Simplification

1. Core Concepts & Theoretical Blueprint

Simplification tests the ability to reduce a complex numerical or algebraic expression to its simplest final value by applying operations in the mathematically correct order, and by recognizing structural shortcuts (identities, common factors, standard fraction patterns) that bypass brute-force computation.

The BODMAS/VBODMAS Rule (absolute operational order, must never be violated):

where Vinculum (bar/overline grouping) is resolved first, then Brackets (in order: small (), curly {}, square []), then Of (meaning multiplication in the specific sense of "a fraction/percentage OF a quantity," resolved before general division/multiplication), then Division, Multiplication, Addition, and finally Subtraction — division and multiplication are performed strictly left-to-right relative to each other when both appear at the same "level," as are addition and subtraction.

Standard Algebraic Identities (the backbone of shortcut-based simplification):

Recurring Decimal to Fraction Conversion (frequently disguised as a simplification sub-task):

• Pure recurring: $0.\overline{ab} = \dfrac{ab}{99}$ (as many 9's as recurring digits)
• Mixed recurring: $0.a\overline{bc} = \dfrac{abc-a}{990}$ (9's for recurring digits, 0's for non-recurring digits after the decimal point)

The Universal Trap: Four traps recur relentlessly:

1. Misordering "Of" and Division — "Of" (as in $\frac{2}{5}$ of 40) must be resolved BEFORE plain division/multiplication in the same expression when both appear ambiguously; students often treat "of" as equal-priority multiplication and process operations in the wrong sequence.
2. Sign errors when removing brackets, especially with a preceding minus sign — $a-(b-c)=a-b+c$, NOT $a-b-c$; distributing a negative sign across a bracket is the single most common simplification error at every level.
3. Applying identities to numbers that don't actually match the pattern — forcing a near-fit onto $a^2-b^2$ or $(a+b)^2$ without verifying the middle term or exact structure produces a confidently wrong answer; always verify the full identity structure before applying it as a shortcut.
4. Losing precision in decimal simplification by rounding too early — intermediate rounding in a multi-step decimal expression compounds error; carry full precision (or convert to fractions) until the final step.

2. Exhaustive Question Typology

                             SIMPLIFICATION
                                   |
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    |             |               |               |               |       |
Type 1:        Type 2:         Type 3:         Type 4:         Type 5:  Type 6:
Basic          Simplification  Simplification  Simplification  Simplifi- Recurring
BODMAS/        of Fractions    Using Algebraic  of Decimals    cation of Decimal to
VBODMAS        (LCD-based)     Identities                      Complex/  Fraction
Application                                                    Nested    Conversion
                                                                Fractions
    |             |               |
Type 7:        Type 8:         Type 9:
"Find x"       Approximation-  Simplification
Equations      Based           Involving Basic
Hidden in      Simplification  Surds/Roots
Simplification (near round     (non-index
               numbers)        level)

Type 1 — Basic BODMAS/VBODMAS application:

• Core Scenario: "Simplify: $18\div3\times2+[15-(6+3)]$."
• Governing Equation: Apply V-B-O-D-M-A-S in strict order, resolving innermost brackets first.

Type 2 — Simplification of fractions (LCD-based operations):

• Core Scenario: "Simplify: $\dfrac{2}{3}+\dfrac{5}{6}-\dfrac{1}{4}$."
• Governing Equation: Convert all fractions to a common denominator (LCM of denominators), then combine numerators.

Type 3 — Simplification using algebraic identities:

• Core Scenario: "Simplify: $\dfrac{(x+y)^2-(x-y)^2}{4xy}$" or "Evaluate $997\times1003$ using identities."
• Governing Equation: Recognize and apply the matching identity from Section 1 (e.g., $997\times1003=(1000-3)(1000+3)=1000^2-3^2$).

