Cattle Market puzzle

[Gabe]

I keep going wrong on a puzzle and I am not sure why. The puzzle is by H.E. Dudeney. Obviously I am missing something but I don't know what it is.

I've found the answer online, but having trouble with the process.

 

Here is the puzzle:

 

AT A CATTLE MARKET.

 

Three countrymen met at a cattle market. "Look here," said Hodge to

Jakes, "I'll give you six of my pigs for one of your horses, and then

you'll have twice as many animals here as I've got." "If that's your

way of doing business," said Durrant to Hodge, "I'll give you fourteen

of my sheep for a horse, and then you'll have three times as many

animals as I." "Well, I'll go better than that," said Jakes to Durrant;

"I'll give you four cows for a horse, and then you'll have six times as

many animals as I've got here."

 

No doubt this was a very primitive way of bartering animals, but it is

an interesting little puzzle to discover just how many animals Jakes,

Hodge, and Durrant must have taken to the cattle market.

 

Now, I get at least this far:

 

2H-15=J

3D-52=H

6J-21=D

 

But here is where I keep stumbling.

 

I basically am substituting on a per variable basis. Take the last equation:

 

6J-21=D

 

Substituting with the first equation I get:

 

6(2H-15)-21=D

12H-90-21=D

12H-69=D

 

Then substituting the second equation for H I get:

 

12(3D-52)-90-21=D

36D-624-90-21=D

36D-513=D

 

But I've already messed up. Where am I going wrong?

[J.C.MacSwell]

I keep going wrong on a puzzle and I am not sure why. The puzzle is by H.E. Dudeney. Obviously I am missing something but I don't know what it is.

I've found the answer online, but having trouble with the process.

 

Here is the puzzle:

 

AT A CATTLE MARKET.

 

Three countrymen met at a cattle market. "Look here," said Hodge to

Jakes, "I'll give you six of my pigs for one of your horses, and then

you'll have twice as many animals here as I've got." "If that's your

way of doing business," said Durrant to Hodge, "I'll give you fourteen

of my sheep for a horse, and then you'll have three times as many

animals as I." "Well, I'll go better than that," said Jakes to Durrant;

"I'll give you four cows for a horse, and then you'll have six times as

many animals as I've got here."

 

No doubt this was a very primitive way of bartering animals, but it is

an interesting little puzzle to discover just how many animals Jakes,

Hodge, and Durrant must have taken to the cattle market.

 

Now, I get at least this far:

 

2H-15=J

3D-52=H

6J-21=D

 

But here is where I keep stumbling.

 

I basically am substituting on a per variable basis. Take the last equation:

 

6J-21=D

 

Substituting with the first equation I get:

 

6(2H-15)-21=D

12H-90-21=D

12H-69=D

 

Then substituting the second equation for H I get:

 

12(3D-52)-90-21=D

36D-624-90-21=D

36D-513=D

 

But I've already messed up. Where am I going wrong?

 

The bolded does not follow. It should be 12H-111=D and 36D-735=D.

[Gabe]

I feel like an idiot for asking this, but why doesn't it follow?

 

Why do we add and not subtract?

 

I feel this is a very rudimentary algebra question and I apologize in advance because I don't want to waste anyone's time, but it feels as though there is some basic rule I am not remembering.

[zapatos]

Actually I think you are the one who added.

 

If you had:

 

CM is Current Money in your pocket

SM is Starting Money in your pocket (say, $100, that you had in your pocket at the beginning of the day)

Then you spend $10 and McDonalds

Then you spend $5 at Subway

 

You could say:

 

CM = SM-10-5

 

Now, do you shorten that to:

 

CM = SM-15

or

CM = SM-5

 

The question being of course, do you have $95 or $85 in your pocket now?

[Gabe]

Actually I think you are the one who added.

 

If you had:

 

CM is Current Money in your pocket

SM is Starting Money in your pocket (say, $100, that you had in your pocket at the beginning of the day)

Then you spend $10 and McDonalds

Then you spend $5 at Subway

 

You could say:

 

CM = SM-10-5

 

Now, do you shorten that to:

 

CM = SM-15

or

CM = SM-5

 

The question being of course, do you have $95 or $85 in your pocket now?

 

 

OK I see now! I wasn't treating, say, 90 or 624 as a negative number!!

 

So for instance - 36D-624-90-21=D

If treating 624 as a negative and subtracting 90 then 21, I would have gotten -735. The adding was confusing me but I see now it gives the same value to subtract from 36D. It is cumulative.

I was subtracting 90 and 21 from the wrong value - I was ignoring its negative status.

 

 

 

[vignesh3790]

Let the number of animals with hari, jaggu and dinanath be x,y, and z respectively..

 

 

 

Case I : Hari gives six pigs to jaggu and gets one horse, then jaggu wil hav twice that of hari's.

 

 

i.e., 2(X-6+1) = Y+6-1

=> 2X-10 = Y+5

=> 2X-Y= 15 ............. eqn (1)

 

 

 

 

Case II :Dinanath gives fourteen sheeps to hari and gets one horse, then hari wil hav thrice that of dinanth's.

 

 

 

 

i.e., 3(Z-14+1) = X+14-1

=> 3Z-39 = X+13

=> X-3Z = -52 ..............eqn(2)

 

 

 

Case III: Jaggu gives four cows to dinanath and gets a horse, then dinanath wil have six times that of jaggu's.

 

 

i.e., 6(Y-4+1) = Z+4-1

=> 6Y-18 = Z+3

=> 6Y-Z= 21 ................eqn(3)

 

 

solving 1 and 2, we get

 

(1)X2=> 2X+Y= 15

(2) 2X-6Z= -104

(-) (+) (+)

............

-Y+6Z= 119 ................eqn (4)

 

 

 

solving (3) and (4)

 

(3)x 6=> 36Y- 6Z = 126

(4)=> -Y+ 6Z = 119

..............

35Y = 245

 

 

=> Y=7 (jaggu)

 

 

 

 

sub Y in eqn (1)

 

2X -7 =15

 

=> 2X = 22

 

=> X = 11 (hari)

 

 

 

 

 

sub X in eqn (2)

 

 

11-3Z= -52

=> -3Z= -63

=> Z= 21 (dinanath)