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Register now and save up to $200 Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code ## Liters of mixture X in the 50% mixture? tagged by: Brent@GMATPrepNow This topic has 5 expert replies and 5 member replies gmattesttaker2 Legendary Member Joined 14 Feb 2012 Posted: 641 messages Followed by: 8 members Upvotes: 11 #### Liters of mixture X in the 50% mixture? 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 Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums! gmattesttaker2 Legendary Member Joined 14 Feb 2012 Posted: 641 messages Followed by: 8 members Upvotes: 11 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 ### GMAT/MBA Expert GMATGuruNY GMAT Instructor Joined 25 May 2010 Posted: 13738 messages Followed by: 1802 members Upvotes: 13060 GMAT Score: 790 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. _________________ Mitch Hunt GMAT Private Tutor GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. GMATinsight Legendary Member Joined 10 May 2014 Posted: 1005 messages Followed by: 21 members Upvotes: 205 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 _________________ Bhoopendra Singh & Sushma Jha - Founder "GMATinsight" Testimonials e-mail: info@GMATinsight.com I Mobile: +91-9999687183 / +91-9891333772 To register for One-on-One FREE ONLINE DEMO Class Call/e-mail One-On-One Private tutoring fee - US$40 per hour & for FULL COURSE (38 LIVE Sessions)-US\$1000

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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
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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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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
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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,

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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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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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,

Cheers,
Brent

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Brent@GMATPrepNow GMAT Instructor
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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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