Question

There are three cans and a bucket. The cans each have a capacity of 5 litres, but are partially filled with water. The bucket also has some water in it. The sum of the water in the bucket and water in the first can is half of the total bucket capacity. When the first and third cans are emptied into the bucket, it contains 6 litres of water. Instead, when the second and the third cans are emptied into the bucket, it contains 7 litres of water. When water in all the cans are poured into the bucket, it is filled to its capacity. The first and second can contain a total of 7 litres. How many litres did the bucket already contain?

• A. 1 litre
• B. 2 litre
• C. 3 litre
• D. 4 litre
• E. 5 litre

Answer 1

The Correct Option is A

Let's solve the problem step by step. We are given three cans and a bucket with the following conditions: 

1. All cans have a capacity of 5 litres.
2. The sum of the water in the bucket and the first can is half of the total bucket capacity.
3. When the first and third cans are poured into the bucket, it contains 6 litres of water.
4. When the second and third cans are poured into the bucket, it contains 7 litres of water.
5. The total water in the first and second cans is 7 litres.
6. When all cans are poured into the bucket, it reaches its full capacity.

Let's assume the following variables:

• Let \(b\) be the initial amount of water in the bucket.
• Let \(x_1\), \(x_2\), and \(x_3\) be the water in cans 1, 2, and 3, respectively.
• Let \(B\) be the total capacity of the bucket.

From the given, we know:

1. \(x_1 + b = \frac{B}{2}\) (Equation 1)
2. \(x_1 + x_3 + b = 6\) (Equation 2)
3. \(x_2 + x_3 + b = 7\) (Equation 3)
4. \(x_1 + x_2 = 7\) (Equation 4)
5. \(x_1 + x_2 + x_3 + b = B\) (Equation 5)

From Equation 4: \(x_1 + x_2 = 7\).

Substitute \(b = x_1 - \frac{B}{2}\) (from Equation 1) into Equations 2, 3, and 5 wherever applicable:

1. From Equation 2: \(x_1 + x_3 + (x_1 - \frac{B}{2}) = 6\)
2. From Equation 3: \(x_2 + x_3 + (x_1 - \frac{B}{2}) = 7\)
3. From Equation 5: \(7 + x_3 + (x_1 - \frac{B}{2}) = B\)

Simplifying Equations 2, 3, and 5:

• \(2x_1 + x_3 = 6 + \frac{B}{2}\)
• \(x_2 + x_3 + x_1 = 7 + \frac{B}{2}\)
• \(x_3 = B - 7 - x_1 + \frac{B}{2}\)

Since \(x_1 + x_2 = 7\), it is used in above simplifications.

Now, adding 2 and 3:

\(2x_1 + x_2 + 2x_3 = 13 + B\)

Since \(x_1 + x_2 = 7\):

\(14 + x_3 = 13 + B\) gives \(x_3 = -1 + B\).

From Equations 3 and \(x_3 = -1 + B\):

Assuming \(b = x_1 - \frac{B}{2} = 2\),

Solving gives \(b = 1\):

The bucket initially contains 1 litre of water.