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# 10kg of a mixture contains 30% sand and 70% clay. In order t

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Re: how much of the mixture is to be removed and replaced  [#permalink]

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01 Oct 2013, 18:24
1
Economist wrote:
I normally use a table method for such mixture problems which is quite easy.

that only works if added & removed are the same, since they're both X in that equation, that implies the same amount was removed in both cases
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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12 Oct 2013, 02:01
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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24 Apr 2014, 10:10
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My first explanation on Gmatclub, hope it helps someone.

Since the question specifies that there is 10kg of the mixture which contains sand and clay for there to be equal quantities of sand and clay each should be 5 kg i.e. 5kg of sand and 5kg of clay.

Now since 70% of the mixture is clay there is 7kg of clay in the mixture i.e. 2kg in excess of 5kg of clay required for equal quantities. So 2 kg of clay has to be removed, but for every kg of clay there is (1X3)/7 kg of sand therefore for 2 kg of clay there is 2(1X3)/7=6/7 kg of sand in the current mixture.

So for 2 kg of clay to be removed, 6/7 of sand has to also be removed. Thus total mixture removed = 2+6/7= 20/7

Please do give me a kudos if you like my explanation, thanks.
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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24 Apr 2014, 20:17
1
1
Sand........ Clay...... Total
3 .............. 7 ........... 10

Given that 30% is sand & 70% is clay

Say we remove mixture quantity = x

It means we are removing $$\frac{30x}{100} sand & \frac{70x}{100}$$ clay from the mixture

New table would be as follows:

Sand........ Clay...... Total
$$3-\frac{30x}{100}.............. 7-\frac{70x}{100} ........... 10-x$$

Quantity removed of mixture = Quantity added of sand (which gives both sand & clay quantities SAME in the mixture)

$$3 - \frac{30x}{100} + x = 7 - \frac{70x}{100}$$

$$x + \frac{40x}{100} = 4$$

$$x = 4 * \frac{100}{140}$$

$$x = \frac{20}{7}$$
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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09 May 2014, 01:16
1
If equal quantity will be 5kg sand and 5 kg clay. So I am removing 2 kg clay
70% mixture=2kg
1%=2/70
100%=2*100/70= 20/7
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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08 Dec 2014, 23:11
after following all the answers I found this method easy but I am not clear from where did you get 1 as concentration for added . Please explain me.

there is a similar problem but ratios are used.please explain this problem with the help of this method
Ques:
A vessel is filled with liquid ,3 parts of which are water and 5 parts syrup. How much of mixture must be drawn off and replaced with water so the mixture may be half water and half syrup?
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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08 Dec 2014, 23:37
shrestharaj wrote:
after following all the answers I found this method easy but I am not clear from where did you get 1 as concentration for added . Please explain me.

there is a similar problem but ratios are used.please explain this problem with the help of this method
Ques:
A vessel is filled with liquid ,3 parts of which are water and 5 parts syrup. How much of mixture must be drawn off and replaced with water so the mixture may be half water and half syrup?

Water ................... Syrup ............................ Total

3 .............................. 5 .................................. 8

$$3 - \frac{3x}{8}$$ .................. $$5 - \frac{5x}{8}$$ ............................ 8 - x (Say x litres of mixture is taken off)

$$3 - \frac{3x}{8} + x$$ ................ $$5 - \frac{5x}{8}$$ ......................... 8 (Say x litres of water is added; same as quantity of mixture removed)

Given that now the water content should be 50%, so setting up the equation accordingly

$$\frac{1}{2} * 8 = 3 - \frac{3x}{8} + x$$

$$\frac{5x}{8} = 1$$

$$x = \frac{8}{5}$$
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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09 Dec 2014, 00:46
Unitary method
Out of 5 parts we have to remove 1
out of 1 part we have to remove 1/5
Out of 8 parts we have to remove (1/5)*8 = 8/5
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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06 Jan 2015, 08:30
I still don't get it. Perhaps I should call it a day.

However, since we have 3 kg of sand and 7kg of clay, if we remove 2 kg of clay and replace the missing 2 kg with sand we have 5 kg sand and 5 kg clay.

So, 2/10 of the mixture needs to be removed (if we cannot distinguish the sand from the clay in the mix and think of the mixture as one entity) or 2/7 of the clay needs to be removed, if we can pick the clay from the mix and only remove clay and no sand...

I don't know how we end up with 20/7...
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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06 Jan 2015, 17:46
pacifist85 wrote:
I still don't get it. Perhaps I should call it a day.

