Algebra — CAT Quantitative Ability

Algebra is CAT's second-largest Quantitative Ability topic, spanning linear and simultaneous equations, quadratics, inequalities and modulus, functions, logarithms, and arithmetic and geometric progressions. Nearly every hard Algebra question is a standard structure wearing an unfamiliar disguise — a quadratic dressed as a word problem, an inequality dressed as a "find the range" question, a progression dressed as a pattern-recognition puzzle. Spotting the disguise is the skill; the algebra underneath is usually short.

1. Linear and simultaneous equations

Two linear equations in two unknowns, $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$, have a unique solution unless their coefficient ratios line up:

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\Rightarrow \text{infinitely many solutions}\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}\Rightarrow \text{no solution}$

A question that asks "for what value of $k$ does this system have infinitely many solutions" is testing exactly this ratio condition, not elimination. Elimination or substitution only applies once you've confirmed a unique solution exists; setting up the ratio test first is faster than attempting to solve and discovering the system is inconsistent.

CAT rarely poses a bare two-variable system — it is usually buried inside an age problem ("A is twice as old as B was five years ago"), a cost problem ("2 pens and 3 pencils cost ₹18; 3 pens and 2 pencils cost ₹17") or a mixture problem.

The translation step — naming two unknowns and writing exactly two independent equations from the words — is the actual difficulty, since the algebra that follows is routine elimination. A system with three unknowns needs three independent equations; a common trap is writing three equations that are not actually independent (one is a combination of the other two), which leaves the system under-determined despite looking complete.

2. Quadratic equations

For $a{x}^2+bx+c=0$, the roots satisfy two relationships that are usually faster than the quadratic formula itself:

$\text{Sum of roots}=-\frac{b}{a},\text{Product of roots}=\frac{c}{a}$

The discriminant $D=b^2-4ac$ decides the nature of the roots: $D>0$ gives two distinct real roots, $D=0$ gives equal real roots, and $D<0$ gives a complex conjugate pair — no real roots.

This lets a question be answered without solving the quadratic at all. "For what values of $k$ does $x^2-6x+k=0$ have real roots?" only needs $D\ge 0$: $36-4k\ge 0\Rightarrow k\le 9$ — the boundary case $k=9$ gives the equal-roots quadratic $x^2-6x+9=(x-3{)}^2=0$, and every $k<9$ opens up two genuinely distinct roots, which is a strictly weaker condition than the question asks for and is worth double-checking whenever a stem says "real and distinct" rather than merely "real."

A quadratic can be built directly from its roots without ever forming $a{x}^2+bx+c$ first: $x^2-(\text{sum})x+(\text{product})=0$. This is the fast route whenever a question states two conditions on the roots rather than the coefficients.

Two quadratics sharing a common root is a distinct and recurring CAT pattern. The common root must satisfy both equations simultaneously, so substitute each equation's own roots into the other equation and check which one fits — there can be more than one value of an unknown parameter that works, corresponding to different choices of which root is shared.

Questions about a symmetric expression in the roots — ${\alpha }^2+{\beta }^2$, ${\alpha }^3+{\beta }^3$, or $(\alpha +1)(\beta +1)$ — never need the individual roots at all. Every symmetric expression can be rewritten purely in terms of the sum $\alpha +\beta$ and product $\alpha \beta$: for instance ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$, and $(\alpha +1)(\beta +1)=\alpha \beta +(\alpha +\beta )+1$. Solving for the roots individually and then combining them is always slower and invites an avoidable arithmetic slip.

Quadratics also arise from word problems disguised as rate or dimension questions — "a rectangle's length exceeds its breadth by 3 cm and its area is 70 cm², find its dimensions" sets up $x(x+3)=70$, i.e. $x^2+3x-70=0$, factoring to $(x+10)(x-7)=0$; only the positive root $x=7$ is physically meaningful, which is why a quadratic word problem's two algebraic roots are not automatically its two possible answers.