Type 4 — Simplification of decimals:

• Core Scenario: "Simplify: $12.5\times0.4+3.6\div0.9$."
• Governing Equation: Convert decimals to fractions where convenient, or align decimal places carefully, then apply BODMAS.

Type 5 — Simplification of complex/nested (continued) fractions:

• Core Scenario: "Simplify: $\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2}}}$."
• Governing Equation: Resolve from the innermost fraction outward, treating each level as a self-contained BODMAS sub-problem before substituting upward.

Type 6 — Recurring decimal to fraction conversion:

• Core Scenario: "Express $0.3\overline{45}$ as a fraction in lowest terms."
• Governing Equation: Mixed recurring formula: $\dfrac{\text{(all digits)}-\text{(non-recurring digits)}}{\underbrace{99\ldots9}<em>{\text{recurring}}\underbrace{00\ldots0}</em>{\text{non-recurring}}}$

Type 7 — "Find x" equations hidden inside simplification-style questions:

• Core Scenario: "If $\dfrac{3}{4}$ of a number is 15 more than $\dfrac{2}{5}$ of the same number, find the number."
• Governing Equation: Translate the words into a linear equation, then solve for x using standard algebraic manipulation (this is simplification's "reverse" direction).

Type 8 — Approximation-based simplification (near round numbers):

• Core Scenario: "Find the approximate value of $\sqrt{4899}\div6.98+11.02\times2.99$."
• Governing Equation: Round each awkward number to its nearest convenient value BEFORE computing (e.g., $\sqrt{4899}\approx70$, $6.98\approx7$, etc.), then apply BODMAS on the rounded values.

Type 9 — Simplification involving basic surds/roots (non-indices level):

• Core Scenario: "Simplify: $\sqrt{50}+\sqrt{18}-\sqrt{8}$."
• Governing Equation: Reduce each surd to its simplest form (factor out perfect squares), then combine LIKE surds as if they were algebraic like-terms.

3. Type-wise Practice MCQs with Full Solutions

Type 1 — Basic BODMAS/VBODMAS Application

MCQ 1. Simplify: $36\div4\times3+15-(8-3)$ (A) 37 (B) 42 (C) 32 (D) 47

Correct Answer: (A) Solution: Brackets first: $8-3=5$. Then D/M left to right: $36\div4=9$; $9\times3=27$. Then A/S: $27+15-5=37$.

MCQ 2. Simplify: $\dfrac{2}{5}$ of $150 \div 15 + 8\times2$ (A) 20 (B) 24 (C) 22 (D) 18

Correct Answer: (A) Solution: "Of" resolves before division: $\dfrac{2}{5}\times150=60$. Then $60\div15=4$. Then $8\times2=16$. Then $4+16=20$.

MCQ 3. Simplify: $[25-{14-(8-3)}]\div2$ (A) 10 (B) 8 (C) 12 (D) 9

Correct Answer: (A) Solution: Innermost bracket: $8-3=5$. Curly: $14-5=9$. Square: $25-9=16$. Then $16\div2=8$. (Recheck arithmetic: this gives 8, not matching marked (A); correcting.)

MCQ 3 (verified). Simplify: $[25-{14-(8-3)}]\div2$ (A) 8 (B) 10 (C) 12 (D) 9

Correct Answer: (A) Solution: As derived: innermost $8-3=5$; curly $14-5=9$; square $25-9=16$; final $16\div2=8$.

Type 2 — Simplification of Fractions

MCQ 1. Simplify: $\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{1}{6}$ (A) $\dfrac{13}{12}$ (B) $\dfrac{11}{12}$ (C) $\dfrac{5}{4}$ (D) $\dfrac{7}{6}$

Correct Answer: (A) Solution: LCD = 12. $\dfrac{8}{12}+\dfrac{9}{12}-\dfrac{2}{12}=\dfrac{8+9-2}{12}=\dfrac{15}{12}=\dfrac{5}{4}$. (Recheck: gives 5/4, matching option C, not A; correcting the marked answer.)

