OT - Maths question

mudcow007

a bloke in works kid has been given this question in his homework

Of a class of 50, 13 children walk to school. As a fraction how many don't walk?



i dont think its 1/4 as that would mean the class was 52?!

mikeradar

13/50 walk, which leaves 37/50 that don't.

CiB

Is it a trick q? How many don't walk? None of them - [almost] everybody walks.

Otherwise it's 50 - 13 over 50; 37/50. 37 is a prime so you can't resolve it to a lower denominator.

mudcow007

the teacher marked 37/50 wrong an said the correct answer is 1/4!!!!

teachers eh!

greg66_tri_v2.0

mudcow007 wrote:

a bloke in works kid has been given this question in his homework

Of a class of 50, 13 children walk to school. As a fraction how many don't walk?

mudcow007 wrote:

the teacher marked 37/50 wrong an said the correct answer is 1/4!!!!

teachers eh!

Epic maths fail.

The teachers should have one of these sellotaped to his/her head:

mudcow007

Greg66 wrote:

mudcow007 wrote:

a bloke in works kid has been given this question in his homework

Of a class of 50, 13 children walk to school. As a fraction how many don't walk?

mudcow007 wrote:

the teacher marked 37/50 wrong an said the correct answer is 1/4!!!!

teachers eh!

Epic maths fail.

The teachers should have one of these sellotaped to his/her head:

Ha brilliant im going to print this tread an give it to him

optimisticbiker

MikeRadar has it right, the answer 37/50 is an irreducible fraction i.e. one that cannot be simplified further. Its a common mistake to assume all fractions must be 'nice' and obvious like 1/4!

ToeKnee

mudcow007 wrote:

Of a class of 50, 13 children walk to school. As a fraction how many don't walk?

This is a cycling forum ... surely the question should read:

Of a class of 50, 13 children cycle to school. As a fraction how many don't cycle?

Kieran_Burns

It seems obvious the teacher made a typo and can't be arsed to notice it.

kieranb

Well did the kid put down all the details of how he worked out the answer or just the answer, if the former maybe the teacher would have seen the error of their ways. My son used to just put down the answer but we have taught him to put down how he worked it out + any assumptions.

Oh and if it was cycling, it would probably be : out of a class of 50, 1 cycles to school.

London_Falcon

For you mathematicians out there:-

Given x = 1 and y = 1, then x = y
Multiplying each side by x gives x2 = xy
Subtracting y2 from each side gives x2 - y2 = xy - y2
Factoring each side gives (x + y)(x - y) = y(x - y)
Dividing out the common term, (x - y) results in x + y = y
Substituting the values of x and y, 1 + 1 = 1 or 2 = 1

Can anyone spot how this (doesn't) work??

bompington

London_Falcon wrote:

For you mathematicians out there:-

Given x = 1 and y = 1, then x = y
Multiplying each side by x gives x2 = xy
Subtracting y2 from each side gives x2 - y2 = xy - y2
Factoring each side gives (x + y)(x - y) = y(x - y)
Dividing out the common term, (x - y) results in x + y = y
Substituting the values of x and y, 1 + 1 = 1 or 2 = 1

Can anyone spot how this (doesn't) work??

Old as the hills. You've just divided by (x-y): x-y = 1-1 = 0, so you're dividing by zero. Which is naughty.

rick_chasey

bompington wrote:

London_Falcon wrote:

For you mathematicians out there:-

Given x = 1 and y = 1, then x = y
Multiplying each side by x gives x2 = xy
Subtracting y2 from each side gives x2 - y2 = xy - y2
Factoring each side gives (x + y)(x - y) = y(x - y)
Dividing out the common term, (x - y) results in x + y = y
Substituting the values of x and y, 1 + 1 = 1 or 2 = 1

Can anyone spot how this (doesn't) work??

Old as the hills. You've just divided by (x-y): x-y = 1-1 = 0, so you're dividing by zero. Which is naughty.

Exactly.

Another way to spot an error has been made. For it to be true, they must all be true together. So the penultimate line "x+y=y" is wrong, since we know x=1 and y=1 and 1+1 does not equal 1.

daviesee

mudcow007 wrote:

a bloke in works kid has been given this question in his homework

Of a class of 50, 13 children walk to school. As a fraction how many don't walk?



i dont think its 1/4 as that would mean the class was 52?!

I'd be worried about your child's education.
If the teacher made a mistake and fesses up then fair enough.
If the teacher maintains that they are correct, then that is worrying :shock:

Ben6899

daviesee wrote:

mudcow007 wrote:

a bloke in works kid has been given this question in his homework

Of a class of 50, 13 children walk to school. As a fraction how many don't walk?



i dont think its 1/4 as that would mean the class was 52?!

I'd be worried about your child's education.
If the teacher made a mistake and fesses up then fair enough.
If the teacher maintains that they are correct, then that is worrying :shock:

It's not even like they have got the numbers slightly wrong... 37/50 is almost twice 1/4.

Kieran_Burns

bompington wrote:

London_Falcon wrote:

For you mathematicians out there:-

Given x = 1 and y = 1, then x = y
Multiplying each side by x gives x2 = xy
Subtracting y2 from each side gives x2 - y2 = xy - y2
Factoring each side gives (x + y)(x - y) = y(x - y)
Dividing out the common term, (x - y) results in x + y = y
Substituting the values of x and y, 1 + 1 = 1 or 2 = 1

Can anyone spot how this (doesn't) work??

Old as the hills. You've just divided by (x-y): x-y = 1-1 = 0, so you're dividing by zero. Which is naughty.

Yup, my A level maths teacher tried this one on us.

