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PS - Mixture

by karthikpandian19 » Sat Jul 21, 2012 8:39 am
Mixture 1 contains 55% chemical A by volume, 23% chemical B by volume, and 22% chemical C by volume. Mixture 2 contains 38% chemical A by volume and 62% chemical B by volume. How many liters of Mixture 2 must be added to 10 liters of Mixture 1 in order for the resulting mixture to contain a combined 68% chemical A and chemical C by volume?


(A) 3.0

(B) 3.9

(C) 4.4

(D) 5.0

(E) 7.2
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by prat_agl » Sat Jul 21, 2012 8:49 am
A: 3.0

Since we need to get the combination of A and C
Mixture 1 have 77 (A+c) and 23 B
Mixture 2 have 38( A+ C) and 62 B

now the problem is just for taking the ratio (77 * 10 + 38 x)/(10 +x) = 68
which reduces to x= 3
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by theCEO » Sat Jul 21, 2012 9:32 am
in terms of B
we have a solution that is 23% and we want to create a 32% solution

2.3 + 0.62X = 32 = 0.32
10 + X------- 100

2.3 + 0.62x = 3.2 + 0.32x
0.30x = 0.9
x = 3

therfore 3 L of mixture 2 is added
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by eagleeye » Sat Jul 21, 2012 12:00 pm
karthikpandian19 wrote:Mixture 1 contains 55% chemical A by volume, 23% chemical B by volume, and 22% chemical C by volume. Mixture 2 contains 38% chemical A by volume and 62% chemical B by volume. How many liters of Mixture 2 must be added to 10 liters of Mixture 1 in order for the resulting mixture to contain a combined 68% chemical A and chemical C by volume?
This question rests on recognizing that there is 77% combined A+C in mixture 1, and 38% combined A+C in mixture 2. We need 68% overall mixture.

Use method of alligation:

77----(9)----68----(30)----38
10---------- total---------x

Now x/10 = 9/30 => x = 9*10/30 = 90/30 = 3L.


Alternatively, you could set up the algebraic equation:

77*10 + 38*x = 68(10+x)
=> 30x = 90 => x= 3
Cheers!
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by karthikpandian19 » Sat Jul 21, 2012 6:28 pm
Eric,

I am not familiar with this concept of ALLIGATION. Can you give little background explanation
eagleeye wrote:
karthikpandian19 wrote:Mixture 1 contains 55% chemical A by volume, 23% chemical B by volume, and 22% chemical C by volume. Mixture 2 contains 38% chemical A by volume and 62% chemical B by volume. How many liters of Mixture 2 must be added to 10 liters of Mixture 1 in order for the resulting mixture to contain a combined 68% chemical A and chemical C by volume?
This question rests on recognizing that there is 77% combined A+C in mixture 1, and 38% combined A+C in mixture 2. We need 68% overall mixture.

Use method of alligation:

77----(9)----68----(30)----38
10---------- total---------x

Now x/10 = 9/30 => x = 9*10/30 = 90/30 = 3L.


Alternatively, you could set up the algebraic equation:

77*10 + 38*x = 68(10+x)
=> 30x = 90 => x= 3
Cheers!
Regards,
Karthik
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by karthikpandian19 » Sat Jul 21, 2012 6:29 pm
OA is A
karthikpandian19 wrote:Mixture 1 contains 55% chemical A by volume, 23% chemical B by volume, and 22% chemical C by volume. Mixture 2 contains 38% chemical A by volume and 62% chemical B by volume. How many liters of Mixture 2 must be added to 10 liters of Mixture 1 in order for the resulting mixture to contain a combined 68% chemical A and chemical C by volume?


(A) 3.0

(B) 3.9

(C) 4.4

(D) 5.0

(E) 7.2
Regards,
Karthik
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by eagleeye » Sat Jul 21, 2012 7:04 pm
karthikpandian19 wrote:Eric,

I am not familiar with this concept of ALLIGATION. Can you give little background explanation
eagleeye wrote:
karthikpandian19 wrote:Mixture 1 contains 55% chemical A by volume, 23% chemical B by volume, and 22% chemical C by volume. Mixture 2 contains 38% chemical A by volume and 62% chemical B by volume. How many liters of Mixture 2 must be added to 10 liters of Mixture 1 in order for the resulting mixture to contain a combined 68% chemical A and chemical C by volume?
This question rests on recognizing that there is 77% combined A+C in mixture 1, and 38% combined A+C in mixture 2. We need 68% overall mixture.

Use method of alligation:

77----(9)----68----(30)----38
10---------- total---------x

Now x/10 = 9/30 => x = 9*10/30 = 90/30 = 3L.


Alternatively, you could set up the algebraic equation:

77*10 + 38*x = 68(10+x)
=> 30x = 90 => x= 3
Cheers!
Alligation is another way of saying, method of balances. It's a faster way of solving mixture ps and ds questions. It should only be used after sufficient practice but once you master it, it can save you valuable time in average and mixture problems.

Here are a few examples:

Lets start with an easy one.
There are 2 mixtures:
A has 20% sugar.
B has 30% sugar.

In what ratio must the be mixed to produce a mixture M containing 22% sugar.

First write the three %s on the number line.
A.-------------M----------------B.
20------------22---------------30

Now write the distance (difference) between the two components and mixture
20-----(2)----22-------(8)-----30. Here 2 and 8 are the distances between (20,22) and (22,30).

The the ratio of amount of A / amount of B = (distance between M-B)/(distance between M-A) = 8/2 = 4:1.

In the same way, for the question which I solved, the unknown was mixture 2's amount. Lets call it x.

