Question

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?

Answer 1

@Loser66
give it a try :)

Answer 2

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.

Answer 3

yes, the tricky part is that it must be solved over "integers" as you cannot buy tickets in fractional amounts..

Answer 4

\(x,y,z\in \mathbb{N}\)

Answer 5

x = 68
y = 22
z = 10

Answer 6

68*25 = 1700
22*50 = 1100
120*10 = 1200
1700+1100+1200 = 2800+1200 = 4000

@ganeshie8 am i correct?

Answer 7

I think to solve it, we need express 1 in terms of 5, 10 and 24, right?

Answer 8

Yep! (68, 22, 10) is one solution,
how do we know thats the only solution ?

Answer 9

@Loser66
that should work, you could also make the work simple by eliminating a variable before doing that..

Answer 10

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.

Answer 11

\[x = 100 - y - z \]\[5x + 10y + 24 z = 800\]\[5(100 - y - z) + 10y + 24 z = 800\]\[5y + 19z = 300\]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.

Answer 12

\[z = 5 \Rightarrow x + y = 95, ~5x+10y + 120 = 800 \Rightarrow x = 54,~ y = 41\]

Answer 13

@ParthKohli can we take z,x or y =0 situations?, if that's the case we already have 3 answers.

Answer 14

\[z = 10 \Rightarrow x+y=90, ~5x+10y + 240 = 800 \Rightarrow \text{@lochana's solution}\]

Answer 15

\[z=15\Rightarrow x + y = 85, 5x + 10y + 360 = 800\Rightarrow x = 82, y = 3 \]

Answer 16

That's nice! yeah z must be a multiple of 5
let me add \(x,y,z > 0\) to the main problem..

Answer 17

wow. that's a good one.

Answer 18

We can do this till \(z = 30\).

Answer 19

z=20 doesn't work though haha

Answer 20

there are exactly 3 solutions...

Answer 21

yes, that's right. at \(z = 5, 10, 15\). afterwards, it's just negative.

Answer 22

Thats it !
@Loser66 are you also getting the same by solving the problem using diophantine method ?

Answer 23

\(x = 100 - y - z\tag{1} \)

\(25x + 50y + 120 z = 4000\)
\(\implies 5x + 10y + 24 z = 800\tag{2}\)

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.

Answer 24

hmm, apparently there's no need to eliminate to see that z is a multiple of five. >_<
the solution could have been made shorter.

Answer 25

Haha right, since \((24, 5)=1\), it follows that \(z\mid 5\).

But two variables are easy to deal with using the diophantine method..

Answer 26

@ganeshie8 how did you say that it has only 3 solutions? any method?

Answer 27

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)

Answer 28

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

Answer 29

@lochana thats the general mehtod to solve any linear diophantine equation...

Answer 30

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 \(y\equiv 3 (mod 19)\)

Answer 31

thats correct

Answer 32

Notice that \(y = 1200 - 19t\) is exactly same as saying \(y\equiv 3\pmod{19}\)

Answer 33

your method is just an alternative way, which avoids euclid gcd algorithm, to find a paritcular solution.

Answer 34

However, you need to solve "two" different congruences to find a particular solution, one for each variable :

\(5y +19 z = 300\)

\[5y\equiv 300\pmod{19}\tag{1} \]\[19z\equiv 300\pmod{5}\tag{2}\]

Answer 35

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.

Answer 36

I meant: if we let minimum y = 21, put back to get
x + z = 79
25x +120z = 2950

Answer 37

you could do that, but that wont reduce the amount of work, it only increases..

Answer 38

we add them together: 26x +121z = 3029
then again, use \(26x \equiv 3029 (mod 121)\) to solve for x

Answer 39

how did u get y = 21 ?

Answer 40

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?

Answer 41

oh, I am tired, it is not 21, it is 22 .

Answer 42

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...

Answer 43

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?

Answer 44

how is this any different from guessing ?

Answer 45

It is not different but "what is the goal of lecture?"

Answer 46

you want to try all y values and pick the ones that work

that doesn't look like a better method

Answer 47

what lecture ?

Answer 48

diophantine equation solving.

Answer 49

diophanitne mehtod is like an algorithm

no guessing is necessary at any step of the procedure..

Answer 50

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.

Answer 51

diophantine method gives you directly the "y" values that work,

we don't need to verify a bunch of y values...

Answer 52

no, the restrain x+y +z=100, doesn't allow us to have many option to verify.

Answer 53

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

Answer 54

Ok.

Answer 55

I'm in 7th grade and I'm gonna try to figure this out. XD

Answer 56

good luck! do let us know your answer when you have it :)

Answer 57

@ganeshie8 I'm so sorry, I forgot all about this. =(