At the fish counter, salmon was selling for $9.60 a pound, whitefish for $7.20 a pound, swordfish for $8.40 a pound, and haddock for $6.90 a pound. Sasha bought 34 lb. of fish for $5.40. What fish did he choose
Let's assume Sasha bought x pounds of salmon, y pounds of whitefish, z pounds of swordfish, and w pounds of haddock.
According to the problem, we have the following information:
x + y + z + w = 34 (Equation 1) ---> Since Sasha bought a total of 34 pounds of fish.
9.60x + 7.20y + 8.40z + 6.90w = 5.40 (Equation 2) ---> Since Sasha spent a total of $5.40.
To solve this system of equations, we can use substitution or elimination methods. In this case, we will use the substitution method.
Let's solve Equation 1 for x:
x = 34 - y - z - w
Now substitute x in Equation 2:
9.60(34 - y - z - w) + 7.20y + 8.40z + 6.90w = 5.40
326.4 - 9.60y - 9.60z - 9.60w + 7.20y + 8.40z + 6.90w = 5.40
317.52 - 2.4y - 1.20z - 2.7w = 5.40
-2.4y - 1.20z - 2.7w = 5.40 - 317.52
-2.4y - 1.20z - 2.7w = -312.12 (Equation 3)
We now have a system of three equations:
x + y + z + w = 34 (Equation 1)
-2.4y - 1.20z - 2.7w = -312.12 (Equation 3)
Let's simplify Equation 3 and divide it by -2.4:
y + 0.5z + w = 130.05 (Equation 4)
Since we have 3 variables (y, z, w) and 3 equations (Equations 1, 3, and 4), we can solve the system.
From Equation 1: y = 34 - x - z - w (Equation 5)
We can plug Equation 5 into Equation 4:
34 - x - z - w + 0.5z + w = 130.05
34 - x + 0.5z = 130.05
0.5z = 130.05 - 34 + x
0.5z = 96.05 + x
z = 192.10 + 2x (Equation 6)
Looking at Equation 6, we can see that z must be an even number. However, none of the given options for fish (salmon, whitefish, swordfish, haddock) have a price that can be divided by 2 without a decimal. Therefore, it appears that there is no solution that satisfies all the conditions of the problem.
To determine what fish Sasha chose, we can set up a system of equations based on the given information.
Let's assign variables to represent the pounds of each type of fish Sasha bought:
Let's say the pounds of salmon Sasha bought is S.
Let's say the pounds of whitefish Sasha bought is W.
Let's say the pounds of swordfish Sasha bought is SW.
Let's say the pounds of haddock Sasha bought is H.
From the given information, we have the following information:
1) The salmon was selling for $9.60 a pound.
2) The whitefish was selling for $7.20 a pound.
3) The swordfish was selling for $8.40 a pound.
4) The haddock was selling for $6.90 a pound.
5) Sasha bought a total of 34 pounds of fish.
6) Sasha spent a total of $5.40.
Based on this information, we can set up the following equations:
1) S + W + SW + H = 34 (Equation 1: Total pounds of fish)
2) 9.60S + 7.20W + 8.40SW + 6.90H = 5.40 (Equation 2: Total cost of fish)
Now, let's solve this system of equations to find the values of S, W, SW, and H.
From Equation 1, we can express one variable in terms of the other variables:
S = 34 - W - SW - H
Substituting this expression for S into Equation 2, we have:
9.60(34 - W - SW - H) + 7.20W + 8.40SW + 6.90H = 5.40
Expanding and simplifying, we have:
326.40 - 9.60W - 9.60SW - 9.60H + 7.20W + 8.40SW + 6.90H = 5.40
Combining like terms, we get:
-2.40W - 1.20SW - 2.70H = -321
Now, we can solve this equation for one variable in terms of the other variables:
-2.40W - 1.20SW = -321 + 2.70H
-2.40W = -321 + 2.70H + 1.20SW
W = (-321 + 2.70H + 1.20SW) / -2.40 (Equation 3: Value of W in terms of H and SW)
From here, we can try different values for H and SW to find possible combinations of pounds for salmon, whitefish, swordfish, and haddock. However, since the total cost is given as only $5.40, it is not possible to find a combination that satisfies all the conditions.
Please check the given information again as there seems to be an inconsistency.