The weight of the container alone is 25% of the container filled with a certain fluid. When some fluid is removed, the weight of the container and remaining fluid is 50% of the original total weight. What fractional part of the liquid has been removed ?
Answers
The weight of the container alone is 25% of the weight of the filled container.
⇒ The weight of the fluid in the container is three times the weight of the container.
Let the weight of the container alone be x units.
⇒ The weight of the fluid in the container is 3x.
Let the weight of the fluid removed be y units.
⇒ The total weight after removal of some quantity of the fluid is 4x−y.
It is given that the weight to of the container and the remaining fluid is 50% if the original total weight.
⇒ The weight to of the container and the remaining fluid is 2x.
⇒4x−y=2x⇒y=2x.
⇒ The fractional part of the liquid that has been removed is 2x/3x=2/3.
Answer:
Step-by-step explanation:
let weight of container with a certain fluid be : 100
given weight of container alone is 25% of weight of container with fluid...means 100*25/100=25
That means the weight of fluid alone is : 100-25=75
let the weight of fluid removed be x
Then the weight of fluid still left in container is 75-x
given that weight of container and remaining fluid is 50% of original total weight That means 25+75-x=100*50/100
therefore by solving above equation we get x=50 which is the weight of fluid removed.
The fractional part of liquid removed is
50/75=2/3 is the answer