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A solution contains 8 parts of water for every 7 parts of
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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
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Originally posted by udaymathapati on 31 Oct 2010, 07:14.
Last edited by Bunuel on 15 Jul 2012, 06:25, edited 2 times in total.
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Re: Lemonade Syrup
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31 Oct 2010, 19:38
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
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Re: Lemonade Syrup
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31 Oct 2010, 08:10
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




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Re: Lemonade Syrup
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31 Oct 2010, 07:48
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
Please explain. \(\frac{water}{syrup}=\frac{8}{7}\); Consider the solution to be 15 liters, so it will contain 8 liters of water and 7 liters syrup. We want to replace \(x\) liters of solution with water so that amount of syrup decreased from 7 liters to 15*0.4=6 liters. So, we should replace (remove) 1 liter of syrup: but with every 1 liter of syrup comes 8/7 liters of water (\(\frac{water}{syrup}=\frac{8}{7}\) > \(\frac{water}{1}=\frac{8}{7}\) > \(water=\frac{8}{7}\)) so \(x=1+\frac{8}{7}\approx{2.14}\) liters. Answer: C.
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Re: Lemonade Syrup
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31 Oct 2010, 19:51
C it is total solution 15, Lemonoid = 7 we want to make it 40% lemonoid >> so 6 parts of lemonoid so we need to remove 1 part of lemonoid each part of solution has 7/15 lemonoid that means for 1 part lemonoid will be in 15/7 = 2.14 prt of solution which will be replaced with water



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Re: Lemonade Syrup
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02 Nov 2010, 21:48
Does anybody else find the question description ambiguous?
"A solution contains 8 parts for every 7 parts of Lemonade syrup" Your interpretation is that syrup and water relate as 7 and 8.
However, it can also mean that syrup takes 7 parts out of total of 8 parts of solution.
Does it make sense?



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Re: Lemonade Syrup
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03 Nov 2010, 04:54
Fijisurf wrote: Does anybody else find the question description ambiguous?
"A solution contains 8 parts for every 7 parts of Lemonade syrup" Your interpretation is that syrup and water relate as 7 and 8.
However, it can also mean that syrup takes 7 parts out of total of 8 parts of solution.
Does it make sense? Yep. When I read the question, I thought that either the question is worded improperly or the word 'water' after '8 parts' is missing. But I figured that if it is 8 parts solution then the options don't match the answer so it must be 8 parts water. Don't fret!
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Re: Lemonade Syrup
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14 Jul 2012, 11:14
Bunuel wrote: 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
Please explain. \(\frac{water}{syrup}=\frac{8}{7}\); Consider the solution to be 15 liters, so it will contain 8 liters of water and 7 liters syrup. We want to replace \(x\) liters of solution with water so that amount of syrup decreased from 7 liters to 15*0.4=6 liters. So, we should replace (remove) 1 liter of syrup: but with every 1 liter of syrup comes 8/7 liters of water (\(\frac{water}{syrup}=\frac{8}{7}\) > \(\frac{water}{1}=\frac{8}{7}\) > \(water=\frac{8}{7}\)) so \(x=1+\frac{8}{7}\approx{2.14}\) liters. Answer: C. Can you explain that why the new ration is w/1 = 8/7 ? I thought it be w/6 =8/7..Thanks



