X = 15000, Y = 50)

=> Z=300 days

 

 

guess nowadays the standard has improved and my math skill has .....

 

[Josho]

OMGGGGGGGGGG Primary 6 using equation? OMGGGGGGGGGG That's a no no for me.. im a D7 Math student... can never pass...

[David]

No wonders the primary students are interesting.

[Parkingidiot]

Let X=initial amount of grass, Y=daily grass growth rate.

 

X + 100Y = 200 * 100 -> Equation 1

X + 150Y = 150 * 150 -> Equation 2

 

Solving above equations give X = 15000, Y = 50

 

If there's 100 sheeps which can be fed in Z days, => X + Y*Z = 100 * Z (X = 15000, Y = 50)

=> Z=300 days

 

 

 

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.

[Tjkbeluga]

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.

OKlah, if you insist,

 

initial growth rate plus daily grass growth rate multiply by hundred is equal to two hundred multiply by one hundred

initial growth rate plus daily grass growth rate multiply by hundred fifty is equal to hundred fifty multiply by hundred fifty

 

 

 

[Route88]

Let X=initial amount of grass, Y=daily grass growth rate.

 

X + 100Y = 200 * 100 -> Equation 1

X + 150Y = 150 * 150 -> Equation 2

 

Solving above equations give X = 15000, Y = 50

 

If there's 100 sheeps which can be fed in Z days, => X + Y*Z = 100 * Z (X = 15000, Y = 50)

=> Z=300 days

 

 

 

 

P6 uses tis equation?

[Grx80]

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

[Leepee]

The X original amount of grass....How come did not factor in the Y growth rate as well?

 

All living grass will grow at Y rate right?

 

But the equation did not factor in the existing grass growth rate!!!

 

It seems that the assumption was made that existing grass will not grow anymore? (Which I think is totally wrong as it goes against nature!)

Meaning as the grass grows longer, the sheeps will still eat it as a unit and not as a proportion to its hunger level! (In simple terms, a long grass or short grass will still be considered as a single unit of grass to satisfy hunger!)

 

I think using grass is a POOR formulation of this questions as the growth rate is a complex thing to fanthom.

 

I think I still like the old skool questions, example 5 painters need 10 days to paint a house, how many day 2 painter use to paint the same house type of question!

 

 

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

Edited March 2, 2009 by Leepee