3. Inequalities and modulus

Polynomial inequalities are solved with the wavy-curve (sign-chart) method, not by inspection. Mark every root of the numerator and denominator on a number line, then track the sign of the expression across each interval, remembering that the sign flips at every root of odd multiplicity and does not flip at a root of even multiplicity. Points where the denominator is zero are always excluded, even when the inequality is non-strict.

$|x-a|<b⟺a-b<x<a+b|x-a|>b⟺x<a-b\text{or}x>a+b$

A modulus equation is solved by casing on the sign of the expression inside the bars, not by squaring both sides indiscriminately — squaring works but often introduces extra terms that must be checked against the original equation's implicit sign conditions, which is exactly why every modulus solution must be verified by substitution back into the original equation.

A quadratic inequality such as $x^2-5x+6<0$ factors to $(x-2)(x-3)<0$, which is the wavy-curve method with only two critical points: the expression is negative exactly between the roots, so the solution is $2<x<3$.

The general rule for a quadratic that opens upward ($a>0$) is that it is negative strictly between its two real roots and positive outside them — a fact worth stating explicitly, because it is the special case of the wavy-curve method candidates reach for automatically, while the version with three or more factors (as in Example 3 below) is where most errors occur.

4. Functions and composition

A function's domain is the set of inputs for which it is defined, and its range is the resulting set of outputs. Composition is not commutative: $f(g(x))$ and $g(f(x))$ are generally different functions, computed by working from the innermost function outward.

A small number of functions are their own inverse under repeated composition, and recognising one saves an entire computation. If $f(x)=\frac{x-1}{x+1}$, then computing $f(f(x))$ directly gives $-\frac{1}{x}$ — a clean simplification rather than a messier nested fraction, and evaluating $f(f(f(f(x))))$ becomes trivial once this pattern is seen, since applying $f$ four times returns to $x$ itself.

An odd function satisfies $f(-x)=-f(x)$ and an even function satisfies $f(-x)=f(x)$; a periodic function repeats its values at a fixed interval. All three properties are usually stated in a stem to shortcut a computation that would otherwise require evaluating the function at an inconvenient point.

Domain restrictions are frequently the entire question in disguise. A rational function $\frac{1}{g(x)}$ excludes every $x$ for which $g(x)=0$; a square root $\sqrt{h(x)}$ requires $h(x)\ge 0$; and a logarithm requires a strictly positive argument. A question that looks like it needs the wavy-curve method on an inequality is sometimes actually asking only for the domain of a function containing a square root or a denominator — read what is actually being asked before reaching for machinery.

A functional equation states a relationship a function must satisfy for all inputs, rather than giving its formula directly — for instance $f(x+y)=f(x)f(y)$ for all real $x,y$ forces $f$ to be (or behave like) an exponential function, since only $f(x)=a^x$ satisfies that multiplicative-to-additive conversion. Recognising the standard functional-equation families (additive, multiplicative, and the self-inverse pattern of Example 5 below) is faster than trying to solve a functional equation from scratch under time pressure.

5. Logarithms

${\log }_a(xy)={\log }_{a}x+{\log }_{a}y,{\log }_a\left(\frac{x}{y}\right)={\log }_{a}x-{\log }_{a}y,{\log }_a(x^n)=n{\log }_{a}x$

${\log }_{a}b=\frac{{\log }_{c}b}{{\log }_{c}a}\text{(change of base)}$

The number of digits in $N$ is $\lfloor {\log }_{10}N\rfloor +1$ — the single most-tested logarithm application on CAT, since it converts an unwieldy power into one multiplication and a floor function. Finding the number of digits in $2^{64}$, given ${\log }_{10}2\approx 0.3010$, needs only $64\times 0.3010=19.264$, floor to $19$, plus $1$, giving $20$ digits — never requiring the actual 20-digit number to be computed.

Trap. ${\log }_{a}x$ is defined only for $x>0$ and $a>0,a\ne 1$. A logarithmic inequality like ${\log }_{a}x<{\log }_{a}y$ flips direction if $0<a<1$, exactly like multiplying an inequality by a negative number — a fact many candidates forget because logarithms "feel" monotonic in only one direction.

