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Question

Assume you have 20 dollars.And you want to buy a total of 20 bottles of beer, wine and coke.The price of a single bottle of beer, wine and coke is 4.00, 0.50 and 0.25 dollars respectively.So how many bottles of each will you buy to get a total bottle of 20 with the 20 dollars??????

Answer 1

This is a Diophantine problem. We have to solve
2 equations in 3 unknowns with the restriction
that all are nonnegative integers less than or equal to 20.
The 2 equations of the problem are

x + y + z = 20
and
4x + y/2 + z/4 = 20.

This yields
x + y + z = 20
16x + 2y + z = 80.

Subtracting,
15x + y = 60
and we find that there are 2 solutions to
the problem:
x = 4 y = 0 z = 16
and
x = 3 y = 15 z = 2

Answer 2

x = beer bottles
y = wine bottles
z = coke bottles

then, x + y + z = 20 --- -1
and
4x + .5y + .25z = 20 ---- 2

Possiblities:

x, y and z are all even

Or, from equation 2, y is odd and z is even, and hence, x is odd

In either cases, z is even and can not be odd

so, z is even and <18

If x, y and z are all even, then,

8x + y + .5z = 40

x + y + z = 20... 2x + 2y + 2z = 40
0r, 7x + y - 1.5z = 0

7x + y = 1.5z

since, x, y and z are all integers, and all are even, for this eq to be satisfied, z must be a multiple of 4...it is either

so, z is either 4, 8, 12 or 16

Z can not be 4 as 1.5z is only 6 and we have 7x

For z = 8, again 7x is minimum 14 as x is even..not possible

For z = 16, this equation is not satisfied as x and y must then be 2 each

For z = 12, x must be 2 and y must be 4 to satisfy the equation... but then x + y + z is not 20

So, x,y and z can not all be even..

So, we go with the second ypothesis that
y is odd and z is even, and hence, x is odd

4x + .5y + .25z = 20
x + y + z = 20
then,

3x -.5y - .75z = 0

or, 5y + 75z = 300x


y + 15z = 60x

Now, z is even ... and greater than equal to 2 and less than equal to 18

y is odd and x is also odd

But this equation is never satisfied if y is odd and z is even.

So, this equation has no solution if all x, y and z have to be positive.


This means, that one of more of the number of bottles is zero...

Now, if z =0
Then solving original equations does not give x and y as inters
(x = 20/3)

If x = 0, then
x = 0, y + z = 20 and .5y + .25z = 20 which is not possible

If y = 0, this gives,
y= 0, x = 4 and z = 16

So, your solution is:

Beer - 4 bottles
Wine - none
Coke - 16 bottles

Answer 3

like doug said but x y z are positive (or zero) integers, so its not quite the solution to multiple simultaneous equaitions so what you can do is
note these facts
x>=0, y>=0,z>=0
x<=20,y<=20,z<=20


4x + .5* y + .25 z = 20
=>
16x + 2y + z = 80
x + y + z = 20
=>
15x + y = 60

now what (positive) integer solutions can you see to the previous solution that satisfies the equation? Find that then back substitute. I see two possible values for (x,y,z)

Answer 4

Let,

x= No. of bottles of wine
y= No. of bottles of beer
z= No. of bottles of coke.

from question, we can form given equations:

x+y+z=20 ........................(1)
4x+(1/2)y+(1/4)z = 20 .......(2)
Simplifying (2), we get:

16x+2y+z=80 ....................(3)

Solving (1) and (3), we get:

15x+y=60 ........................ (4)
Now, when we plot the given equation (4), the only values of x and y between 1 and 20 is at the point (3,15).

Hence, we conclude the following:
x ( no. of bottles of wine) = 3
y (no. of bottles of beer) =15
Substitutingthese values in (1), we get,
z=2.

Hence, I would buy 3 bottles of wine, 15 bottles of beer and 2 bottles of coke with 20 dollars.

i hope this satisfies you.......

Answer 5

Let x= number of bottles of beer, y= number of bottles of wine, and z= number of bottles of coke.

From the problem
4x+.5y+.25z = 20 and
x+y+z = 20

Now solve it just like any other system of 3 simultaneous linear equations. (If you don't know how to do that, it's time to hit the books☺)


Doug

Answer 6

Let x= Beer bottles
y = wine bottles
z = coke bottles
As per given rates
4x +0.50 y +0.25 z = 20..............................i
and total bottles are 20, therefore
x+y+ z =20..................................................ii
Multiply ii by 4
4x +4y +4z =80.............................................iii
Subtract i from iii
3.5 y + 3.75z = 60........................................iv
Let y=0 then
(7/2)(0) +(15/4)z =60
(15/4)z = 60
z = 16..............................................v
When y=0, z=16 and x= 4

Answer 7

what grade are u in im in 8th grade