ganeshie8
  • ganeshie8
The price of concert tickets was $25 for local students, $50 for local nonstudents, and $120 for foreigners. There are 100 tickets sold for a total of $4000. It is also known that at least one tick of each type was sold. How many tickets of each type were sold?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
@Loser66 give it a try :)
Loser66
  • Loser66
it looks liked algebra with the equations 25x + 50y + 120 z = 4000 x+y+z = 100 but they are have 2 equations with 3 unknown.
ganeshie8
  • ganeshie8
yes, the tricky part is that it must be solved over "integers" as you cannot buy tickets in fractional amounts..

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More answers

ganeshie8
  • ganeshie8
\(x,y,z\in \mathbb{N}\)
lochana
  • lochana
x = 68 y = 22 z = 10
lochana
  • lochana
68*25 = 1700 22*50 = 1100 120*10 = 1200 1700+1100+1200 = 2800+1200 = 4000 @ganeshie8 am i correct?
Loser66
  • Loser66
I think to solve it, we need express 1 in terms of 5, 10 and 24, right?
ganeshie8
  • ganeshie8
Yep! (68, 22, 10) is one solution, how do we know thats the only solution ?
ganeshie8
  • ganeshie8
@Loser66 that should work, you could also make the work simple by eliminating a variable before doing that..
lochana
  • lochana
yes. that's what i was about to ask. however, you need $120 to be 0 in its 10^2 base. like 1200 , 2400 etc. otherwise. we can't combine multiplication of $120 with multiplication of $50 and $25 to be made $4000. So I started with 10 tickets of 120$. and then got two equations with two variables.
ParthKohli
  • ParthKohli
\[x = 100 - y - z \]\[5x + 10y + 24 z = 800\]\[5(100 - y - z) + 10y + 24 z = 800\]\[5y + 19z = 300\]This is only possible when \(z \) is a multiple of five. Let's plug that into the original equation and try stuff.\[5x + 10y + 24z = 800\]Case I: \(z = 0\), meaning \(x + y = 100\). \(5x + 1000 - 10x = 800\Rightarrow x = 40, y = 60\). Found another one.
ParthKohli
  • ParthKohli
\[z = 5 \Rightarrow x + y = 95, ~5x+10y + 120 = 800 \Rightarrow x = 54,~ y = 41\]
lochana
  • lochana
@ParthKohli can we take z,x or y =0 situations?, if that's the case we already have 3 answers.
ParthKohli
  • ParthKohli
\[z = 10 \Rightarrow x+y=90, ~5x+10y + 240 = 800 \Rightarrow \text{@lochana's solution}\]
ParthKohli
  • ParthKohli
\[z=15\Rightarrow x + y = 85, 5x + 10y + 360 = 800\Rightarrow x = 82, y = 3 \]
ganeshie8
  • ganeshie8
That's nice! yeah z must be a multiple of 5 let me add \(x,y,z > 0\) to the main problem..
lochana
  • lochana
wow. that's a good one.
ParthKohli
  • ParthKohli
We can do this till \(z = 30\).
ParthKohli
  • ParthKohli
z=20 doesn't work though haha
ganeshie8
  • ganeshie8
there are exactly 3 solutions...
ParthKohli
  • ParthKohli
yes, that's right. at \(z = 5, 10, 15\). afterwards, it's just negative.
ganeshie8
  • ganeshie8
Thats it ! @Loser66 are you also getting the same by solving the problem using diophantine method ?
ganeshie8
  • ganeshie8
\(x = 100 - y - z\tag{1} \) \(25x + 50y + 120 z = 4000\) \(\implies 5x + 10y + 24 z = 800\tag{2}\) eliminating \(x\) gives \(5(100 - y - z) + 10y + 24 z = 800\) \(5y + 19z = 300\tag{3}\) This is a linear diophantine equation in two variables.
ParthKohli
  • ParthKohli
hmm, apparently there's no need to eliminate to see that z is a multiple of five. >_< the solution could have been made shorter.
ganeshie8
  • ganeshie8
Haha right, since \((24, 5)=1\), it follows that \(z\mid 5\). But two variables are easy to deal with using the diophantine method..
lochana
  • lochana
@ganeshie8 how did you say that it has only 3 solutions? any method?
ganeshie8
  • ganeshie8