MCQ 1 (verified). Correct Answer: (C) $\dfrac{5}{4}$ Solution: As derived: $\dfrac{8+9-2}{12}=\dfrac{15}{12}=\dfrac{5}{4}$.

MCQ 2. Simplify: $2\dfrac{1}{3}\times1\dfrac{1}{2}\div\dfrac{7}{4}$ (A) 2 (B) $\dfrac{3}{2}$ (C) 3 (D) $\dfrac{7}{2}$

Correct Answer: (A) Solution: Convert: $2\frac13=\frac73$, $1\frac12=\frac32$. $\frac73\times\frac32=\frac{21}{6}=\frac72$. Then $\frac72\div\frac74=\frac72\times\frac47=\frac{28}{14}=2$.

MCQ 3. Simplify: $\dfrac{3}{5}\div\dfrac{9}{10}\times\dfrac{3}{4}$ (A) $\dfrac{1}{2}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{3}{4}$ (D) $\dfrac{5}{6}$

Correct Answer: (A) Solution: $\dfrac{3}{5}\div\dfrac{9}{10}=\dfrac{3}{5}\times\dfrac{10}{9}=\dfrac{30}{45}=\dfrac{2}{3}$. Then $\dfrac{2}{3}\times\dfrac{3}{4}=\dfrac{6}{12}=\dfrac{1}{2}$.

Type 3 — Simplification Using Algebraic Identities

MCQ 1. Find the value of $997\times1003$ using algebraic identities. (A) 999991 (B) 998991 (C) 999891 (D) 997991

Correct Answer: (A) Solution: $997\times1003=(1000-3)(1000+3)=1000^2-3^2=1000000-9=999991$.

MCQ 2. Simplify: $\dfrac{(59)^2-(41)^2}{18}$ (A) 100 (B) 90 (C) 110 (D) 80

Correct Answer: (A) Solution: $a^2-b^2=(a+b)(a-b)=(59+41)(59-41)=100\times18=1800$. Then $\dfrac{1800}{18}=100$.

MCQ 3. If $x+y=12$ and $xy=27$, find the value of $x^2+y^2$. (A) 90 (B) 108 (C) 144 (D) 81

Correct Answer: (A) Solution: $x^2+y^2=(x+y)^2-2xy=12^2-2\times27=144-54=90$.

Type 4 — Simplification of Decimals

MCQ 1. Simplify: $12.5\times0.4+3.6\div0.9$ (A) 9 (B) 8 (C) 10 (D) 7

Correct Answer: (A) Solution: $12.5\times0.4=5$; $3.6\div0.9=4$. Sum $=5+4=9$.

MCQ 2. Simplify: $0.75\times0.75+0.75\times0.5+0.25\times0.25...$ let's use a cleaner standard identity-based decimal MCQ.

MCQ 2 (restated). Simplify: $2.5\times2.5-1.5\times1.5$ (A) 4 (B) 3 (C) 5 (D) 6

Correct Answer: (A) Solution: Recognize $a^2-b^2$ form: $(2.5+1.5)(2.5-1.5)=4\times1=4$.

MCQ 3. Simplify: $(0.6)^3+(0.4)^3$ using the sum-of-cubes identity, given $0.6+0.4=1$. (A) 0.28 (B) 0.216 (C) 0.064 (D) 0.352

Correct Answer: (A) Solution: $a^3+b^3=(a+b)(a^2-ab+b^2)=(1)\times(0.36-0.24+0.16)=1\times0.28=0.28$.

Type 5 — Simplification of Complex/Nested Fractions

MCQ 1. Simplify: $\cfrac{1}{2+\cfrac{1}{3}}$ (A) $\dfrac{3}{7}$ (B) $\dfrac{7}{3}$ (C) $\dfrac{2}{7}$ (D) $\dfrac{3}{2}$

Correct Answer: (A) Solution: Inner: $2+\dfrac13=\dfrac73$. Then $\dfrac{1}{7/3}=\dfrac37$.

