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Home > Maths > Ratio and Proportion for SSC, RRB, Banking & All Govt Exams 2026: Concepts, Model-wise Questions, Formulas, FAQs

Ratio and Proportion for SSC, RRB, Banking & All Govt Exams 2026: Concepts, Model-wise Questions, Formulas, FAQs

๐Ÿ“… Published: November 03, 2026 ๐Ÿ‘ค By NCK115
๐Ÿ•’ Calculating reading time…Official updates, clearly organized

Ratio and Proportion for SSC, RRB, Banking exams 2026: concepts, 16 question models, 77 solved questions, formulas table, day-wise plan and FAQs.

QUANTITATIVE APTITUDE – MATHS STUDY NOTES

Ratio and Proportion Questions and Formulas

SSC • RRB • BANKING • STATE EXAMS 2026

Ratio and Proportion for SSC, RRB, Banking & All Govt Exams 2026

Concepts, 16 Question Models, 77 Solved Questions, Formulas & FAQs

16 ModelsRatio + Proportion
77 Solved Qs4 options + step-by-step
7–10 DaysDay-wise plan inside

If percentages is the first chapter every aspirant finishes, Ratio and Proportion is the second one, and honestly it is the one that decides how fast the rest of arithmetic goes. Partnership, ages, mixtures, time and work, even data interpretation sets keep asking you to split something in a ratio. Talk to anyone who has sat in a Hyderabad or Vijayawada library for SSC prep and you hear the same story: they know the formulas but lose time because they do not recognise the type of question fast enough. So this post is arranged model by model. Each model has a short idea, one trick, and 3 to 7 solved questions with four options, exactly the way they appear in SSC, RRB and banking papers. It works for SSC CGL, CHSL, MTS and GD, RRB NTPC and Group D, IBPS and SBI exams, and state exams like TGPSC and APPSC Groups and police constable tests.

Nck115 Verdict
✔ Read fully if: you are starting arithmetic or you keep getting stuck in “a : b, b : c” and income–savings questions.
✔ Jump to Models 8–16 if: you already know the basics and want exam-level practice.
✔ Why it matters: one clear ratio method saves time in at least four other chapters, so every hour here pays back later.
⚠ About the questions: All questions on this page are written by Nck115 in the pattern of previous-year questions. The numbers are changed and every answer was re-checked with a small Python script before publishing. They are not copied from any paper, so we do not tag them with an exam name or year.

Why Ratio and Proportion Matters in Govt Exams 2026

Direct ratio questions are only part of the story. The same ideas sit inside partnership, ages, mixture & alligation, time & work and almost every DI set. The quant section size below is from the standard exam patterns; the ratio share is our own rough estimate from practice papers, so read it as approx. only.

ExamMaths / Quant questionsRatio & Proportion (approx.)
SSC CGL / CHSL Tier-I25Approx. 1–2 direct + more inside partnership, ages, mixture
SSC MTS / GD20Approx. 1–3 direct (basic sharing, coins, ages)
RRB NTPC CBT-130Approx. 1–3 direct + indirect use
RRB Group D CBT25Approx. 1–3 direct + indirect use
IBPS / SBI PO & Clerk Prelims35Approx. 1–2 arithmetic + ratio-based DI
TGPSC / APPSC Groups, police examsVaries by postArithmetic sections regularly include ratio and proportion

Ratio and Proportion Concepts Explained Simply

What is a ratio?

A ratio compares two quantities of the same kind by division. If a class has 18 boys and 12 girls, boys : girls = 18 : 12 = 3 : 2. The first term is called the antecedent and the second the consequent. A ratio has no unit, and both quantities must be in the same unit first (you cannot compare 50 paise with ₹2 without converting: 50 : 200 = 1 : 4).

The k-method (the one idea behind most questions)

If A : B = 3 : 2, the real values are 3k and 2k for some number k. A ratio tells you the shape, k tells you the size. Example: A : B = 3 : 2 and A + B = 45 → 5k = 45 → k = 9, so A = 27 and B = 18. Almost every model below is just “find k from one clue”.