The digit-counting trick generalises into two related ideas worth keeping separate. $\lfloor {\log }_{10}N\rfloor$ alone is called the characteristic of ${\log }_{10}N$, and it is exactly one less than the number of digits in $N$ for any $N\ge 1$; the fractional part (the mantissa) determines the leading digits but is not needed for a pure digit-count question.

Confusing "number of digits" with "the characteristic itself" (forgetting the $+1$) is the single most common slip on this topic, and it is worth a dedicated final check on every digit-count answer.

6. Arithmetic and geometric progressions

$\text{AP: }{T}_n=a+(n-1)d,S_n=\frac{n}{2}[2a+(n-1)d]$

$\text{GP: }{T}_n=a{r}^{n-1},S_n=\frac{a(r^n-1)}{r-1}(r\ne 1),S_{\infty }=\frac{a}{1-r}(|r|<1)$

An AP question that gives two specific terms is really giving two linear equations in $a$ and $d$ — write both term equations, subtract to eliminate $a$ first, and the remaining term follows in one more step.

The infinite sum of a GP exists only when $|r|<1$, and applying the formula outside that range silently produces a meaningless number rather than an error, so checking $|r|<1$ before using $S_{\infty }$ is not optional.

The AM-GM inequality is CAT's most useful non-progression optimisation tool, and it appears constantly inside algebra questions asking for a minimum or maximum:

$\frac{x+y}{2}\ge \sqrt{xy}\text{for }x,y>0,\text{equality iff }x=y$

For an expression like $x+\frac{4}{x}$ with $x>0$, AM-GM gives a minimum of $2\sqrt{x\cdot \frac{4}{x}}=2\sqrt{4}=4$, attained exactly when $x=\frac{4}{x}$, i.e. $x=2$ — a one-line answer to a question that calculus would also solve, but far more slowly under exam time pressure.

A harmonic progression (HP) is simply a sequence whose reciprocals form an AP, so any HP question is solved by inverting every term, solving it as an AP, and inverting the answer back at the end — there is no separate HP formula worth memorising beyond this one translation. The three means relate by a fixed, always-true inequality for positive numbers:

$\text{AM}\ge \text{GM}\ge \text{HM}$

with equality throughout exactly when all the numbers are equal. This chain is what justifies the AM-GM shortcut used above, and it is also the fastest way to bound a quantity when a question gives an average (AM) and asks for a bound on a product (related to GM) or a rate (related to HM), without setting up the full optimisation from scratch.

Worked Examples

Example 1 (linear equations — easy). For what value of $k$ does the system $2x+3y=5$ and $4x+ky=10$ have infinitely many solutions?

The ratio condition requires $\frac{2}{4}=\frac{3}{k}=\frac{5}{10}$. Since $\frac{2}{4}=\frac{5}{10}=\frac{1}{2}$, we need $\frac{3}{k}=\frac{1}{2}\Rightarrow k=6$.

Example 2 (quadratic — common root, hard). The equations $x^2-5x+6=0$ and $x^2+kx-6=0$ have a common root. Find all possible values of $k$.

The roots of the first equation are $2$ and $3$. If the common root is $r=2$: substituting into the second equation gives $4+2k-6=0\Rightarrow k=1$. If the common root is $r=3$: $9+3k-6=0\Rightarrow k=-1$. Both are valid, giving $k=1$ (shared root $2$) or $k=-1$ (shared root $3$) — a question asking for "the value of $k$" in the singular is testing whether a candidate checks both roots rather than stopping at the first.

Example 3 (inequality, wavy curve — hard). Solve $\frac{(x-1)(x-3)}{x-2}\ge 0$.

