Ratio and Proportion

1. Core Concepts & Theoretical Blueprint

A Ratio compares two quantities of the SAME kind by division ($a:b=a/b$), while a Proportion states that two ratios are EQUAL ($a:b::c:d$, meaning $a/b=c/d$) — nearly every problem in this chapter reduces to setting up a correct ratio or proportion equation and solving via cross-multiplication.

Absolute Core Definitions:

Fourth, Third, and Mean Proportional:

Compounded, Duplicate, and Triplicate Ratios:

Componendo-Dividendo (a powerful algebraic shortcut for ratio equations):

The Universal Trap: Four persistent traps:

1. Adding/subtracting the SAME absolute number to both terms of a ratio and assuming the ratio changes by a simple proportional amount — ratios do NOT scale linearly under addition/subtraction of a constant; the new ratio must be recomputed as a fresh fraction, never estimated.
2. Confusing "in the ratio a:b" with "a is a fraction of b" — a ratio 3:5 means the quantities are $3x$ and $5x$ for some common multiplier x, not that one quantity equals $3/5$ of a fixed total.
3. Misapplying compounded/duplicate ratio formulas — duplicate ratio ($a^2:b^2$) is NOT the same as "ratio doubled" ($2a:2b$, which is actually identical to the original ratio); this naming confusion causes frequent errors.
4. Forgetting to verify the ratio is in LOWEST TERMS before comparing or combining with another ratio — un-simplified ratios can lead to incorrect combined ratios when merging multiple ratio relationships (e.g., A:B and B:C) if the B-terms aren't first equalized via LCM.

2. Exhaustive Question Typology

                          RATIO AND PROPORTION
                                  |
    -----------------------------------------------------------------------
    |            |              |               |               |          |
Type 1:       Type 2:        Type 3:        Type 4:         Type 5:     Type 6:
Basic Ratio   Finding a      Dividing a     Mean            Third       Compounded/
Simplification Quantity      Quantity in    Proportional    Proportional Duplicate/
              Given Ratio    Given Ratio                                Triplicate
              and One Value                                             Ratio
              (Fourth Prop.)
    |            |              |
Type 7:       Type 8:        Type 9:
Componendo-   Continued      Ratio Changes
Dividendo     Proportion     (If x is
Application   (3+ Terms,     Added/
              A:B:C form)    Subtracted
                              from Both
                              Terms)

Type 1 — Basic ratio simplification:

• Core Scenario: "Simplify the ratio 45:75 to its lowest terms."
• Governing Equation: Divide both terms by their HCF.

Type 2 — Finding a quantity given ratio and one value (fourth proportional):

• Core Scenario: "If A:B = 3:5 and A = 27, find B," or "find the fourth proportional to 4, 6, 10."
• Governing Equation: Cross-multiplication: $\dfrac{a}{b}=\dfrac{c}{x}\Rightarrow x=\dfrac{bc}{a}$

Type 3 — Dividing a quantity in a given ratio:

• Core Scenario: "Divide ₹1,200 among A, B, C in the ratio 2:3:5."
• Governing Equation: Each share $=\dfrac{\text{individual ratio part}}{\text{sum of all parts}}\times\text{Total}$

Type 4 — Mean proportional:

• Core Scenario: "Find the mean proportional between 9 and 25."
• Governing Equation: $x=\sqrt{ab}$

Type 5 — Third proportional:

• Core Scenario: "Find the third proportional to 8 and 12."
• Governing Equation: $x=\dfrac{b^2}{a}$

Type 6 — Compounded/duplicate/triplicate ratio:

• Core Scenario: "Find the compounded ratio of 2:3 and 4:5," or "find the duplicate ratio of 3:4."
• Governing Equation: Compounded: $ac:bd$; Duplicate: $a^2:b^2$; Sub-duplicate: $\sqrt a:\sqrt b$

Type 7 — Componendo-Dividendo application:

• Core Scenario: "If $\dfrac{x}{y}=\dfrac35$, find the value of $\dfrac{x+y}{x-y}$."
• Governing Equation: $\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}$ whenever $\dfrac ab=\dfrac cd$

Type 8 — Continued proportion (3+ terms, A:B:C form):

• Core Scenario: "If A:B = 2:3 and B:C = 4:5, find A:B:C."
• Governing Equation: Equalize the common term (B here) via LCM, then scale both ratios accordingly before combining.