However, since we have 3 kg of sand and 7kg of clay, if we remove 2 kg of clay and replace the missing 2 kg with sand we have 5 kg sand and 5 kg clay.

So, 2/10 of the mixture needs to be removed (if we cannot distinguish the sand from the clay in the mix and think of the mixture as one entity) or 2/7 of the clay needs to be removed, if we can pick the clay from the mix and only remove clay and no sand...

I don't know how we end up with 20/7...

Its a 10kg mixture. Agreed that "pure 2kg" clay has to be removed and "pure 2kg" sand has to added to make it 50-50.

However, in this case, as its a mixture, we cannot separate clay & sand. A variable has to be take to treat this problem further. Please refer my earlier related post
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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20 Jun 2015, 03:53
Economist wrote:
I normally use a table method for such mixture problems which is quite easy.

Hi Economist,

The amount in that table will be total amount ? If we're considering "Sand" then why not use the amount of Sand in the mixture which is 3 kg in this case? Please give some idea.Thanks
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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20 Jun 2015, 20:28
Thanks for the detailed explanation!
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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20 Jun 2015, 20:32
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davesinger786 wrote:
Thanks for the detailed explanation!

Tradition of "Thanking" someone on GMAT CLUB community is followed by pressing the button "+1KUDOS" for the post that helped you
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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20 Jun 2015, 20:57
Alright,haha.Thanks for the reminder.Kudos to you!
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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09 May 2016, 05:58
A super quick way of solving mixture questions is using the rule of alligation
(In this case, it took me < 1 minute)

Difference between the diagonals is the numbers in red

Current proportion of sand _______________ 3/10____________ 5/10 ______ (10/10-5/10)

Desired proportion of sand ________________________ 5/10

Proportion of sand in the mix to be added ___10/10___________ 2/10 ______ (5/10-3/10)
(pure sand)

Proportion of current mix to pure sand should be red number (5:2)
In other words, 2/7 of the mix must be pure sand.
2/7 * 10 kg = 20/7
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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14 Jun 2016, 06:54
havoc7860 wrote:
My first explanation on Gmatclub, hope it helps someone.

Since the question specifies that there is 10kg of the mixture which contains sand and clay for there to be equal quantities of sand and clay each should be 5 kg i.e. 5kg of sand and 5kg of clay.

Now since 70% of the mixture is clay there is 7kg of clay in the mixture i.e. 2kg in excess of 5kg of clay required for equal quantities. So 2 kg of clay has to be removed, but for every kg of clay there is (1X3)/7 kg of sand therefore for 2 kg of clay there is 2(1X3)/7=6/7 kg of sand in the current mixture.

So for 2 kg of clay to be removed, 6/7 of sand has to also be removed. Thus total mixture removed = 2+6/7= 20/7

Please do give me a kudos if you like my explanation, thanks.

Everyone else has pretty much said the same thing as you have, yet yours is the one that made it click for me. Thank you +1000000 Kudos!
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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14 Jun 2016, 09:29
let x=amount of mixture to be removed
.3*10-.3x+x=.5*10
x=20/7
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10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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16 Apr 2017, 16:08
1
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?

OA:
20/7

My fav. method to solve replacement problems:
Consider only sand: We remove 'x%' of the mixture and replace it with pure sand so that the mixture has 50% sand in it.
Mixture (percent sand)--------Average-----------------------------Sand(pure)
30%-----------------------------------50%-------------------------------------100%
$$\frac{(100-50)}{(50-30)} = \frac{Mixture}{Sand}$$
$$\frac{5}{2}= \frac{Mixture}{Sand}$$
$$\frac{Sand}{Mixture} = \frac{2}{5}$$:This means, 2 parts of pure sand was mixed with 5 parts of the mixture to give 7 parts of final mixture
PURE Sand is 2/7th of new mixture. So can we not say 2/7th of 10kg was replaced?
Ans = 20/7
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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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16 Apr 2017, 21:39
Net sand gain when 1 kg of mixture is removed and pure sand is added=(-0.3+1)+0.7= 1.4
Difference between Clay and Sand (since we are looking for a 50-50 mixture)=4kg, so 4/1.4 or 40/14=20/7

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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t  [#permalink]

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08 Jul 2020, 06:52
1
I have used allegations here.
30% 100%(as we are replacing with pure sand)
50%
50% 20%
Which means 50% of the mixture has to be removed and replaced with 20% of pure sand.

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Re: 10kg of a mixture contains 30% sand and 70% clay. In order t   [#permalink] 08 Jul 2020, 06:52

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