Maths Problem of Simultaneous Equations: "Each week the junior civil servant in Downing Street wheels a suitcase...."

[Gavin300]

I keep trying to find the solutions for V by elimination but still can't find the answer. Any help would be much appreciated:

Each week the junior civil servant in Downing Street wheels a suitcase to the local supermarket to buy alcohol. He always spends the same amount of money.

In week 1 he buys 10 bottles of gin, 5 bottles of vodka and 9 bottles of wine.

In week 2 he buys 4 bottles of gin, 10 bottles of vodka and 12 bottles of wine.

In week 3 he buys 2 bottles of gin and 27 bottles of wine.

In week 4 the Cabinet Secretary decrees that only vodka can be purchased. How many bottles of vodka can he buy?

I've written this as:

10G + 5V + 9W = A

4G + 10V + 12V = A

2G + 27W = A

 

[stapel]

I keep trying to find the solutions for V by elimination but still can't find the answer. Any help would be much appreciated:

Each week the junior civil servant in Downing Street wheels a suitcase to the local supermarket to buy alcohol. He always spends the same amount of money.

In week 1 he buys 10 bottles of gin, 5 bottles of vodka and 9 bottles of wine.

In week 2 he buys 4 bottles of gin, 10 bottles of vodka and 12 bottles of wine.

In week 3 he buys 2 bottles of gin and 27 bottles of wine.

In week 4 the Cabinet Secretary decrees that only vodka can be purchased. How many bottles of vodka can he buy?

I've written this as:

10G + 5V + 9W = A

4G + 10V + 12V = A

2G + 27W = A

When I approached this exercise, I starting with the following steps:

-5R_3 + R_1 -> new R_1

-2R_3 + R_2 -> new R_2

-2R_1 + R_2 -> new R_2

(1/120)R_2 -> new R_2

After another two steps, this gave me the amount of the total A that is the cost of one bottle of vodka. From this I could determine the answer to the question.

Please reply showing your work for at least one of your attempts, so we can see what you're trying and where things are going astray.

Thank you!

 

[JeffM]

I keep trying to find the solutions for V by elimination but still can't find the answer. Any help would be much appreciated:

Each week the junior civil servant in Downing Street wheels a suitcase to the local supermarket to buy alcohol. He always spends the same amount of money.

In week 1 he buys 10 bottles of gin, 5 bottles of vodka and 9 bottles of wine.

In week 2 he buys 4 bottles of gin, 10 bottles of vodka and 12 bottles of wine.

In week 3 he buys 2 bottles of gin and 27 bottles of wine.

In week 4 the Cabinet Secretary decrees that only vodka can be purchased. How many bottles of vodka can he buy?

I've written this as:

10G + 5V + 9W = A

4G + 10V + 12V = A

2G + 27W = A

You cannot find V as a NUMBER by normal algebraic methods. Why? You have four unknowns, A, G, V, and W but only three equations.

It is a sensible first step to go

[math] 10G + 5V + 9W = A\\ 4G + 10V + 12W = A\\ 2G + 27W = A. [/math]
Notice that you did not quite do that first step correctly. But, if you do it correctly, you will then see that there are four unknowns and only three equations. Most such problems do not have a unique solution. There are, however, several methods to find REASONABLE answers to many problems of this type.

Here is one such method. We first note that A is assumedly positive (or things have changed a lot since Churchill was Prime Minister). Moreover, G, V, and W must be non-negative integers. And we are interested in small integers because 10 Downing Street is (presumably at least) not the site of perpetual drunken revels.

So, we are interested in small, non-negative integers that satisfy the (correct) equations. These are known as Diophantine equations. Here is a way to solve some such problems..

[math] \text{Let } = p = G/A, \ q = V/A, \text { and } r = W/A. [/math]
Instead of looking for integers we are looking for ratios.

[math] 2G + 27W = A \implies \dfrac{2G}{A} + \dfrac{27W}{A} = \dfrac{A}{A} \implies\\ 2(G/A) + 27(W/A) = 1 \implies 2p + 27r = 1.\\ \text {Similarly, } 4p + 10q + 12r = 1 \text { and } 10p + 5q + 9r = 1. [/math]
Now that system of equations is almost certainly soluble (3 equations and 3 unknowns). What do you get?

So what is a reasonable answer to the problem?

 

[Steven G]

I keep trying to find the solutions for V by elimination but still can't find the answer. Any help would be much appreciated:

Each week the junior civil servant in Downing Street wheels a suitcase to the local supermarket to buy alcohol. He always spends the same amount of money.

In week 1 he buys 10 bottles of gin, 5 bottles of vodka and 9 bottles of wine.

In week 2 he buys 4 bottles of gin, 10 bottles of vodka and 12 bottles of wine.

