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

Question

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?

(A) 2.0
(B) 2.9
(C) 3.4
(D) 4.0
(E) 5.0

(A) 2.0
(B) 2.9
(C) 3.4
(D) 4.0
(E) 5.0

MartyMurray · Accepted Answer

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?

(A) 2.0
(B) 2.9
(C) 3.4
(D) 4.0
(E) 5.0

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 that way we need to handle only removal of humus whereas, if we focus on the sand, we'll have to consider both subtracting and adding sand.

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

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 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:  B

Guntabulla · 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

Kinshook · Answer

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

sayan640 · Answer

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

KarishmaB · Answer

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 weighted averages and mixtures here: https://anaprep.com/arithmetic-weighted-averages/ https://anaprep.com/arithmetic-mixtures/ Check Replacement here: https://anaprep.com/arithmetic-replacement-in-mixtures/

kevinhirose · Answer

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

kevinhirose · Answer

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?

Ranu2024 · Answer

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

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Purnank · Answer

In the end we wanted to make a mixture that will have 5kg of humus. So to do that we need to remove that extra 2 kg of humus that was already present in 10kg.
Then how much we need to remove that will make sure we have removed 2kg of humus.
lets assume that as x,

now  \frac{7}{10} * x = 2

on solving  x=2.857  appx 2.9

Answer is B.

MBAchild · Answer

Is it possible to do it with Teeter-tottor method? if yes can somebody post it here plz?

Dheeraj_079 · Answer

The total mixture is 10kg -----------> containing 3kg sand and 7kg humus
Now when x kg of mixture is removed, sand and humus are removed in ratio 3:7
So, the 10 - x kg mixture hax 3- 3x/10 sand and 7-7x/10 kg humus

Now we are adding x kg of pure sand to bring it up to 10 kg
So, the mixture now has 3-3x/10+x kg sand and 7-7x/10 kg humus

So, 3+0.7x = 7-0.7x
Solving for x we get 2.85 ~ 2.9 Kg
Answer is Option B

sanebeyondone · Answer

Just want to give my own method out there:
(rule of three in arithmetic to rule them all!)
10 kg mixture / 7 kg of humus = X kg of mixture / 2 kg of humus

X = (10/7)*2 = approx 2.9 kg

My mistakes:
1. At first I didn't know that the question asks how many kilogram of mixture. I thought the question asks for how many kilogram of humus. So I went ahead for 2 kg.
2. With my instinct, whenever I saw mixture problem, I promptly make a table and/or C1V1*C2V2. It didn't help anything for this question (mixture removal question)

sutharpralhad · Answer

We can solve it by using allegations : 
We have a 30% sand and 70% humus mixture which translates to 3kg and 7 kg respectively . We have to add pure sand (100% ) 0% humus and remove the uniform mixture .

we have to reach a 50% for sand . Let say we add x kg of pure sand and remove 10-x kg of mixture of 30% sand

so allegation ratio will be

x/20 = (10-x)/50 -> x=20/7 ->2.9 (B)

zlishz · Answer

10 kg soil -> 30% sand, 70% humus.
Ratio of Sand to Humus is 3:7
We want to make it 5:5 by removing some of the soil and adding pure sand back to it.

Let the sand removed be 'x' kgs from 10 kgs.
Removal Ratio =   x/10   
Thus, Leftover Ratio =   (10-x)/10

We will proceed with equating humus on both sides of the equation as it is the one that does not change throughout the process of removal and replacement.
Earlier, we had   7/10   Humus out of the entire soil.
We want humus to be   5/10   now.
Thus,

7/10   *   (10-x)/10   =   5/10   
  (10-x)/10   =   5/7   
  x = 20/7  
  x = 2.9  
 Option B.

hfell · Answer

Yes, it is possible draw a line from 30 to 100 with 50 in the middle. So, that gives us a ratio of 50:20 meaning for every 5 parts of the 30% mixture we need 2 parts of pure sand. So, 2/7 of the final mixture needs to be pure sand, multiply that number by 10 and we get 2.85 which is about 2.9.

VikramMarkiv · Answer

To solve this problem, we need to calculate how much of the original soil mixture must be removed and replaced with pure sand to achieve a 10-kilogram mixture containing 50% sand and 50% humus.

Step 1: Analyse the current mixture
The current 10-kilogram mixture is:
 
 30% sand = 0.3×10=3 kg 
 70% humus = 0.7×10=7 kg

Step 2: Let x represent the kilograms of the current mixture removed and replaced by pure sand.
If x kilograms of the mixture are removed:
 
 The amount of sand removed is 0.3x (as the mixture contains 30% sand; if x kg of the mixture is removed, we are removing 30% of x = 0.3x of sand) and similarly 
 The amount of humus removed is 0.7x

These x kilograms are replaced with pure sand:
 
 The total sand added is x 
 The new amounts of sand and humus in the mixture will be: 
 Sand: (3−0.3x+x)=(3+0.7x) Humus: (7−0.7x)

Step 3: Set up the condition for 50% sand and 50% humus.
For the new mixture to have 50% sand and 50% humus in a total of 10 kilograms:
Sand: (3+0.7x)=0.5×10=5 kg
Humus: (7−0.7x)=0.5×10=5 kg

Step 4: Solve for x.
From the sand equation:
3 + 0.7x = 5 
0.7x=2
 x=2/0.7≈2.86 kg

Approximately  2.9 kilograms  of the original mixture should be removed and replaced with pure sand.

Nabeela199Nasar · Answer

from where are we getting 100 below 50? Please explain

gmatbeauty · Answer

Let "x" be the amount of the mixture we remove (and the amount of pure sand that we add). Since we need a 10 kg mixture that is 50/50, there will be 5 kg of pure sand. The original mixture has 3 kg of pure sand (ie. 30%). So,

3 + x - 0.3x = 5
simplify/solve
x=2.9

This is a tough question, but when you think about converting the text into algebra, it becomes easier imo.

DanTheGMATMan · Answer

Classic mixture of hummus and sand, very tasty:
 https://www.youtube.com/watch?v=fWmeBAM3sXo

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

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