Type 9 — Ratio changes (if x is added/subtracted from both terms):

• Core Scenario: "Two numbers are in the ratio 3:5. If 4 is added to both, the ratio becomes 5:7. Find the numbers."
• Governing Equation: Represent original numbers as $3k,5k$; set up the new ratio equation $\dfrac{3k+4}{5k+4}=\dfrac57$; solve for k.

3. Type-wise Practice MCQs with Full Solutions

Type 1 — Basic Ratio Simplification

MCQ 1. Simplify the ratio 45:75 to its lowest terms. (A) 3:5 (B) 9:15 (C) 4:5 (D) 3:4

Correct Answer: (A) Solution: HCF(45,75)=15. $45/15:75/15=3:5$.

MCQ 2. Simplify the ratio 0.75:1.25. (A) 3:5 (B) 4:5 (C) 3:4 (D) 5:6

Correct Answer: (A) Solution: Multiply by 100: $75:125$. Simplify by HCF(25): $3:5$.

MCQ 3. Simplify the ratio $\dfrac23:\dfrac34$. (A) 8:9 (B) 2:3 (C) 3:4 (D) 9:8

Correct Answer: (A) Solution: $\dfrac23\div\dfrac34=\dfrac23\times\dfrac43=\dfrac89$, giving ratio $8:9$.

Type 2 — Finding a Quantity Given Ratio and One Value

MCQ 1. If A:B = 3:5 and A = 27, find B. (A) 45 (B) 40 (C) 50 (D) 35

Correct Answer: (A) Solution: $\dfrac3{27}=\dfrac5B$? More directly: since $A=3k=27\Rightarrow k=9$; $B=5k=45$.

MCQ 2. Find the fourth proportional to 4, 6, 10. (A) 15 (B) 12 (C) 18 (D) 20

Correct Answer: (A) Solution: $x=\dfrac{bc}{a}=\dfrac{6\times10}{4}=15$.

MCQ 3. If x:y = 5:7 and y = 42, find x. (A) 30 (B) 35 (C) 28 (D) 25

Correct Answer: (A) Solution: $y=7k=42\Rightarrow k=6$. $x=5k=30$.

Type 3 — Dividing a Quantity in a Given Ratio

MCQ 1. Divide ₹1,200 among A, B, C in the ratio 2:3:5. (A) A=240, B=360, C=600 (B) A=200,B=400,C=600 (C) A=300,B=300,C=600 (D) A=250,B=350,C=600

Correct Answer: (A) Solution: Total parts$=10$. A$=\dfrac2{10}\times1200=240$; B$=\dfrac3{10}\times1200=360$; C$=\dfrac5{10}\times1200=600$.

MCQ 2. A sum of ₹2,850 is divided among P, Q, R such that P:Q=3:4 and Q:R=5:6. Find R's share. (A) ₹1200 (B) ₹1000 (C) ₹1100 (D) ₹1300

Correct Answer: (A) Solution: Combine ratios: P:Q=3:4=15:20 (scaled by 5); Q:R=5:6=20:24 (scaled by 4). Combined P:Q:R=15:20:24, total=59 parts. R's share$=\dfrac{24}{59}\times2850\approx1159$. (Recheck: not clean; recalibrate total sum for exam-standard clean figures.)

MCQ 2 (verified, clean version). A sum of ₹5,900 is divided among P, Q, R such that P:Q=3:4 and Q:R=5:6. Find R's share. (A) ₹2400 (B) ₹2000 (C) ₹2200 (D) ₹2600

Correct Answer: (A) Solution: Combined ratio P:Q:R=15:20:24 (total 59 parts). R's share$=\dfrac{24}{59}\times5900=24\times100=2400$.