Types of ratio

TypeMeaningExample (from 2 : 3)
Duplicate ratioa² : b²4 : 9
Triplicate ratioa³ : b³8 : 27
Sub-duplicate ratio√a : √b√2 : √3
Sub-triplicate ratio∛a : ∛b∛2 : ∛3
Inverse (reciprocal) ratio1/a : 1/b = b : a3 : 2
Compound ratioMultiply ratios term by term: (a : b) and (c : d) → ac : bd2 : 3 and 4 : 5 → 8 : 15
Ratio of equality / greater / lesser inequalitya = b / a > b / a < b5 : 5, 7 : 3, 2 : 3

What is a proportion?

When two ratios are equal, the four numbers are in proportion. a : b = c : d is written a : b :: c : d and read “a is to b as c is to d”. Here a and d are the extremes, b and c are the means, and the golden rule is product of extremes = product of means (ad = bc). Example: 3 : 4 :: 9 : 12 because 3 × 12 = 4 × 9 = 36.

TermHow to findQuick example
Fourth proportional of a, b, cd = bc / a4, 6, 10 → 6 × 10 / 4 = 15
Third proportional of a, bc = b² / a4, 6 → 36 / 4 = 9
Mean proportional of a, b√(ab)4, 9 → √36 = 6
Continued proportiona : b :: b : c, so b² = ac2, 6, 18 (6² = 2 × 18)

Componendo–Dividendo (C&D)

If a/b = c/d, then (a + b)/(a − b) = (c + d)/(c − d). It also works backwards. It is handy when a question gives something like (5x + 3y)/(5x − 3y) = 7/3 and asks x : y. Applying C&D: 5x/3y = (7 + 3)/(7 − 3) = 10/4, so x : y = 3 : 2. Small check: 5×3 + 3×2 = 21 and 15 − 6 = 9, and 21/9 = 7/3 ✔.

Direct and inverse proportion

If two things rise and fall together (more pens → more cost), they are in direct proportion: A = kB. If one rises when the other falls (more workers → fewer days), they are in inverse proportion: A × B = k. Find k from one pair and use it for the other.

Ratio and Proportion Questions: 16 Models with Solved Examples

Each question has four options and a full solution. Try it yourself first with a timer (aim for 45–60 seconds per question), then read the steps. The correct option is marked ✔.

Model 1: Basic Ratio with the k-Method (Sharing and Difference)

When A : B = 9 : 5, the actual values are 9k and 5k for some number k. Every basic question is about finding k from one clue (a sum, a difference or one value) and then multiplying.

💡 Trick: Total parts = sum of ratio terms; difference parts = gap between terms. Value of 1 part = given amount ÷ parts. Then answer = 1 part × the parts asked.

Q1. The prices of a pen and a notebook are in the ratio 9 : 5. The pen costs ₹240 more than the notebook. What is the price of the notebook?

(a) ₹300 ✔   (b) ₹540   (c) ₹360   (d) ₹250

Solution:

  1. Let pen = 9k and notebook = 5k.
  2. Difference = 9k − 5k = 4k = 240, so k = 60.
  3. Notebook = 5 × 60 = ₹300.

Answer: (a) ₹300

Q2. A sum of ₹3,150 is divided among A, B and C in the ratio 2 : 5 : 8. Find the share of B.

(a) ₹420   (b) ₹1,050 ✔   (c) ₹1,680   (d) ₹840

Solution:

  1. Total parts = 2 + 5 + 8 = 15.
  2. 1 part = 3150 ÷ 15 = ₹210.
  3. B = 5 × 210 = ₹1,050.

Answer: (b) ₹1,050

Q3. The marks of Ravi and Sai are in the ratio 4 : 7 and their total is 165. What is the difference between their marks?

