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Liters of mixture X in the 50% mixture?

This topic has 5 expert replies and 5 member replies
gmattesttaker2 Legendary Member Default Avatar
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Liters of mixture X in the 50% mixture?

Post Sat Nov 02, 2013 1:05 pm
Hello,

Can you please assist with this:

Mixture X is 40% oil and mixture Y is 80% oil, if the two mixtures are put together to form 44 litres of a mix which is 50% oil, how many liters of mixture X are in the 50% mixture?

My approach:

Let mixture X be 100. So, oil is 40
Let mixture Y be x. So, oil is (80/100)x

Mixtures X + Y = (100 + x) which has 50% oil

So, 50/100(100 + x) = 40 + 80/100x

Simplifying, x = 100/3

However, I am not able to proceed beyond this point. I dont know where this 44 liters would fit in here. Can you please assist?

Thanks,
Sri

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gmattesttaker2 Legendary Member Default Avatar
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Post Sat Nov 02, 2013 4:32 pm
Brent@GMATPrepNow wrote:
Hey Sri,

We can assign arbitrary values (like 100) when we aren't required to find an actual number.

For example, a question might ask us to find some PERCENTAGE in which it may be okay to assign some arbitrary values in your solution. Likewise, as you've shown in this last question, we're asked to find the RATIO of solution volumes. So, in that case, assigning arbitrary values helped.

However, in the original question in this thread, we're asked to determine the NUMBER of liters of mixture X in the new mixture. As such, we can't assign arbitrary values.

I hope that helps.

Cheers,
Brent
Hello Brent,

Thanks again for the explanation. I tried to solve the following problem as follows:

Michael mixed a liters of a 10-percent alcohol solution with b liters of a 20-percent
alcohol solution to get c liters of a 14-percent alcohol solution, what is the value of
b?
(1) a = 12
(2) c = 20


My approach:

10/100(a) + 20/100(b) = 14/100(c) (Given)

Since, c = a + b
10/100(a) + 20/100(b) = 14/100(a+b) - Eq. 1

Also I have,
10/100(a) + 20/100(b) = 30/100(a+b) - Eq. 2

I was wondering if the way I am reasoning Eq. 2 is correct here? I am thinking that 10% alcohol in "a" liters + 20% alcohol in "b" liters = 30% alcohol in "a+b" liters

From 1 and 2:
14/100(a+b) = 30/100(a+b)
Simplifying, b = 4/6a

1) a = 12 => b = 8
2) c (i.e. a+b = 20). Simplifying, b = 8

Ans: D

I was also wondering how we can check that Eq. 1 and Eq. 2 are correct. When I plug in the values for a and b in Eq. 1 I get:

10/100(a) + 20/100(b) = 14/100(a+b) - Eq. 1

Left Hand Side: 10/100(12) + 20/100(8) = 14/5
Right Hand Side: 14/100(20) = 14/5

I was wondering though how can we prove that Eq. 2 is correct?

Thanks again for all your help.

Best Regards,
Sri

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Post Sat Nov 02, 2013 5:28 pm
gmattesttaker2 wrote:
Mixture X is 40% oil and mixture Y is 80% oil, if the two mixtures are put together to form 44 litres of a mix which is 50% oil, how many liters of mixture X are in the 50% mixture?
The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the percentages for X and Y on the ends and the percentage for the mixture in the middle.
X 40%-----------50%-----------80% Y

Step 2: Calculate the distances between the percentages.
X 40%----10-----50%----30-----80% Y

Step 3: Determine the ratio in the mixture.
The required ratio of X to Y is equal to the RECIPROCAL of the distances in red.
X:Y = 30:10 = 3:1.

Since X:Y = 3:1, of every 4 liters, X=3 liters and Y=1 liter.
Thus, X is equal to 3/4 of the 44 liters of mixture:
(3/4) * 44 = 33.

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Post Fri Jul 25, 2014 2:43 am
gmattesttaker2 wrote:
Hello,

Mixture X is 40% oil and mixture Y is 80% oil, if the two mixtures are put together to form 44 litres of a mix which is 50% oil, how many liters of mixture X are in the 50% mixture?

Let, the Quantity of X used is X

The Quantity of Oil in the Mixture can be calculated as

(40/100)X +(80/100)Y = (50/100)(X+Y)

i.e. 40X + 80Y = 50X + 50Y
i.e. 10X = 30Y
i.e. X = 3Y
But Given : X+Y = 44
therefore, (3Y) + Y = 44
i.e. 4Y = 44
i.e. Y = 11 and X = 44-11 = 33

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Post Fri Jul 25, 2014 1:52 am
Get comfortable with the approach(es) above, then think about it this way when your proficient at mixture problems.
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gmattesttaker2 Legendary Member Default Avatar
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Post Sat Nov 02, 2013 4:01 pm
Brent@GMATPrepNow wrote:
Hey Sri,

We can assign arbitrary values (like 100) when we aren't required to find an actual number.

For example, a question might ask us to find some PERCENTAGE in which it may be okay to assign some arbitrary values in your solution. Likewise, as you've shown in this last question, we're asked to find the RATIO of solution volumes. So, in that case, assigning arbitrary values helped.

However, in the original question in this thread, we're asked to determine the NUMBER of liters of mixture X in the new mixture. As such, we can't assign arbitrary values.

I hope that helps.

Cheers,
Brent
Hello Brent,

Thank you very much for clarifying and for your excellent explanations.

Best Regards,
Sri

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Post Sat Nov 02, 2013 3:55 pm
Hey Sri,

We can assign arbitrary values (like 100) when we aren't required to find an actual number.

