In a dairy farm, there are two drums, one full of milk and the other full of water. There are a certain number of cans, each of capacity 35 litres. The total liquid in the two drums is exactly sufficient to fill all the cans. If each can hold exactly one-sixth of the milk along with one-eighth of the water, what is the number of cans in the dairy farm?
Question
In a dairy farm, there are two drums, one full of milk and the other full of water. There are a certain number of cans, each of capacity 35 litres. The total liquid in the two drums is exactly sufficient to fill all the cans. If each can hold exactly one-sixth of the milk along with one-eighth of the water, what is the number of cans in the dairy farm?
Solution
To solve this problem, we need to set up a system of equations based on the information given in the problem.
Let's denote: M = the total amount of milk W = the total amount of water C = the number of cans
From the problem, we know that each can has a capacity of 35 litres, and the total liquid (milk and water) is exactly enough to fill all the cans. This gives us our first equation:
M + W = 35C ...(1)
We also know that each can holds exactly one-sixth of the milk and one-eighth of the water. This gives us our second equation:
M/6 + W/8 = 35 ...(2)
Now we have a system of two equations, and we can solve it step by step.
First, let's multiply the second equation by 48 to get rid of the fractions:
8M + 6W = 1680 ...(3)
Then, multiply the first equation by 8:
8M + 8W = 280C ...(4)
Subtract equation (3) from equation (4):
2W = 280C - 1680
Solve for W:
W = 140C - 840
Substitute W into equation (1):
M + 140C - 840 = 35C
Solve for M:
M = -105C + 840
Since M and W must be positive (you can't have negative amounts of milk or water), the maximum value for C that satisfies both equations is 8.
So, there are 8 cans in the dairy farm.
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