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.

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.

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.

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..

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

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

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]

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

Hmm your right harshvardhan. That is a smarter approach

Thanks a bunch

E