MCQ 2. Simplify: $\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2}}}$ (A) $\dfrac{3}{5}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{5}{3}$ (D) $\dfrac{3}{2}$

Correct Answer: (A) Solution: Innermost: $1+\dfrac12=\dfrac32$. Next level: $1+\dfrac{1}{3/2}=1+\dfrac23=\dfrac53$. Outer: $\dfrac{1}{5/3}=\dfrac35$.

MCQ 3. Simplify: $\cfrac{2+\cfrac{3}{4}}{1-\cfrac{1}{4}}$ (A) $\dfrac{11}{3}$ (B) $\dfrac{7}{3}$ (C) $\dfrac{5}{3}$ (D) $\dfrac{13}{3}$

Correct Answer: (A) Solution: Numerator: $2+\dfrac34=\dfrac{11}{4}$. Denominator: $1-\dfrac14=\dfrac34$. Result: $\dfrac{11/4}{3/4}=\dfrac{11}{3}$.

Type 6 — Recurring Decimal to Fraction Conversion

MCQ 1. Convert $0.\overline{36}$ to a fraction in lowest terms. (A) $\dfrac{4}{11}$ (B) $\dfrac{36}{99}$ (C) $\dfrac{9}{25}$ (D) $\dfrac{7}{22}$

Correct Answer: (A) Solution: $0.\overline{36}=\dfrac{36}{99}=\dfrac{4}{11}$ (dividing numerator and denominator by their HCF, 9).

MCQ 2. Convert $0.4\overline{5}$ to a fraction. (A) $\dfrac{41}{90}$ (B) $\dfrac{45}{99}$ (C) $\dfrac{9}{20}$ (D) $\dfrac{5}{11}$

Correct Answer: (A) Solution: Mixed recurring, 1 non-recurring digit (4), 1 recurring digit (5): $\dfrac{45-4}{90}=\dfrac{41}{90}$.

MCQ 3. Convert $2.3\overline{18}$ to a fraction. (A) $\dfrac{2315}{990}$ (B) $\dfrac{2318}{999}$ (C) $\dfrac{2295}{990}$ (D) $\dfrac{2318}{990}$

Correct Answer: (A) Solution: All digits after decimal (non-recurring + recurring, as one block): "318"; non-recurring part: "3". Denominator: 2 nines (for 2 recurring digits "18") + 1 zero (for 1 non-recurring digit "3") = 990. Numerator = 318 − 3 = 315, then add the integer part scaled: total fraction $=2+\dfrac{315}{990}=\dfrac{2\times990+315}{990}=\dfrac{1980+315}{990}=\dfrac{2295}{990}$. (Recheck against options — this matches option C's numerator style; correcting the marked correct option.)

MCQ 3 (verified). Correct Answer: (C) $\dfrac{2295}{990}$ Solution: As derived, the full fraction (integer part included) is $\dfrac{2295}{990}$, which simplifies further to $\dfrac{51}{22}$ in lowest terms.

Type 7 — "Find x" Equations Hidden in Simplification

MCQ 1. If $\dfrac{3}{4}$ of a number exceeds $\dfrac{2}{5}$ of the same number by 34, find the number. (A) 40 (B) 45 (C) 50 (D) 60

Correct Answer: (A) Solution: Let the number be x. $\dfrac{3}{4}x-\dfrac{2}{5}x=34$. LCD=20: $\dfrac{15x}{20}-\dfrac{8x}{20}=\dfrac{7x}{20}=34\Rightarrow x=\dfrac{34\times20}{7}=\dfrac{680}{7}\approx97.1$. (Recheck: this doesn't give a clean integer; adjust the problem's constant for a clean textbook answer.)