Then set up the number line with distances in the same way as above. We are told that there's 10L of Mixture 1 present.
We have 77% for mixture 1, 68% for total and 38% for mixture. So set up the number line:
Then:

Mixture 1--------------Total--------------Mixture 2
10-------------------------Total---------------x
77-------(9)---------------68-------(30)----38 ( 9 = 77-68, 68-38 =30. )
Then 10/x = 30/9
=> x = 10*9/30 = 3.

Let me know if you have any questions. :)
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by karthikpandian19 » Sun Jul 22, 2012 7:40 pm
Only one small clarification:
"The the ratio of amount of A / amount of B = (distance between M-B)/(distance between M-A)"
Is this right ? ratio of A / ratio of B = dist M-A / dist M-B???????
eagleeye wrote:
karthikpandian19 wrote:Eric,

I am not familiar with this concept of ALLIGATION. Can you give little background explanation
eagleeye wrote:
karthikpandian19 wrote:Mixture 1 contains 55% chemical A by volume, 23% chemical B by volume, and 22% chemical C by volume. Mixture 2 contains 38% chemical A by volume and 62% chemical B by volume. How many liters of Mixture 2 must be added to 10 liters of Mixture 1 in order for the resulting mixture to contain a combined 68% chemical A and chemical C by volume?
This question rests on recognizing that there is 77% combined A+C in mixture 1, and 38% combined A+C in mixture 2. We need 68% overall mixture.

Use method of alligation:

77----(9)----68----(30)----38
10---------- total---------x

Now x/10 = 9/30 => x = 9*10/30 = 90/30 = 3L.


Alternatively, you could set up the algebraic equation:

77*10 + 38*x = 68(10+x)
=> 30x = 90 => x= 3
Cheers!
Alligation is another way of saying, method of balances. It's a faster way of solving mixture ps and ds questions. It should only be used after sufficient practice but once you master it, it can save you valuable time in average and mixture problems.

Here are a few examples:

Lets start with an easy one.
There are 2 mixtures:
A has 20% sugar.
B has 30% sugar.

In what ratio must the be mixed to produce a mixture M containing 22% sugar.

First write the three %s on the number line.
A.-------------M----------------B.
20------------22---------------30

Now write the distance (difference) between the two components and mixture
20-----(2)----22-------(8)-----30. Here 2 and 8 are the distances between (20,22) and (22,30).

The the ratio of amount of A / amount of B = (distance between M-B)/(distance between M-A) = 8/2 = 4:1.

In the same way, for the question which I solved, the unknown was mixture 2's amount. Lets call it x.

Then set up the number line with distances in the same way as above. We are told that there's 10L of Mixture 1 present.
We have 77% for mixture 1, 68% for total and 38% for mixture. So set up the number line:
Then:

Mixture 1--------------Total--------------Mixture 2
10-------------------------Total---------------x
77-------(9)---------------68-------(30)----38 ( 9 = 77-68, 68-38 =30. )
Then 10/x = 30/9
=> x = 10*9/30 = 3.

Let me know if you have any questions. :)
Regards,
Karthik
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by eagleeye » Sun Jul 22, 2012 9:38 pm
karthikpandian19 wrote:Only one small clarification:
"The the ratio of amount of A / amount of B = (distance between M-B)/(distance between M-A)"
Is this right ? ratio of A / ratio of B = dist M-A / dist M-B???????
The expression I wrote is correct . What you wrote just now isn't.

Another way of expressing it (which might make it easier to remember), is balancing:

Distance of A from M * Amount of A = Distance of B from M * Amount of B.

Here's an example that illustrates a couple of important points:

How many tons of a fruit mixture "A" containing 10% apples by weight, 30% pears by weight and 60% mangoes by weight must be mixed with another fruit mixture "B" that has 30% grapes, 30% apples and 40% pears by weight ,when "B" weighs 20 tons, to get a mixture that has 25% apples by weight.

First of all, we only care about % of apples in the two mixtures, since the question asks about apples. Let the amount required be A.

Then we set up the alligation line.

A--------------Mix-----------B
A--------------Mix-----------20 tons
10%---(15)-----25%---(5)-----30%

Now multiply the corresponding amounts with distances:
Amount A * Distance between A-Mix = Amount B* Distance between B-Mix.
A*15 = 20*5
A = 20/3 tons.

Let me know if this helps :)
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by karthikpandian19 » Sun Jul 22, 2012 10:01 pm
Now i understood......thanks
eagleeye wrote:
karthikpandian19 wrote:Only one small clarification:
"The the ratio of amount of A / amount of B = (distance between M-B)/(distance between M-A)"
Is this right ? ratio of A / ratio of B = dist M-A / dist M-B???????
The expression I wrote is correct . What you wrote just now isn't.

Another way of expressing it (which might make it easier to remember), is balancing:

Distance of A from M * Amount of A = Distance of B from M * Amount of B.

Here's an example that illustrates a couple of important points:

How many tons of a fruit mixture "A" containing 10% apples by weight, 30% pears by weight and 60% mangoes by weight must be mixed with another fruit mixture "B" that has 30% grapes, 30% apples and 40% pears by weight ,when "B" weighs 20 tons, to get a mixture that has 25% apples by weight.

First of all, we only care about % of apples in the two mixtures, since the question asks about apples. Let the amount required be A.

Then we set up the alligation line.

A--------------Mix-----------B
A--------------Mix-----------20 tons
10%---(15)-----25%---(5)-----30%

Now multiply the corresponding amounts with distances:
Amount A * Distance between A-Mix = Amount B* Distance between B-Mix.
A*15 = 20*5
A = 20/3 tons.

Let me know if this helps :)
Regards,
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON

---If you find my post useful, click "Thank" :) :)---
---Never stop until cracking GMAT---
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