Mark the critical points $1,2,3$ on a number line ($x=2$ excluded, since the expression is undefined there). Testing one point in each interval shows the expression is negative for $x<1$, positive for $1<x<2$, negative for $2<x<3$, and positive for $x>3$. Including the zeros of the numerator ($x=1,3$, where the expression equals zero and the inequality is non-strict) and excluding $x=2$: the solution is $x\in [1,2)\cup [3,\infty )$.

Example 4 (modulus — medium). Solve $|2x-3|=x+4$.

Case 1: $2x-3=x+4\Rightarrow x=7$. Check: $x+4=11\ge 0$ and $|2(7)-3|=11$. Valid.

Case 2: $-(2x-3)=x+4\Rightarrow -2x+3=x+4\Rightarrow x=-\frac{1}{3}$. Check: $x+4=\frac{11}{3}\ge 0$ and $\left|2\left(-\frac{1}{3}\right)-3\right|=\frac{11}{3}$. Valid.

Both $x=7$ and $x=-\frac{1}{3}$ satisfy the original equation.

Example 5 (functions, self-inverse composition — hard). If $f(x)=\frac{x-1}{x+1}$, find $f(f(x))$, and hence evaluate $f(f(f(f(5))))$ without further computation.

$f(f(x))=\frac{f(x)-1}{f(x)+1}=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=\frac{\frac{(x-1)-(x+1)}{x+1}}{\frac{(x-1)+(x+1)}{x+1}}=\frac{-2}{2x}=-\frac{1}{x}$

So $f(f(x))=-\frac{1}{x}$, and applying $f\circ f$ twice more gives $f(f(f(f(x))))=-\frac{1}{-1/x}=x$ — four applications of $f$ return to the original input. Hence $f(f(f(f(5))))=5$ directly, with no fraction arithmetic needed at all.

Example 6 (logarithms — medium). Find the number of digits in $2^{64}$, given ${\log }_{10}2\approx 0.3010$.

Number of digits $=\lfloor 64\times 0.3010\rfloor +1=\lfloor 19.264\rfloor +1=19+1=20$.

Example 7 (AP — easy). In an arithmetic progression, the 5th term is 17 and the 12th term is 38. Find the 20th term.

$a+4d=17$ and $a+11d=38$. Subtracting: $7d=21\Rightarrow d=3$, so $a=17-12=5$. The 20th term is $a+19d=5+57=62$.

Example 8 (AM-GM optimisation — hard). Find the minimum value of $x+\frac{4}{x}$ for $x>0$, and state where it occurs.

By AM-GM, $\frac{x+\frac{4}{x}}{2}\ge \sqrt{x\cdot \frac{4}{x}}=\sqrt{4}=2$, so $x+\frac{4}{x}\ge 4$. Equality holds when $x=\frac{4}{x}\Rightarrow x^2=4\Rightarrow x=2$ (taking the positive root, since $x>0$). The minimum value is $4$, attained at $x=2$ — verified directly: $2+\frac{4}{2}=2+2=4$.

Summary

Check the coefficient-ratio condition before solving a simultaneous system — it decides unique, infinite, or no solution in one step.

Sum and product of roots, $-b/a$ and $c/a$, are usually faster than the quadratic formula; a shared root between two quadratics must be checked against each equation's own roots individually, since more than one parameter value can work.

Solve polynomial inequalities with the wavy-curve method across marked critical points, excluding any point where the expression is undefined; solve modulus equations by casing on sign, and always verify each case's solution against the original equation.

Composition is not commutative, and a function that is its own inverse under two applications ($f(f(x))=-1/x$ for $f(x)=(x-1)/(x+1)$) turns a repeated-composition question into a one-line pattern check.

The number of digits in $N$ is $\lfloor {\log }_{10}N\rfloor +1$ — the highest-yield single logarithm fact on the exam.

An AP question giving two terms is two linear equations in $a$ and $d$; a GP's infinite sum exists only for $|r|<1$. AM-GM, $\frac{x+y}{2}\ge \sqrt{xy}$, solves most "minimum of $x$ plus something over $x$" questions in one line, with equality exactly when the two terms are equal.