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#1 Re: Jokes » Experimental Mathematics Troll » Today 06:23:44

I did do it by hand.  If the 'correct' answer is 90, then please give one element that is missing from my set.


#2 Re: Jokes » Experimental Mathematics Troll » Today 05:56:35

Ok.  Here's my solution set:

G1 C1 G2 C2 G3 C3
G1 C1 G2 G3 C2 C3
G1 C1 G2 G3 C3 C2

G1 C1 G3 C3 G2 C2
G1 C1 G3 G2 C2 C3
G1 C1 G3 G2 C3 C2

G1 G2 C1 C2 G3 C3
G1 G2 C1 G3 C1 C3
G1 G2 C1 G3 C3 C1

G1 G2 C2 C1 G3 C3
G1 G2 C2 G3 C1 C3
G1 G2 C2 G3 C3 C1

G1 G3 C1 C3 G2 C2
G1 G3 C3 G2 C1 C2
G1 G3 C3 G2 C2 C3

G1 G3 C1 C3 G2 C2
G1 G3 C3 G2 C1 C2
G1 G3 C3 G2 C2 C3

G2 C2 G1 C1 G3 C3
G2 C2 G1 G3 C1 C3
G2 C2 G1 G3 C3 C1

G2 C2 G3 C3 G1 C1
G2 C2 G3 G1 C1 C3
G2 C2 G3 G1 C3 C1

G2 G1 C2 C1 G3 C3
G2 G1 C2 G3 C1 C3
G2 G1 C2 G3 C3 C1

G2 G1 C1 C2 G3 C3
G2 G1 C1 G3 C2 C3
G2 G1 C1 G3 C3 C2

G2 G3 C2 C3 G1 C1
G2 G3 C2 G1 C1 C3
G2 G3 C2 G1 C3 C1

G2 G3 C3 C2 G1 C1
G2 G3 C3 G1 C1 C2
G2 G3 C3 G1 C2 C1

G3 C3 G1 C1 G2 C2
G3 C3 G1 G2 C1 C2
G3 C3 G1 G2 C2 C1

G3 C3 G2 C2 G1 C1
G3 C3 G2 G1 C1 C2
G3 C3 G2 G1 C2 C1

G3 G1 C3 C1 G2 C2
G3 G1 C3 G2 C1 C2
G3 G1 C3 G2 C2 C1

G3 G1 C1 C3 G2 C2
G3 G1 C1 G2 C2 C3
G3 G1 C1 G2 C3 C2

G3 G2 C3 C2 G1 C1
G3 G2 C3 G1 C1 C2
G3 G2 C3 G1 C2 C1

G3 G2 C2 C3 G1 C1
G3 G2 C2 G1 C1 C3
G3 G2 C2 G1 C3 C1


#3 Re: Jokes » Experimental Mathematics Troll » Today 04:24:51

I'm doing that now.  But I've only got 48.  Have to take a break and then have another look.


#4 Re: Jokes » Experimental Mathematics Troll » Today 03:31:41

Yes, But it just goes to show my devotion to MIF.  smile


ps.  As it's only 54, I'm tempted to write out all the solutions.  Then you'll have something to agree / disagree with.

#5 Re: Jokes » Experimental Mathematics Troll » Today 01:37:41


I'm getting 54. 

Like this:

(On paper I keep muddling Gs and Cs because of my writing so I'm using A for the guardians and C for the children.)

First interview must be a guardian, let's say Ax.  3 ways.

Hereafter I'll list the interview sequence:

Second interview

AxCx (3 x 1)
AxAy (3 x 2).

