10kg of a mixture contains 30% sand and 70% clay. In order t

The mixture contains 3kg sand and 7 kg clay.
For the mixture to be in equal quantities, there should be 2kg of clay removed.
clay and sand are in the ratio 7:3
So part of sand to be removed = 2*3/7 = 6/7
So total mixture to be removed = 2 + 6/7 = 20/7

Alee's way is probably faster but here's another way of thinking about it.

You have 7kg clay and 3kg sand. For every 1kg of mixture you remove you lose 7/10kg clay and 3/10kg sand. You then give that 1kg back to sand, so the net effective is gaining -3/10kg + 1kg = 7/10kg sand.

So,

7-(7/10)x
divided by
3+(7/10)x

= 1.

Solve for x => 20/7

actually i'm having trouble understanding alee's method even though it's clearly correct. if you just solve for 2 grams of clay removal (fixing the ratio at 1:1), and then independently solve for sand, doesn't that imply that the ratio of sand/clay then changes from 1:1 to something else?

Let the mixture to be replaced by pure sand be x
we want the sand/ clay ratio to be 1:1
therefore after replacement


sand/clay = (3 - 0.3x + x)/(7 - 0.7x) = 1/1

removing x kg of mixture means removing 3/10*x kg of sand(as x kg mixture has 30% sand) from the existing sand that is 3 kg,also we are adding x kg pure sand(replacement!) so +x in numerator, same goes for the clay, x kg of mixture has 7/10*x kg clay...so after replacement clay left is 7 - 0.7x (no addition here). the required ratio is 1:1 so....

solving for x gives x = 20/3


so here is the "detailed" explanation ....hope that helps

bipolarbear, even I found it somewhat difficult to understand
but I got it...heres how

here we are trying to remove 2 kg of clay from the mixture so that removing 2 kg clay leaves 5 kg clay and adding 2 kg sand leaves 5 kg sand..giving a 1:1 mixture
sand:clay=3:7
there is 7 kg clay for every 3 kg sand
for 1 kg clay, sand =3/7
for 2 kg clay, sand =3/7*2

so removing 2 kg clay from the mixture means also removing 21/2 kg sand along with it.

so the total mixture removed is 2kg (clay for which we calculated) + sand (according to the ratio)
= 2 + 6/7 = 20/7

PROOF
from the mixture 6/7 kg sand has been removed which leaves us 3 - 6/7= 15/7 kg sand
but we are also adding the same amount of pure sand which is equal to the mixture removed(20/7)

so... checking the answer: remaining sand+new sand
= 15/7 + 20/7
= 5 kg

so now the sand is 5kg and the clay is also 5kg
ratio 1:1

actually ,never seen a mixture problem solved like this....I too was doubtful abt alee's method
bit eccentric

wow, thats a simple and quick one...I did considering clay with this method and it works... thanks

O- R+ A=D
7-.7x+0=5
x=20/7

Hummm I hate mixture problems...

You have 7kg of C and 3 kg of S

Your final mixture has to have 5kg of C and 5kg of S.

But if you remove Clay, you will remove sand as well.
So you have to take into account this.

The question is: how much of (S+C) will you have to remove in order to remove 2kg of C?

The answer is:
0.7*X=2
X = 2/0.7

This is not an easy question.

This is an interesting problem. The best and fastest way I'd suggest is to derive an equation based on the given statements.

I remove X kg of mixture and add X kg of pure sand. so the total weight of the mixture should remain constant at 10kg. I am not considering adding pure sand to 70% of clay because, it complicates the second equation.

(10 - X)(30/100) + X = 10(50/100)
X = 20/7 kg.

Required: 5kg to remain for both sand & clay
let x = quantity remaining after some mixture have been removed.
70% of x = 5; x=50/7
sand to be replaced = 10 - 50/7 = 20/7

yes it will be 20/7 only

as 2+6/7 = 20/7 , the 6/7 of sand which is removed will be replaced with pure sand, thus it is nullified.

We should remove 2 kilograms of clay, thus we need to remove 20/7 kilograms of mixture: 7/10*20/7=2.

that only works if added & removed are the same, since they're both X in that equation, that implies the same amount was removed in both cases

My first explanation on Gmatclub, hope it helps someone.

