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Mixtures

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by newton9 » Wed Mar 31, 2010 10:24 am
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?
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by harshavardhanc » Wed Mar 31, 2010 11:07 am
newton9 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?
7 Kgs of clay
3 Kgs of sand

make them equal. i.e 5Kgs each.

OR remove 2 Kgs of clay which will come from 20/7 kgs of the mixture.
Last edited by harshavardhanc on Wed Mar 31, 2010 11:41 am, edited 1 time in total.
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by newton9 » Wed Mar 31, 2010 11:08 am
Harsha,

You cannot directly take out clay or sand. That's the trick.

You have to take out the mixture and replace it with pure sand.
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by harshavardhanc » Wed Mar 31, 2010 11:47 am
newton9 wrote:Harsha,

You cannot directly take out clay or sand. That's the trick.

You have to take out the mixture and replace it with pure sand.
correct!

I was in hurry as my ride had come and I jumped the gun with that solution! :)

Anyway, yes that's the trick. you can only take out the mixture.

The correct answer will be that quantity of mixture which will have 2 kgs of clay.
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by pops » Wed Mar 31, 2010 10:11 pm
suppose you take out x kg of mixture and replace it with pure sand

so, amount of sand in x kg of mixture = 0.3x
and, amount of clay in x kg of mixture = 0.7x
add, x kg od sand in the mixture

so net sand = 3-0.3x+x
net clay=7-0.7x
hence, 3+0.7x = 7-0.7x
or, 1.4x=4
x=4/1.4=2.857..
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by eaakbari » Wed Mar 31, 2010 10:29 pm
There is an interesting formula I use that works with mixtures and weighted averages.

% of higher ratio - % of mean : % of mean - %of lower = lower :higher


Here we have 1 part containing 100% sand
1 part containing 30 % sand
what we want is 50 % sand
substitute in formula

100 - 50 : 50 - 30 = lower:higher
5:2 = lower :higher
That means the ratio of 30%mixture to pure sand is 5:2

The total weight has to be 10 kgs
5x+2x=7x
7x=10 ; x=10/7
Hence 5 x = 50/7
Hence we can find how many kgs would a 10 kg misture contain of the 30%misture

= 50/7

Therefore amount removed = 10 -50/7 = 20/7
which is the answer
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by harshavardhanc » Wed Mar 31, 2010 11:50 pm
eaakbari wrote:There is an interesting formula I use that works with mixtures and weighted averages.

% of higher ratio - % of mean : % of mean - %of lower = lower :higher


Here we have 1 part containing 100% sand
1 part containing 30 % sand
what we want is 50 % sand
substitute in formula

100 - 50 : 50 - 30 = lower:higher
5:2 = lower :higher
That means the ratio of 30%mixture to pure sand is 5:2

The total weight has to be 10 kgs
5x+2x=7x
7x=10 ; x=10/7
Hence 5 x = 50/7
Hence we can find how many kgs would a 10 kg misture contain of the 30%misture

= 50/7

Therefore amount removed = 10 -50/7 = 20/7
which is the answer
indeed, it's a good formula.

in cases like this one, however, you can save some more time by having a "birds eye view".

look at the percentages. one is 30 and the other one is 70. You want to make them equal. or 50-50.

Or you have to remove that 20% of clay.

Now, we know that 70% clay comes from 100% of mixture. Therefore, 20% will come from (20 * 100)/70.

or 200/7 % of the mixture OR 2/7 of the mixture.

Hence, the answer will be 2/7th of 10 = 20/7 . in fact it is even possible to do it orally :)
Regards,
Harsha
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by sanju09 » Thu Apr 01, 2010 12:02 am
newton9 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?
If x kg of the mixture is removed and replaced with pure sand, then the sand contents in the new 10 kg of mixture are [0.3 (10 - x) + x] kg and the clay contents in that are 0.7 (10 - x) kg, such that

[0.3 (10 - x) + x] = 0.7 (10 - x)

3 + 0.7 x = 7 - 0.7 x

1.4 x = 4

[spoiler]x = 20/7 kg[/spoiler]
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by sanju09 » Thu Apr 01, 2010 12:05 am
harshavardhanc wrote:
newton9 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?
7 Kgs of clay
3 Kgs of sand

make them equal. i.e 5Kgs each.

OR remove 2 Kgs of clay which will come from 20/7 kgs of the mixture.
to me, it's even better
The mind is everything. What you think you become. -Lord Buddha



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by eaakbari » Thu Apr 01, 2010 12:53 am
Hmm your right harshvardhan. That is a smarter approach

Thanks a bunch

E
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