Question

packets are formed?
A shopkeeper 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

Answer:

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

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 shopkeeper 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

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

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