For example, a question might ask us to find some PERCENTAGE in which it may be okay to assign some arbitrary values in your solution. Likewise, as you've shown in this last question, we're asked to find the RATIO of solution volumes. So, in that case, assigning arbitrary values helped.

However, in the original question in this thread, we're asked to determine the NUMBER of liters of mixture X in the new mixture. As such, we can't assign arbitrary values.

I hope that helps.

Cheers,
Brent

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gmattesttaker2 Legendary Member Default Avatar
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Post Sat Nov 02, 2013 3:47 pm
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote:
Hello,

Can you please assist with this:

Mixture X is 40% oil and mixture Y is 80% oil, if the two mixtures are put together to form 44 litres of a mix which is 50% oil, how many liters of mixture X are in the 50% mixture?

My approach:

Let mixture X be 100. So, oil is 40 (are you saying we have 100 liters of mixture X? We don't know how many liters of mixture X are combined. In fact, this is what the question is asking us to find)
Let mixture Y be x. So, oil is (80/100)x (it might be confusing to use x when referring to mixture Y)

Mixtures X + Y = (100 + x) which has 50% oil (I'm not really sure what this means)

So, 50/100(100 + x) = 40 + 80/100x

Simplifying, x = 100/3

However, I am not able to proceed beyond this point. I dont know where this 44 liters would fit in here. Can you please assist? If we have 44 liters that is 50%, we know that there are 22 liters of oil in the mix

Thanks,
Sri
Hey Sri,

There are a few problems with your solution. I added my comments in green.

Cheers,
Brent
Hello Brent,

Thanks for your excellent and detailed explanation. When I solve these kinds of mixture problems I thought it was OK to set one of the unknowns to 100. For example in this problem:

A 25% alcohol solution is mixed with a 40% alcohol solution to get a mixture which is 30% alcohol, what is the ratio of the 25% alcohol solution to the 40% alcohol solution in the mixture?

This is how I try to solve:

Let the 1st solution be 100 liters. So it has 25 liters alcohol.
Let the 2nd solution be y liters. So it has (40/100)y liters alcohol.

The mixture is (100 + y) liters and it has 30/100 (100 + y) liters alcohol

=> 25 + (40/100)y = 30/100(100 + y)
Solving, y = 50

Hence, Ratio = 100/y = 100/50 = 1:2


Here, when I set the 1st solution to be x liters, I have (25/100)x liters of alcohol.
and when the 2nd solution is y liters, I have (40/100)y liters alcohol.

So I get 2 unknowns in my final equation. So I was just wondering if in such cases it is OK to set one of the unknowns to 100?

Thanks again for all your help.

Best Regards,
Sri

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Post Sat Nov 02, 2013 1:38 pm
gmattesttaker2 wrote:
Mixture X is 40% oil and mixture Y is 80% oil, if the two mixtures are put together to form 44 litres of a mix which is 50% oil, how many liters of mixture X are in the 50% mixture?
We can also solve the question using 1 variable.

We'll let the number of liters of mixture X = x
Since the COMBINED mixture has 44 liters, the number of liters of mixture Y = 44 - x

So, we get the following diagram:
ttp://postimg.org/image/6y8b5uoj9/" target="_blank">

Now, if we focus solely on the oil, we can write 0.4x + 0.8(44 - x) = 22
Expand: 0.4x + 35.2 - 0.8x = 22
Simplify: - 0.4x = -13.2
Solve: x = 33

Cheers,
Brent

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Post Sat Nov 02, 2013 1:31 pm
gmattesttaker2 wrote:
Hello,

Can you please assist with this:

Mixture X is 40% oil and mixture Y is 80% oil, if the two mixtures are put together to form 44 litres of a mix which is 50% oil, how many liters of mixture X are in the 50% mixture?

My approach:

Let mixture X be 100. So, oil is 40 (are you saying we have 100 liters of mixture X? We don't know how many liters of mixture X are combined. In fact, this is what the question is asking us to find)
Let mixture Y be x. So, oil is (80/100)x (it might be confusing to use x when referring to mixture Y)

Mixtures X + Y = (100 + x) which has 50% oil (I'm not really sure what this means)

So, 50/100(100 + x) = 40 + 80/100x

Simplifying, x = 100/3

However, I am not able to proceed beyond this point. I dont know where this 44 liters would fit in here. Can you please assist? If we have 44 liters that is 50%, we know that there are 22 liters of oil in the mix

Thanks,
Sri
Hey Sri,

There are a few problems with your solution. I added my comments in green.

Cheers,
Brent

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Post Sat Nov 02, 2013 1:27 pm
gmattesttaker2 wrote:
Mixture X is 40% oil and mixture Y is 80% oil, if the two mixtures are put together to form 44 litres of a mix which is 50% oil, how many liters of mixture X are in the 50% mixture?

I like to sketch the mixtures with their individual components separated.
Also, for this question, I'll use 2 variables, so that things are clear.
ttp://postimg.org/image/fykwbp9pz/" target="_blank">

We can now write two equations with 2 variables.

First, if x liters of mixture X added to y liters of mixture Y gives us a resulting mixture of 44 liters, we can write x + y = 44

Second, if we focus on the oil only, we can write 0.4x + 0.8y = 22

We'll now solve the system . . .
x + y = 44
0.4x + 0.8y = 22
. . . for x.

Multiply both sides of the red equation by 4 to get 4x + 4y = 176
Multiply both sides of the blue equation by 5 to get 2x + 4y = 110

Subtract the blue equation from the red equation to get 2x = 66
So, x = 33

Cheers,
Brent

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