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Re: Mixture Problems Made Easy [#permalink]
28 Oct 2010, 06:54

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?

Re: Mixture Problems Made Easy [#permalink]
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:

Re: Mixture Problems Made Easy [#permalink]
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.

Re: Mixture Problems Made Easy [#permalink]
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:

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 [ 655.36 KiB | Viewed 3728 times ]

Re: Mixture Problems Made Easy [#permalink]
20 Dec 2013, 02:40

KillerSquirrel wrote:

Mixture problems made Easy

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?

Re: Mixture Problems Made Easy [#permalink]
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 )

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?

Re: Mixture Problems Made Easy [#permalink]
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:

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]
25 Nov 2014, 05:34

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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