Question

A merchant has 120 litres of oil of one kind and 240 litres of third kind .He wants to sell the oil by filling the three kind of oil in tins of equal capacity. What should be the greatest capacity of such a tin​

Answer 1

ɢɪᴠᴇɴ:

• Quantity of oil A = 120 liters
• Quantity of oil B = 180 liters
• Quantity of oil C = 240 liters

ᴛᴏ ғɪɴᴅ:

• The greatest capacity of such a tin?

sᴏʟᴜᴛɪᴏɴ:

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

So,

HCF = 22 × 3 × 5

= 4 × 3 × 5

= 60 litres

The greatest capacity of tin = 60 litres.

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

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

$\rm\orange{Thanks}$