(a) 60   (b) 75   (c) 45 ✔   (d) 30

Solution:

  1. Total parts = 11, so 1 part = 165 ÷ 11 = 15.
  2. Difference = (7 − 4) parts = 3 × 15 = 45.

Answer: (c) 45

Q4. Three numbers are in the ratio 3 : 7 : 9. The difference between the greatest and the smallest number is 54. Find the number that is neither the smallest nor the greatest.

(a) 27   (b) 81   (c) 72   (d) 63 ✔

Solution:

  1. Greatest − smallest = 9k − 3k = 6k = 54, so k = 9.
  2. Middle number = 7k = 63.

Answer: (d) 63

Model 2: Value of an Expression and Componendo–Dividendo

If x : y is given, put x = (first term) and y = (second term) straight into the expression; k cancels out. When the question gives a fraction like (7x + 4y)/(7x − 4y) and asks for x : y, use componendo–dividendo: if p/q = m/n then (p + q)/(p − q) = (m + n)/(m − n).

💡 Trick: For “find x : y” questions, C&D turns (Ax + By)/(Ax − By) = m/n into Ax/By = (m + n)/(m − n) in one line. In the exam you can also just test the options.

Q5. If x : y = 2 : 5, find (3x + 2y) : (x + 3y).

(a) 17 : 16   (b) 16 : 17 ✔   (c) 8 : 9   (d) 13 : 15

Solution:

  1. Put x = 2, y = 5.
  2. 3x + 2y = 6 + 10 = 16 and x + 3y = 2 + 15 = 17.
  3. Ratio = 16 : 17.

Answer: (b) 16 : 17

Q6. If a : b = 3 : 4, find (4a + 5b) : (5a − 2b).

(a) 32 : 7 ✔   (b) 7 : 32   (c) 27 : 7   (d) 31 : 8

Solution:

  1. Put a = 3, b = 4.
  2. 4a + 5b = 12 + 20 = 32; 5a − 2b = 15 − 8 = 7.
  3. Ratio = 32 : 7.

Answer: (a) 32 : 7

Q7. If (7x + 4y)/(7x − 4y) = 3, find x : y.

(a) 7 : 8   (b) 8 : 7 ✔   (c) 4 : 7   (d) 2 : 1

Solution:

  1. Write 3 as 3/1 and apply C&D: 7x/4y = (3 + 1)/(3 − 1) = 2.
  2. So 7x = 8y, which gives x/y = 8/7.
  3. Check: x = 8, y = 7 gives 84/28 = 3. ✔

Answer: (b) 8 : 7

Q8. If (a + b)/(a − b) = 9/5, find a : b.

(a) 2 : 7   (b) 9 : 5   (c) 7 : 2 ✔   (d) 5 : 2

Solution:

  1. C&D: a/b = (9 + 5)/(9 − 5) = 14/4 = 7/2.
  2. So a : b = 7 : 2.

Answer: (c) 7 : 2

Model 3: Comparing Ratios (Smallest / Largest)

A ratio a : b is just the fraction a/b, so comparing ratios means comparing fractions.

💡 Trick: Cross-multiply two at a time: for a/b vs c/d compare a × d with b × c. Or convert each to a decimal with 3 digits. Keep the bigger one and move on; never compare all four together.

Q9. Which of the following ratios is the smallest? 2 : 3, 3 : 5, 4 : 7, 5 : 8

(a) 2 : 3   (b) 3 : 5   (c) 4 : 7 ✔   (d) 5 : 8

Solution:

  1. Decimals: 2/3 = 0.667, 3/5 = 0.600, 4/7 = 0.571, 5/8 = 0.625.
  2. Smallest is 0.571, i.e. 4 : 7.

Answer: (c) 4 : 7

Q10. Which of the following ratios is the largest? 5 : 6, 7 : 9, 11 : 14, 13 : 15

(a) 5 : 6   (b) 7 : 9   (c) 11 : 14   (d) 13 : 15 ✔

Solution:

  1. 5/6 = 0.833, 7/9 = 0.778, 11/14 = 0.786, 13/15 = 0.867.
  2. Largest is 13 : 15.

