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10 kg of soil mixture X consists of 20% of sand

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10 kg of soil mixture X consists of 20% of sand

by 4meonly » Tue Aug 26, 2008 9:07 am
10 kg of soil mixture X consists of 20% of sand and 70% of clay. By replacing how many kiligrams of soil with the same kilograms of pure sand, the new soil will consist of 50 percent of sand?
a. 7/5
b. 5/7
c. 20/7
d. 10/7
e. 20/3

OA C

I think OA is not correct...
Moreover, I am very confused with replacing kiligrams of soil.
Usually mixture problems are with adding.

May be someone from Instructors will help?
Last edited by 4meonly on Thu Aug 28, 2008 5:58 am, edited 1 time in total.
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Re: 10 kg of soil mixture X consists of 20% of sand

by sudhir3127 » Tue Aug 26, 2008 9:20 am
4meonly wrote:10 kg of soil mixture X consists of 20% of sand and 70% of clay. By replacing how many kiligrams of soil with the same kilograms of pure sand, the new soil will consist of 50 percent of sand?
a. 7/5
b. 5/7
c. 20/7
d. 10/7
e. 20/3

OA C
Are u sure abt the OA.. i am getting 20/3 E
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by hengirl03 » Tue Aug 26, 2008 9:29 am
E can't be the answer. Here is why:

20/3 = 6.6667 kg

Add 6.666.7 to 2kg and you get 8.6667kg of sand. Therefore, the sand ratio is more than 80% of the 10kg soil mixture.
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Solution

by mayur00 » Tue Aug 26, 2008 10:40 am
*I think there is an error in the question. Mixture X should have 30% sand and 70% clay.

Using this the solution will be as follows:

Amount of sand in original mixture = 3 kg

if A kg of soil is removed and A kg of sand is added, then

The amount of sand removed = 30% of A = (3/10)A

Amount of sand added = A

Therefore total sand in the final mixture = 3 - (3/10)A + A = 3 + (7/10)A

Since we know that amount of sand in the final mixture = 5 kg (50% of 10kg)


Therefore, 3 + (7/10)A = 5

or, A = 20/7
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by kshankker » Wed Aug 27, 2008 3:40 am
Iam Totally confused about Mixture problems... Guys pls help me in this...pls say me the procedure to solve mixture problems.....
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by smackmartine » Sat Jun 25, 2011 11:41 am
Posting the correct question... Some people don't even care to edit the wrong question!!!

10 kg of soil mixture X consists of 30% of sand and 70% of clay. By replacing how many kiligrams of soil with the same kilograms of pure sand, the new soil will consist of 50 percent of sand?
a. 7/5
b. 5/7
c. 20/7
d. 10/7
e. 20/3
Smack is Back ...
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by GMATGuruNY » Sat Jun 25, 2011 2:14 pm
smackmartine wrote:Posting the correct question... Some people don't even care to edit the wrong question!!!

10 kg of soil mixture X consists of 30% of sand and 70% of clay. By replacing how many kiligrams of soil with the same kilograms of pure sand, the new soil will consist of 50 percent of sand?
a. 7/5
b. 5/7
c. 20/7
d. 10/7
e. 20/3
Percentage of sand in X = 30.
Percentage of sand in the pure sand = 100.
Percentage of sand in the mixture = 50.

We can use alligation to determine the ratio of the elements in the mixture.
Alligation dictates the following:

The proportion of each element in the mixture = the distance between the percentage attributed to the other element in the mixture and the percentage attributed to the mixture.

Thus:
Proportion of X in the mixture = |100-50| = 50.
Proportion of sand in the mixture = |30-50| = 20.
X : Sand = 50:20 = 5:2.

Thus, for every 5+2=7 grams of mixture, 2 grams must be pure sand. In other words, the pure sand must be 2/7 of the mixture.
(2/7)*10 = 20/7.

The correct answer is C.
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