Question

A trader has three different types of oils of volume \(870 \text{ l}\), \(812 \text{ l}\) and \(638 \text{ l}\). Find the least number of containers of equal size required to store all the oil without getting mixed.

Hint:

In problems asking for "least number of items" based on a capacity/size, you are looking for the Highest Common Factor (HCF) of the given quantities. Dividing the total amount by this HCF gives the minimal count.

Answer 1

Step 1: Understanding the Concept:
To find the least number of containers of equal size, each container must have the maximum possible capacity that can exactly divide the given volumes.
This maximum capacity is the Highest Common Factor (HCF) of the three volumes.
Step 2: Key Formula or Approach:
1. Find the HCF of 870, 812, and 638.
2. Total number of containers = \(\frac{\text{Volume}_1}{\text{HCF}} + \frac{\text{Volume}_2}{\text{HCF}} + \frac{\text{Volume}_3}{\text{HCF}}\).
Step 3: Detailed Explanation:
First, we find the prime factorization of each number:
\[ 870 = 2 \times 3 \times 5 \times 29 \]
\[ 812 = 2^2 \times 7 \times 29 \]
\[ 638 = 2 \times 11 \times 29 \]
The common factors are 2 and 29.
\[ \text{HCF}(870, 812, 638) = 2 \times 29 = 58 \text{ litres} \]
This is the capacity of one container.
Now, calculate the number of containers for each type of oil:
- For \(870 \text{ l}\): \(\frac{870}{58} = 15\)
- For \(812 \text{ l}\): \(\frac{812}{58} = 14\)
- For \(638 \text{ l}\): \(\frac{638}{58} = 11\)
Total number of containers = \(15 + 14 + 11 = 40\).
Step 4: Final Answer:
The least number of containers required is 40.