Since the question specifies that there is 10kg of the mixture which contains sand and clay for there to be equal quantities of sand and clay each should be 5 kg i.e. 5kg of sand and 5kg of clay.

Now since 70% of the mixture is clay there is 7kg of clay in the mixture i.e. 2kg in excess of 5kg of clay required for equal quantities. So 2 kg of clay has to be removed, but for every kg of clay there is (1X3)/7 kg of sand therefore for 2 kg of clay there is 2(1X3)/7=6/7 kg of sand in the current mixture.

So for 2 kg of clay to be removed, 6/7 of sand has to also be removed. Thus total mixture removed = 2+6/7= 20/7

Please do give me a kudos if you like my explanation, thanks.

Sand........ Clay...... Total
3 .............. 7 ........... 10

Given that 30% is sand & 70% is clay

Say we remove mixture quantity = x

It means we are removing \(\frac{30x}{100} sand & \frac{70x}{100}\) clay from the mixture

New table would be as follows:

Sand........ Clay...... Total
\(3-\frac{30x}{100}.............. 7-\frac{70x}{100} ........... 10-x\)

Quantity removed of mixture = Quantity added of sand (which gives both sand & clay quantities SAME in the mixture)

\(3 - \frac{30x}{100} + x = 7 - \frac{70x}{100}\)

\(x + \frac{40x}{100} = 4\)

\(x = 4 * \frac{100}{140}\)

\(x = \frac{20}{7}\)

If equal quantity will be 5kg sand and 5 kg clay. So I am removing 2 kg clay
70% mixture=2kg
1%=2/70
100%=2*100/70= 20/7

Hi Economist,

The amount in that table will be total amount ? If we're considering "Sand" then why not use the amount of Sand in the mixture which is 3 kg in this case? Please give some idea.Thanks

Don't ever try to find a Easier/Shortcut way unless you are clear about the basic method

The basic method goes like this (Making an equation and solving it further)

Let, x is the amount of mixture to be replaced with Pure Sand. The please understand that since we are replacing Mixture with sand so
1) the amount of Sand that goes out with x kg Mixture will come back with the pure sand
2) the amount of Clay that goes out with x kg Mixture will be replaced by pure sand

i.e. Total Amount of Sand will increase by the amount of clay that is removed while removing x kg of mixture

Please know that Sand in the mixture as of now = (30/100)*10 = 3 kg.
Clay as of now = 10 - 3 = 7kg
Clay that goes out in x kg of mixture = (70/100)*x = (7x/10) kg.

METHOD-1
Then Total Sand after the mixture is replaced with pure sand = 3+ (7x/10) kg.

But now Sand and Clay should be equal in the final mixture
i.e. Sand = 50% of total Mixture amount
i.e. 3+ (7x/10) = (50/100)*10
i.e. 3+ (7x/10) = 5
i.e. (7x/10) = 2
i.e. x = 20/7 kg

METHOD-2
Same equations can also be made on the quantity of clay after the x kg Mixture is replaced with x kg pure Sand

i.e. Clay that goes out in x kg of mixture = (70/100)*x = (7x/10) kg.
i.e. Clay after the Mixture is Replaced with Pure Sand = 7 - (7x/10) kg.

But now Sand and Clay should be equal in the final mixture
i.e. Clay = 50% of total Mixture amount
i.e. 7 - (7x/10) = (50/100)*10
i.e. 7 - (7x/10) = 5
i.e. (7x/10) = 2
i.e. x = 20/7 kg

I hope it helps!

My fav. method to solve replacement problems:
Consider only sand: We remove 'x%' of the mixture and replace it with pure sand so that the mixture has 50% sand in it.
Mixture (percent sand)--------Average-----------------------------Sand(pure)
30%-----------------------------------50%-------------------------------------100%
\(\frac{(100-50)}{(50-30)} = \frac{Mixture}{Sand}\)
\(\frac{5}{2}= \frac{Mixture}{Sand}\)
\(\frac{Sand}{Mixture} = \frac{2}{5}\):This means, 2 parts of pure sand was mixed with 5 parts of the mixture to give 7 parts of final mixture
PURE Sand is 2/7th of new mixture. So can we not say 2/7th of 10kg was replaced?
Ans = 20/7