In week 3 he buys 2 bottles of gin and 27 bottles of wine.

In week 4 the Cabinet Secretary decrees that only vodka can be purchased. How many bottles of vodka can he buy?

I've written this as:

10G + 5V + 9W = A

4G + 10V + 12W = A

2G + 27W = A

Why is there no equation for week 4?
Let k = the number of bottles of vodka that can be purchased in week 4. Then kV=A
Now can you solve this system below?? You need to use all ALL the information that is given!

10G + 5V + 9W = A

4G + 10V + 12W = A

2G + 27W = A

kV = A

If you replace A with kV, then you get

10G + 5V + 9W = kV ==> 10G + (5-k)V + 9W = 0

4G + 10V + 12W = kV==> 4G + (10-k)V + 12W = 0

2G + 27W = kV======> 2G -kV + 27W =0

Now elimate G from the 3 equations. This take two steps and will yield two equations in terms of V and W. Then solve for V and W.
Then solve for G.

Now V, W and G will be in terms of k. At this point any value k>0 that can make V, W AND G all positive integers will be solutions for V, W and G.

You can solve for G, V and W in terms of k and then use JeffM comment about (in this k) k being small. Just make sure that G, V and W are all integers.

 

[Dr.Peterson]

Just make sure that G, V and W are all integers.

Why do the prices have to be integers? (I don't think anyone has yet stated what the variables mean!) Why can't there be pennies involved?

And how does smallness enter into it?

I solved it, uniquely, by taking A to be 1, and defining G, V, and W to be the fractions of the daily amount that each costs. (This is equivalent to others solving for G/A, V/A, and W/A.) This is another useful technique: Rescale the problem using a convenient unit.

Then the three variables each turn out to be nice fractions, and the solution is just 1/V.

 

[Steven G]

Why do the prices have to be integers? (I don't think anyone has yet stated what the variables mean!) Why can't there be pennies involved?

And how does smallness enter into it?

I solved it, uniquely, by taking A to be 1, and defining G, V, and W to be the fractions of the daily amount that each costs. (This is equivalent to others solving for G/A, V/A, and W/A.) This is another useful technique: Rescale the problem using a convenient unit.

Then the three variables each turn out to be nice fractions, and the solution is just 1/V.

Yes, yes, I was thinking that V, G and W was the number of bottles purchased. You are correct that if the variables were defined then I would not have made this mistake.

 

[JeffM]

Why is there no equation for week 4?
Let k = the number of bottles of vodka that can be purchased in week 4. Then kV=A
Now can you solve this system below?? You need to use all ALL the information that is given!

Steven wrote me a private message asking if I objected to his proposed method of solving. The answer is "No."

Perhaps others will find his question as interesting as I do so I am making my more extensive answer public.

Being super-pedantic, I do not think his approach is perfect because it requires assuming that the amount spent per week is an integer multiple of the cost of a bottle of vodka. We are not given that information. But this is mere pettifogging, and I am perfectly happy to admit that his approach is just as good as mine for practical purposes..

Pedagogically, however, Steven's method does not address the Diophantine nature of the problem, which I suspect is where the student had difficulty. We add a fourth equation, but simultaneously add a fifth variable, namely k. We still have more variables than equations.

This is is a good problem for students. In practical applications, we frequently have situations where the number of equations is less than the number of variables, but there are other constraints, such as non-negativity or restriction to integer or rational solutions, that permit solution or solutions. In that regard, I disagree with Dr. Peterson that the criterion of "smallness" is irrelevant. An answer of 74,513 bottles of vodka per week (even for Bojo, who seemed to like partying) seems like an answer that would get a zero mark.

 

[Steven G]

Jeff, as Dr Peterson pointed out, k doesn't have to be an integer. I was mistaken with that comment. As Dr Peterson pointed out, no one defined the OP's variables so mistakes are made from there and I surely did!

 

[Steven G]

According to WA:

k=25, V=4G/5 and W=2G/3

 

[Dr.Peterson]

I disagree with Dr. Peterson that the criterion of "smallness" is irrelevant. An answer of 74,513 bottles of vodka per week (even for Bojo, who seemed to like partying) seems like an answer that would get a zero mark.

You're misreading the problem:

Moreover, G, V, and W must be non-negative integers. And we are interested in small integers because 10 Downing Street is (presumably at least) not the site of perpetual drunken revels.

We already know the number of bottles in each scenario (except the last); G, V, and W are not the number of bottles, but the prices. These don't have to be integers, and large values would mean drinks are expensive, not that they drink a lot.

Moreover, as I indicated before, the problem has a unique solution without making any such assumptions; they are not needed, even if they were valid. It is not a Diophantine problem at all. And that's why smallness is irrelevant. You take the answer you get.