Question

A shopkeeper has 120 liters of petrol, 180 litres of diesel and 240 litres of kerosene. He wants to sell oil by filling the three kinds of oils in tins of equal capacity. What should be the greatest capacity of such a tin?

Answer 1

Answer:We need to find the HCF or GCD that is Greatest Common Divisor  

120=2  

3

×3×5

180=2  

2

×3  

2

×5

240=2  

4

×3×5

GCD=2  

2

×3×5=60

The greatest capacity = 60 liters

So the merchant needs to fill 60 liters of all types of oils  

Hence option C is correct

Step-by-step explanation:

Answer 2

Quantity of oil A = 120 liters

Quantity of oil A = 120 liters Quantity of oil B = 180liters

Quantity of oil A = 120 liters Quantity of oil B = 180liters Quantity of oil C = 240liters

Quantity of oil A = 120 liters Quantity of oil B = 180liters Quantity of oil C = 240liters We want to fill oils A, B and C in tins of the same capacity ∴ The greatest capacity of the tin chat can hold oil

Quantity of oil A = 120 liters Quantity of oil B = 180liters Quantity of oil C = 240liters We want to fill oils A, B and C in tins of the same capacity ∴ The greatest capacity of the tin chat can hold oil. A, B and C = HCF of 120, 180 and 240 By fundamental theorem of arithmetic 120 = 23 × 3 × 5 180 = 22 × 32 × 5 240 = 24 × 3 × 5 HCF = 22 × 3 × 5 = 4 × 3 × 5 = 60 litres

The greatest capacity of tin = 60 litres