Question

A merchant has 120 litres of oil of one kind 180 litres of another kind and 240 litres of third kind hewants to sell the oil by filling the three kinds of oil in the tins of equal capacity what should be the greatest capacity of such a tin

Answer 1

Step-by-step explanation:

half of 120 =60

60 is divisible by 120

60 is divisible by 180

60 is divisible by 240

greatest capacity of such a tin =60 litre

Answer 2

$\huge \fbox \red{Solution:}$

Given:

The merchant has 3 different oils:

Capacity Of 1 oil= 120 litres

Capacity Of 2 oil= 180 litres

Capacity Of 3 oil= 240 litres

So, the greatest capacity of the tin for filling three different types of oil can be found out by simply finding the H.C.F. of the three quantities 120,180 and 240.

Solve:

$\fbox \blue{Apply Euclid’s division lemma on 180 and 120.}$

180 = 120 x 1 + 60

120 = 60 x 2 + 0 (here the remainder becomes zero in this step)

Since the divisor at the last step is 60, the HCF (120, 180) = 60.

Now, let’s find the H.C.F of 60 and the third quantity 240.

Applying Euclid’s division lemma, we get

240 = 60 x 4 + 0

And here, since the remainder is 0, the HCF (240, 60) is 60

$\fbox \blue{Therefore, the tin should be of 60 litres.}$

$\rm\orange{Thanks}$