Question:
A trader has three different types of oils of volume 870 l, 812 l and 638 l. Find the least number of containers of equal size required to store all the oil without getting mixed.
A trader has three different types of oils of volume 870 l, 812 l and 638 l. Find the least number of containers of equal size required to store all the oil without getting mixed.
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An easy way to find the quotients after dividing by the HCF is to look at the leftover prime factors in your factorization:
- For 870: \(3 \times 5 = 15\)
- For 812: \(2 \times 7 = 14\)
- For 638: \(11\)
This saves you from doing long division calculations!
- For 870: \(3 \times 5 = 15\)
- For 812: \(2 \times 7 = 14\)
- For 638: \(11\)
This saves you from doing long division calculations!
Updated On: Jul 9, 2026
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Solution and Explanation
Step 1: Understanding the Question:
The topic of this question is Real Numbers, specifically the practical application of the Highest Common Factor (HCF).
A trader has three different volumes of oil: 870 liters, 812 liters, and 638 liters.
To find the least number of containers of equal size, each container must have the maximum possible capacity that divides each volume exactly.
This maximum capacity is the Highest Common Factor (HCF) of the three given volumes.
Step 2: Key Formula or Approach:
1. Calculate the HCF of 870, 812, and 638 using prime factorization.
2. The HCF gives the capacity of one container.
3. Calculate the number of containers required for each type of oil by dividing the respective volumes by the HCF.
4. Sum these values to find the total least number of containers required.
Step 3: Detailed Explanation:
• Find the prime factorization of 870:
\[ 870 = 2 \times 435 \] \[ 870 = 2 \times 3 \times 145 \] \[ 870 = 2 \times 3 \times 5 \times 29 \]
• Find the prime factorization of 812:
\[ 812 = 2 \times 406 \] \[ 812 = 2 \times 2 \times 203 \] \[ 812 = 2^2 \times 7 \times 29 \]
• Find the prime factorization of 638:
\[ 638 = 2 \times 319 \] \[ 638 = 2 \times 11 \times 29 \]
• Calculate the HCF of 870, 812, and 638:
Identify the common prime factors:
The common prime factors are 2 and 29.
\[ \text{HCF} = 2 \times 29 = 58 \] So, the capacity of each container is 58 liters.
• Calculate the number of containers required for each volume:
- For 870 liters:
\[ \text{Containers}_1 = \frac{870}{58} = 15 \] - For 812 liters:
\[ \text{Containers}_2 = \frac{812}{58} = 14 \] - For 638 liters:
\[ \text{Containers}_3 = \frac{638}{58} = 11 \]
• Calculate the total least number of containers:
\[ \text{Total Containers} = 15 + 14 + 11 = 40 \]
Step 4: Final Answer:
The least number of containers of equal size required to store all the oil is 40.
The topic of this question is Real Numbers, specifically the practical application of the Highest Common Factor (HCF).
A trader has three different volumes of oil: 870 liters, 812 liters, and 638 liters.
To find the least number of containers of equal size, each container must have the maximum possible capacity that divides each volume exactly.
This maximum capacity is the Highest Common Factor (HCF) of the three given volumes.
Step 2: Key Formula or Approach:
1. Calculate the HCF of 870, 812, and 638 using prime factorization.
2. The HCF gives the capacity of one container.
3. Calculate the number of containers required for each type of oil by dividing the respective volumes by the HCF.
4. Sum these values to find the total least number of containers required.
Step 3: Detailed Explanation:
• Find the prime factorization of 870:
\[ 870 = 2 \times 435 \] \[ 870 = 2 \times 3 \times 145 \] \[ 870 = 2 \times 3 \times 5 \times 29 \]
• Find the prime factorization of 812:
\[ 812 = 2 \times 406 \] \[ 812 = 2 \times 2 \times 203 \] \[ 812 = 2^2 \times 7 \times 29 \]
• Find the prime factorization of 638:
\[ 638 = 2 \times 319 \] \[ 638 = 2 \times 11 \times 29 \]
• Calculate the HCF of 870, 812, and 638:
Identify the common prime factors:
The common prime factors are 2 and 29.
\[ \text{HCF} = 2 \times 29 = 58 \] So, the capacity of each container is 58 liters.
• Calculate the number of containers required for each volume:
- For 870 liters:
\[ \text{Containers}_1 = \frac{870}{58} = 15 \] - For 812 liters:
\[ \text{Containers}_2 = \frac{812}{58} = 14 \] - For 638 liters:
\[ \text{Containers}_3 = \frac{638}{58} = 11 \]
• Calculate the total least number of containers:
\[ \text{Total Containers} = 15 + 14 + 11 = 40 \]
Step 4: Final Answer:
The least number of containers of equal size required to store all the oil is 40.
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