Third interview:

AxCxAy  (3 x 1 x 2)
AxAyCx  (3 x 2 x 1)
AxAyCy  (3 x 2 x 1)

Fourth interview:

AxCxAyCy  (3 x 1 x 2 x 1)
AxCxAyAz  (3 x 1 x 2 x 1)

AxAyCxCy  (3 x 2 x 1 x 1)
AxAyCxAz  (3 x 2 x 1 x 1)

AxAyCyCx  (3 x 2 x 1 x 1)
AxAyCyAz  (3 x 2 x 1 x 1)

Fifth and sixth interview:

AxCxAyCyAzCz is forced (3 x 1 x 2 x 1 x 1 x 1 = 6)
AxCxAyAzCyCz  (3 x 1 x 2 x 1 x  1 x 1 = 6)
AxCxAyAzCzCy  (3 x 1 x 2 x 1 x 1 x 1= 6)

AxAyCxCyAzCz is forced (3 x 2 x 1 x 1 x 1 x 1 = 6)
AxAyCxAzCyCz  (3 x 2 x 1 x 1 x 1 x 1 = 6)
AxAyCxAzCzCy  (3 x 2 x 1 x 1 x 1 x 1 = 6)

AxAyCyCxAzCz is forced (3 x 2 x 1 x 1 x 1 x 1 =6)
AxAyCyAzCxCz  (3 x 2 x 1 x 1 x 1 x 1 = 6)
AxAyCyAzCzCx  (3 x 2 x 1 x 1 x 1 x 1 = 6)

Total = 9 x 6 = 54


#6 Re: Jokes » Experimental Mathematics Troll » Yesterday 20:20:26

hi Agnishom,

Thank you for clearing up a puzzle for me.  When I try to start a 'chat' session with Amazon there is always one person ahead of me in the queue and their problem seems to take ages to resolve.  Now I know why. 

You could only use EM to get an estimate on delivery by repeating the order many times and collecting data.  To avoid the high cost of the experiment here are some bounds for you, obtained by theory rather than EM:

Further information on this can be found at

Hope that helps.  smile


#7 Re: Help Me ! » An equal problem » Yesterday 20:00:17

hi Arna

I also am confused by what you have written.  Maybe you are using the wrong symbols?

=  equal

Two numbers are equal if they are the same.


: is used to show a ratio.

eg.  If there are three times as many children in a room as adults then    children : adults = 3 : 1

you wrote:


This line is incomplete.  You have said that y equals z implies the ratio of y to z is ?????

you wrote:


Substituting x, y, and z

If the numbers are equal then you cannot have different numbers for x y and z.

I'll go back to the original problem.

curved area : total area = 1:2

implies curved area : curved area + top and bottom = 1 : 2

implies curved area : top and bottom = 1 : 1

This means they are equal so


#8 Re: Science HQ » collision » 2014-10-21 19:49:41

You are welcome.  smile

If you ever find out how this question was expected to be done, please let me know.


#9 Re: Help Me ! » some hard algebra problems » 2014-10-21 19:47:32

Eeeekkk!  Don't do that.  Without your answer, I wouldn't have thought I could do the problem. 

Teamwork!  smile


#10 Re: Help Me ! » some hard algebra problems » 2014-10-20 23:57:30

hi phro,

Funnily enough it was your answer of 45 that made me think properly about this one.  It's got loads of irrationals from the roots, but somehow they must all cancel out and leave a simple number.  So I started to think seriously about how that might happen.

If you times each term by a fraction thus

then the denominators simplify to just 2.  Thus the roots are moved to the numerators and with a minus sign now instead of a plus.  It's a bit like rationalising a surd.

Now put all over the common denominator of 2, and you get a load of cancelling like this

The only terms that don't cancel are -root(100) at the start and + root(10000) at the end.  As both have integer roots that does it.  smile


#11 Re: Help Me ! » some hard algebra problems » 2014-10-20 19:30:02

hi thedarktiger

If pq + qr + rp = 20 (correct so far) you need to divide by three for the average.


#12 Re: Help Me ! » some hard algebra problems » 2014-10-20 05:27:43

1)  Use the difference of two squares thus:

Do that for every term, cancel all but the first and last, and it comes out (as 45, thanks Phro smile )

4)Let that expression = N


Using the quadratic formula you want b^2 - 4ac to be a perfect square.

So you just need to find the biggest t (under 10000) that gives such a perfect square.


