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Mixture Problems Made Easy

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Re: Mixture Problems Made Easy [#permalink] New post 06 Dec 2010, 10:07
Thanks for sharing
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Re: Mixture Problems Made Easy [#permalink] New post 16 Mar 2011, 23:34
thanks for sharing
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Re: Mixture Problems Made Easy [#permalink] New post 17 Mar 2011, 23:16
Always have trouble with mixture problems, thanks!
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Re: Mixture Problems Made Easy [#permalink] New post 20 Mar 2011, 16:35
great post! thanks
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Re: Mixture Problems Made Easy [#permalink] New post 20 Jul 2011, 20:32
Thanks alot for this, its very helpful
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Re: Mixture Problems Made Easy [#permalink] New post 27 Jul 2011, 21:18
KillerSquirrel wrote:
Mixture problems made Easy

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

see attachment

Thanks

:-D


Awesome material Bro.

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

-Raghu
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Re: Mixture Problems Made Easy [#permalink] New post 15 Aug 2011, 23:08
good one...!!!!
Could someone please explain the 3rd and 5th example in detail.?
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Re: Mixture Problems Made Easy [#permalink] New post 07 Nov 2011, 02:01
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?



Could anyone answer this question?
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

I'm not confident about my answer at all. is it right?
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Re: Mixture Problems Made Easy [#permalink] New post 07 Nov 2011, 02:02
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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Re: Mixture Problems Made Easy [#permalink] New post 29 Feb 2012, 15:06
I'm struggling using this method... may be because I'm not very strong with ratios!
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Re: Mixture Problems Made Easy [#permalink] New post 23 Apr 2012, 23:45
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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Re: Mixture Problems Made Easy [#permalink] New post 05 Aug 2012, 06:22
Great post, thanks.
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Re: Mixture Problems Made Easy [#permalink] New post 21 Oct 2013, 01:23
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?



Could anyone answer this question?
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

I'm not confident about my answer at all. is it right?


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
m&a_analysis.png [ 655.36 KiB | Viewed 2510 times ]

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Re: Mixture Problems Made Easy [#permalink] New post 20 Dec 2013, 02:40
KillerSquirrel wrote:
Mixture problems made Easy

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

see attachment

Thanks

:-D


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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Re: Mixture Problems Made Easy [#permalink] New post 20 Dec 2013, 02:45
Expert's post
soneesingh wrote:
KillerSquirrel wrote:
Mixture problems made Easy

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

see attachment

Thanks

:-D


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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Re: Mixture Problems Made Easy [#permalink] New post 21 Dec 2013, 04:06
radioguy wrote:
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?



Could anyone answer this question?
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

I'm not confident about my answer at all. is it right?


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
Re: Mixture Problems Made Easy   [#permalink] 21 Dec 2013, 04:06
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