A 3rd grade Western math question

[WarpSpeed]

My 3rd grader came home with this math question I do not believe has a 3rd grade answer. While neither my wife nor I claim to be mathematicians we aren't slouches in the math department either and there is no apparent 3rd grade answer to this question IOO, maybe we're missing something? If so we're all eyes. Basically since there is already a total number of miles mentioned that doesn't evenly divide by 3rds you have to get the total number of miles the other 2 girls ran first but then you have to add the deficit of Sara's miles to that total which then changes her final total of miles run. Sort of chicken before the egg so to speak.

Regardless of how you work it out, (if you do) to do this math question, if one gets an answer is far ahead of a 3rd graders skills IMO, other techniques ending up leaving with a mile left over and that then needs to be broken down into 3rds as well..

Jan, Mya and Sara ran a total of 64 miles last week.

Jan and Mya ran the same number of miles.

Sara ran 8 miles less than Mya.

How many miles did Sara run?

There is no doubt an answer, but please all you genius's out there, keep in mind this is supposed to be a 3rd grade question before you slam the thread or our math skills..

Edited April 26, 2014 by WarpSpeed

[partington]

If the number of miles Jan and Mya ran is x, then the equation for the total number of miles is

Jan ran x miles, Mya ran x miles, Sara ran x-8 miles

Put together total number of miles run by all:

2x + (x-8) = 64

which means

2x + x = 64 + 8

So 2x + x = 72

so 3x = 72

so x = 24

therefore if Jan and Myra ran x miles they ran 24 miles each.

Sara ran 24 -8 = 16 miles.

Total Jan ran 24 + Mya ran 24 + Sara ran 16 = 64 miles

Edited April 26, 2014 by partington

[sustento]

Jan and Mya ran 24 miles, Sara ran 16. How old is a 3rd grader?

[muchogra]

If you don't set up the equations, observant people may say Sara ran zero mile which is the correct answer. Who cares if Sara ran 8 miles less than Mya!

Correct approach is:

j+m+s=64

j+m=64

s=m-8

3 equations and 3 unknowns!

Actually, you can see from equation 1 and 2 that s is zero.

[muchogra]

For a third grader, or even a high schooler, many would miss.

Glad that I was educated in the old British system where math and science took precedence rather than teaching kids about "debate" which they ca!l "critical thinking"!

How can one have critical thinking when fundamentals are not learnet first?

Edited April 26, 2014 by muchogra

[partington]

If you don't set up the equations, observant people may say Sara ran zero mile which is the correct answer. Who cares if Sara ran 8 miles less than Mya!

Correct approach is:

j+m+s=64

j+m=64

s=m-8

3 equations and 3 unknowns!

Actually, you can see from equation 1 and 2 that s is zero.

Nonsense. The original premise says the total ran was 64 miles. There is only one unknown, x.

Fail!

[muchogra]

If you don't set up the equations, observant people may say Sara ran zero mile which is the correct answer. Who cares if Sara ran 8 miles less than Mya!

Correct approach is:

j+m+s=64

j+m=64

s=m-8

3 equations and 3 unknowns!

Actually, you can see from equation 1 and 2 that s is zero.

Nonsense. The original premise says the total ran was 64 miles. There is only one unknown, x.

Fail!

Really? Are you one who classifies yourself as a critical thinker? I bet you do!

[partington]

If you don't set up the equations, observant people may say Sara ran zero mile which is the correct answer. Who cares if Sara ran 8 miles less than Mya!

Correct approach is:

j+m+s=64

j+m=64

s=m-8

3 equations and 3 unknowns!

Actually, you can see from equation 1 and 2 that s is zero.

Nonsense. The original premise says the total ran was 64 miles. There is only one unknown, x.

Fail!

Really? Are you one who classifies yourself as a critical thinker? I bet you do!

!deleted! off

Edited April 26, 2014 by partington

[muchogra]

To elaborate further even when I'm drunk. These are the answers:

s=0

m=8

j=56

Dig?

[sustento]

To elaborate further even when I'm drunk. These are the answers:

s=0

m=8

j=56

Dig?

Jan and Mya ran the same number of miles. Dig?

[WarpSpeed]

Ok so I'm not losing my mind, the question is way beyond the league of a 9 year old which is 3rd grade.. I'm still not certain I've seen the correct answer yet as I have a pretty good headache going now but there is something wrong with the first one posted for some reason I just can't recall at the mo? But it's the same approach I took initially and somehow something jumped out at me like, where the deficit 8 miles went to? It had to be added to the previous two runners total miles and that can't be done until you have done the math of the total and it will then change Sara's total mileage being the she ran 8 miles shorter..

[muchogra]

So easy a problem but people seem to want debating about it!

