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A solution contains 8 parts of water for every 7 parts of

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Re: A solution contains 8 parts of water for every 7 parts of  [#permalink]

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New post 23 Mar 2015, 04:49
VeritasPrepKarishma wrote:
GauravSolanky wrote:
VeritasPrepKarishma wrote:
Though Bunuel and soumanag have already explained the solution well, I will add the method I like the most.

In replacement questions, focus on the thing that decreases. When solution is removed, water decreases but then water is added. While when solution is removed, lemonade decreases and does not get added later. So we will work with lemonade concentration.

The fraction of lemonade in the solution is 7/15
We need to get this fraction down to 2/5 (to make it 40%)
Let us say, we remove a fraction 'f' of the solution.
Then 7/15 - f * (7/15) = 2/5
f = 1/7
So (1/7)th of the solution has to be removed. But we want the answer in terms of parts (how many of the 15 parts have to be removed)
So we need to remove (1/7) * 15 = 2.14 parts




Hi Karishma,

Thanks for explaining it so well. I got it wrong as I thought we have to give answer in terms of lemonade.
So, here can we say that replacing 1 unit of lemonade and 1.14 units of water will serve the purpose ?

Regards,

Gaurav :-D


Yes, you are removing a total of 2.14 units, of which (7/15)*2.14 = 1 unit is lemonade and rest 1.14 units is water.

Note that saying "replace 1 unit of lemonade and 1.14 units of water" is not very logical since you cannot remove the two separately. They are mixed together so you need to remove the solution only. You cannot remove 1 unit of lemonade alone since water will come along with it. So it will be logical to say that we must remove 2.14 units of solution of which 1 unit will be lemonade and rest will be water since solutions are assumed homogeneous.



I stated that way as I just wanted to put more stress on quantities separately. But yes, you are correct that was illogical.

Thank Again. :)
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A solution contains 8 parts of water for every 7 parts of  [#permalink]

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New post Updated on: 10 Jun 2017, 20:45
x=fraction of solution replaced
8/15-8x/15+x=9/15
x=1/7
1/7*15=2.14 parts of solution replaced
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Originally posted by gracie on 21 Dec 2015, 15:26.
Last edited by gracie on 10 Jun 2017, 20:45, edited 1 time in total.
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Re: A solution contains 8 parts of water for every 7 parts of  [#permalink]

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New post 29 Dec 2015, 13:50
VeritasPrepKarishma wrote:
VSabc wrote:
It seems I took a completely different tangent here, can someone please help me:
Let there be total 15l of solution implying 8l water and 7l lemonade. Acc. to the problem, let's take off 'w'l from solution and add 'w'l of water implying:
(8+w)/(15-w)=0.6 solving which gives w=0.525
What's wrong here?


Here is the problem with your equation.

When you took out 'w' lt of solution, the water left is less than 8 lt. So how can total water after replacement be (8 + w) lts? It will 'something less than 8 + w' lts.
Also, the new solution after you replace with water is again 15 lts. So why would you have (15 - w) in the denominator?

Your equation should be

\(\frac{8 - (8/15)*w + w}{15} = 0.6\) (the fraction of water removed will be (8/15) of w)

\(8 + (7/15)*w = 9\)

\(w = 15/7\)


HI Karishma,

I am just wondering if we can apply scale method to this problem?! The reason is - we know desired %, we have relevant proportions of water & lemonade which can be converted into %
Or Is my approach not right?
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Re: A solution contains 8 parts of water for every 7 parts of  [#permalink]

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New post 29 Dec 2015, 23:34
2
madhusudhan237 wrote:
VeritasPrepKarishma wrote:
VSabc wrote:
It seems I took a completely different tangent here, can someone please help me:
Let there be total 15l of solution implying 8l water and 7l lemonade. Acc. to the problem, let's take off 'w'l from solution and add 'w'l of water implying:
(8+w)/(15-w)=0.6 solving which gives w=0.525
What's wrong here?


Here is the problem with your equation.

When you took out 'w' lt of solution, the water left is less than 8 lt. So how can total water after replacement be (8 + w) lts? It will 'something less than 8 + w' lts.
Also, the new solution after you replace with water is again 15 lts. So why would you have (15 - w) in the denominator?

Your equation should be

\(\frac{8 - (8/15)*w + w}{15} = 0.6\) (the fraction of water removed will be (8/15) of w)

\(8 + (7/15)*w = 9\)

\(w = 15/7\)


HI Karishma,

I am just wondering if we can apply scale method to this problem?! The reason is - we know desired %, we have relevant proportions of water & lemonade which can be converted into %
Or Is my approach not right?


You can apply scale method but it will be a two step method and will need you to understand the logic properly.

