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Re: How many liters of water must be evaporated from 50 liters of a 3-perc [#permalink]

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26 Nov 2014, 08:06

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How many liters of water must be evaporated from 50 liters of a 3-percent sugar solution to get a 10-percent solution?

3% of a 50 liter solution is 1.5L. So you are trying to determine how many liters must a solution be for the 1.5L to represent 10% of the solution. Set up an inequality and solve for x:

1.5/x = 1/10

x = 15

Since you need a 15L solution, you must evaporate 35 of the original 50L solution to get a 10% solution.

Re: How many liters of water must be evaporated from 50 liters of a 3-perc [#permalink]

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26 Nov 2014, 11:51

We start with 1.5L of sugar and 48.5L of water. This is a 3% sugar solution. To get to a 10% sugar solution, we need 1.5L sugar per 15L of total volume or 13.5L of water. Just evaporate 48.5-13.5=35L of water.

Re: How many liters of water must be evaporated from 50 liters of a 3-perc [#permalink]

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01 Dec 2014, 00:09

PareshGmat wrote:

Answer = A) 35

Sugar should constitute 10% of the solution post water removal \(\frac{10}{100} (50-x) = 1.5\)

x = 50-15 = 35

Hey , could anyone please tell me why shouldnt we substract \(x\) FROM 1.5 as well here , since a part of solution is removed from main solution. When we remove and replace the solution, we remove that part ofsolution from LHS as well as RHS of the equation.

Kind Regards, Vardhaman
_________________

ITS NOT OVER , UNTIL I WIN ! I CAN, AND I WILL .PERIOD.

Re: How many liters of water must be evaporated from 50 liters of a 3-perc [#permalink]

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01 Dec 2014, 00:20

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vards wrote:

PareshGmat wrote:

Answer = A) 35

Sugar should constitute 10% of the solution post water removal \(\frac{10}{100} (50-x) = 1.5\)

x = 50-15 = 35

Hey , could anyone please tell me why shouldnt we substract \(x\) FROM 1.5 as well here , since a part of solution is removed from main solution. When we remove and replace the solution, we remove that part ofsolution from LHS as well as RHS of the equation.

Kind Regards, Vardhaman

1.5 is 100% sugar concentrate. We cannot remove "only sugar" from the liquid mixture, so it remains as it is.

The equation setup is for the concentration to evaluate value of x.

Yes, you can see in the table that from the total of 50, "x" was removed (Here LHS/RHS is taken into account)
_________________

How many liters of water must be evaporated from 50 liters of a 3-perc [#permalink]

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01 Dec 2014, 02:56

PareshGmat wrote:

vards wrote:

PareshGmat wrote:

Answer = A) 35

Sugar should constitute 10% of the solution post water removal \(\frac{10}{100} (50-x) = 1.5\)

x = 50-15 = 35

Hey , could anyone please tell me why shouldnt we substract \(x\) FROM 1.5 as well here , since a part of solution is removed from main solution. When we remove and replace the solution, we remove that part ofsolution from LHS as well as RHS of the equation.

Kind Regards, Vardhaman

1.5 is 100% sugar concentrate. We cannot remove "only sugar" from the liquid mixture, so it remains as it is.

The equation setup is for the concentration to evaluate value of x.

Hi Paresh. Thanks for the reply.

Yes, you can see in the table that from the total of 50, "x" was removed (Here LHS/RHS is taken into account)

Hi Paresh. Thanks for the reply.

Yes i understand, but the solution contains sugar. So, when the solution evaporates, the sugar within the solution should also evaporate.

I mean , suppose=100 litres of water has 10 percent concentration of alcohol. so when 50 percent of the solution evaporates, it means 50 litres of solution will evaporate which will contain 5 litres of alcohol. In the same context here, i thought sugar should also evaporate from solution in equal proportion.

Please clear my doubt. Many Thanks.
_________________

ITS NOT OVER , UNTIL I WIN ! I CAN, AND I WILL .PERIOD.

Re: How many liters of water must be evaporated from 50 liters of a 3-perc [#permalink]

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01 Dec 2014, 03:17

1

This post received KUDOS

vards wrote:

vards wrote:

PareshGmat wrote:

Answer = A) 35

Sugar should constitute 10% of the solution post water removal \(\frac{10}{100} (50-x) = 1.5\)

x = 50-15 = 35

Hey , could anyone please tell me why shouldnt we substract \(x\) FROM 1.5 as well here , since a part of solution is removed from main solution. When we remove and replace the solution, we remove that part ofsolution from LHS as well as RHS of the equation.

Kind Regards, Vardhaman

1.5 is 100% sugar concentrate. We cannot remove "only sugar" from the liquid mixture, so it remains as it is.

The equation setup is for the concentration to evaluate value of x.

Hi Paresh. Thanks for the reply.

Yes, you can see in the table that from the total of 50, "x" was removed (Here LHS/RHS is taken into account)

Hi Paresh. Thanks for the reply.

Yes i understand, but the solution contains sugar. So, when the solution evaporates, the sugar within the solution should also evaporate.

I mean , suppose=100 litres of water has 10 percent concentration of alcohol. so when 50 percent of the solution evaporates, it means 50 litres of solution will evaporate which will contain 5 litres of alcohol. In the same context here, i thought sugar should also evaporate from solution in equal proportion.

Please clear my doubt. Many Thanks.

Sugar cannot evaporate. Only "water content" in the solution evaporates.

100 Litres of 10% alcohol solution = 90 Litres of water + 10 Litres of Alcohol (Pure alcohol, 100%; NO water)

50% solution evaporates; so resultant would be

(100-50) Litres of solution = (90-50) Litres of water + 10 Litres of Alcohol

50 Litres of solution = 40 Litres of water + 10 Litres of Alcohol

As quantity of alcohol remains same, its concentration is now 20% in the solution
_________________

Re: How many liters of water must be evaporated from 50 liters of a 3-perc [#permalink]

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15 Oct 2017, 09:06

50L of solution contains 3% sugar, i.e 1.5L The rest(97%) is water, 50-1.5 = 48.5L

In order to get a target solution that is 90% water, we need to remove some water. Let this be 'X' liters. Hence, the equation becomes: 48.5 - x = 0.9(50-x) 3.5=0.1x x = 35 L
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