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A tank holds x gallons of a saltwater solution that is 20% [#permalink]
22 Feb 2008, 15:33

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00:00

A

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E

Difficulty:

75% (hard)

Question Stats:

52% (02:55) correct
48% (03:12) wrong based on 66 sessions

A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?

A. 37.5 B. 75 C. 100 D. 150 E. 175

Can you show me how to solve this without backsolving, ie the real algabraic way?

Re: Salt Storage Tank [#permalink]
22 Feb 2008, 18:13

4

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Manbehindthecurtain wrote:

neelesh wrote:

Is it e. 175 ?

Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.

Original mixture has 20% salt and 80% water. Total = x Out of which Salt = 0.2x and water = 0.8x Now, 1/4 water evaporates and all salt remains. So what remains is 0.2x salt and 0.6x water. Now 20 gallons salt is added and 10 gallons of water is added. So salt now becomes -> (0.2x + 20) and water --> (0.6x+10) Amount of salt is 33.33% of total. So amount of water is 66.66%. So salt is half of the volume of water. So (0.2x+20) = (0.6x+10)/2, Solving, x = 150.

Re: A tank holds x gallons of a saltwater solution that is 20% [#permalink]
18 Apr 2014, 13:01

1

This post received KUDOS

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Understand the logic of the flow of question.

Consider X initial solution.

0.2X = Salt 0.8X = Water

1/4(0.8X)= Evaporated Leaving 0.6X=water

0.6X+10=Water 0.2X+20=Salt

Salt concentration now becomes= 1/3 of the Solution

(0.2X+20)/(0.8X+30) = 1/3

X=150 _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: Salt Storage Tank [#permalink]
22 Feb 2008, 19:21

rgajare14 wrote:

Manbehindthecurtain wrote:

neelesh wrote:

Is it e. 175 ?

Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.

Original mixture has 20% salt and 80% water. Total = x Out of which Salt = 0.2x and water = 0.8x Now, 1/4 water evaporates and all salt remains. So what remains is 0.2x salt and 0.6x water. Now 20 gallons salt is added and 10 gallons of water is added. So salt now becomes -> (0.2x + 20) and water --> (0.6x+10) Amount of salt is 33.33% of total. So amount of water is 66.66%. So salt is half of the volume of water. So (0.2x+20) = (0.6x+10)/2, Solving, x = 150.

Outstanding response. thank you. Can you describe how you started unpeeling this? I found the question wording really confusing.

Also, what level difficulty would you say this is?

Re: Salt Storage Tank [#permalink]
22 Feb 2008, 19:53

Quote:

Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.

Original mixture has 20% salt and 80% water. Total = x Out of which Salt = 0.2x and water = 0.8x Now, 1/4 water evaporates and all salt remains. So what remains is 0.2x salt and 0.6x water. Now 20 gallons salt is added and 10 gallons of water is added. So salt now becomes -> (0.2x + 20) and water --> (0.6x+10) Amount of salt is 33.33% of total. So amount of water is 66.66%. So salt is half of the volume of water. So (0.2x+20) = (0.6x+10)/2, Solving, x = 150.

Quote:

Outstanding response. thank you. Can you describe how you started unpeeling this? I found the question wording really confusing.

Also, what level difficulty would you say this is?

Thanks. Actually , I did not solve it backwards. I solved it as per the flow given in the problem question. Honestly, I dont think this was more than a 650 level question.

Re: A tank holds x gallons of a saltwater solution that is 20% [#permalink]
15 Feb 2014, 16:40

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