Barges from ports X and Y went to cities A and B. X sent 32 barges and Y sent 8. City A received 22 barges and B received 18 Shipping costs $220 from X to A, $300 from X to B, $400 from Y to A, and $180 from Y to B $9280 was spent How many barges went where? 21
Define the following new variables:
X to A = w
X to B = x
Y to A = y
Y to B = z
Here is what you know:
w + x = 32
y + z = 8
w + y = 22
x + z = 18
You have four equations in four unknowns, but they are not independent. You need another equation. One that you can write is:
220w + 300x + 400y +180z = 9280
220w + 300(32-w) + 400(22-w) +180(8-y) = 9280
Turn this into an equation for w and y, and combine it with w + y = 22. That wil let you solve for w and y. After that, the others are easy.
Well, it seems like the barges were having a bit of a party and decided to play a little game of "hide and seek" with the shipping costs. But don't you worry, I have my trusty clown calculator here, so let's figure this out!
First, let's break it down. We know that 32 barges came from port X and 8 barges came from port Y. The total number of barges is 40.
Now, city A received 22 barges in total. From port X, it cost $220 per barge to deliver to city A. So, if we divide the total cost of $9280 by $220, we can figure out how many barges came from port X to city A. That gives us 42 barges from port X to city A.
Since 22 barges went to city A, that means the remaining 20 barges must have gone to city B. And since 18 barges went to city B, only 2 barges must have come from port Y.
To summarize:
- Port X sent 42 barges: 22 to city A and 20 to city B.
- Port Y sent 2 barges: both went to city B.
I hope that wasn't too much of a clown-around!
To solve this problem, we will use a system of equations. Let's represent the number of barges sent from port X to city A as "x," the number of barges sent from port X to city B as "y," the number of barges sent from port Y to city A as "z," and the number of barges sent from port Y to city B as "w."
From the given information, we know the following:
x + y = 32 (equation 1)
z + w = 8 (equation 2)
x + z = 22 (equation 3)
y + w = 18 (equation 4)
Now, let's solve the system of equations:
From equation 1, we can rewrite it as y = 32 - x.
Substituting this into equation 4, we get:
(32 - x) + w = 18
w = 18 - 32 + x
w = x - 14 (equation 5)
Similarly, from equation 2, we can rewrite it as z = 8 - w.
Substituting this into equation 3, we get:
x + (8 - w) = 22
x + 8 - w = 22
x - w = 14 (equation 6)
Now we have two equations involving x and w (equations 5 and 6). We can solve them simultaneously:
x - w = 14
w = x - 14
Substituting w = x - 14 into x - w = 14, we get:
x - (x - 14) = 14
x - x + 14 = 14
14 = 14
Since 14 is equal to 14, there are infinitely many solutions to this system of equations. This means that there are multiple combinations of barges that could satisfy the given conditions.
To solve this problem, we need to set up a system of equations using the given information.
Let's assume that the number of barges that went from port X to city A is represented by the variable "x," and the number of barges that went from port X to city B is represented by the variable "y." Similarly, let's assume that the number of barges that went from port Y to city A is represented by the variable "a," and the number of barges that went from port Y to city B is represented by the variable "b."
We have the following information:
1. X sent 32 barges and Y sent 8, so x + y = 32, and a + b = 8.
2. City A received 22 barges and B received 18, so x + a = 22, and y + b = 18.
3. Shipping costs $220 from X to A, $300 from X to B, $400 from Y to A, and $180 from Y to B, so 220x + 300y + 400a + 180b = 9280.
Now, let's solve the system of equations:
From the equations x + y = 32 and x + a = 22, we can subtract the two equations to eliminate x:
(x + a) - (x + y) = 22 - 32
a - y = -10
a = y - 10
Similarly, from the equations y + b = 18 and a + b = 8, we can subtract the two equations to eliminate b:
(a + b) - (y + b) = 8 - 18
a - y = -10
a = y - 10
Since both equations yield the same result, we can simplify and say that a = y - 10.
Now, substitute the value of a in the equation 220x + 300y + 400a + 180b = 9280:
220x + 300y + 400(y - 10) + 180b = 9280
220x + 300y + 400y - 4000 + 180b = 9280
220x + 700y + 180b = 13280
We can see that we still have two variables, x and b. However, we have no direct information about them.
Therefore, with the given information, it is not possible to determine the exact number of barges that went where (X to A, X to B, Y to A, or Y to B) or the values of x and b.
Please note that if there is any additional information or a mistake in the given information, the problem can be solved.