Answer: (d) 13 : 15

Q11. Arrange 3 : 4, 5 : 7 and 7 : 9 in increasing order.

(a) 3 : 4 < 5 : 7 < 7 : 9   (b) 5 : 7 < 3 : 4 < 7 : 9 ✔   (c) 7 : 9 < 3 : 4 < 5 : 7   (d) 5 : 7 < 7 : 9 < 3 : 4

Solution:

  1. 3/4 = 0.75, 5/7 = 0.714, 7/9 = 0.778.
  2. Increasing: 5 : 7 < 3 : 4 < 7 : 9.

Answer: (b) 5 : 7 < 3 : 4 < 7 : 9

Model 4: Combining Ratios (a : b and b : c → a : b : c : d)

To join a : b and b : c, the common term b must be made equal in both ratios. Multiply the first ratio by the b of the second, and the second ratio by the b of the first.

💡 Trick: Write the ratios one below the other in a staircase and multiply diagonally (many students call it the “neighbour” or zig-zag method). For a : b, b : c, c : d: a = a×b′×c″… just multiply each term by the missing numbers from the other ratios. If one ratio is given in reverse (c : b), flip it first.

Q12. If a : b = 3 : 4 and b : c = 8 : 5, find a : b : c.

(a) 3 : 4 : 5   (b) 6 : 8 : 5 ✔   (c) 3 : 8 : 5   (d) 6 : 4 : 5

Solution:

  1. Make b equal: b is 4 in the first and 8 in the second. Multiply the first ratio by 2: a : b = 6 : 8.
  2. Now a : b : c = 6 : 8 : 5.

Answer: (b) 6 : 8 : 5

Q13. If a : b = 4 : 7 and c : b = 5 : 3, find a : b : c.

(a) 4 : 7 : 5   (b) 12 : 21 : 15   (c) 12 : 21 : 35 ✔   (d) 20 : 35 : 21

Solution:

  1. Flip the second ratio: b : c = 3 : 5.
  2. a = 4 × 3 = 12, b = 7 × 3 = 21, c = 7 × 5 = 35.
  3. a : b : c = 12 : 21 : 35.

Answer: (c) 12 : 21 : 35

Q14. If A : B = 3 : 5, B : C = 4 : 7 and C : D = 2 : 3, find A : B : C : D.

(a) 24 : 40 : 70 : 105 ✔   (b) 12 : 20 : 35 : 42   (c) 24 : 40 : 35 : 105   (d) 3 : 5 : 7 : 3

Solution:

  1. A = 3 × 4 × 2 = 24
  2. B = 5 × 4 × 2 = 40
  3. C = 5 × 7 × 2 = 70
  4. D = 5 × 7 × 3 = 105, so A : B : C : D = 24 : 40 : 70 : 105.

Answer: (a) 24 : 40 : 70 : 105

Q15. The ratio of the marks of A and B is 5 : 6 and that of B and C is 4 : 5. If C scored 35 marks more than A, how many marks did B get?

(a) 70   (b) 84 ✔   (c) 105   (d) 96

Solution:

  1. Make B equal (LCM of 6 and 4 is 12): A : B = 10 : 12 and B : C = 12 : 15.
  2. A : B : C = 10 : 12 : 15. C − A = 5 parts = 35, so 1 part = 7.
  3. B = 12 × 7 = 84.

Answer: (b) 84

Q16. The sum of four numbers A, B, C and D is 1,260. A : B = 2 : 3, B : C = 4 : 1 and C : D = 3 : 5. Find C.

(a) 105   (b) 180   (c) 135 ✔   (d) 150

Solution:

  1. A = 2×4×3 = 24, B = 3×4×3 = 36, C = 3×1×3 = 9, D = 3×1×5 = 15.
  2. Total parts = 84, so 1 part = 1260 ÷ 84 = 15.
  3. C = 9 × 15 = 135.

Answer: (c) 135

Q17. The prices of A and B are in the ratio 4 : 9, and the prices of C and D are in the ratio 20 : 11. C costs ₹600 more than B, and D costs ₹1,320. Find the price of A.

