­A 10-kilogram soil mixture containing 30 percent sand and 70 percent

­A 10-kilogram soil mixture containing 30 percent sand and 70 percent humus, by weight, is thorougly mixed together. In order to create a 10-kilogram mixture that contains 50 percent of each component, approximately how many kilograms of the soil mixture should be removed and replaced by pure sand?

It's easy to get the impression that we should focus on the sand since the question involves replacing some of the mixture with pure sand. However, it's much easier to find the answer by focusing on the humus since we need to handle only removal of humus whereas, if we focus on the sand, we'll have to handle first subtracting and then adding sand.

We need to create a 10-kilogram mixture such that each component makes up 50 percent of the mixture. So, the final mixture will contain 10 × 0.5 = 5 kilograms of humus. If it contains 5 kilograms of humus, it will contain 5 kilograms of sand as well.

So, since the mixture currently has 10 × 0.7 = 7 kilograms of humus, we need to remove enough of the current mixture to remove 2 kilograms of humus.

The soil mixture is "thoroughly mixed." So, the humus is evenly distributed throughout the mixture. So, to remove 2 kilograms of humus, we remove 2/7 of the entire mixture.

1/7 is a little over 0.14. So, 2/7 is about 0.29.

0.29 × 10 = 2.9

(In case you didn't know the decimal equivalent of 1/7, you could quickly see that 2/7 is going to be a little less 0.3 because 21/7 = 3.)

(A) 2.0

(B) 2.9

(C) 3.4

(D) 4.0

(E) 5.0

Correct answer:­­­­­­
_________________

Amount of sand in mixture = 3 kgs
Amount needed = 5 kgs
Say we remove x amount of the mixture. Then that mixture would be 3x/10 sand. We then add x amount of pure sand.
Therefore 3 - 3x/10 + x =5
On solving we get x=20/7 which is approx 2.9 since 21/7 is 3

AnkurGMAT20
Given: A 10-kilogram soil mixture containing 30 percent sand and 70 percent humus, by weight, is thorougly mixed together.
Asked: In order to create a 10-kilogram mixture that contains 50 percent of each component, approximately how many kilograms of the soil mixture should be removed and replaced by pure sand?
In 10 kg soil mixture: - 
Sand = 3 kg
Humus = 7 kg

Let us assume that x kg of soild mixture should be removed and replaced by pure sand.

Sand = 3 - .3x + x = 5; .7x = 2; x = 2/.7 = 20/7 = 2.9 approx
Humus = 7 - .7x = 5; .7x = 2; x = 2/.7 = 20/7 = 2.9 approx

IMO B

KarishmaB I solved it using conventional method. But can this question be solved using alligation method/ any quicker one ?

 

­It is a replacement question. There are multiple ways of doing it quickly and efficiently. Here are a couple:

We have 10 kg of mixture containing 70% humus. Some of it is removed. Say now we have M kg of mixture. Next we add some sand to get 10 kg of 50% humus mixture.

Method 1: Use \(C_1*V_1 = C_2*V_2\)

\(0.7*M = 0.5*10\)
M = 7.1 kg approx

This means 2.9 kg of mixture was removed and 2.9 kg of sand was added. 

Method 2: Use weighted averages.

\(\frac{w1}{w2} = \frac{(0 - 50)}{(50 - 70)} = \frac{5}{2}\)
So M and the weight of sand added was in the ratio 5:2. 
So sand added is 2/7 of the new 10 kg mixture i.e. about 2.9 kg.

Answer (B)

Check Replacement here: https://anaprep.com/arithmetic-replacement-in-mixtures/
_________________

I am confused. Why do you say we remove 2/7 of the entire mixture, wouldn't it be 2/10?

­

I am not understanding how 0.3x and x are used as the same variable in the equation. Doesn't x represent the entire mixture? So wouldn't we need two variables, one to represent the entire mixture and one to represent just sand? 

­

Another way :- For any replacement qs( where same amount is removed and replaced)

S:H = 3:7

Now we have to make 50% each that means 1:1

Here the catch is we have to make total should be same (3+7=10)

So now we have to make the other ratio as sum as 10 and since it's given as 1:1 so equal amount should be added. Hence (5:5)

Now, 3:7 and 5:5 now, since we have to remove the humus and replace with pure sand that means 7 is decreasing to 5 , that is 2 unit is reduced from 7 so it becomes 2/7 and this proportion is also valid for original mix, so 2/7(10)= 2.85 ~ 2.9

Posted from my mobile device