I visit the GMAT Club forum regularly and discuss some ideas, some methodologies there. The weighted averages method I discussed in my previous two posts is one of my most highly appreciated inputs on the forum. People love how easily they can solve some of the most difficult questions by just drawing a scale or using a ratio. If you are not a Quant jock, I am sure you feel a chill run down your spine every time you see a mixtures problem. But guess what, they are really simple if you just use the same weighted average concepts we discussed in the previous two posts. Let’s look at a mixtures question in detail*:*

Mixture A is 20 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 15 gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?

(A) 3 gallons

(B) 4 gallons

(C) 5 gallons

(D) 10 gallons

(E) 12 gallons

Let’s try and imagine what is taking place here:

Imagine two tumblers, one containing Mixture A and the other containing Mixture B. The mixtures from the two tumblers are poured into a new big empty tumbler. The first thing that we can conclude from this step is that V1 + V2 gallons = 15 gallons because the total volume when you combined the mixtures turned out to be 15 gallons (in the big tumbler).

Can you say anything about the relative value of V1 and V2? Can you say which one will be greater? We discussed in the previous post that when we find the weighted average of two quantities, the average is closer to the one which has higher weight because that quantity ‘pulls’ the average toward itself. So if the new mixture contains 30% alcohol, which of the two mixtures – A (20% alcohol) or B (50% alcohol), will have higher volume? Obviously A because it pulled the average toward itself. The average, 30% alcohol, is closer to 20% than to 50%. Now, let’s calculate exactly how much of the 15 gallons was mixture A and how much was mixture B.

*w1/w2 = (A2 – Aavg)/(Aavg – A1)*

w1/w2 = (50 – 30)/(30 – 20) = 2/1

Ratio of volume of mixture A : volume of mixture B = 2:1.

So out of a total of 15 gallons, 10 gallons will be mixture A and 5 gallons will be mixture B. (Check out the post on Ratios if you are not sure how we arrived at this.)

Let’s try the same thing using the scale method:

V1 : V2 = 2:1

So out of a total of 15 gallons, 10 gallons will be mixture A and 5 gallons will be mixture B. It wasn’t very bad, was it?

And now, let’s try a quick example that looks ominous. People end up making tons of equations with ‘x’ and ‘y’ to solve this one but we can do it in a moment!

*There are 2 bars of copper-nickel alloy. One bar has 2 parts of copper to 5 parts of nickel. The other has 3 parts of copper to 5 parts of nickel. If both bars are melted together to get a 20 kg bar with the final copper to nickel ratio of 5:11. What was the weight of the first bar?*

(A) 1 kg

(B) 4 kg

(C) 6 kg

(D) 14 kg

(E) 16 kg

This question might look different from the question above but it is actually very similar to it. Focus on any one of the two elements, say Copper (or Nickel if you fancy it more. Either way, you will get the same answer). Forget about the other one. If you notice, in the question above there are two elements in each mixture too (obviously, that is why they are called mixtures!) – Alcohol and Water. 20% Alcohol means that the rest 80% is Water. We just work with the concentration of Alcohol because it is given explicitly to us. We could as well say that mixture A has 80% water and mixture B has 50% water and the combined mixture has 70% water. Still the ratio obtained will be 2:1. But, still, somehow, giving the ratio of both Copper and Nickel in this question throws people off.

Let me quote one of my favorite GMAT instructors here: “And this is one of many reasons why McDonald’s, the greatest marketing organization on earth, is a lot like the GMAT, the test that can help you get into the greatest marketing MBA programs on earth. They know how to make those tiny tweaks that make massive differences, at least in perception to the end consumer.” (Check out the complete(-ly) delicious article that discusses Shamrock Shakes here.)

Anyway, let’s get back to this question. I am keen to show you that it is, in fact, just like the previous question.

First bar is 2/7^{th} copper (2 parts copper and 5 parts nickel to get a total of 7 parts). Second bar is 3/8^{th} copper and the combined alloy is 5/16^{th} copper.

Now, let’s calculate the ratio of weights of the two bars.

