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.
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.
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.
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;
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.
(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?
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
And by extension
………………… eq2
……………………………………. using the identity –
[a/b] =[ c/d] =[ (a+c) /(b+d)]
Let us look at the question now armed with eq
2
So we have by virtue of Eq
2
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.
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 .
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
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
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
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
I can be reached on ranjan.rahul885@gmail.com