#13 Re: Help Me ! » some hard algebra problems » 2014-10-20 04:04:15

hi thedarktiger

Wow!  I see your algebra has come on a lot.  smile

1) and 4) are beyond my powers but others will hopefully jump in on those and give you (and me) an idea what to do.

2)  I don't have a super number cruncher for this, so I'd have to slog through elimination.  Do-able but tedious.

3)  Arhh! One I can contribute to.  If you expand (p+q+r)^2 you'll find you can re-arrange this so that pq+qr+rp is the subject and the other side of your equation can be substituted with numbers.  So it's easy from there.


Might have an idea for 2).  I'll be back if it works.  smile

LATER EDIT:  Yes it does.

Add all five equations and simplify to get the value of a+b+c+d+e

Now subtract the first from the second, substitute in for  a+b+c+d+e and you'll get the value of e.

Do similar tricks with subtractions of other pairs of equations to get other letters.

#14 Re: Science HQ » collision » 2014-10-19 03:05:23

hi Niharika,

When I was taught this it was called Newton's experimental law.  If two balls collide with velocities u1 and u2 before the impact and v1 and v2 afterwards, then the coefficient of restitution is given by

and the law of conservation of momentum is

In real life 0 < e < 1.  But the kinetic theory of gases assumes that for molecules e = 1.

For the impact of the ball with the ground, e must be 1 for otherwise the ball will not climb back to its original height.  As we are not told the value of e for the impact with the block, I am assuming it, also, must be 1.  There's no way to do the problem without knowing e, because it affects the velocity that the ball bounces off the block.

By taking e = 1 we get

Substituting the given velocities (and taking right movement as positive):

So the ball bounces back with a horizontal component of velocity of 3v/2.

Alternatively, what happens if e = 0.5, say.

So the ball bounces back with a horizontal component of velocity of 7v/8.

So you can see the value of e is critical.

That's the best I can do with the information given in the problem.


#15 Re: Help Me ! » What Are X And Y Stand For? » 2014-10-19 02:45:01

hi Dawn0388

Welcome to the forum.

The x-y system of coordinates was invented by the mathematician Descartes.

But you don't have to use x and y at all.  For example to draw a graph of velocity changes with time, you might have 't' across and 'v' up.

You'll find more about coordinates at … nates.html


#16 Re: Help Me ! » Quick question on tangent » 2014-10-18 19:30:06

hi Al-Allo

Yes, those answers look good to me.  smile


#17 Re: Science HQ » collision » 2014-10-18 04:18:32

hi Niharika,

Any information about 'coefficient of restitution' ?  ie.  How 'elastic' are the collisions  ?

I could assume it's the same for the block and the floor and hope it will cancel out I suppose.

More thinking:  If e = 1 for perfectly elastic collisions, then the ball would come back to its starting position without A moving at all.  As A is moving, this suggests that the energy lost during the collisions is exactly the same as the energy imparted to the ball by the block.  Hhhmmmm.  dizzy


Ok.  I've thought about it.  Let's just think about the vertical motion first.

The initial velocity is zero and the ball accelerates under gravity.  When the ball hits the wall the vertical velocity is unaffected because it is parallel to the plain of the block and perpendicular to the motion of the block.  So we can just treat the vertical motion without worrying about the block at all.

So use

to get the time to the bottom.

Now the ball will only rise up to the start point if no energy is lost on impact with the ground, so the collision must be perfectly elastic (e=1).  This means the time for the ball to rise up is the same as the time to drop, so you can get the total time by doubling.

Now for the horizontal motion.  At impact with the block, the velocity is still V and the block is moving at -V/4.  We aren't told the mass of the block so no momentum equation is possible.  But we can use the elasticity equation.  We'll have to assume e=1  again as we aren't told anything else.  So

You'll need to be careful with negatives here, but you should be able to get the velocity of the ball after the impact (assume block is still -v/4).

So now you can get the total time for the two horizontal motions and hence get an equation for x in terms of everything else.


#18 Re: Help Me ! » no. of ways of dividing in a couple » 2014-10-15 06:33:08

hi bobbym,

It looks like you've got the same woman dancing with two men?