No wonder the world is at a discord where "critical thinkers" are debating to death to no avail!

Sustento, my post in Jinthing where I praised you got deleted even when I received 2 "like". I guess I can't beat the trend with all the pc and progressive people who think I am abnormal rather than the other way round. Can you dig it?

Edited April 26, 2014 by muchogra

[sustento]

It's a simple algebra question as Partington pointed out.

If x is the number of miles that Jan ran then Mya ran x miles too and Sara ran x-8 miles.

Altogether they ran 64 miles

Thus x+x+x-8 = 64

Thus 3x=64+8 = 72

Thus x = 72/3 = 24

So Jan ran 24 miles, Mya ran 24 miles and Sara ran 24-8 = 16 miles.

[WarpSpeed]

To elaborate further even when I'm drunk. These are the answers:

s=0

m=8

j=56

Dig?

I'd say you're not drunk enough yet, but thanks for participating..

[partington]

Ok so I'm not losing my mind, the question is way beyond the league of a 9 year old which is 3rd grade.. I'm still not certain I've seen the correct answer yet as I have a pretty good headache going now but there is something wrong with the first one posted for some reason I just can't recall at the mo? But it's the same approach I took initially and somehow something jumped out at me like, where the deficit 8 miles went to? It had to be added to the previous two runners total miles and that can't be done until you have done the math of the total and it will then change Sara's total mileage being the she ran 8 miles shorter..

For God's sake man, the answer adds up correctly. What can you be worrying about?

If Mya ran 24 and Jan ran 24 and Sara ran 16, it all adds up correctly. Why then do you doubt it? They add up to 64, the figure for Mya and Jan are the same and Sara's figure is eight less.

What is your problem with this solution????

Really can you explain?

[WarpSpeed]

It's a simple algebra question as Partington pointed out.

If x is the number of miles that Jan ran then Mya ran x miles too and Sara ran x-8 miles.

Altogether they ran 64 miles

Thus x+x+x-8 = 64

Thus 3x=64+8 = 72

Thus x = 72/3 = 24

So Jan ran 24 miles, Mya ran 24 miles and Sara ran 24-8 = 16 miles.

Except algebra is not a 3rd grade subject.. So not so simple..And still your answer is questionable as you don't account for the deficit 8 miles.. Jan and Mya didn't run 24 miles they ran 28 and that leaves Sara's mileage an unknown, because you can't divide it into equal thirds given that there is an extra mile and she ran 8 miles less. You have to apply the 8 miles to Jan and Mya's mileage throwing off the entire equation....

[nidieunimaitre]

To elaborate further even when I'm drunk. These are the answers:

s=0

m=8

j=56

Dig?

I hope that tomorrow morning your hangover will be strong enough to make you forget you posted this.

[WarpSpeed]

Ok so I'm not losing my mind, the question is way beyond the league of a 9 year old which is 3rd grade.. I'm still not certain I've seen the correct answer yet as I have a pretty good headache going now but there is something wrong with the first one posted for some reason I just can't recall at the mo? But it's the same approach I took initially and somehow something jumped out at me like, where the deficit 8 miles went to? It had to be added to the previous two runners total miles and that can't be done until you have done the math of the total and it will then change Sara's total mileage being the she ran 8 miles shorter..

For God's sake man, the answer adds up correctly. What can you be worrying about?

If Mya ran 24 and Jan ran 24 and Sara ran 16, it all adds up correctly. Why then do you doubt it? They add up to 64, the figure for Mya and Jan are the same and Sara's figure is eight less.

What is your problem with this solution????

Really can you explain?

First off chill out please, it's a debate and discussion, I don't want to get the thread deleted due to aggro..Secondly my point is quite well supported by the debate and confusion, it is not a 3rd grade question, it's complicated algebra.

Finally there is not a true answer, read my postulation, you did not account for the deficited 8 miles in your equation, it has to be added to the previous runners totals first that then throws off Sara's total miles run.

Edited April 26, 2014 by WarpSpeed

[partington]

It's a simple algebra question as Partington pointed out.

If x is the number of miles that Jan ran then Mya ran x miles too and Sara ran x-8 miles.

Altogether they ran 64 miles

Thus x+x+x-8 = 64

Thus 3x=64+8 = 72

Thus x = 72/3 = 24

So Jan ran 24 miles, Mya ran 24 miles and Sara ran 24-8 = 16 miles.

Except algebra is not a 3rd grade subject.. So not so simple..And still your answer is questionable as you don't account for the deficit 8 miles.. Jan and Mya didn't run 24 miles they ran 28 and that leaves Sara's mileage an unknown, because you can't divide it into equal thirds given that there is an extra mile and she ran 8 miles less. You have to apply the 8 miles to Jan and Mya's mileage throwing off the entire equation....