*w1/w2 = (A2 – Aavg)/(Aavg – A1)*

w1/w2 = (3/8 – 5/16)/(5/16 – 2/7) = 7/3

So weight of first bar: weight of second bar = 7:3

Out of a total of 20 kgs, the weight of the first bar was 14 kg and the weight of the second bar was 6 kg.

Or we can use the scale method too.

So the ratio of their weights is 1:3/7 or 7:3

Out of a total of 20 kgs, the weight of the first bar was 14 kg and the weight of the second bar was 6 kg.

Hope these questions don’t look very intimidating now. Use this method for your mixture problems; they won’t feel like problems any more!

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep in Detroit, Michigan, and regularly participates in content development projects such as this blog!*

## Quarter Wit, Quarter Wisdom: Don't Get Mixed-Up in Mixtures

*Posted on April 11, 2011, filed in: GMAT, Quarter Wit, Quarter Wisdom*

Related posts:

Hi Karishma,

Awesome. Thank you..

how to solve this mixture problem with the equation

there is a 40 litre solution of milk and water in which milk forms 72%.How much water must be added to the solution to make it a solution in which milk forms 60%?

A mixture of spirit and water measure 80 gallons. It contains 20% water. How much water should be added to it so that water may be 25%?

For this question:

A mixture of spirit and water measure 80 gallons. It contains 20% water. How much water should be added to it so that water may be 25%?

v1=80, v2 to be figured out.

try out the scale method taking water percentages. %1=20, avg%=25, %2=100 (since its only water, no spirit)

it comes out to be – v1:v2=75:5=15:1

v2=80/15 liters of water

First Question:

Solution 1 has 72% milk.

Solution 2 has 0% milk. (You are mixing solution 1 with pure water.)

Mixture should have 60% milk.

w1/w2 = (0 – 60)/(60 – 72) = 5/1

So volume of solution 1 : volume of water = 5:1

Solution 1 is 40 litres, so 8 litres of water must be added.

Now, can you figure out how to solve the second question using this formula?

Hi karishma

I tried to solve this taking water instead of milk , somehow I am making a mistake somewhere , please can you show me the right way

solution one : 28 percent water ( 72 %milk )

solution 2 : 100 percent water ( No milk )

Final mixture : 40 % water ( 60% milk )

so 28…..40…..x ( where x = liters of water added )

ratio of their weights (x-40)/ 12 = 40 /x

is this equation correct, where am I making the mistake ?

Solution 1: 28% water

Solution 2: 100% water

Mixture: 40% water

Ratio of weights = (100 – 40)/(40 – 28) (Why x? You already know the concentration of water in both the solutions)

Ratio of weights = 60/12 = 5:1

Volume of solution 1: Volume of solution 2 i.e. water) = 5:1

Now same as above.

Thank you got it now,

One point I was making a mistake is that if we are considering percentages ( concentration ) then equation should be in the form of concentration’s only

like 28…….40……….100

I was taking x as the volume of water added and forming the equation as

28…….40…….x which is wrong as the first two quantities are concentration( percentages ) and x is the volume of water added not the concentration..so this second equation is wrong.

all the three quantities should be the same kind ( either concentration’s , or volume or ratio’s ) ,

right ?

Yes, think about it – you are taking the avg of 2 things. They must have the same units and their avg will have the same units too.

Karishma,

How do we tackle this with your approach?

How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol?

I get A:B as 15 :1

Yes, you are right. You get w1/w2 = 15:1

For every 15 parts of 20% alcohol solution, pure alcohol added must be 1 part.

For 100 lts of 20% solution, pure alcohol added must be 100/15 = 6.667 lts

Please let me know how was 15:1 was arrived.

You have 20% alcohol solution and 100% alcohol (pure alcohol). You want to mix them to get 25% alcohol. Use the formula:

w1/w2 = (100 – 25)/(25 – 20) = 75/5 = 15/1

So you should mix 20% alcohol and 100% alcohol in the ratio 15:1

Hi Karishma,

Can the following question using the formula or the time scale. I’m not able to get the equation right.

Solution Y is 30 percent liquid X and 70 percent water. If 2 kilograms of water evaporates from 8 kilograms of solution Y and 2 kilograms of liquid X are added to the remaining 6 kilograms of solution Y, what percent of this new solution is liquid X?

