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how much of the mixture is to be removed and replaced [#permalink]
 14 Aug 2009, 17:56
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10kg of a mixture contains 30% sand and 70% clay. In order to make the mixture contain equal quantities of clay and sand. how much of the mixture is to be removed and replaced with pure sand? Detailed explanation please.
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Re: how much of the mixture is to be removed and replaced [#permalink]
14 Aug 2009, 18:25
The mixture contains 3kg sand and 7 kg clay.
For the mixture to be in equal quantities, there should be 2kg of clay removed.
clay and sand are in the ratio 7:3
So part of sand to be removed = 2*3/7 = 6/7
So total mixture to be removed = 2 + 6/7 = 20/7
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Re: how much of the mixture is to be removed and replaced [#permalink]
14 Aug 2009, 19:53
cant write the fractions in symbols 
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Last edited by joebloggs on 14 Aug 2009, 20:13, edited 1 time in total.
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Re: how much of the mixture is to be removed and replaced [#permalink]
14 Aug 2009, 20:07
Alee's way is probably faster but here's another way of thinking about it.
You have 7kg clay and 3kg sand. For every 1kg of mixture you remove you lose 7/10kg clay and 3/10kg sand. You then give that 1kg back to sand, so the net effective is gaining -3/10kg + 1kg = 7/10kg sand.
So,
7-(7/10)x
divided by
3+(7/10)x
= 1.
Solve for x => 20/7
actually i'm having trouble understanding alee's method even though it's clearly correct. if you just solve for 2 grams of clay removal (fixing the ratio at 1:1), and then independently solve for sand, doesn't that imply that the ratio of sand/clay then changes from 1:1 to something else?
Last edited by bipolarbear on 14 Aug 2009, 20:14, edited 2 times in total.
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Re: how much of the mixture is to be removed and replaced [#permalink]
14 Aug 2009, 20:12
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Let the mixture to be replaced by pure sand be x we want the sand/ clay ratio to be 1:1 therefore after replacement sand/clay = (3 - 0.3x + x)/(7 - 0.7x) = 1/1 removing x kg of mixture means removing 3/10*x kg of sand(as x kg mixture has 30% sand) from the existing sand that is 3 kg,also we are adding x kg pure sand(replacement!) so +x in numerator, same goes for the clay, x kg of mixture has 7/10*x kg clay...so after replacement clay left is 7 - 0.7x (no addition here). the required ratio is 1:1 so.... solving for x gives x = 20/3so here is the "detailed" explanation ....hope that helps 
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Re: how much of the mixture is to be removed and replaced [#permalink]
14 Aug 2009, 20:36
bipolarbear wrote: Alee's way is probably faster but here's another way of thinking about it.
You have 7kg clay and 3kg sand. For every 1kg of mixture you remove you lose 7/10kg clay and 3/10kg sand. You then give that 1kg back to sand, so the net effective is gaining -3/10kg + 1kg = 7/10kg sand.
So,
7-(7/10)x
divided by
3+(7/10)x
= 1.
Solve for x => 20/7
actually i'm having trouble understanding alee's method even though it's clearly correct. if you just solve for 2 grams of clay removal (fixing the ratio at 1:1), and then independently solve for sand, doesn't that imply that the ratio of sand/clay then changes from 1:1 to something else? bipolarbear, even I found it somewhat difficult to understand but I got it...heres how here we are trying to remove 2 kg of clay from the mixture so that removing 2 kg clay leaves 5 kg clay and adding 2 kg sand leaves 5 kg sand..giving a 1:1 mixture sand:clay=3:7 there is 7 kg clay for every 3 kg sand for 1 kg clay, sand =3/7 for 2 kg clay, sand =3/7*2 so removing 2 kg clay from the mixture means also removing 21/2 kg sand along with it. so the total mixture removed is 2kg (clay for which we calculated) + sand (according to the ratio) = 2 + 6/7 = 20/7 PROOFfrom the mixture 6/7 kg sand has been removed which leaves us 3 - 6/7= 15/7 kg sand but we are also adding the same amount of pure sand which is equal to the mixture removed(20/7) so... checking the answer: remaining sand+new sand = 15/7 + 20/7 = 5 kg so now the sand is 5kg and the clay is also 5kg ratio 1:1 actually ,never seen a mixture problem solved like this....I too was doubtful abt alee's method bit eccentric 
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Re: how much of the mixture is to be removed and replaced [#permalink]
14 Aug 2009, 20:53
I know my solution is bit diff, but I generally take less time using this methodology for solving mixture problems. So only I use this. In GMAT ultimately, it is only the final answer which counts not how we solve it
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Re: how much of the mixture is to be removed and replaced [#permalink]
14 Aug 2009, 20:55
Haha I am still confused... however while thinking about it I guess I sort of understand. I have to do it a more roundabout way though. You need to remove 2 grams of clay to get a 5:5 split. However, if you remove 2 grams of pure clay, you are actually removing 2*(10/7) total grams and replacing that with sugar. So the sugar concentration is now 3+20/7= 41/7. That is 6/7 higher than the 5 you were aiming for so you must subtract it. The total is then 6/7 + 2 = 20/7. I think I'll stick with our original method which seems significantly easier and much less abstract 
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Re: how much of the mixture is to be removed and replaced [#permalink]
15 Aug 2009, 00:16
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I normally use a table method for such mixture problems which is quite easy.
