Question

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. What should be the greatest capacity of such a tin?

Answer 1

$\boxed{\fcolorbox{black}{pink}{Answer}}$

It will be 60 litres.

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

Answer 2

$\large \boxed{\sf{Correct \: Question:}}$

A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres 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?

$\large \boxed{ \sf{Required \: Answer:}}$

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.

To find:

• LCM of 120, 180 and 240

Solution:

$\boxed{\sf {\red{Apply\: Euclid’s\: division\: lemma\: on \:180 \:and \:120.}}}$

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

$\underline{\boxed{\sf{\blue{Therefore, the\: tin \:should\: be\: of\: 60 \:litres.}}}}\pink\bigstar$

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