A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?
A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?
The merchant has 3 different oils of 120 liters, 180 liters and 240 liters respectively.
So the greatest capacity of the tin for filling three different types of oil is given by the H.C.F. of 120,180 and 240.
So first we will calculate H.C.F of 120 and 180 by Euclid’s division lemma.
180 = (120) (1) + 60
120 = (60) (2) + 0
The divisor at the last step is 60. So the H.C.F of 120 and 180 is 60.
Now we will find the H.C.F. of 60 and 240,
240 = (60) (4) + 0
The divisor at the last step is 60. So the H.C.F of 240 and 60 is 60.
Therefore, the tin should be of 60 liters.
The merchant has 3 different oils of 120 liters, 180 liters and 240 liters respectively.
So the greatest capacity of the tin for filling three different types of oil is given by the H.C.F. of 120,180 and 240.
So first we will calculate H.C.F of 120 and 180 by Euclid’s division lemma.
180 = (120) (1) + 60
120 = (60) (2) + 0
The divisor at the last step is 60. So the H.C.F of 120 and 180 is 60.
Now we will find the H.C.F. of 60 and 240,
240 = (60) (4) + 0
The divisor at the last step is 60. So the H.C.F of 240 and 60 is 60.
Therefore, the tin should be of 60 liters.