MCQ 1 (verified, clean version). If $\dfrac{3}{4}$ of a number exceeds $\dfrac{2}{5}$ of the same number by 21, find the number. (A) 60 (B) 45 (C) 50 (D) 40

Correct Answer: (A) Solution: $\dfrac{7x}{20}=21\Rightarrow x=\dfrac{21\times20}{7}=3\times20=60$.

MCQ 2. A number is such that when 24 is subtracted from $\dfrac23$ of it, the result equals $\dfrac13$ of the number. Find the number. (A) 72 (B) 60 (C) 48 (D) 96

Correct Answer: (A) Solution: $\dfrac23x-24=\dfrac13x\Rightarrow\dfrac23x-\dfrac13x=24\Rightarrow\dfrac13x=24\Rightarrow x=72$.

MCQ 3. Simplify to find x: $\dfrac{x}{5}+\dfrac{x}{4}=9$ (A) 20 (B) 18 (C) 16 (D) 24

Correct Answer: (A) Solution: LCD=20: $\dfrac{4x}{20}+\dfrac{5x}{20}=\dfrac{9x}{20}=9\Rightarrow x=20$.

Type 8 — Approximation-Based Simplification

MCQ 1. Find the approximate value of $\sqrt{4899}\div7$ (nearest whole number). (A) 10 (B) 9 (C) 11 (D) 12

Correct Answer: (A) Solution: $\sqrt{4899}\approx\sqrt{4900}=70$. Then $70\div7=10$.

MCQ 2. Find the approximate value of $11.98\times5.02+29.97\div6.01$ (A) 65 (B) 60 (C) 55 (D) 70

Correct Answer: (A) Solution: Round: $11.98\approx12$, $5.02\approx5$, $29.97\approx30$, $6.01\approx6$. $12\times5=60$; $30\div6=5$. Sum $=60+5=65$.

MCQ 3. Find the approximate value of $(20.02)^2-(19.98)^2$ (A) 1.6 (B) 2 (C) 1.2 (D) 0.8

Correct Answer: (A) Solution: Apply $a^2-b^2=(a+b)(a-b)$ directly (no rounding error introduced this way): $(20.02+19.98)(20.02-19.98)=40\times0.04=1.6$.

Type 9 — Simplification Involving Basic Surds

MCQ 1. Simplify: $\sqrt{50}+\sqrt{18}-\sqrt{8}$ (A) $6\sqrt2$ (B) $4\sqrt2$ (C) $8\sqrt2$ (D) $5\sqrt2$

Correct Answer: (A) Solution: $\sqrt{50}=5\sqrt2$; $\sqrt{18}=3\sqrt2$; $\sqrt8=2\sqrt2$. Combine: $5\sqrt2+3\sqrt2-2\sqrt2=6\sqrt2$.

MCQ 2. Simplify: $\sqrt{75}\div\sqrt{3}$ (A) 5 (B) 25 (C) $5\sqrt3$ (D) 15

Correct Answer: (A) Solution: $\sqrt{75}\div\sqrt3=\sqrt{75/3}=\sqrt{25}=5$.

MCQ 3. Simplify: $3\sqrt{2}\times2\sqrt{8}$ (A) 24 (B) 12 (C) 16 (D) 48

Correct Answer: (A) Solution: $3\sqrt2\times2\sqrt8=6\times\sqrt{2\times8}=6\times\sqrt{16}=6\times4=24$.

4. High-Yield Speed Tricks & Shortcut Mental Models

Shortcut 1 — The $(1000\pm a)(1000\pm b)$ Identity Recognition Reflex

• Application: Any multiplication of two numbers close to a round base (100, 1000, etc.), like $997\times1003$ or $98\times102$.
• Mental Model: Instantly rewrite both numbers as (round base ± small deviation), then apply $a^2-b^2$ or the direct expansion — this converts a large, error-prone multiplication into a small-number squaring and subtraction, executable almost entirely mentally.

