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Manager
Joined: 29 Oct 2010
Posts: 89

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06 Dec 2010, 11:07
Thanks for sharing

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Intern
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17 Mar 2011, 00:34
thanks for sharing

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Manager
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18 Mar 2011, 00:16
Always have trouble with mixture problems, thanks!

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Intern
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20 Mar 2011, 17:35
great post! thanks

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Intern
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20 Jul 2011, 21:32
Thanks alot for this, its very helpful

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Manager
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WE 1: 2.5 yrs in manufacturing

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27 Jul 2011, 22:18
KillerSquirrel wrote:

solving mixture problems fast ! - (30 day money back guarantee )

see attachment

Thanks

Awesome material Bro.

Thanks a lot for sharing. Now I am a lot more confident on these mixture problems.

-Raghu
_________________

ARISE AWAKE AND REST NOT UNTIL THE GOAL IS ACHIEVED

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Intern
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16 Aug 2011, 00:08
good one...!!!!
Could someone please explain the 3rd and 5th example in detail.?

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Manager
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07 Nov 2011, 03:01
tkp wrote:
gr8 post!

but can anyone show me how to solve this problem , with the method given here

Question: 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?

Do we need to know the weight of sand or clay?
My approach is:
Based on this new method:

30% ------------------------ X%
--------------50%
(X-50)% -------------------- 20%

so in new mixture
$$\frac{20}{100}=\frac{weight}{10}$$

weight = 2

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Manager
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07 Nov 2011, 03:02
Thank you. it's a perfect approach. but could you plz elaborate the last example in PDF file? i didn't understand that one.
Thanks

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29 Feb 2012, 16:06
I'm struggling using this method... may be because I'm not very strong with ratios!

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Intern
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24 Apr 2012, 00:45
Hi everyone,

I, too, am struggling with understanding the first example on this pdf.

I believe I understand the diagram, but where I'm struggling is the final step: so it appears that the 30 corresponds to the 15%, and the 50 corresponds with the 5%. So I understood this ratio to mean the 30% solution compared to the 50% solution has a ratio of 15:5 or 3:1. Based on this, I thought the answer should be 7.5 (but I know logically it should actually be 2.5). Would appreciate any insight on this.

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Intern
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05 Aug 2012, 07:22
Great post, thanks.

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Intern
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21 Oct 2013, 02:23
shahideh wrote:
tkp wrote:
gr8 post!

but can anyone show me how to solve this problem , with the method given here

Question: 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?

Do we need to know the weight of sand or clay?
My approach is:
Based on this new method:

30% ------------------------ X%
--------------50%
(X-50)% -------------------- 20%

so in new mixture
$$\frac{20}{100}=\frac{weight}{10}$$

weight = 2

Hi ,

I think there's something you missed here which is that since this being a homogenous mixture, when we take out 2kg of the mixture.it'll be in the ratio of 3:7 i.e. 600gm of sand and 1400 gms of clay will be taken out.This is how the 2kg taken out will be divided as.Now since we need to equate the quantities of clay and sand,so in essence the total quantity to be added will be - 2kg of pure clay + 600 gms (of the 2kg that was taken out).So the answer will be 2.6kg.
Attachments

File comment: Please refer the attachment for more clarity.Let me know if it isn't clear.

m&a_analysis.png [ 655.36 KiB | Viewed 5750 times ]

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Intern
Joined: 14 Sep 2013
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20 Dec 2013, 03:40
KillerSquirrel wrote:

solving mixture problems fast ! - (30 day money back guarantee )

see attachment

Thanks

Can we solve this quest using ur technique?

Bob just filled his car's gas tank with 20 gallons of gasohol, a mixture consisting of 5% ethanol and 95% gasoline. If his car runs best on a mixture consisting of 10% ethanol and 90% gasoline, how many gallons of ethanol must he add into the gas tank for his car to achieve optimum performance?

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Math Expert
Joined: 02 Sep 2009
Posts: 41890

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20 Dec 2013, 03:45
soneesingh wrote:
KillerSquirrel wrote:

solving mixture problems fast ! - (30 day money back guarantee )

see attachment

Thanks

Can we solve this quest using ur technique?

Bob just filled his car's gas tank with 20 gallons of gasohol, a mixture consisting of 5% ethanol and 95% gasoline. If his car runs best on a mixture consisting of 10% ethanol and 90% gasoline, how many gallons of ethanol must he add into the gas tank for his car to achieve optimum performance?

Several ways to solve this question are discussed here: bob-just-filled-his-car-s-gas-tank-with-20-gallons-of-45790.html

Hope this helps.
_________________

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Manager
Joined: 04 Oct 2013
Posts: 162

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Location: India
GMAT Date: 05-23-2015
GPA: 3.45

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21 Dec 2013, 05:06
shahideh wrote:
tkp wrote:
gr8 post!

but can anyone show me how to solve this problem , with the method given here

Question: 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?

Do we need to know the weight of sand or clay?
My approach is:
Based on this new method:

30% ------------------------ X%
--------------50%
(X-50)% -------------------- 20%

so in new mixture
$$\frac{20}{100}=\frac{weight}{10}$$

weight = 2

Hi ,

I think there's something you missed here which is that since this being a homogenous mixture, when we take out 2kg of the mixture.it'll be in the ratio of 3:7 i.e. 600gm of sand and 1400 gms of clay will be taken out.This is how the 2kg taken out will be divided as.Now since we need to equate the quantities of clay and sand,so in essence the total quantity to be added will be - 2kg of pure clay + 600 gms (of the 2kg that was taken out).So the answer will be 2.6kg.

Let X be the amount of mixture that is removed and replaced with pure sand to get 50% sand in the 10 kg mixture.
Thus, the algebraic equation we may write as
((3/10)(10-X) + X )/10 = 0.5
Solving we get X = (20/7) kg

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Intern
Joined: 25 Mar 2014
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25 Nov 2014, 06:34
cheetarah1980 wrote:
Quick question.

In the Example 5, the explanation says that the volume of the new solution is x/2. If one amount of solution replaces the same amount, how is the volume of the new solution cut in half? If I take out 1 liter of solution and replace it with 1 liter of another solution, my overall volume stays the same. Could someone please explain the assumption that the volume is x/2.

Thanks!

The result is 1:1, therefore, you will need a "half-half"solution. Hence the answer is 1/2

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Intern
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25 Nov 2014, 06:34
I did not get from where the $$\frac{1}{3}$$ came from on example 3!

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Math Expert
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25 Nov 2014, 07:06
plaverbach wrote:
I did not get from where the $$\frac{1}{3}$$ came from on example 3!

That question is discussed here: one-fourth-of-a-solution-that-was-10-by-weight-was-replaced-149134.html

Hope it helps.
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Intern
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24 Dec 2014, 09:46
KillerSquirrel wrote:

solving mixture problems fast ! - (30 day money back guarantee )

see attachment

Thanks

i did not understand even a single strategy .. would you please simplify any one of them with the image structure??

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