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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > A vessel is filled with liquid, 3 parts of which are...
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A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

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Suppose 80 liters of liquid we have 30 l water and 50 l syrup.we take out 'x' units of liquid from the vessel. So we draw 3x/8 units of water and 5x/8 units of syrup from the vessel. Remaining quantities of water and syrup left are now 30-(3x/8) and 50-(5x/8) units respectively. We now add 'x' units...
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Suppose 80 liters of liquid we have 30 l water and 50 l syrup.we take out 'x' units of liquid from the vessel. So we draw 3x/8 units of water and 5x/8 units of syrup from the vessel. Remaining quantities of water and syrup left are now 30-(3x/8) and 50-(5x/8) units respectively. We now add 'x' units of water. Now quantities of water and syrup are 30-(3x/8)+x and 50-(5x/8) respectively. As there must be equal quantities of water and syrup now, we have 30-(3x/8)+x = 50-(5x/8) Solve to get x=16 units which means you need to draw 20% or 1/5th of the liquid from the vessel. read less
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Maths, Physics & Chemistry, IIT Kanpur & IIT Bombay graduate

Answer: 1/5th of the vessel volume. Let's assume the capacity of vessel be (3+5=) 8 ltrs. Now, the basic principle of all such questions is that when you remove any amount of the mixture, it removes both the liquids in the same ratio as the mixture itself. So, let's suppose we take x ltrs out of the...
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Answer: 1/5th of the vessel volume. Let's assume the capacity of vessel be (3+5=) 8 ltrs. Now, the basic principle of all such questions is that when you remove any amount of the mixture, it removes both the liquids in the same ratio as the mixture itself. So, let's suppose we take x ltrs out of the vessel. This means 5x/8 amount of syrup and 3x/8 amount of water gets removed. which means the remaining mixture has (3-(3x/8)) ltrs of water & (5-(5x/8)) syrup. Now when we add x ltrs of water to this mixture, the amount of water and syrup becomes equal. Thus, (x+3-(3x/8)) = (5-(5x/8)) => x=1.6 ltrs. Since we assumed that vessel is 8 ltrs. in volume=> 1.6/8 = 1/5th of the mixture in the vessel needs to be replaced with water. Hope it helps. :) read less
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Let x litres of this liquid be replaced with water Let x litres of this liquid be replaced with water. Quantity of syrup in new mixture = (5 -5x/8)litres 3- 3x/8 + x= 5-5x/8 24- 3x+8x= 40-5x 10x=16 X=1.6 lits
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Specialization in Organic Chemistry. Teaching Chemistry has always been my passion.

40% of the mixture shall be drawn and replaced with water.
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Let total volume =v unit, then W= 3/8V, Syrup=5/8V, Let x qty is drawn, then water= 3/5(V-x), Syrup= 5/8(V-x) let y qty water added, 3/5(V-x)+y=V/2, 5/8(V-x)=V/2 then on solving, y=V/5=1/5(volume of syrup)
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passion for teaching mathematics

Let total volume =v unit, then W= 3/8V, Syrup=5/8V, Let x qty is drawn, then water= 3/5(V-x), Syrup= 5/8(V-x) let y qty water added, 3/5(V-x)+y=V/2, 5/8(V-x)=V/2 then on solving, y=V/5=1/5(volume of Mix)
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One fifth......it is like if l lts of content is assumed, 3l/8 will b water and 5l/8 will be syrup. Now if a lts of content is removed and a lts of water is added , then water content in new mixture will be 3(l-a)/8+a which must b equal to syrup in new mixture, 5(l-a)/8. Equating them, you'd get a=l/5....
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One fifth......it is like if l lts of content is assumed, 3l/8 will b water and 5l/8 will be syrup. Now if a lts of content is removed and a lts of water is added , then water content in new mixture will be 3(l-a)/8+a which must b equal to syrup in new mixture, 5(l-a)/8. Equating them, you'd get a=l/5. That means 1/5th of original content needs to b removed and replaced with water read less
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Tutor,Mentor,Consultant and Professional drives through result oriented approach.

1/8 part
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Mathematics is an epic

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