Three tankers contain 385l 415l and 505l of petrol respectively. Find ...
**Problem Statement**
Three tankers contain 385 L, 415 L, and 505 L of petrol, respectively. We need to find the minimum capacity of a container that can measure the petrol of the three tankers an exact number of times.
**Approach**
To find the minimum capacity of a container that can measure the petrol of the three tankers an exact number of times, we need to find the common factors of the three capacities and then determine the minimum among them.
**Step-by-Step Solution**
1. Find the factors of each tanker's capacity.
- The factors of 385 L are: 1, 5, 7, 11, 35, 55, 77, 385.
- The factors of 415 L are: 1, 5, 83, 415.
- The factors of 505 L are: 1, 5, 101, 505.
2. Identify the common factors among the three capacities.
The common factors of 385 L, 415 L, and 505 L are: 1 and 5.
3. Determine the minimum common factor.
The minimum common factor among 1 and 5 is 1.
4. Calculate the minimum capacity of the container.
The minimum capacity of the container that can measure the petrol of the three tankers an exact number of times is the product of the minimum common factor and the sum of the capacities of the three tankers.
Minimum capacity of the container = 1 * (385 + 415 + 505) = 1305 L.
Therefore, the minimum capacity of the container that can measure the petrol of the three tankers an exact number of times is 1305 L.
Three tankers contain 385l 415l and 505l of petrol respectively. Find ...
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