Simplification, Surds and Indices — SSC CGL Quantitative Aptitude

Simplification is where marks are lost not to difficulty but to sequence: doing addition before division, or squaring a surd carelessly. Fix the VBODMAS order once, learn the index laws cold, and memorise a few surd patterns (rationalising, $\sqrt{a\pm 2\sqrt{b}}$, $a^3\pm b^3$). Then these become the fastest guaranteed marks on the page.

1. What SSC actually asks

Tier 1: 2–3 Q · Tier 2: 2–4 Q. Formats: long BODMAS expressions with fractions and 'of', find-the-value power questions, rationalising a surd, simplifying a nested radical, and Tier-2 approximation ("find the nearest value"). All are mechanical once the rules are automatic.

2. VBODMAS — the operation hierarchy

Evaluate strictly in this order:

Vinculum (a bar over terms, e.g. $\overline{7-3}$) → Brackets ( ), { }, [ ] → Of → Division → Multiplication → Addition → Subtraction.

• 'of' means multiply but ranks above division: $\frac{1}{2}$ of $8÷2=4÷2=2$, not $\frac{1}{2}\times 4$.
• Division and multiplication share a rank — work left to right: $6÷2\times 3=9$, never $1$.
• A vinculum is an invisible bracket: simplify what's under the bar first.

3. Index laws (memorise, don't re-derive)

$a^m\cdot a^n=a^{m+n},\frac{a^m}{a^n}=a^{m-n},(a^m{)}^n=a^{mn}$ $a^0=1,a^{-n}=\frac{1}{a^n},a^{1/n}=\sqrt[n]{a},a^{m/n}=\sqrt[n]{a^m}$

• Reduce fractional powers by writing the base as a prime power: $16^{3/4}=(2^4{)}^{3/4}=2^3=8$.
• Equal bases ⇒ equal exponents: $3^{x+1}=81=3^4\Rightarrow x=3$.
• ${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n$ — a negative power just flips the fraction.

4. Surds — the four moves

1. Simplify: pull out perfect squares — $\sqrt{72}=\sqrt{36\cdot 2}=6\sqrt{2}$.
2. Rationalise: multiply by the conjugate. $\frac{1}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}+\sqrt{2}}{3-2}=\sqrt{3}+\sqrt{2}$.
3. Nested radical $\sqrt{a\pm 2\sqrt{b}}$: find $x,y$ with $x+y=a$ and $xy=b$; then it equals $\sqrt{x}\pm \sqrt{y}$. So $\sqrt{7+4\sqrt{3}}=\sqrt{7+2\sqrt{12}}=\sqrt{4}+\sqrt{3}=2+\sqrt{3}$.
4. Compare surds: raise to a common power or square both — never approximate blindly under time pressure.

5. Algebraic identities that shortcut simplification

$a^2-b^2=(a+b)(a-b),\frac{a^3+b^3}{a^2-ab+b^2}=a+b,\frac{a^3-b^3}{a^2+ab+b^2}=a-b$

The decimal version is a stock SSC question: $\frac{0.6^3+0.4^3}{0.6^2-0.6\cdot 0.4+0.4^2}=0.6+0.4=1$. Spotting the identity turns a page of arithmetic into one addition.

6. Approximation (Tier 2)

When the question says "approximately", round to friendly numbers before computing: $39.99\%$ of $1299\approx 40\%$ of $1300=520$. Keep rounding consistent (round both up or estimate the net drift) so errors cancel rather than compound.

7. Solved PYQ-style examples

Q1. Simplify $12+6÷2\times 3-4$. Solution. Division/multiplication left to right: $6÷2=3$, $\times 3=9$; then $12+9-4=$ 17.

Q2. Value of $16^{3/4}$? Solution. $16=2^4\Rightarrow (2^4{)}^{3/4}=2^3=$ 8.

Q3. Simplify $\sqrt{7+4\sqrt{3}}$. Solution. $7+4\sqrt{3}=7+2\sqrt{12}$ with $4+3=7,4\times 3=12$ → $\sqrt{4}+\sqrt{3}=$ $2+\sqrt{3}$.

Q4. Evaluate $\frac{0.6^3+0.4^3}{0.6^2-0.6\times 0.4+0.4^2}$. Solution. Identity $\frac{a^3+b^3}{a^2-ab+b^2}=a+b=0.6+0.4=$ 1.

Q5. If $3^{x+1}=81$, find $x$. Solution. $81=3^4\Rightarrow x+1=4\Rightarrow x=$ 3.

Q6. Simplify $2+\frac{1}{2+\frac{1}{2}}$. Solution. Inner $2+\frac{1}{2}=\frac{5}{2}$; reciprocal $\frac{2}{5}$; total $2+\frac{2}{5}=$ $\frac{12}{5}$.

8. Exam protocol

1. Write VBODMAS mentally and obey it — most simplification errors are ordering errors.
2. Division and multiplication: left to right, same rank; 'of' outranks both.
3. Reduce every power to a prime base before touching fractional exponents.
4. Rationalise with the conjugate; for $\sqrt{a\pm 2\sqrt{b}}$ find the $x+y=a,xy=b$ pair.
5. Scan for $a^2-b^2$ and $a^3\pm b^3$ patterns — they collapse the whole expression.
6. "Approximately"? Round first, compute second, keep the rounding consistent.