Algebra is nothing but the expression of number patterns using variables. Importance of variable based questions in CAT is increasing year after year. On an average around 50% of the questions are from algebra over the last 5 years. The best thing about algebra is that almost all questions can be solved using some very innovative and easy to use strategies as shown below. I have included a few questions from actual CAT papers also.

ASSUMPTION method

There are at least 10 questions out of 25 in the CAT 2007 paper where you can apply this technique. This involves assuming simple values for the variables in the questions, and substituting in answer options based on those values. Assumption helps in tremendously speeding up the process of evaluating the answer as shown below.

1) Let ‘x’ be the arithmetic mean and y,z be the two geometric means between any two positive numbers. The value of  (y³ + z³) / (xyz) is

a) 2                  b) 3                  c) 1/2               d) 3/2

Assume a GP  1        2          4          8 . This implies that the 2 GMs between 1 and 8 are y=2 and z=4.

Arithmetic Mean, x =(8+1)/2 = 4.5 .Substitute in (y³ + z³) / (xyz)

Answer = (2³+4³)/(2x4x4.5) = 2.

NOTE:- Assume a GP 1    1    1    1  and            AP 1    1          1,

then x=1, y=1 z=1. Answer on substitution=2, which will make the calculation even faster, Half of the problems in Algebra can be solved using assumption. Note that this is not direct substitution. In the next example see how you can use the same technique in an equation question.

ADVANTAGE 1: The moment you get values for x,y,z you are in a position to solve any question on x,y,z and not just the above one.

ADVANTAGE 2 :You can assume any GP and solve the question. You will get the same answer. You are not supposed to get two different answers. If you get two answers, there are only 2 possibilities

1) There is an option “Cannot be Determined”

2) The question is wrong!

That’s the beauty of algebra.

Let’s look at another question we can solve using a similar approach from the topic of equations

2) k &  2k² are the two roots of the equation x² – px + q. Find q + 4q² + 6pq =

a) q²                 b) p³                 c) 0                  d) 2p³

Solution:

Assume an equation with roots 1&2 (k=1)

=>p (sum of roots)= 3 and q(product of roots)=2. Now you can solve any question in p & q

Substitute in q + 4q² + 6pq = 54.  Look in the answer options for 54 on substituting values of p=3 and q=2. We get 2p³ = 54.=> Ans = 2p³ .

Even though the above example is a question from an entirely different topic, for us it is one and the same. i.e the method which we used for solving is assumption.

SOME ACTUAL QUESTIONS FROM CAT-2007 SOLVED USING THIS TECHNIQUE

3) Consider the set S={2,3,4……2n+1), where n is a positive integer larger than 2007. Define X as the average of odd integers in S and Y as the average of the even integers in S. What is the value of X-Y?

a) 1                 b) n/2                        c)(n+1)/3n                  d) 2008                      e) 0

Solution:

The question is independent of n, which is shown below.

Take n=2. Then S= {2, 3, 4, 5). X= 4 and Y=3. X-Y =1,

Take n=3. Then S={2,3,4,5,6,7}. X=5 and Y=4. X-Y=1

Hence you can directly mark the answer option (a) .You can solve the question in less than 60 seconds.

4) Let a₁=p and b₁=q, p and q are positive quantities. Define

a_n=pb_n-1, b_n=qb_n-1 for even n>1

a_n=pa_n-1, b_n=qa_n-1 for odd n>1.

which of the following best describes a_n+b_n for even n.

a) qp ^(n/2)(p+q)                               b) q^(n/2)(p+q)ⁿ        

c) q^(n/2)(p+q)^n/2-1                            d)q(pq)^(n/2)-1(p+q)

 Solution:

Since the answer options involve square roots, we will assume

 a₁=p=4 and b₁ =q=4 (when n=1)

Now for n=2,we will get  a₂= 4*4=16 ; b₂=4*4=16

In our assumption, a₂+b₂=32. Check where you are getting 8 among answer options.  Answer is option (d) q(pq)^(n/2)-1(p+q)= 4*8=32. This won’t take more than more than two minutes.

5) A GP consists of 500 terms. Sum of the terms occupying  the odd places is P₁

and the sum of the terms occupying the even places is P2. Find the common ratio?

a) P₂ / P₁        b) P₁ / P₂        c) P₂ + (P₂/ P₁)                          d) P₂ + (P₁ / P₂)

Solution:

We can solve this question easily by assuming a GP with just 2 terms, e.g.: 1, 2

In our assumption P₁ =1  and P₂ = 2 and common ratio = 2

Substitiute P₁ = 1 and P₂ = 2 and check among answer options where you are getting 2.

Answer in 30 seconds. Option (a)

CAT -08 question solved using a similar approach

There were more questions which could be solved using similar strategies. The methods given above clearly show that for someone with good conceptual knowledge and right strategies the quant section is a cakewalk. The following question is straight out of the CAT-2008 paper.

6)√[1+(1/1²)+(1/2²)+ √(1+1/2²+1/3²) +………..√[1+(1/2007²)+(1/2008²) ]

a) 2008- (1/2008)   b) 2007-(1/2008)    c) 2008 – (1/2009)    d) 2008- (1/2007)

Solution:

Assume an extreme case, i.e. only the first term. We can get the answer for a seemingly difficult question in a completely unconventional method as given below.

Value of the first term = 3/2

Among the options we can substitute 2008 = n = 2 and 2007 = n-1 = 1 and 2009 = n+1 = 3

We get option (a) = 2 – (1/2) = 3/2. In other options we are not getting 3/2. So answer is option (a). Unbelievably easy!