Attachments
mix.JPG [ 17.11 KiB | Viewed 5509 times ]
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Re: how much of the mixture is to be removed and replaced [#permalink]
15 Aug 2009, 05:29
Economist wrote: I normally use a table method for such mixture problems which is quite easy. wow, thats a simple and quick one...I did considering clay with this method and it works... thanks O- R+ A=D 7-.7x+0=5 x=20/7 
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Re: how much of the mixture is to be removed and replaced [#permalink]
28 Oct 2009, 18:46
ummm...Ok MAJOR querie..Why is the answer not 4?! We need sand and clay to be 50% each.We have clay 7kgs.This represents 50%.So we get sand as 7 kgs. 7-3=4 kgs..
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Re: how much of the mixture is to be removed and replaced [#permalink]
09 Nov 2009, 11:56
tejal777 wrote: ummm...Ok MAJOR querie..Why is the answer not 4?!
We need sand and clay to be 50% each.We have clay 7kgs.This represents 50%.So we get sand as 7 kgs. 7-3=4 kgs.. Because you cannot remove the clay alone - you can only remove a mix of clay and sand. Think of it this way: Imagine you removed everything from the container and started pouring the mixture and pure sand back in. In what ratio of sand:mixture would you put them back in to get to 50-50? The ratio would be 5:2 [i.e. 5 mixture to 2 pure sand]. Now, since you have 10kg, the total amount of pure sand would be (2/7)*10 = 20/7. The amount of pure sand is what the question was looking for in the first place.
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Re: how much of the mixture is to be removed and replaced [#permalink]
22 Nov 2009, 02:34
Very useful and simple.
Thanks the Economist
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Re: how much of the mixture is to be removed and replaced [#permalink]
22 Nov 2009, 13:01
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Hummm I hate mixture problems...
You have 7kg of C and 3 kg of S
Your final mixture has to have 5kg of C and 5kg of S.
But if you remove Clay, you will remove sand as well.
So you have to take into account this.
The question is: how much of (S+C) will you have to remove in order to remove 2kg of C?
The answer is:
0.7*X=2
X = 2/0.7
This is not an easy question.
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Re: how much of the mixture is to be removed and replaced [#permalink]
22 Nov 2009, 15:59
tejal777 wrote: 10kg of a mixture contains 30% sand and 70% clay. In order to make the mixture contain equal quantities of clay and sand. how much of the mixture is to be removed and replaced with pure sand?
Detailed explanation please. This is an interesting problem. The best and fastest way I'd suggest is to derive an equation based on the given statements. I remove X kg of mixture and add X kg of pure sand. so the total weight of the mixture should remain constant at 10kg. I am not considering adding pure sand to 70% of clay because, it complicates the second equation. (10 - X)(30/100) + X = 10(50/100) X = 20/7 kg.
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Re: how much of the mixture is to be removed and replaced [#permalink]
25 Jan 2010, 00:23
the method described for approaching averages works for this too. It's listed under "Strategies for solving problems with Average" on the Math Tips page.
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Re: how much of the mixture is to be removed and replaced [#permalink]
28 Mar 2010, 14:33
tejal777 wrote: 10kg of a mixture contains 30% sand and 70% clay. In order to make the mixture contain equal quantities of clay and sand. how much of the mixture is to be removed and replaced with pure sand?
Detailed explanation please. Required: 5kg to remain for both sand & clay let x = quantity remaining after some mixture have been removed. 70% of x = 5; x=50/7 sand to be replaced = 10 - 50/7 = 20/7
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Re: how much of the mixture is to be removed and replaced [#permalink]
16 Apr 2010, 13:54
yes it will be 20/7 only as 2+6/7 = 20/7 , the 6/7 of sand which is removed will be replaced with pure sand, thus it is nullified.
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Re: how much of the mixture is to be removed and replaced [#permalink]
11 Jul 2010, 05:45
I have gone through almost all the methods of solving mixture problem. However, i am not able to apply any of them to any of the mixture problem. It is getting toooooo discouraging now  Can someone pleaseeeee explain this problem once again....
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Re: how much of the mixture is to be removed and replaced [#permalink]
11 Jul 2010, 12:43
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bibha wrote: I have gone through almost all the methods of solving mixture problem. However, i am not able to apply any of them to any of the mixture problem. It is getting toooooo discouraging now Can someone pleaseeeee explain this problem once again.... Initially, Sand = 30% of 10 = 3kg Clay = 70% of 10 = 7 kg Suppose x kg of the material is removed. in x kg 30% of x will be sand and 70% of x will be clay thus now quantity of Sand left = 3-0.3x and Clay = 7-0.7x since equal amount is replaced => x kg of sand is now added As per the given condition after the replacement both sand and clay are equal in quantity, => 3-0.3x + x = 7-0.7x => x=20/7
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Re: how much of the mixture is to be removed and replaced
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