Shortcut 2 — "Of" Before "Division/Multiplication" Reflex Check

• Application: Every expression containing the word "of" mixed with division or multiplication symbols.
• Mental Model: Train a reflex pause whenever "of" appears alongside ÷ or ×: resolve "of" FIRST regardless of left-to-right position in the expression, since it functionally behaves like an implicit bracket around the two quantities it connects — treating it as equal-priority with plain multiplication is the single most common BODMAS-order error.

Shortcut 3 — Convert Awkward Decimals to Fractions Before Multi-Step Operations

• Application: Any Type 4/5 problem with repeating or multi-step decimal chains (e.g., $0.375\times0.8$).
• Mental Model: Recognize common decimal-fraction equivalents instantly (0.25=1/4, 0.125=1/8, 0.375=3/8, 0.75=3/4, 0.2=1/5, 0.4=2/5) and convert before multiplying — fraction multiplication with cancellation is almost always faster and more accurate than chained decimal multiplication.

5. Deep-Dive: Most Frequently Asked Questions

Problem 1 (SSC/RRB Standard): Simplify: $\dfrac{2}{5}$ of $\left(\dfrac{3}{4}\times240\right)+\dfrac{1}{3}$ of $150-45$

Traditional Method (Slow): Step 1: $\dfrac34\times240=180$. Step 2: $\dfrac25\times180=72$. Step 3: $\dfrac13\times150=50$. Step 4: $72+50-45=77$. (Each step written out and computed sequentially with full fraction multiplication — ~35-40 seconds.)

Exam Shortcut (Fast): Mentally cancel before multiplying: $\dfrac34\times240$: cancel 240/4=60, then $60\times3=180$ (single mental step). $\dfrac25\times180$: cancel 180/5=36, then $36\times2=72$ (single mental step). $\dfrac13\times150=50$ (instant, since 150/3 is a clean division). Final: $72+50-45=77$. Answer: 77, reached via the same operations but executed as instant cancel-then-multiply mental steps rather than written-out fraction arithmetic — under 15 seconds.

Problem 2 (UPSC/Banking Advanced): If $a+b+c=15$ and $a^2+b^2+c^2=83$, find the value of $ab+bc+ca$. Hence, if $a,b,c$ are also known to satisfy $abc=105$, verify using the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ what $a^3+b^3+c^3$ equals.

Step-by-Step Breakdown:

1. Use the identity $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$.
2. Substitute known values: $15^2=83+2(ab+bc+ca)\Rightarrow225=83+2(ab+bc+ca)$.
3. $2(ab+bc+ca)=225-83=142\Rightarrow ab+bc+ca=71$.
4. Now apply the sum-of-cubes-minus-3abc identity: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.
5. Compute the bracket: $a^2+b^2+c^2-ab-bc-ca=83-71=12$.
6. So $a^3+b^3+c^3-3abc=15\times12=180$.
7. Given $abc=105$: $a^3+b^3+c^3=180+3\times105=180+315=495$.
8. Answer: $ab+bc+ca=71$, and $a^3+b^3+c^3=495$. This demonstrates the chapter's advanced ceiling: chaining two separate standard identities (the square-of-sum expansion, then the sum-of-cubes identity) rather than solving for a, b, c individually — a technique that appears repeatedly in UPSC/banking-level "find the symmetric expression" simplification questions.

6. Chapter Checklist for Students

• I apply V-B-O-D-M-A-S in exact order every time, never treating "of" as equal-priority with plain multiplication/division.
• I correctly distribute a negative sign across an entire bracket when removing it, never dropping just the first term's sign.
• I verify an expression's full structure before applying an algebraic identity as a shortcut — never force-fit a near-match.
• I have memorized common decimal-fraction equivalents (0.125, 0.25, 0.375, 0.5, 0.75, 0.2, 0.4, 0.6, 0.8) to skip decimal multiplication chains.
• I use the $(1000\pm a)(1000\pm b)$-style near-round-number identity reflex for any multiplication of two numbers close to a clean base.