Follow up question:

Can all the mixture questions be solved using either the equation or the scale method?

This formula/scale method is useful when you add two mixtures together. Removal/replacement needs a different approach. I might discuss it in my posts down the line. As for this question, since only one component evaporates, it’s best to work on the components individually.

Y(8 kg) = X(2.4 kg) + Water(5.6 kg)

Water evaporates -> Y(6 kg) = X(2.4 kg) + Water (3.6 kg)

Add 2kg X -> Y(8 kg) = X(4.4 kg) + Water (3.6 kg)

% of X in new solution = 4.4/8 * 100 = 55%

Alternatively, you see that X = 30% and Water = 70%

Water decreases by 25% of total solution (2kg out of 8kg solution evaporates)

So Water leftover will be 45%. Therefore, the rest of the 55% in the new 8 kg solution must be X (after X was added to bring the solution back to 8 kg)

Hi Karishma,

I just wanted to thank you for posting this. I’ve been having the hardest time understanding mixtures and this instantly cleared up all the confusion and I have been able to solve many problems now. Thank you so much for your help!

S

Hi Karishma,

Is there an approach for removal/replacement of a solution in a mixture problem?

Thanks,

Prat

Yes, I have been meaning to discuss removal/replacement but I have been lost in P and C for the past many weeks! I will take up removal/replacement next week and continue with probability thereafter.

Great, Thanks!!

Hi Karishma,

Trying to solve a mixture problem but not sure if I have solved it correctly. I have not been able to use the scale or weighted average method. Can either method be used to solve this:

A Food and Drug lab has two new samples: a 240 gram cup of drip coffee, which contains 124 mg of caffeine, and a 60 gram cup of espresso, containing 160 mg of caffeine. If a technician were to create a new 120 gram cup sample that contained 50% coffee and 50% espresso, how many mg of caffeine would the new drink contain?

The way I solved this was as follows:

of the 120 gms- 50% (60 gms) is espresso so it includes 160 mg of caffeine

– 50% (60 gms)is drip coffee, so 124/4= 31 mg of caffeine

So total is 191 gms of caffeine in the new 120 gram sample.

Is this correct? I don’t hope it is a fluke..

Thanks,

Pratap.

Sorry for the delayed reply! I missed replying somehow. The solution is perfectly fine. It is what I would use instead of any other method. It is the most straight forward solution. Good work!

Hi Karishma,

If it’s already published, can you please share the link for solving mixture problems containing removal/replacement ?

I really liked your method and I’m going to use it for adding mixture problems now onwards.

Thanks,

Abhishek.

Here you go:

http://www.veritasprep.com/blog/2012/01/quarter-wit-quarter-wisdom-removalreplacement-in-mixtures/

Two glasses contain milk and water. In the first glass the ratio of milk to water is 3 : 2 and in the second glass the ratio of milk and water is reverse of that in first glass. What part of the mixture from first glass should be drawn out and added to the second glass, so that milk and water will be in equal quantities in the second glass?

A. 1/3

B. 1/4

C. 2/5

D. 1/2

E. cannot be determined

How to solve this using allegations ( Scale method)

Using the scale method, you can easily find the ratio in which the two solutions should be mixed:

w1/w2 = (3/5 – 1/2)/(1/2 – 2/5) = 1:1

In fact, you can just reason it out too. One solution has excess of one quantity. Another has excess of the other quantity (in same measure). To get equal amounts of both quantities, you must mix equal amounts of solutions.

The problem here is we don’t know whether the two glasses are identical, whether they are completely full/half full etc. All we know that is that both solutions should be mixed in equal quantities but we don’t know what fraction of the first solution needs to be removed since we don’t know how much total solution there is in the first glass. We also need to know the amount of solution in the second glass because the same amount of solution 1 needs to be drawn.

Hence you cannot answer this question with the given information.

How do you get 5/16th copper?

We are given that ‘the final copper to nickel ratio of 5:11′ which means copper is 5/16th of the final mixture. Hence, we know have the Aavg.

Hello Karishma. Thanks for your effort into writing this guide. It definitely makes things easier. One small question.in the scale method how do you decide which one is W1 and which one is W2.