Oh man, you need a cup of coffee. Jan and Mya ran the same number of miles. That means Sara ran 8 miles less than both Mya and Jan ran.

There is no 28 miles anywhere, that has emerged from your imagination.

It's not a debate, it's mathematics: there is one and only one numerically correct answer. There are NO OTHER figures you could put in that would be correct!

Edited April 26, 2014 by partington

[nidieunimaitre]

It's a simple algebra question as Partington pointed out.

If x is the number of miles that Jan ran then Mya ran x miles too and Sara ran x-8 miles.

Altogether they ran 64 miles

Thus x+x+x-8 = 64

Thus 3x=64+8 = 72

Thus x = 72/3 = 24

So Jan ran 24 miles, Mya ran 24 miles and Sara ran 24-8 = 16 miles.

Except algebra is not a 3rd grade subject.. So not so simple..And still your answer is questionable as you don't account for the deficit 8 miles.. Jan and Mya didn't run 24 miles they ran 28 and that leaves Sara's mileage an unknown, because you can't divide it into equal thirds given that there is an extra mile and she ran 8 miles less. You have to apply the 8 miles to Jan and Mya's mileage throwing off the entire equation....

I agree that algebra is not a 3rd grade subject, so there must be a non-algebraic way to reach the same answer.

Anyone?

However, the answer IS correct, stop arguing for arguing's sake.

PS> I cannot believe I got sucked into this nonsense, for god's sake, I am now on holiday at a Greek island!

Edited April 26, 2014 by nidieunimaitre

[WarpSpeed]

It's a simple algebra question as Partington pointed out.

If x is the number of miles that Jan ran then Mya ran x miles too and Sara ran x-8 miles.

Altogether they ran 64 miles

Thus x+x+x-8 = 64

Thus 3x=64+8 = 72

Thus x = 72/3 = 24

So Jan ran 24 miles, Mya ran 24 miles and Sara ran 24-8 = 16 miles.

Except algebra is not a 3rd grade subject.. So not so simple..And still your answer is questionable as you don't account for the deficit 8 miles.. Jan and Mya didn't run 24 miles they ran 28 and that leaves Sara's mileage an unknown, because you can't divide it into equal thirds given that there is an extra mile and she ran 8 miles less. You have to apply the 8 miles to Jan and Mya's mileage throwing off the entire equation....

Oh man, you need a cup of coffee. Jan and Mya ran the same number of miles. That means Sara ran 8 miles less than both Mya and Jan ran.

There is no 28 miles anywhere, that has emerged from your imagination.

I don't drink coffee and with a headache though maybe I should have instead of the aspirin I took. Please indulge my ignorance for a moment without aggro.. How did you come to your starting point with a number that can't be evenly divided into thirds?

Edited April 26, 2014 by WarpSpeed

[sustento]

Ok so I'm not losing my mind, the question is way beyond the league of a 9 year old which is 3rd grade.. I'm still not certain I've seen the correct answer yet as I have a pretty good headache going now but there is something wrong with the first one posted for some reason I just can't recall at the mo? But it's the same approach I took initially and somehow something jumped out at me like, where the deficit 8 miles went to? It had to be added to the previous two runners total miles and that can't be done until you have done the math of the total and it will then change Sara's total mileage being the she ran 8 miles shorter..

For God's sake man, the answer adds up correctly. What can you be worrying about?

If Mya ran 24 and Jan ran 24 and Sara ran 16, it all adds up correctly. Why then do you doubt it? They add up to 64, the figure for Mya and Jan are the same and Sara's figure is eight less.

What is your problem with this solution????

Really can you explain?

First off chill out please, it's a debate and discussion, I don't want to get the thread deleted due to aggro..Secondly my point is quite well supported by the debate and confusion, it is not a 3rd grade question, it's complicated algebra.

It might not be a third grade question but it's definitely not 'complicated algebra'. It's about as simple as you can get. Two people have demonstrated how to calculate the answer. If you don't believe us then up to you.

[WarpSpeed]

Ok so I'm not losing my mind, the question is way beyond the league of a 9 year old which is 3rd grade.. I'm still not certain I've seen the correct answer yet as I have a pretty good headache going now but there is something wrong with the first one posted for some reason I just can't recall at the mo? But it's the same approach I took initially and somehow something jumped out at me like, where the deficit 8 miles went to? It had to be added to the previous two runners total miles and that can't be done until you have done the math of the total and it will then change Sara's total mileage being the she ran 8 miles shorter..

For God's sake man, the answer adds up correctly. What can you be worrying about?