MCQ 3. Divide 720 into three parts such that the first is twice the second, and the second is thrice the third. (A) First=480, Second=240, Third=80 (B) First=400,Second=240,Third=80 (C) First=500,Second=200,Third=20 (D) First=450,Second=225,Third=45

Correct Answer: (A) Solution: Let third=x, second=3x, first=6x. Sum$=6x+3x+x=10x=720\Rightarrow x=72$. First$=432$,Second$=216$,Third$=72$. (Recheck: gives 432,216,72; doesn't match option A; correcting.)

MCQ 3 (verified). Correct Answer: (E)/restated as First=432, Second=216, Third=72 Solution: As derived: $x=72$; First=6x=432, Second=3x=216, Third=x=72.

Type 4 — Mean Proportional

MCQ 1. Find the mean proportional between 9 and 25. (A) 15 (B) 17 (C) 13 (D) 12

Correct Answer: (A) Solution: $x=\sqrt{9\times25}=\sqrt{225}=15$.

MCQ 2. Find the mean proportional between 4 and 64. (A) 16 (B) 32 (C) 20 (D) 24

Correct Answer: (A) Solution: $x=\sqrt{4\times64}=\sqrt{256}=16$.

MCQ 3. The mean proportional between two numbers is 12, and one of the numbers is 8. Find the other. (A) 18 (B) 16 (C) 20 (D) 24

Correct Answer: (A) Solution: $12^2=8\times x\Rightarrow144=8x\Rightarrow x=18$.

Type 5 — Third Proportional

MCQ 1. Find the third proportional to 8 and 12. (A) 18 (B) 16 (C) 20 (D) 14

Correct Answer: (A) Solution: $x=\dfrac{12^2}{8}=\dfrac{144}{8}=18$.

MCQ 2. Find the third proportional to 5 and 10. (A) 20 (B) 15 (C) 25 (D) 18

Correct Answer: (A) Solution: $x=\dfrac{10^2}{5}=\dfrac{100}{5}=20$.

MCQ 3. The third proportional to 6 and x is 24. Find x. (A) 12 (B) 10 (C) 14 (D) 16

Correct Answer: (A) Solution: $24=\dfrac{x^2}{6}\Rightarrow x^2=144\Rightarrow x=12$.

Type 6 — Compounded/Duplicate/Triplicate Ratio

MCQ 1. Find the compounded ratio of 2:3 and 4:5. (A) 8:15 (B) 6:8 (C) 8:8 (D) 6:15

Correct Answer: (A) Solution: Compounded ratio$=2\times4:3\times5=8:15$.

MCQ 2. Find the duplicate ratio of 3:4. (A) 9:16 (B) 6:8 (C) 3:4 (D) 12:16

Correct Answer: (A) Solution: Duplicate ratio$=3^2:4^2=9:16$.

MCQ 3. Find the sub-duplicate ratio of 16:25. (A) 4:5 (B) 8:10 (C) 16:25 (D) 2:5

Correct Answer: (A) Solution: Sub-duplicate ratio$=\sqrt{16}:\sqrt{25}=4:5$.

Type 7 — Componendo-Dividendo Application

MCQ 1. If $\dfrac{x}{y}=\dfrac35$, find the value of $\dfrac{x+y}{x-y}$. (A) −4 (B) 4 (C) 8 (D) −8

Correct Answer: (A) Solution: $\dfrac{x+y}{x-y}=\dfrac{3+5}{3-5}=\dfrac{8}{-2}=-4$.

MCQ 2. If $\dfrac{a}{b}=\dfrac74$, find the value of $\dfrac{a+b}{a-b}$. (A) 11/3 (B) 3/11 (C) 7/4 (D) 4/7

Correct Answer: (A) Solution: $\dfrac{a+b}{a-b}=\dfrac{7+4}{7-4}=\dfrac{11}{3}$.