solving \(5y + 19z = 300\) over integers : step1 : finding gcd of 5 and 19 19 = 5*3 + 4 5 = 4*1 + 1 step2 : expressing gcd as combination of 5 and 19 1 = 5 - 4*1 = 5 - (19 - 5*3)*1 = 5(1+3) - 19*1 = 5(4) + 19(-1) step3 : particular solution 1 = 5(4) + 19(-1) multiplying 300 through out gives 300 = 5(1200) + 19(-300) so (1200, -300) is one particular solution to the equation 5y+19z = 300 step4 : null solution (-19, 5) is a solution to the equation 5y + 19z = 0 step5 : all the solutions (y, z) = (1200, -300) + t(-19, 5)
ganeshie8
  • ganeshie8
y = 1200 - 19t z = -300 + 5t since we want the solutions only in positive integers : 1200 - 19t > 0 -300 + 5t > 0 solving gives \(t \in \{61, 62, 63\}\) we can plugin these \(t\) values to get the corresponding \(y, z\) values @Loser66
ganeshie8
  • ganeshie8
@lochana thats the general mehtod to solve any linear diophantine equation...
Loser66
  • Loser66
Question: is it not that if 5y +19 z = 300, then \(5y\equiv 300\equiv 15 (mod 19)\) and (5,9) =1, we can have \(y\equiv 3 (mod 19)\)
ganeshie8
  • ganeshie8
thats correct
ganeshie8
  • ganeshie8
Notice that \(y = 1200 - 19t\) is exactly same as saying \(y\equiv 3\pmod{19}\)
ganeshie8
  • ganeshie8
your method is just an alternative way, which avoids euclid gcd algorithm, to find a paritcular solution.
ganeshie8
  • ganeshie8
However, you need to solve "two" different congruences to find a particular solution, one for each variable : \(5y +19 z = 300\) \[5y\equiv 300\pmod{19}\tag{1} \]\[19z\equiv 300\pmod{5}\tag{2}\]
Loser66
  • Loser66
Another question, when we get y, like above, can we put it back to original one to eliminate y, then we have just 2 unknowns for 2 equations. Then we again, use substitute to find x, z.
Loser66
  • Loser66
I meant: if we let minimum y = 21, put back to get x + z = 79 25x +120z = 2950
ganeshie8
  • ganeshie8
you could do that, but that wont reduce the amount of work, it only increases..
Loser66
  • Loser66
we add them together: 26x +121z = 3029 then again, use \(26x \equiv 3029 (mod 121)\) to solve for x
ganeshie8
  • ganeshie8
how did u get y = 21 ?
Loser66
  • Loser66
Actually, we don't need to reduce work to get the answer. As long as we can apply the lecture to solve the problem, we are ok?
Loser66
  • Loser66
oh, I am tired, it is not 21, it is 22 .
ganeshie8
  • ganeshie8
we are not ok, y can take infinitely many integer values. we're not achieving anything useful by plugging just one value in the equation...
Loser66
  • Loser66
Not that, if we take value of x, y, z too large, the solution is not ok since they are restricted by the amount of money 4000. We need be back to original problem to solve, right?
ganeshie8
  • ganeshie8
how is this any different from guessing ?
Loser66
  • Loser66
It is not different but "what is the goal of lecture?"
ganeshie8
  • ganeshie8
you want to try all y values and pick the ones that work that doesn't look like a better method
ganeshie8
  • ganeshie8
what lecture ?
Loser66
  • Loser66
diophantine equation solving.
ganeshie8
  • ganeshie8
diophanitne mehtod is like an algorithm no guessing is necessary at any step of the procedure..
Loser66
  • Loser66
Like what lochana posted, at the first look, I thought that is guessing, then parkholi used algebra to solve. Surely we can apply any method to solve but that is not good for me.
ganeshie8
  • ganeshie8
diophantine method gives you directly the "y" values that work, we don't need to verify a bunch of y values...
Loser66
  • Loser66
no, the restrain x+y +z=100, doesn't allow us to have many option to verify.
ganeshie8
  • ganeshie8
what would you do if x + y + z = 100000000 ? it doesn't matter if they are few or plenty, you're still verifying and rejecting values that don't work
Loser66
  • Loser66
Ok.
Willie579
  • Willie579
I'm in 7th grade and I'm gonna try to figure this out. XD
ganeshie8
  • ganeshie8
good luck! do let us know your answer when you have it :)
Willie579
  • Willie579
@ganeshie8 I'm so sorry, I forgot all about this. =(

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