If Mya ran 24 and Jan ran 24 and Sara ran 16, it all adds up correctly. Why then do you doubt it? They add up to 64, the figure for Mya and Jan are the same and Sara's figure is eight less.

What is your problem with this solution????

Really can you explain?

First off chill out please, it's a debate and discussion, I don't want to get the thread deleted due to aggro..Secondly my point is quite well supported by the debate and confusion, it is not a 3rd grade question, it's complicated algebra.

It might not be a third grade question but it's definitely not 'complicated algebra'. It's about as simple as you can get. Two people have demonstrated how to calculate the answer. If you don't believe us then up to you.

Ok here we go with the pedantic out of perspective comments. Let me emphasize again shall we?? From a 3rd graders perspective it is complicated algebra... Therefore I'm trying to find the simple math that makes it a 3rd graders question, get it??

Edited April 26, 2014 by WarpSpeed

[Tywais]

Partington and susteno are both correct. There is no missing or extra mile that you mention or fractions. Just reverse the procedure and plug the numbers in and they work correctly.

J = M = 24 (They ran the same distance)

J + M = 48

(J + M) + S = 64

48 + S = 64

S = 64 - 48 = 16

So M+J+S = 24 + 24 + 16 = 64

As for 3rd grade, depends on if it is Prathom 1-6 (the lower grades) or Maythom 1-6 (the upper grades). But sounds like the OP is referring to Prathom 3.

Now have to figure it out in the perspective of a 9 year old.

[partington]

Ok here we go with the pedantic out of perspective comments. Let me emphasize again shall we?? From a 3rd graders perspective it is complicated algebra... Therefore I'm trying to find the simple math that makes it a 3rd graders question, get it??

It isn't explicable any more simply than the two explanations you already have.

(The poster muchogra is either deranged or deliberately fooling around for his own amusement.)

Honestly if you haven't been able to understand the reasoning behind the answers given, I for one don't see any other way that it could be explained. Sorry.

[WarpSpeed]

Surely, "A 3rd grade Western math question" could not be understood by many Westerners tells a lot about their countries!

To me, it's declining due to people talk (debate) too much when they've got it all wrong!

Are you drunk?

He already said he was, now can you try to answer the question without the complicated algebra you used, in a 3rd graders fashion? That's why my question still stands, if you can't then my post is still relevant as to the missing mileage because basic, 3rd grade elementary math a third grader would use does not explain it and this is also directed towards Tywais..

[WarpSpeed]

Now have to figure it out in the perspective of a 9 year old.

Yes this is and has been my question from the onset, how to do it without the complicated algebra and that is what my questions postulate, it can't be done without it in 3 grade level which requires the approach I've applied so essentially not an answerable question. It's not prathum either it's 3rd grade western elementary in the states.

[nidieunimaitre]

OK, here we go, straight from a Greek island, and non-algebraic, so that even 3rd grade educated westerners that hate modern teaching can understand it.

64 has to be divided into 3 parts, of wich 2 are equal, 1 is 8 down.

Let us say the answer is 27 / 27 / 10

This is not correct, since the difference is not 8

Let us say the answer is 26 / 26 / 12

This is not correct, since the difference is not 8

So we try 25 / 25 / 14

.

Next try 24 / 24 / 16

Halelujah! We found the answer!

And without algebra!

[WarpSpeed]

Surely, "A 3rd grade Western math question" could not be understood by many Westerners tells a lot about their countries!

To me, it's declining due to people talk (debate) too much when they've got it all wrong!

Are you drunk?

I think the answer to that is as obvious as the answer to Warpspeed's algebra problem

Except I'm not asking an algebra problem, nor has anyone answered the problem I am asking about without algebra. I'm asking a 3rd grade problem, and it seems to be even harder for those here to grasp that point and respond accordingly with an intelligent, non-aggro response..

Edited April 26, 2014 by WarpSpeed

[WarpSpeed]

OK, here we go, straight from a Greek island, and non-algebraic, so that even 3rd grade educated westerners that hate modern teaching can understand it.

64 has to be divided into 3 parts, of wich 2 are equal, 1 is 8 down.

Let us say the answer is 27 / 27 / 10

This is not correct, since the difference is not 8

Let us say the answer is 26 / 26 / 12

This is not correct, since the difference is not 8

So we try 25 / 25 / 14

.

Next try 24 / 24 / 16

Halelujah! We found the answer!

And without algebra!

That seems a winner! Props, obviously I wasn't the only fooled. Why did it seem so complicated? I guess it was how you began to get to those numbers? What did you use as a starting point for example? Why 27? Or any number between 1 and 64?Never mind, now after the fact it is logical to have started with some combination of numbers that essentially splits with an uneven balance left and count from there.

Edited April 26, 2014 by WarpSpeed