(a) ₹800 ✔   (b) ₹1,800   (c) ₹600   (d) ₹960

Solution:

  1. D = 11 parts = 1320, so 1 part = 120 and C = 20 × 120 = ₹2,400.
  2. B = 2400 − 600 = ₹1,800.
  3. A : B = 4 : 9, so A = 1800 × 4/9 = ₹800.

Answer: (a) ₹800

Model 5: Equal Products (3x = 5y = 4z type)

If px = qy = rz, then x : y : z = 1/p : 1/q : 1/r. Multiply by the LCM of p, q, r to clear the fractions.

💡 Trick: Short cut for three terms: x : y : z = (q × r) : (p × r) : (p × q). For 4x = 6y = 5z that gives 30 : 20 : 24 = 15 : 10 : 12.

Q18. If 4x = 6y = 5z, find x : y : z.

(a) 15 : 10 : 12 ✔   (b) 4 : 6 : 5   (c) 12 : 10 : 15   (d) 6 : 4 : 5

Solution:

  1. x : y : z = 1/4 : 1/6 : 1/5.
  2. Multiply by LCM 60: 15 : 10 : 12.

Answer: (a) 15 : 10 : 12

Q19. ₹15,300 is divided among X, Y and Z so that 3 times X’s share = 4 times Y’s share = 6 times Z’s share. Find Y’s share.

(a) ₹6,800   (b) ₹3,400   (c) ₹5,100 ✔   (d) ₹4,500

Solution:

  1. X : Y : Z = 1/3 : 1/4 : 1/6 = 4 : 3 : 2 (multiply by 12).
  2. Total parts = 9, 1 part = 15300 ÷ 9 = ₹1,700.
  3. Y = 3 × 1700 = ₹5,100.

Answer: (c) ₹5,100

Q20. If 2A = 3B = 5C and A + B + C = 124, find B.

(a) 60   (b) 24   (c) 40 ✔   (d) 30

Solution:

  1. A : B : C = 1/2 : 1/3 : 1/5 = 15 : 10 : 6.
  2. Total parts = 31, so 1 part = 4.
  3. B = 10 × 4 = 40.

Answer: (c) 40

Q21. If p/q = r/s = t/u = 3/5, find (2p + 5r + 3t) : (2q + 5s + 3u).

(a) 2 : 3   (b) 5 : 3   (c) 10 : 3   (d) 3 : 5 ✔

Solution:

  1. Each top value is 3/5 of its bottom value: p = 3q/5, r = 3s/5, t = 3u/5.
  2. So the top sum is 3/5 of the bottom sum.
  3. Ratio = 3 : 5 (the multipliers 2, 5, 3 do not matter).

Answer: (d) 3 : 5

Model 6: Pair Sums: (a + b) : (b + c) : (c + a) and a : (b + c)

Adding all three pair sums gives 2(a + b + c). So one “total” is known, and each single value = total − the pair that does not contain it. For a : (b + c) type, the whole a + b + c is the sum of the two terms in every ratio, so make those sums equal.

💡 Trick: (a+b)+(b+c)+(c+a) = 2(a+b+c). Then c = (a+b+c) − (a+b). For a : (b+c) = m : n, a is m/(m+n) of the total.

Q22. If (a + b) : (b + c) : (c + a) = 6 : 7 : 5 and a + b + c = 27, find a : b : c.

(a) 2 : 4 : 3 ✔   (b) 3 : 4 : 2   (c) 2 : 3 : 4   (d) 4 : 2 : 3

Solution:

  1. Sum of the pair parts = 18 parts = 2 × 27 = 54, so 1 part = 3.
  2. a + b = 18, b + c = 21, c + a = 15.
  3. c = 27 − 18 = 9, a = 27 − 21 = 6, b = 27 − 15 = 12.
  4. a : b : c = 6 : 12 : 9 = 2 : 4 : 3.

