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April 11th, 2018, 05:10 AM   #1
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Question Coins question

I pick out all the coins in my pocket and discover I have 30 coins, of total value £1.70. I only have two pence, five pence and ten pence coins and i have more ten pence coins than I have two pence coins.

1) How many five pence coins do I have?

2) Show there is only one solution to the problem

Please help!
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April 11th, 2018, 05:34 AM   #2
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It is hard to help when we know nothing about your mathematical knowledge.

Do you know any algebra?
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April 11th, 2018, 05:38 AM   #3
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Quote:
It is hard to help when we know nothing about your mathematical knowledge.

Do you know any algebra?


Yes, I do know algebra.
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April 11th, 2018, 06:00 AM   #4
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Originally Posted by Student2018 View Post
Yes, I do know algebra.
OK

Start by writing down the names of things AS ALWAYS.

$\text {number of two pence } = x.$

$\text {number of five pence } = y.$

$\text {number of ten pence } = z.$

Next, write down in mathematical form what you know about your unknowns AS ALWAYS.

$x + y + z = 30.$

$2x + 5y + 10z = 170.$ (Unless you are working with 240 pennies to the pound.)

$z > x.$

Now that is not enough information to find a unique solution. But you have other clues.

$x,\ y,\ z \in \mathbb Z \text { and } x \ge 0 \le y.$

Furthermore x must be a multiple of 5 and y must be a multiple of 2. Now what?
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April 13th, 2018, 02:21 AM   #5
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Quote:
x, y, z∈ℤ

What does the last two symbols mean?
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April 13th, 2018, 03:57 AM   #6
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This means that x, y, z belong to the set of integers. This is clear because they represent amounts of coins.
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April 13th, 2018, 04:46 AM   #7
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Quote:
Originally Posted by Student2018 View Post
What does the last two symbols mean?
Oh, that is mathematical symbolism for a MAJOR constraint on the solution to THIS problem.

You know, I presume that there are many different kinds of numbers.

$\in \text { means "is a member of the set of." }$

$\mathbb Z \text { means "integers."}$ So

$x,\ y,\ z \in \mathbb Z \text { means "} x,\ y,\ \text { and } z \text { are all integers.}$

Any questions on what that means and why it is true?

You are correctly taught that n real unknown quantities can be determined with n independent and consistent equations in those n unknowns. But sometimes n unknown quantities can be determined with fewer than n consistent equations if additional information is provided in the form of constraints on what the unknowns can be.

So in this problem you have 3 unknowns and only 2 equations, but you have additional constraints, namely three inequations plus a restriction to integers plus the knowledge that two of the unknowns are evenly divisible by 5 and 2 respectively. These can be summarized as:

$z > x,\ x \ge 0,\ y \ge 0,\ 5 \ | \ x,\ 2 \ | \ y,\ \text { and } x,\ y,\ z \in \mathbb Z.$

Thus x = 0 might be a solution. Try it.

$0 + y + z = 30 \implies y = 30 - z \implies$

$2 * 0 + 5(30 - z) + 10z = 170 \implies 5z = 20 \implies$

$z = 4,\ y = 26,\ x = 0.$

$26 + 4 = 30 \text { and } 5 * 26 + 10 * 4 = 130 + 40 = 170.$

Now this is not a unique solution so I am guessing that the problem really means that he had at least one of each kind of coin. If that is what the problem means then the inequalities are

$x \ge 5,\ y \ge 2,\ \text { and } z > x.$

Have a go.
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April 13th, 2018, 06:42 AM   #8
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Math Focus: Geometry
I think it's you again! How much can you ask these British questions?
Do you post this question on all the resources?
I'm ready to argue that you asked this question on mathsisfun.com, mymathforum.com, study daddy and freemathhelp.com Answer: you have coins in your pocket and you can not buy anything for these £1.70
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April 13th, 2018, 07:24 AM   #9
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No.

It is a capital mistake to theorise before you have all the evidence. It biases the judgement.


-Sir Arthur Conan Doyle
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April 13th, 2018, 08:17 AM   #10
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Quote:
Originally Posted by ScottyBlack228 View Post
I think it's you again! ...
Do you post this question on all the resources?
study daddy
Can't we just ban this spammer?
Thanks from Student2018
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