Algebra is one of the most important topics when it comes to the Quant portion of the CAT syllabus. The majority of the questions in this section, are either form algebra or from geometry. Keeping in mind the vast syllabus of number system in CAT, we will cover only the concepts and tricks which are less known yet important.

### 1. Maxima of Quadratic Rational Function

The functions of kind (ax2 + bx + c) / (mx2 + nx +q) are called quadratic rational functions. To find maximum (or minimum) value of such functions, we equate it with y, i.e. (ax2 + bx + c) / (mx2 + nx +q) = y.

ax2 + bx + c = ymx2 + ynx + q

(a – ym) x2 + (b – yn) x + c – q = 0

For real roots, D 0, i.e. (b – yn)2 – 4(a – ym)(c – q) 0

Now solving this equality gives us the maximum (or minimum) value of y, or the quadratic rational function.

Example: Find out the least value of (x2 – 6x + 5) / (x2 + 2x + 1).

Answer: Let (x2 – 6x + 5) / (x2 + 2x + 1) = y.

(1 – y)x2 – (6 + 2y) + 5 – y =0

D ≥ 0 => 4(6 + 2y)2 – 4(1 – y)(5 – y) ≥ 0

=> 36 + 24y + 4y2 – (4y2 – 24y + 20) ≥ 0

=> 48y + 16 ≥ 0 or y ≥ -1/3

Thus minimum value of the given function is -1/3.

### 2. Smallest value of maximum function

To solve such questions, all you need to do it to find out the point of intersection of the individual values. To find the smallest value of g(x) = max (5 − x, x + 2), we equate 5 – x with x + 2.

That gives us x = 3/2, and the value of g(x) comes out to be 7/2, which is its smallest value.

### 3. Number of Integral Solutions

#### A. Type Ax + By + C = 0 (in lowest reducible form)

If A and B are not co-prime, there is no solution.

Find any one integral solution, and the rest solutions can be found easily. The value x and y will change by a coefficient of the other variable. An increase in x will cause decrease in y, and vice versa.

#### B. Type: x1 + x2 + ⋯ + xr = n

Number of positive integral solutions is n-1Cr-1.

To find number of non-negative integral solution, replace x1 by x1‘ – 1, x2 by x2‘ – 1 and so on. Now x1‘, x2‘ … are positive integers, hence the above formula can be applied.

#### C. Type: |x| + |y| = n

Let |x| = a and |y| = b. Then, a + b = n.

Number of positive integral solution is n-1C2-1 = n – 1. As (x, y) can take four values corresponding to each (a, b) solution, Total number of non negative solutions is 4(n – 1).

You can add solutions where a = 0 or b = 0, if total integer solutions is asked in the question.

#### D. x2 – y2 = N

Simplify the equation (x – y)(x + y) = N, and find factors of N. Now N can be one of these four types-

Case 1: N is an odd number and not a perfect square

Total positive integral solution is (Total number of factors of N) / 2.

Case 2: N is an odd number and a perfect square

Total positive integral solution is ((Total number of factors of N) – 1) / 2

Case 3: N is an even number and not a perfect square

Total positive integral solution is  ((Total number of factors of N/4)) / 2.

Case 4: N is an even number and a perfect square

Total positive integral solution is  (Total number of factors of N/4) – 1) / 2.

PS: Comment below if you need explanation of any the formulae given or more examples of any of the concepts explained.

You can find more such blogs here- Advance concepts and tricks of quantitative aptitude.

I would recommend you to practice solving questions from this section at Prepleaf – CAT Portal.