This is what I did (before I researched it).

Get all 30 into the ballroom.  Expel everyone else. (except the band of course).

Line the men up against the wall.  (So they cannot escape if they don't like their partner smile )

Man one can have any of 15 partners.  Once chosen, man two can choose from 14, man three from 13 etc etc and man fifteen gets the last woman.

Let's simplify.

Say three men Al, Bob and Colin and three women, Di, Elaine and Fiona.

Al-Di; Bob-Elaine; Colin-Fiona
Al-Di; Bob-Fiona; Colin-Elaine
Al-Elaine; Bob-Di; Colin-Fiona
Al-Elaine; Bob-Fiona; Colin-Di
Al-Fiona; Bob-Di; Colin-Elaine
Al-Fiona; Bob-Elaine; Colin-Di

That's 3 x 2 x 1 = 6 possibilities.



#19 Re: Help Me ! » Linear programming question » 2014-10-15 06:19:53

Sorry, that took a while.  I had to find a four variable solver on-line.  I eventually found

which gave b = 0, w = 350, m = 0 and e = 150.

How does that seem?


#20 Re: Help Me ! » Linear programming question » 2014-10-15 06:02:48

hi Fiona,

Yes, those look good to me.  So give me a moment to get an solution.  smile


#21 Re: Help Me ! » no. of ways of dividing in a couple » 2014-10-15 05:57:56

hi jacks and bobbym,

Am I missing something here?  Why isn't it 15! example 2.


#22 Re: Help Me ! » Linear programming question » 2014-10-15 05:30:26

hi fiona1110,

Welcome to the forum.  smile

i'm stuck on part b

So what have you got for part a?


#23 Re: Help Me ! » Urgent - Linear Programming » 2014-10-15 05:23:30

hi anson,

I see you have become a member.  Welcome.  smile

I suggest you copy my post and re-post it back with your answers to the ???.  Then I can see if you are doing it correctly.

I've not used Management Scientist but I looked it up and it should handle this problem.  Have you managed to use it for a simpler linear programming problem?

If yes, then please post that one and your solution.  Thanks.

What I am trying to find out  is "What can you do already"?  So far I am unclear about this.


#24 Re: Help Me ! » Urgent - Linear Programming » 2014-10-14 21:06:36

hi anson,

When there's a lot of text like this, I like to summarise the key information in a table.  Here's mine:


Initially I thought there would be three variables, x y and z, for the three grades.  But as I created the table, I realised that there are only two choices: how much raw material is given the stage one processing, and then how much of the resulting grade 2 is given the stage two processing.  So that fixes just two variables, so I've used x for the first stage processing and y for the second.

So let's say that x is the amount of raw material that is given the first stage process to convert it into grade 1 and grade 2 product.  I'll measure x in 100s of litres.

And y is the amount of grade 2 that is given the second stage process to convert it into grade 3 and grade 1 product.  I'll measure y in 100s of litres.

So let's now make a model for the stage one process.  I'll take x of raw and convert it:

cost = 6x + 3x
time = 3x
grade 1 produced = ????
grade 2 produced = 0.4x

Now let's model the stage two process.  I'll take y of the grade 2 and convert it:

cost = y
time = ???
grade 1 produced = 0.4y
grade 3 produced = 0.6y

And the final amounts are now:

grade 1 = ???? + 0.4y
grade 2 = 0.4x - y
grade 3 = ????

total time = 3x + y

total cost = 9x + y

total sale value = ????

I've left a few ??? so you get to put in some of the answers yourself.

One constraint will be that y cannot be bigger than the amount of grade 2 available.  ie.  y ≤ 0.4x

There will be a time constraint and three constraints due to the maximum amounts that can be made.

The profit will be sales - costs. 

Is that enough for you to complete this?

Please post back with your progress.  smile


#25 Re: Help Me ! » Urgent - Linear Programming » 2014-10-14 03:42:24

hi Anson,

So have you managed to get any of the constraint equations ?


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