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Manager
Joined: 29 Oct 2010
Posts: 92
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Kudos [?]: 9 [0], given: 10

Thanks for sharing
Intern
Joined: 08 Mar 2011
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Kudos [?]: 12 [0], given: 4

thanks for sharing
Manager
Joined: 10 Jul 2010
Posts: 196
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Kudos [?]: 18 [0], given: 12

Always have trouble with mixture problems, thanks!
Intern
Joined: 12 Feb 2011
Posts: 2
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Kudos [?]: 0 [0], given: 0

great post! thanks
Intern
Joined: 29 Jun 2011
Posts: 38
Location: Yugoslavia
GMAT 1: 700 Q48 V38
GPA: 3.72
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Kudos [?]: 5 [0], given: 3

Thanks alot for this, its very helpful
Manager
Joined: 11 Jan 2006
Posts: 230
Location: Arkansas, US
WE 1: 2.5 yrs in manufacturing
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Kudos [?]: 13 [0], given: 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

Intern
Joined: 04 May 2011
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Kudos [?]: 2 [0], given: 11

good one...!!!!
Could someone please explain the 3rd and 5th example in detail.?
Manager
Joined: 25 May 2011
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Kudos [?]: 46 [0], given: 71

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

Manager
Joined: 25 May 2011
Posts: 158
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Kudos [?]: 46 [0], given: 71

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
Manager
Joined: 02 Feb 2012
Posts: 195
Location: United States
Schools: HBS '18
GPA: 3.08
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Kudos [?]: 4 [0], given: 104

I'm struggling using this method... may be because I'm not very strong with ratios!
Intern
Joined: 17 Apr 2012
Posts: 5
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Kudos [?]: 0 [0], given: 2

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.
Intern
Joined: 28 May 2012
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Kudos [?]: 0 [0], given: 1

Great post, thanks.
Intern
Joined: 06 Feb 2012
Posts: 16
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Kudos [?]: 21 [0], given: 7

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 3715 times ]

Intern
Joined: 14 Sep 2013
Posts: 10
Followers: 0

Kudos [?]: 2 [0], given: 14

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?
Math Expert
Joined: 02 Sep 2009
Posts: 29108
Followers: 4725

Kudos [?]: 49727 [0], given: 7403

Expert's post
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.
_________________
Manager
Joined: 04 Oct 2013
Posts: 162
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Followers: 2

Kudos [?]: 75 [0], given: 54

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
Intern
Joined: 25 Mar 2014
Posts: 33
Followers: 0

Kudos [?]: 18 [0], given: 38

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
Intern
Joined: 25 Mar 2014
Posts: 33
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Kudos [?]: 18 [0], given: 38

I did not get from where the $$\frac{1}{3}$$ came from on example 3!
Math Expert
Joined: 02 Sep 2009
Posts: 29108
Followers: 4725

Kudos [?]: 49727 [0], given: 7403

Expert's post
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.
_________________
Intern
Joined: 08 May 2014
Posts: 8
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Kudos [?]: 13 [0], given: 1

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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