It doesn’t matter. Take one of them as w1 and A1 and the other as w2 and A2. The only thing to remember is that when you say that one of them is A1, then the weight of that thing must be w1 (and not w2). Do not mix them up.

What I do prefer though is that I take the smaller one (the one which is less than the average) as w1 and put it on the left on the number line since it is more intuitive that way.

Hi Karishma, Thanks for the tips on mixtures. I am applying these concepts on mixture problems, and able to get to the answer without any confusion now.

Here is one question, for which I am not able to apply the concept you have provided above, though I am able to get the correct answer through substitution in terms of x and y. Could you please provide your approach to this?

“A rabbit on a controlled diet is fed daily 300 grams of a mixture of two foods, food X and food Y. Food X contains 10% protein and food Y contains 15% protein. If the rabbit’s diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture. ”

Thanks,

Srini

Sure.

Food X – 10% protein

Food Y – 15% Protein

Mixture that rabbit gets – 38/300 * 100 = (38/3)% protein

wX/wY = (15 – 38/3)/(38/3 – 10) = 7/8

So, mixture contains 7 parts X and 8 parts Y.

This is equal to (7/15)*300 = 140 gms of food X in the 300 gms mixture.

Karishma,

Great Post!

Regarding one of the questions above ” Yes, you are right. You get w1/w2 = 15:1

For every 15 parts of 20% alcohol solution, pure alcohol added must be 1 part.

For 100 lts of 20% solution, pure alcohol added must be 100/15 = 6.667 lts ”

How is it that you are considering 100 lts instead of 80 ltrs.

According to my calculations:

w1 / w2 = 15:1 same as yours.

Then, since w1 = 80 we will get 80/15 = 16/3 —-> Can you please help.

Exactly where am I wrong?

Hi Karishma,

Regarding the 1st example.

Mixture A is 20 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 15 gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?

The resulting mixture is 15 gallons containing 30% alcohol which means that rest 70% is some other liquid say water. Now we have got the ratio of 2:1 but we are equating this ratio to the whole 15 gallons. Indirectly we are saying that 15 gallons of solution contains only alcohol. Should we equate the ratio of 2:1 to 30% of 15 gallons i.e. to 4.5 gallons rather than the entire 15 gallons??

Please let me know if i missed something.

The ratio 2:1 is the ratio in which we mixed mixture A and mixture B (the two components of the combined mixture). We have already taken their alcohol content into account by considering that they have 20% and 50% alcohol respectively.

When you mix 10 lts of 20% alcohol solution with 5 lts of 50% alcohol solution, you get 15 lts of 30% alcohol (4.5 lts in the 15 lts) solution.

Ur shower of concepts have ultimately fertiled my barren terrains of mathematics…nywz even I love alliteration.:P..thank u karishma..ur blog is a panacea for tha aspirants like us who cant visit coachings due to myriad reasons..

Hello,

w1/w2 = (3/8 – 5/16)/(5/16 – 2/7) = 7/3

how can we calculate this the fastest way?

w1/w2 = (3/8 – 5/16)/(5/16 – 2/7) = 7/3

w1/w2 = (6/16 – 5/16)/(35/16*7 – 32/7*16)

w1/w2 = (1/16)/(3/16*7)

w1/w2 = 1*16*7/16*3 = 7/3

Hi Karishma,

Could you explain me how do you get 1:3/7, and then 7:3 in the scale method for the second problem?

Thanks,

Dan

The weights will be given by the lengths of the green and red lines.

Length of green line = 5/16 – 2/7 = 5*7/16*7 – 2*16/7*16 (we made the denominators same) = (35 – 32)/7*16 = 3/7*16

Length of red line = 3/8 – 5/16 = 3*2/16 – 5/16 = 1/16

So the lengths of lines are 3/7*16 and 1/16

The ratio is (3/7*16)/(1/16)

Since 3/7*16 is divided by 1/16, it can be multiplied by 16 instead. You get (3*16)/(7*16) = 3/7

The ratio of the weight is the inverse of this 7/3.

I would suggest you to check out the formula instead: w1/w2 = (A2 – Aavg)/(Aavg – A1)

A lot of these calculations are simplified in that.