MCQ 3. If $\dfrac{x+y}{x-y}=\dfrac32$, find the value of $\dfrac{x}{y}$ (using componendo-dividendo in reverse). (A) 5 (B) 4 (C) 3 (D) 6

Correct Answer: (A) Solution: By componendo-dividendo (reverse form): if $\dfrac{x+y}{x-y}=\dfrac{p}{q}$, then $\dfrac{x}{y}=\dfrac{p+q}{p-q}=\dfrac{3+2}{3-2}=\dfrac51=5$.

Type 8 — Continued Proportion (3+ Terms)

MCQ 1. If A:B = 2:3 and B:C = 4:5, find A:B:C. (A) 8:12:15 (B) 2:4:5 (C) 6:12:15 (D) 8:3:5

Correct Answer: (A) Solution: Equalize B: A:B=2:3=8:12 (×4); B:C=4:5=12:15 (×3). Combined: A:B:C=8:12:15.

MCQ 2. If P:Q = 3:4, Q:R = 6:7, find P:Q:R. (A) 9:12:14 (B) 3:6:7 (C) 18:24:28 (D) 9:4:7

Correct Answer: (A) Solution: Equalize Q: P:Q=3:4=9:12 (×3); Q:R=6:7=12:14 (×2). Combined: P:Q:R=9:12:14.

MCQ 3. If A:B:C = 2:3:4 and B:C:D = 6:8:9, find A:B:C:D. (A) 4:6:8:9 (B) 2:3:4:9 (C) 6:9:12:9 (D) 4:6:9:12

Correct Answer: (A) Solution: Equalize C (common term): A:B:C=2:3:4=4:6:8 (×2); B:C:D=6:8:9 stays with C=8 (already matching after scaling A:B:C by 2). Combined: A:B:C:D=4:6:8:9.

Type 9 — Ratio Changes (If x is Added/Subtracted from Both Terms)

MCQ 1. Two numbers are in the ratio 3:5. If 4 is added to both, the ratio becomes 5:7. Find the numbers. (A) 6, 10 (B) 9, 15 (C) 3, 5 (D) 12, 20

Correct Answer: (A) Solution: Let numbers$=3k,5k$. $\dfrac{3k+4}{5k+4}=\dfrac57\Rightarrow7(3k+4)=5(5k+4)\Rightarrow21k+28=25k+20\Rightarrow8=4k\Rightarrow k=2$. Numbers$=6,10$.

MCQ 2. Two numbers are in the ratio 4:5. If 6 is subtracted from each, the ratio becomes 3:4. Find the numbers. (A) 24, 30 (B) 20, 25 (C) 16, 20 (D) 28, 35

Correct Answer: (A) Solution: Let numbers$=4k,5k$. $\dfrac{4k-6}{5k-6}=\dfrac34\Rightarrow4(4k-6)=3(5k-6)\Rightarrow16k-24=15k-18\Rightarrow k=6$. Numbers$=24,30$.

MCQ 3. The ratio of two numbers is 5:6. If 8 is added to each, the ratio becomes 7:8. Find the smaller number. (A) 20 (B) 24 (C) 18 (D) 22

Correct Answer: (A) Solution: Let numbers$=5k,6k$. $\dfrac{5k+8}{6k+8}=\dfrac78\Rightarrow8(5k+8)=7(6k+8)\Rightarrow40k+64=42k+56\Rightarrow8=2k\Rightarrow k=4$. Smaller number$=5\times4=20$.

4. High-Yield Speed Tricks & Shortcut Mental Models

Shortcut 1 — The Single Multiplier Variable k for Every Ratio Problem

• Application: Virtually every problem in this chapter, especially Types 2, 3, 8, 9.
• Mental Model: Whenever a ratio is given, immediately represent the quantities as $ak, bk$ (or $ak, bk, ck$ for three terms) using ONE variable k — never introduce separate independent variables for each quantity in a ratio relationship. This is the same principle used throughout Partnership, Ages, and Chain Rule, and recognizing this UNIFIED technique across chapters is the single highest-leverage mental model in the entire syllabus.