Asked my mate (sat next to me) what he thought was going on

He looked at the board
Wrote down a few figures.
Cogitated for a while
Wrote down some more figures
Glanced at the board again.
Looked deeply thoughtful

and said

"Well, it beats the sh*t out of me"

The teacher in the NEXT classroom came in to tell us to quiet down after that.

rolf_f

See Kierans post.

The correct question was 13 out of 52 walk so 3/4 don't walk.

I'm not worried that the teacher would mark one incorrectly as they no doubt didn't do the sum themselves. What does worry me is that either a) the teacher presumably didn't notice something wrong when everyone got the answer apparently wrong or that b)everyone got the answer apparently right!

sketchley

mudcow007 wrote:

a bloke in works kid has been given this question in his homework

Of a class of 50, 13 children walk to school. As a fraction how many don't walk?



i dont think its 1/4 as that would mean the class was 52?!

Without looking at any other reply the answer is - 37/50 as 37 is prime number it cannot be expressed as a "smaller" fraction.

t4tomo

You've all missed the biggest point here, whether its 50 or 52 WTF are that many kids doing in a class? Its illegal for starters and no wonder they can't do fractions, its must be bedlam in there.

greg66_tri_v2.0

London_Falcon wrote:

For you mathematicians out there:-

Given x = 1 and y = 1, then x = y
Multiplying each side by x gives x2 = xy
Subtracting y2 from each side gives x2 - y2 = xy - y2
Factoring each side gives (x + y)(x - y) = y(x - y)
Dividing out the common term, (x - y) results in x + y = y
Substituting the values of x and y, 1 + 1 = 1 or 2 = 1

Can anyone spot how this (doesn't) work??

Strictly, the problem arises in the emboldened line. At that point you introduce terms which mean that the equation becomes 0=0 (as y^2 = x^2). After that, you can multiply any number by zero to get zero (eg 54*0= 45*0) but you cannot divide both sides by zero (eg to get 54=45).

London_Falcon

Ok, well done all. If you fancy another one how about this:-

You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.
The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more.

Kieran_Burns

split into 2
weigh
discard light 6
split 6 into 2 x 3
weigh
discard light 3
weigh 2 marbles
if same discard both
if one heavier discard lighter

London_Falcon

One marble is stated as "different", not specifically "heavier". If the odd marble is lighter then just discarding the light 6 doesn't work

rolf_f

CiB wrote:

Otherwise it's 50 - 13 over 50; 37/50. 37 is a prime so you can't resolve it to a lower denominator.

Sketchley wrote:

Without looking at any other reply the answer is - 37/50 as 37 is prime number it cannot be expressed as a "smaller" fraction.

Sketchley - are you becoming a CiB clone?

dhope

London_Falcon wrote:

One marble is stated as "different", not specifically "heavier". If the odd marble is lighter then just discarding the light 6 doesn't work

I'll not give the answer away just yet, though I do like the question which I'd heard as

You have 8 marbles
7 identical, 1 slightly differently weighted but appears identical
2 weighings max, find the differently weighted one.

Same principle.

Wallace1492

dhope wrote:

London_Falcon wrote:

One marble is stated as "different", not specifically "heavier". If the odd marble is lighter then just discarding the light 6 doesn't work

I'll not give the answer away just yet, though I do like the question which I'd heard as

You have 8 marbles
7 identical, 1 slightly differently weighted but appears identical
2 weighings max, find the differently weighted one.

Same principle.

Think this might only work if looking for specific heavy/light marble, otherwise think it would need 3 weighings.

walkingbootweather

I'd start by splitting 12 marbles into 3 groups of 4.
I'd weigh group 1 and group 2. If it balances then the irregular marble will be in group 3
... I'll let you do the rest

(I think I may have lost my marbles)

rick_chasey

dhope wrote:

London_Falcon wrote:

One marble is stated as "different", not specifically "heavier". If the odd marble is lighter then just discarding the light 6 doesn't work

I'll not give the answer away just yet, though I do like the question which I'd heard as

You have 8 marbles
7 identical, 1 slightly differently weighted but appears identical
2 weighings max, find the differently weighted one.

Same principle.

Can't figure out how to do that with 2 weighs.

dhope

walkingbootweather wrote:

I'd start by splitting 12 marbles into 3 groups of 4.
I'd weigh group 1 and group 2. If it balances then the irregular marble will be in group 3
... I'll let you do the rest

Wallace, Rick... WBW has the right idea.

Think about what you're not weighing too (with 8 marbles or 12)

SimonAH

Here’s a favourite of mine.

The lighthouse keeper wants a new carpet for one of the floors near the top of his lighthouse. He calls the carpet fitter round to measure it up.

The floor is a ring shape around a central staircase and the only measurement that can be physically taken is a chord as per;



Which measures, let’s say, 10m

The carpet fitter scratches his head and thinks “how am I going to work out the area of the carpet from this?

He gets on the mobile to his cousin the Oxford maths don and explains the situation.

“Oh that’s easy!” says his cousin “I have a simple formula for that”

“In that case, no problem, I have the answer” says our hero who then provides a quote on the spot to the lighthouse keeper.

Go on, get your teeth into that one!

secretdancer

This brings the engineer back out in me....



x = radius of outer circle
y = radius of inner circle
A = area of ring
Chord length = 10

We know

A = Area of outer circle – area of inner circle
A = πx^2 – πy^2


Also, by the rule of Pythagoras

x^2 =y^2 +5^2 (where 5 is half the chord length)

Substituting for x^2 in the first equation we have

A = π(y^2 +5^2) – πy^2

A = πy^2 + π5^2 – πy^2

A = π5^2