Step1:
w1/w2 = (0 - 2/5)/(2/5 - 7/15) = (6/15)/(1/15) = 6/1
So say 1ml water is added to 6 ml previous solution. This means that initially, the solution was a total of 7 ml out of which 1 ml was replaced so (1/7)th of the solution was replaced.

Step 2:
But the initial solution was actually 15 parts.
(1/7) * 15 = 2.14 parts
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Re: A solution contains 8 parts of water for every 7 parts of  [#permalink]

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New post 10 Nov 2016, 04:30
VeritasPrepKarishma wrote:
Though Bunuel and soumanag have already explained the solution well, I will add the method I like the most.

In replacement questions, focus on the thing that decreases. When solution is removed, water decreases but then water is added. While when solution is removed, lemonade decreases and does not get added later. So we will work with lemonade concentration.

The fraction of lemonade in the solution is 7/15
We need to get this fraction down to 2/5 (to make it 40%)
Let us say, we remove a fraction 'f' of the solution.
Then 7/15 - f * (7/15) = 2/5
f = 1/7
So (1/7)th of the solution has to be removed. But we want the answer in terms of parts (how many of the 15 parts have to be removed)
So we need to remove (1/7) * 15 = 2.14 parts


Responding to a pm:
Quote:
I did exactly what you did and arrived at the answer. But I have question regarding the wording. Is using "parts" here legit? I could easily arrive at f = 1/7. And then I thought to myself - "Okay, so 1 in every 7 total parts should be replaced", and then I was kind of blank for a while and then tried using 15 parts originally provided to solve it. Please explain..


Sure, the use of "parts" is fine. All measurements are in terms of the same "part". That part could 1 ml, 10 ml or 100 ml etc. There are total 15 of these parts in the solution.
When we get f = 1/7, f is the fraction of the solution removed. (1/7)th divides the whole solution into 7 equal parts and removes 1. These parts are not the same as the parts above. The total solution has 15 of the "parts" discussed above. Of these 15 parts, we remove 1/7th which is 2.14 "parts".
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Re: A solution contains 8 parts of water for every 7 parts of  [#permalink]

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New post 10 Nov 2016, 13:59
udaymathapati wrote:
A solution contains 8 parts of water for every 7 parts of Lemonade syrup. How many parts of the solution should be removed and replaced with water so that the solution will now contain 40% lemonade syrup?

A. 1.5
B. 1.75
C. 2.14
D. 2.34
E. 2.64


use C final = C initial (Volume original / volume final ) , final concentration is 40/100 , initial concentration is 7/15 , volume original is volume of the solution after removing part of it and before adding water to dilute it , V final is the total volume after replacing of x parts of the solution by x parts of water and thus total volume remains the same ( 15)

40/100 = 7/15 ( 15-x / 15 ) thus x = 15/7 = 2.14
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Re: A solution contains 8 parts of water for every 7 parts of  [#permalink]

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New post 15 Jan 2017, 20:00
y = amount removed from original = amount replaced with water
x = original amount

(8/15)x - (8/15)y + y = (3/5)x
(8/15)x + (7/15)y = (3/5)x
8x+7y = 9x
7y = x --> y = (1/7)x

There are 15 parts in the original --> so... 15/7 is our answer

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A solution contains 8 parts of water for every 7 parts of  [#permalink]

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New post 10 Jun 2017, 19:55
udaymathapati wrote:
A solution contains 8 parts of water for every 7 parts of Lemonade syrup. How many parts of the solution should be removed and replaced with water so that the solution will now contain 40% lemonade syrup?

A. 1.5
B. 1.75
C. 2.14
D. 2.34
E. 2.64

1. (Initial quantity of solution- quantity of solution removed)* strength of solution + (Quantity of water added *strength of water)/ Initial quantity= 40/100
2. Let x be the initial quantity and y be the quantity removed
3.( (x-y)*7/15 + y*0) / x = 0.4, x/y=7/1
4. If initial quantity is 7 parts, 1 part is removed, if 15 parts 15/7=2.14
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Re: A solution contains 8 parts of water for every 7 parts of  [#permalink]

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New post 16 Jul 2018, 04:49
soumanag wrote:
Let the total solution is 150 L with 80 L water & 70 L syrup.

To make 40% syrup solution, the result solution must have 90 L syrup and 60 L syrup.

Therefore we are taking 10 L of syrup from initial solution and replacing with water.

using urinary method:
70 L syrup in 150 L solution
10 L syrup in 21.4 L solution

We started by multiplying 10
Now to get to the result we need to divide by 10 => amount of solution to be replaced with water = (21.4/10) = 2.14.

Correct option : C

Great solution. But I think you meant unitary method. ;)
Re: A solution contains 8 parts of water for every 7 parts of &nbs [#permalink] 16 Jul 2018, 04:49

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