A merchant has $120$ litres of oil of one kind, $180$ litres of another kind, and $240$ litres of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. Find the maximum capacity of such a tin, and the total number of tins required to store all the oil.
Step-by-Step Solution
Key Concept: The maximum capacity of a tin is $\text{HCF}(120, 180, 240)$. The total number of tins is the sum of quantities divided by the HCF.
Stepwise Solution:
To find the maximum capacity of each tin, we calculate $\text{HCF}(120, 180, 240)$. [0.5 Mark]
Prime factorisation:
$120 = 2^3 \times 3 \times 5$
$180 = 2^2 \times 3^2 \times 5$
$240 = 2^4 \times 3 \times 5$. [1.5 Marks]
$\text{HCF}(120, 180, 240) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$ litres. So maximum capacity of a tin $= 60$ litres. [1.5 Marks]
Number of tins of first kind $= \dfrac{120}{60} = 2$.
Number of tins of second kind $= \dfrac{180}{60} = 3$.
Number of tins of third kind $= \dfrac{240}{60} = 4$. [1.0 Mark]
Total number of tins required $= 2 + 3 + 4 = 9$ tins. [0.5 Mark]
Marking Scheme:
• Identifying HCF as maximum capacity: 0.5 Mark
• Correct prime factorisation of 120, 180, 240: 1.5 Marks
• Correct computation of HCF (60 litres): 1.5 Marks
• Calculating number of tins for each kind: 1.0 Mark
• Total count of tins (9 tins): 0.5 Mark
Correct Answer: