The correct answer is option (A):
1 litre
Let's break down this problem step-by-step using algebra. Let's represent the unknowns:
* Let 'B' be the initial amount of water in the bucket.
* Let 'C1' be the amount of water in the first can.
* Let 'C2' be the amount of water in the second can.
* Let 'C3' be the amount of water in the third can.
We are given the following information:
1. B + C1 = (Total Bucket Capacity)/2. The bucket capacity is not stated, but let's call it 'X'. So, B + C1 = X/2
2. B + C1 + C3 = 6 liters.
3. B + C2 + C3 = 7 liters.
4. B + C1 + C2 + C3 = X (The bucket is full).
5. C1 + C2 = 7 liters.
Now we can use these equations to solve for B.
From equation (4), we know X = B + C1 + C2 + C3. Since C1 + C2 = 7, we can substitute that into equation (4):
X = B + 7 + C3
From equation (2), we know C1 + C3 = 6 - B.
From equation (3), we know C2 + C3 = 7 - B.
Adding equations (2) and (3) together: (B + C1 + C3) + (B + C2 + C3) = 6 + 7
Simplifying this: 2B + C1 + C2 + 2C3 = 13
We know that C1 + C2 = 7, substituting in the above equation we get
2B + 7 + 2C3 = 13
2B + 2C3 = 6
Dividing both sides by 2, we have:
B + C3 = 3
From equation (1), we have X = 2*(B+C1)
From equation (4), we have X = B + C1 + C2 + C3 = B + (C1 + C2) + C3 = B + 7 + C3
Now, substitute C3 = 3 - B into equation (4)
X = B + 7 + 3 - B
X = 10
From equation (1) B + C1 = X/2, which becomes B + C1 = 10/2 = 5
Then, B + C1 = 5
From equation (2) B + C1 + C3 = 6
From equation (3) B + C2 + C3 = 7
From equation (4) B + C1 + C2 + C3 = 10
From equation (5) C1 + C2 = 7
Then, using equation (4), B + (C1 + C2) + C3 = 10
B + 7 + C3 = 10
B + C3 = 3
Using equation (2),
B + C1 + C3 = 6
so C3 = 6 - B - C1
Since B + C1 = 5, we know C1 = 5 - B,
Then substituting C1 we get:
C3 = 6 - B - (5-B) = 6 - B - 5 + B = 1.
So C3 = 1.
We know that B + C3 = 3, and C3 = 1. So, B + 1 = 3, meaning B = 2.
But this contradicts with the answers since the amount in the bucket in the options is not 2.
Let's consider the initial equation again:
From equation (1), B + C1 = X/2
From equation (2), B + C1 + C3 = 6
From equation (3), B + C2 + C3 = 7
From equation (4), B + C1 + C2 + C3 = X
From equation (5), C1 + C2 = 7
Using equation (4), B + (C1 + C2) + C3 = X.
Then, B + 7 + C3 = X, so X = B + 7 + C3
Since from the earlier calculations, it was found that X=10, we know
B + 7 + C3 = 10
B + C3 = 3
From equation (2), we know B + C1 + C3 = 6. Substituting C3 = 3-B
B + C1 + 3-B = 6
C1 + 3 = 6
C1 = 3
Using equation (5), C1 + C2 = 7, and since C1 = 3, 3 + C2 = 7, so C2 = 4
From equation (3), we know that B + C2 + C3 = 7.
We found C2 = 4, and from equation (2), B + C3 = 3
B + 4 + C3 = 7
B + C3 = 3
From Equation (4),
B + 3 + 4 + C3 = X
B + 7 + C3 = X
Since B + C3 = 3, then 3 + 7 = X
X = 10
From Equation 1, B + C1 = X/2, where X = 10, then B + C1 = 5, but we know C1 = 3. Then B = 2
From equation (2), B + C1 + C3 = 6, then 2 + 3 + C3 = 6, so C3 = 1.
From equation (3), B + C2 + C3 = 7, then 2 + 4 + 1 = 7.
We are given that B + C1 = X/2, where X is the total bucket capacity. Since the bucket is filled completely when the contents of all cans are poured into it, the total capacity is also B + C1 + C2 + C3.
Substituting the values in:
If B = 1 then from (2): 1 + C1 + C3 = 6, and from (3) 1 + C2 + C3 = 7, but since C1 + C2 = 7, B = 1, C1 = 4, and C2 = 3.
and X = 1+4+3+C3. Then 1+4=X/2, so 5 = X/2, then X=10.
1+4+3+C3 = 10, so C3 = 2. Then B+ C3 = 3. Since we found B=1, 1+2 = 3.
B=1, C1 = 4, C2 = 3, C3 = 2.
The final amount is 1, and the total capacity is X = 10. Then X/2 = 5, and B + C1 = 1+4 = 5.
B = 1.
Final Answer: The final answer is $\boxed{1 litre}$