Question

144L of liquid P and 216L of liquid Q are to be packed in contaier of the same size.the minimum number ofcontainers required are​

Answer 1

Answer:

The minimum number of containers required are​ 72.

Step-by-step explanation:

Given:

144L of liquid P and 216L of liquid Q are to be packed in container of the same size.

To find:

the minimum number of containers required  

Solution:

The total sum of liquids = 144 + 216 = 360

HCF of 144 and 216 = 360

The number of containers required = $\frac{360}{5}$ = 72

Result:

The minimum number of containers required are​ 72.

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Answer 2

It required 5 containers.

Step-by-step explanation:

Since we have given that  

Quantity of liquid P = 144 L

Quantity of liquid Q = 216 L

So, We need to find the minimum number of containers.

So, We will find L.C.M. of 144 and 216.

lcm of 144 and 216 = 432.

So, it will require 432 L of container.

So, Number of container for liquid P would be

$\dfrac{432}{144}=3$

Number of container for liquid Q would be

$\dfrac{432}{216}=2$

Hence, it required 3+2 =5 containers.

# learn more:

: Two tankers contain 850 litres and 680 litres of petrol respectively. Find the max. capacity of a container, which can measure the petrol of either tanker in exact number

https://brainly.in/question/1194591