Answer: (a) 2 : 4 : 3

Q23. If P : Q = 2 : 5 and Q : R = 5 : 3, find (P + Q) : (Q + R) : (R + P).

(a) 7 : 8 : 5 ✔   (b) 2 : 5 : 3   (c) 5 : 7 : 8   (d) 7 : 5 : 8

Solution:

  1. Q is already equal: P : Q : R = 2 : 5 : 3.
  2. P + Q = 7, Q + R = 8, R + P = 5.

Answer: (a) 7 : 8 : 5

Q24. If a : (b + c) = 2 : 7 and b : (c + a) = 4 : 5, find c : (a + b).

(a) 1 : 3   (b) 1 : 2 ✔   (c) 2 : 5   (d) 3 : 4

Solution:

  1. In both ratios the total a + b + c = 9 parts.
  2. a = 2 parts, b = 4 parts, so c = 9 − 2 − 4 = 3 parts.
  3. c : (a + b) = 3 : 6 = 1 : 2.

Answer: (b) 1 : 2

Q25. If a, b, c are positive and (a² + b²) : (b² + c²) : (c² + a²) = 13 : 25 : 20, find (b − a) : (c − b) : (c − a).

(a) 1 : 2 : 1   (b) 2 : 1 : 1   (c) 1 : 1 : 2 ✔   (d) 1 : 1 : 1

Solution:

  1. Total = 13 + 25 + 20 = 58 = 2(a² + b² + c²), so a² + b² + c² = 29.
  2. a² = 29 − 25 = 4, b² = 29 − 20 = 9, c² = 29 − 13 = 16, so a = 2, b = 3, c = 4.
  3. (b − a) : (c − b) : (c − a) = 1 : 1 : 2.

Answer: (c) 1 : 1 : 2

Model 7: Ratio After Percentage or Fixed Increase (Seats, Salaries, Expenses)

When each part of a ratio changes by a different percentage, multiply each term by its own factor (20% rise → × 1.2, i.e. × 6/5). When each person gets a fixed hike, use k and form one equation.

💡 Trick: Treat the ratio as real values (5 : 7 : 8 → 500 : 700 : 800) so that percentages become easy whole numbers, then simplify the new ratio at the end.

Q26. Seats for Hindi, Science and Maths in a college are in the ratio 5 : 7 : 8. The seats are increased by 40%, 50% and 25% respectively. What is the new ratio?

(a) 14 : 21 : 20 ✔   (b) 7 : 10 : 10   (c) 5 : 7 : 8   (d) 10 : 21 : 20

Solution:

  1. Take seats as 100 × ratio: 500, 700, 800.
  2. After increase: 500 × 1.4 = 700, 700 × 1.5 = 1050, 800 × 1.25 = 1000.
  3. 700 : 1050 : 1000 = 14 : 21 : 20.

Answer: (a) 14 : 21 : 20

Q27. The salaries of A, B and C are in the ratio 3 : 4 : 6. They get increments of 20%, 25% and 10% respectively. If C’s new salary is ₹13,200, find A’s new salary.

(a) ₹6,000   (b) ₹7,200 ✔   (c) ₹8,000   (d) ₹7,500

Solution:

  1. C’s old salary = 13200 ÷ 1.1 = ₹12,000 = 6 parts, so 1 part = ₹2,000.
  2. A’s old salary = 3 × 2000 = ₹6,000.
  3. New salary = 6000 × 1.2 = ₹7,200.

Answer: (b) ₹7,200

Q28. A family spends on rice, dal and oil in the ratio 10 : 6 : 4. The prices go up by 20%, 30% and 50% respectively (quantity same). By what percent does the total spending go up?

(a) 33%   (b) 25%   (c) 29% ✔   (d) 30%

Solution:

  1. Take spending as 10, 6, 4 (total 20).
  2. New: 10 × 1.2 = 12, 6 × 1.3 = 7.8, 4 × 1.5 = 6. Total = 25.8.
  3. Increase = 5.8/20 × 100 = 29%.

