Question

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?

Answer 1

SOLUTION :  

We have to find the HCF of 120, 80 and 240 litres to find the greatest capacity of the tin.

Given : Oil of one kind : 120 l

Another kind : 180 l

Third kind = 240 l

First we will find the HCF of 180 &120

Here, 180 > 120

Let a = 180 and b = 120

180 = 120 × 1 + 60

[By applying division lemma, a = bq + r]

Here, remainder = 60 ≠ 0, so take new dividend as 120 and new divisor as 60.

Let a = 120 and b= 60

120 = 60 × 2 + 0

[By applying division lemma, a = bq + r]

Here, remainder is zero and divisor is 60.

So ,H.C.F. of 180 and 120 is 60.

Now, we find the HCF of 60 and 240

Here, 240 > 60

Let a = 240 & b= 60

240 = 60 ×4 + 0

[By applying division lemma, a = bq + r]

Here, remainder is zero and divisor is 60.

So ,H.C.F. of 240 and 60 is 60.

Hence, the greatest capacity of the tin should be of 160 liters.

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Answer 2

Answer:

From the question, it’s given that the merchant has 3 different oils of 120 litres, 180 litres and

240 litres respectively.

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.

Firstly, 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.

Therefore, the tin should be of 60 litres.

Step-by-step explanation: