Money in wallet problem (haha)

N

nanase

Junior Member
Joined
Aug 8, 2019
Messages
232
can somebody help me with this problem?
Thank you in advance

Mon Q.jpeg
 
Last edited by a moderator:
I tried to setting up simultaneous equations
x for number of $5 notes
y for number of $10 notes
is there a third variable for the total?
how do I find total possible combinations?
is using a graph helpful at all?
money ans.jpeg
 
nanaseailie said:
I tried to setting up simultaneous equations
x for number of $5 notes
y for number of $10 notes
is there a third variable for the total?
how do I find total possible combinations?
is using a graph helpful at all?
View attachment 33768
Click to expand...
If you have x $5 notes and y $10 notes, what is their total value?

The graph may help a little in imagining the problem, but since x and y must be positive integers, it will not solve the problem by itself.

Do you know anything about Diophantine equations? You're look for all the integer solutions within the triangle.
 
Dr.Peterson said:
If you have x $5 notes and y $10 notes, what is their total value?

The graph may help a little in imagining the problem, but since x and y must be positive integers, it will not solve the problem by itself.

Do you know anything about Diophantine equations? You're look for all the integer solutions within the triangle.
Click to expand...
5x +10y = ? it will still be 5x +10y right?
I have no clue about diophantine equations , I am still 16 years old, not a university student... is this supposed to be advance?
 
nanaseailie said:
5x +10y = ? it will still be 5x +10y right?
I have no clue about diophantine equations , I am still 16 years old, not a university student... is this supposed to be advance?
Click to expand...
No, I just asked because knowing your level of knowledge can be very important, and because it's a relevant term; I don't have any advanced methods in mind.

But now we need to know what you have learned, either in the present context or elsewhere. What ideas do you have that might be relevant to finding integer solutions?

What you have now is that xy = 5x + 10y, and that x + y must be less than 50. One approach you could take is just to solve for one variable in terms of the other, and then try every value less than 50 to see if the other variable is an integer and the sum is small enough. There are other tricks you could use to reduce the number of trials you need, and you might discover those while you start the tedious process. (You'll find yourself saying, "Oh, I didn't need to check that because ...", and that will provide useful experience!)
 
I can solve simultaneous equations using elimination & substitution method, but i can't seem to do both here, can you give me more hint? or simpler explanation? should i just try brute force in trial & error?
I did try many values of x and y in multiples of 5, but still unable to find the values that will fit xy=5x+10y, any hint on the process?
 
Can you solve for one variable in terms of the other in xy=5x+10y as requested by Dr Peterson?
 
nanaseailie said:
I can solve simultaneous equations using elimination & substitution method, but i can't seem to do both here, can you give me more hint? or simpler explanation? should i just try brute force in trial & error?
I did try many values of x and y in multiples of 5, but still unable to find the values that will fit xy=5x+10y, any hint on the process?
Click to expand...
Multiples of 5 is the main "trick" I had in mind to save work. That makes it a little better than mere brute force. (I assume you mean that you recognized that at least one of the variables has to be a multiple of 5.)

Solving for x or y will do more to make the work easier. This is not a system of equations.
 
so it is still trial and error? no mathematical algebra to solve the answer?
 
can somebody really help me in getting the combination, i am still stuck honestly despite trying several combinations
 
nanaseailie said:
can somebody really help me in getting the combination, i am still stuck honestly despite trying several combinations
Click to expand...
Share the combinations you have tried.

Did you write an expression for 'y' - in terms of 'x'?
 
I tried multiple of 5 on the right side, but never match
 

Attachments

  • sss.jpeg
    sss.jpeg
    88.1 KB · Views: 1
oh wait i put them in an organised table
first column is x and second column is y
then i have shorter possible list of multiples of 5
i check that x=20 and y=10 is the possible answer
they match, is it correct?
 
nanaseailie said:
oh wait i put them in an organised table
first column is x and second column is y
then i have shorter possible list of multiples of 5
i check that x=20 and y=10 is the possible answer
they match, is it correct?
Click to expand...

Does it match the conditions given in the original problem?

1661086084726.png
If those do match - you are most probably correct.
 
nanaseailie said:
oh wait i put them in an organised table
first column is x and second column is y
then i have shorter possible list of multiples of 5
i check that x=20 and y=10 is the possible answer
they match, is it correct?
Click to expand...
Yes, that is one of several answers; you have to find all of them.

I did what you did, using the formula to make it easier to create the table, using only multiples of 5 for x.

But I also had to solve for x in terms of y, and repeat the table in the other direction, because it isn't always x that is a multiple of 5. In all, I made 15 rows to check (though some didn't need to be actually calculated.
 
Dr.Peterson said:
Yes, that is one of several answers; you have to find all of them.

I did what you did, using the formula to make it easier to create the table, using only multiples of 5 for x.

But I also had to solve for x in terms of y, and repeat the table in the other direction, because it isn't always x that is a multiple of 5. In all, I made 15 rows to check (though some didn't need to be actually calculated.
Click to expand...
oh bummer, I just realised that I got one answer only out of several possible answers.... so to find the rest, i really have to "observe and match" from a list or a table? there is really no algebraic or other mathematical methods available?
 
nanaseailie said:
oh bummer, I just realised that I got one answer only out of several possible answers.... so to find the rest, i really have to "observe and match" from a list or a table? there is really no algebraic or other mathematical methods available?
Click to expand...
Yes. There is nothing wrong with trying all cases, as long as you have used reasoning to reduce their number. This is not at all uncommon even after you learn about Diophantine equations and methods for solving them!

I haven't thought of another way.
 
You must log in or register to reply here.
Top