Shortcut 2 — Direct Componendo-Dividendo Substitution (Skip Cross-Multiplication)

• Application: Every Type 7 problem, and any question asking for $(a+b)/(a-b)$-style expressions given $a/b$.
• Mental Model: The moment you see a ratio $a/b=p/q$ and a request for $(a+b)/(a-b)$ or similar, skip setting up and cross-multiplying entirely — directly substitute the given ratio's numerator and denominator into $(p+q)/(p-q)$. This is a direct plug-in formula, not something to be re-derived via algebra each time.

5. Deep-Dive: Most Frequently Asked Questions

Problem 1 (SSC/RRB Standard): The ratio of the ages of A and B is 5:7. Eighteen years hence, the ratio of their ages will be 3:4. Find A's present age.

Traditional Method (Slow): Let present ages$=5k,7k$. After 18 years: $5k+18,7k+18$, ratio$=3:4$. $\dfrac{5k+18}{7k+18}=\dfrac34\Rightarrow4(5k+18)=3(7k+18)\Rightarrow20k+72=21k+54\Rightarrow k=18$. A's present age$=5\times18=90$. (Note: this age value is unrealistically large, signaling a likely need to double check the problem's given ratios — but the ALGEBRAIC METHOD itself, treated as the exam skill, involves setting up the single-variable ratio equation and cross-multiplying — ~30-35 seconds.)

Exam Shortcut (Fast): Recognize this is structurally IDENTICAL to a Type 9 "ratio changes" problem (from THIS chapter) merged with the Ages chapter's ratio-multiplier convention — apply the single unified $5k,7k\to$cross-multiply reflex immediately without treating it as a "new" problem type requiring fresh derivation. $4(5k+18)=3(7k+18)\Rightarrow20k+72=21k+54\Rightarrow k=18\Rightarrow A=90$. Answer: 90 years (using the given numbers as stated; the key exam-speed insight is recognizing the CROSS-CHAPTER pattern reuse — ages, partnership shares, and ratio-change problems all use the identical single-multiplier-plus-cross-multiplication technique), reached via one direct substitution — under 20 seconds once the pattern is recognized as familiar rather than novel.

Problem 2 (UPSC/Banking Advanced): If $x:y:z=2:3:5$ and $2x+3y-z=90$, find the value of $x+y+z$.

Step-by-Step Breakdown:

1. Since $x:y:z=2:3:5$, represent $x=2k, y=3k, z=5k$ for some common multiplier k.
2. Substitute into the given linear equation: $2(2k)+3(3k)-5k=90$
3. $4k+9k-5k=90\Rightarrow8k=90\Rightarrow k=11.25$
4. Compute $x+y+z=2k+3k+5k=10k=10\times11.25=112.5$.
5. Answer: $x+y+z=112.5$. This demonstrates the standard advanced technique for problems combining a GIVEN ratio with an ADDITIONAL independent linear equation: always substitute the ratio's multiplier-based representation ($2k,3k,5k$) directly into the extra equation, solve for the single unknown k, then use that same k to compute whatever final combined expression (here, the simple sum) the question asks for — this pattern generalizes to any ratio-plus-linear-constraint problem, regardless of how many terms the ratio involves or how complex the additional equation is.

6. Chapter Checklist for Students

• I represent every ratio using a single multiplier variable k (e.g., $ak,bk$), never independent variables for each quantity.
• I never assume a ratio changes proportionally when the SAME constant is added to or subtracted from both terms — I always set up and solve the resulting equation freshly.
• I correctly distinguish duplicate ratio ($a^2:b^2$) from a ratio that has simply been "doubled" (which remains unchanged as a ratio).
• I equalize the common term (via LCM) before combining two separate ratios into a single continued proportion (A:B:C form).
• I apply the componendo-dividendo shortcut $(a+b)/(a-b)=(p+q)/(p-q)$ as a direct substitution, without re-deriving it via cross-multiplication each time.