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Re: Lemonade Syrup
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15 Jul 2012, 06:35
farukqmul wrote: Bunuel wrote: 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
Please explain. \(\frac{water}{syrup}=\frac{8}{7}\); Consider the solution to be 15 liters, so it will contain 8 liters of water and 7 liters syrup. We want to replace \(x\) liters of solution with water so that amount of syrup decreased from 7 liters to 15*0.4=6 liters. So, we should replace (remove) 1 liter of syrup: but with every 1 liter of syrup comes 8/7 liters of water (\(\frac{water}{syrup}=\frac{8}{7}\) > \(\frac{water}{1}=\frac{8}{7}\) > \(water=\frac{8}{7}\)) so \(x=1+\frac{8}{7}\approx{2.14}\) liters. Answer: C. Can you explain that why the new ration is w/1 = 8/7 ? I thought it be w/6 =8/7..Thanks 8/7 is not the new ratio. Again: we want to remove 1 liter of syrup, but with every 1 liter of syrup comes 8/7 liters of water (because if s=1 then w from w/s=8/7 becomes 8/7), hence in order to remove 1 liter of syrup we should remove total of 1+8/7 liters of mixture. Hope it's clear.
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Re: 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
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17 Jul 2013, 01:57
1. Let the number of parts of the original solution be 30. 2.The original solution contains 14 parts lemonade and 16 parts water 3. The new solution contains 12 parts lemonade and 18 parts water 4. Let x parts of solution be replaced. Water replaced is 16x/30 5. x parts of water is added 6. From (2) and (3) we see the net increase in water is 1816=2 parts 7. From (4) and (5) we see the net increase in water is x(16x/30) parts 8. Equating (6) and (7) we have x=4.28 9. If we take the total number of parts as 30, the solution removed is 4.28 parts. Therefore for a total of 15 parts, the solution removed is 2.14 parts
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Re: A solution contains 8 parts of water for every 7 parts of
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13 Aug 2013, 22:46
Quote: Why we take 2/5 as 40% ..we need to take 40 % of 7/15 ?
The question tells you that in the final solution, lemonade syrup is 40% of the solution i.e. there is 2 parts lemonade syrup for 3 parts of water It does not imply that the concentration of lemonade syrup is 40% of its initial concentration. The final concentration is actually 40% i.e. 2/5.
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Re: A solution contains 8 parts of water for every 7 parts of
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14 Aug 2013, 09:34
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 My solution with explanation:
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Re: A solution contains 8 parts of water for every 7 parts of
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08 Feb 2014, 12:59
Alternative approach, give kudos if you like
Let's say total of 15 liters, 8 of water and 7 of lemonade Therefore, if we need lemonade to be 60% we need the ratio to be 3:2
Thus we have
2(8+x8/15x) = 3(77/15x)
Solving for 'x' we get 15/7=2.14 (C)
Just to elaborate a little more on this. 2:3 are of course the ratios, in some problems we are asked so that there's an equal amount of both, then we don't need to care about the 2 and 3.
Next, we are basically replacing quantities so if we put x we take away x of the solution. But the 'x' liters of the solution contain part of both lemonade and water, therefore, that's why we use the respective ratios.
Hope its clear now
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Re: A solution contains 8 parts of water for every 7 parts of
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08 Feb 2014, 21:25
Def took longer to rationalize than I would have allowed on test day...anyway, here's my take: Replacing one part of the solution with water will take away 8/15 of water and 7/15 of Lemonade (and replace it with 15/15 water)...essentially just switching out 7/15 of Lemonade with 7/15 Water for every part removed... So, from a Lemonade composition perspective: [7  (7x/15)]/15 = 4/10 x=15/7
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Re: A solution contains 8 parts of water for every 7 parts of
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22 Mar 2015, 18:29
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)/(15w)=0.6 solving which gives w=0.525 What's wrong here?



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Re: A solution contains 8 parts of water for every 7 parts of
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22 Mar 2015, 18:45
Hi VSabc, This question is a little tougher than a typical "mixture" question. The prompt tells us to REPLACE some of the existing mixture with pure water (with the goal of turning the new mixture into a 40% syrup mix. To start, we have 15 total liters >a mixture that is 8 liters water and 7 liters syrup. If we pour 1 liter of this mixture into a glass, we would have a liquid that is 7/15 syrup (so a little less than half syrup). For the mixture to be 15 total liters and 40% syrup, we need the mixture to be 9 liters water and 6 liters syrup. In basic math terms, we need to pour out enough of the mixture that we remove 1 full liter of syrup; when we pour an equivalent amount of water back in, we'll have 15 total liters (and 6 of them will be syrup). Since each liter is 7/15 syrup...... We need to remove 15/7 liters and replace them with 15/7 liters of pure water. 15/7 is a little more than 2 liters (about 2.14 liters) Final Answer: GMAT assassins aren't born, they're made, Rich
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A solution contains 8 parts of water for every 7 parts of
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22 Mar 2015, 23:08
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)/(15w)=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) \(8 + (7/15)*w = 9\) \(w = 15/7\)
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Re: A solution contains 8 parts of water for every 7 parts of
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22 Mar 2015, 23:17
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



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Re: A solution contains 8 parts of water for every 7 parts of
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23 Mar 2015, 00:52
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 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.
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