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Averages and Mixtures Demystified - Road to GMAT Q51

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Averages and Mixtures Demystified

Most of us must have at least once in our lives spent some time on the see-saw with our friends. How about using them to solve the problems of Mixtures and Alligation, and Averages. Yes, you heard it right. Math is fun after all. Let us rewind our lives to the point when we used to have fun on the seesaws.

image

I was quite skinny when I was young (don’t ask me what I look like nowJ). Consequently, I had to use the long handle to balance the “normal” guy on the other side.
image

Fig In fig 1,

 a)   A is heavier than B (that’s me).

 b)   L1=L2

Clearly the see-saw is unbalanced.  In order to balance the seesaw. Either B must go further away from the support or fulcrum, which essentially means going out the frame, or A must come closer to fulcrum. See fig 2. 

image

What is the underlying physics behind this? The equation that governs where one needs to sit in order to balance the seesaw about the support is

(a)  x (L1) = (b)  x  (L2)…………….  Eq (1)

Where

     A, B= weights of A and B respectively

     L1 and L2= Distance of A and B from the fulcrum. 

Now, you can think of the fulcrum or the support as the point where the average weights of A and B is balanced. Let’s call the average weight as ‘M’. If either the length or the weight of our guy A is increased. The equilibrium is disturbed. B must move further from M to maintain M at the same place, or increase his weight in some way to maintain Eq 1.

Now what does this all have to do with averages or mixtures?

Well, we can use Eq 1 and its implications to solve problems of averages and mixtures. Let’s focus on averages first.

    Consider this problem

Q1.) What is the average weight of a fruit bag containing 1 orange of 100gms and 1 apple of 200gms? (Consider the weight of the bag negligible)

It is easy to solve average M= (100+200) /2 = 150gms 

Let us use the graphical implications of the problem. M=150;

image

And the equation= L1xA=L2xB

                                        Or

                                   50x1=50x2;

What if we didn’t know the value of M? can we still use Eq 1. Well that’s the plan!

                                           (M-150)x1= (200-M)x1

=> 2M=300

=>   M=150

Are we making matters difficult for ourselves? No, we are making it easier. How?Let’s me add complications to the question we just discussed.

Q2.) What is the average weight of a fruit bag containing 10 orange of 100gms and 5 apple of 200gms? (Consider the weight of the bag negligible)

Using Eq 1.  

image

(Why 200-M and not M-100 you may ask? Well Average can’t be greater than the maximum value of an individual element)

  ð  (M-100)x10= (200-M)x5

  ð  15M=2000

  ð  M=133.33

Now, how do we use this to solve slightly difficult problems? Let’s take a slightly difficult problem.

Q3)  A batsman scores with an average of 49 runs in 99 innings. How many should he score in his 100th match to take his career average to 50 runs?

image

To solve this question let us introduce two more variables in place of 100gms and 200gms used in Q2 and Q3. Let’s call them n1 and n2

So

L1=M-n1

L2=n2-M

The eq1 becomes

L1xA=L2xB

=>            (M-n1)x A = (n2-M)xB

 

image

 

And by extension

 

image

…………………eq2

 

……………………………………. using the identity – [a/b]=[c/d] =[(a+c)/(b+d)]

 

Let us look at the question now armed with eq 2 

 

image

 

So we have by virtue of Eq 2

image

Solving, you can get n2=149;

Another way to look at this is the extra runs, n2-49, has to be distributed among all the innings (99+1) in such a way that that the average is 50. So he needs 1 more run for each of his 99 innings to compensate for the average. And 50 more in the last inning to maintain the average. Or he needs to score

                              99 +50 =149.

There are many variants of Eq 2 that you may find in literature. However, I like to write them down in a full equation and solve.

 

Mixtures

 

How can we use the concept developed for averages in mixtures problems? Well they are not much different to be honest.

 

Let’s look at a 600 level (according to GMATClub) problem to understand this

 

Q1) Krisp cereal is 10% sugar by weight. Brano cereal is 2% sugar by weight. If you wanted to make the mixture 4% sugar by weight what should the ratio of Krisp to Brano.

image

so we have 

image

 

Or 6/A=2/B

A/B = 3:1

 

Note- An interesting thing to note here is that, if A=B, then the average will be 6%. But in the question the average is 4%. 4% is closer to n1 or 2%. So A must be greater than B to balance out. Recall the see-saw, where to counter my heavier friend I needed a longer handle (or Torque for the engineers out there!). Consequently A/B will be greater than 1. If the question asked us to find the ratio if the average % was 8% (closer to n2). Then A/B would have been less than 1. Actually A/B = 1:3. (You can imagine the central point or the strength of the solution as the center or the fulcrum of the see-saw and imagine the equilibrium conditions). This is a very important observation. If a question like this appears on the test. Chances are that you will get half of the ratios less than one and half greater than one. You can eliminate half of the options using this simple analysis. Saves time! J

Let’s take a 700 level question

Q2) A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?
A. 37.5
B. 75
C. 100
D. 150
E. 175

 

                                 20% salt implies that = 20 units salt and 80 units water. If ¼ of water evaporates, that is (80/4=) 20 units of water is lost. (Don’t try to evaporate salt please J) You are level with 60 units water. So new strength of the solution after evaporation= 20/ (20 + 60) 0r Y gallons of 1/4 (25%) strength. (It is no longer X gallons. Since part of the water evaporated, right). 

                                 10 gallons of water + 20 gallons of salt = 30 gallons of 2/3 (66.67 %) strength.

So our figure becomes

 

image

 

image

 

Or

Y= 120 gallons

 

So we have 1/4 (25%) strength 120 gallons solution.

Or 30 gallon of salt and 90 gallon water. But 90 gallon is 3/4 of water before evaporation. Or the total water before evaporation

90*4/3=120

 

So, X= 150 

Note that the question can be solved differently in many ways. One of them is by pure algebra.

x= salt + water

  = 0.2x + 0.8x

When 1/4 of 0.8x evaporates. You are left with 0.6x

Adding 10 gallons of water and 20 gallons of salt. New solution becomes

 

 

image

 

Solving, X=150

 

The Idea is to introduce the concept using a fun filled yet powerful analogy, so that the questions can be solved intuitively. Besides, it is a very good way to visualize the problem.

CheersJ

 

Posted Yesterday by rahul ranjan

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