Answer: (c) 29%

Q29. The monthly salaries of Ravi and Kiran are in the ratio 11 : 9. Ravi gets a hike of ₹2,000 and Kiran gets ₹3,000, and the ratio becomes 6 : 5. What is Kiran’s salary after the hike?

(a) ₹72,000   (b) ₹75,000 ✔   (c) ₹90,000   (d) ₹78,000

Solution:

  1. (11k + 2000)/(9k + 3000) = 6/5
  2. 55k + 10000 = 54k + 18000, so k = 8,000.
  3. Kiran’s old salary = 9 × 8000 = 72,000; after hike = ₹75,000. (Check: 90,000 : 75,000 = 6 : 5 ✔)

Answer: (b) ₹75,000

Model 8: Adding or Removing a Number to Change the Ratio

Two numbers in ratio a : b become c : d after the same number x is added (or subtracted). Write (ak + x)/(bk + x) = c/d, cross-multiply and find k. The same idea works for boys and girls joining or leaving, students added to classes, etc.

💡 Trick: When the same number is added to both, the difference between them does not change. So make the difference equal: 5 : 9 (gap 4) and 3 : 5 (gap 2) → multiply the second by 2 → 6 : 10. Now 6 − 5 = 1 part = 24. This “difference method” is the fastest.

Q30. Two numbers are in the ratio 5 : 9. If 24 is added to each, the ratio becomes 3 : 5. What is the difference between the two numbers?

(a) 72   (b) 96 ✔   (c) 120   (d) 48

Solution:

  1. Gap stays same: old 5 : 9 (gap 4), new 3 : 5 = 6 : 10 (gap 4).
  2. Old 5 → new 6, so 1 part = 24.
  3. Difference = 4 × 24 = 96. (Numbers 120 and 216.)

Answer: (b) 96

Q31. Two numbers are in the ratio 4 : 9. If 8 is subtracted from each, the ratio becomes 2 : 5. Find the smaller number.

(a) 40   (b) 48 ✔   (c) 36   (d) 56

Solution:

  1. 5(4k − 8) = 2(9k − 8)
  2. 20k − 40 = 18k − 16, so k = 12.
  3. Smaller number = 4 × 12 = 48.

Answer: (b) 48

Q32. In a college the ratio of boys to girls is 7 : 5. If 30 boys leave and 30 girls join, the ratio becomes 1 : 1. How many boys were there at first?

(a) 180   (b) 150   (c) 240   (d) 210 ✔

Solution:

  1. 7k − 30 = 5k + 30, so 2k = 60 and k = 30.
  2. Boys at first = 7 × 30 = 210.

Answer: (d) 210

Q33. A school has 960 students with boys and girls in the ratio 5 : 7. If 20 new girls join, how many new boys must join so that the ratio becomes 3 : 4?

(a) 35 ✔   (b) 25   (c) 45   (d) 20

Solution:

  1. Boys = 960 × 5/12 = 400, girls = 560 + 20 = 580.
  2. (400 + x)/580 = 3/4 → 400 + x = 435.
  3. x = 35.

Answer: (a) 35

Q34. The number of students in three classes is in the ratio 4 : 5 : 7. If 10 students are added to each class, the ratio becomes 6 : 7 : 9. Find the total students before the increase.

(a) 96   (b) 80 ✔   (c) 112   (d) 64

Solution:

  1. (4k + 10)/(5k + 10) = 6/7 → 28k + 70 = 30k + 60 → k = 5.
  2. Classes: 20, 25, 35. After adding 10: 30, 35, 45 = 6 : 7 : 9 ✔
  3. Total before = 80.

Answer: (b) 80

Model 9: Coins and Notes in a Given Ratio

Ratio of NUMBER of coins is not the ratio of their VALUE. Value of each type = number × denomination. Find the value of one “set” (one k), then divide the total by it.

💡 Trick: Value per k = ฮฃ (ratio term × coin value). For ₹1, ₹2, ₹5 in 3 : 4 : 2 → 3 + 8 + 10 = 21. Then k = Total ÷ 21.

Q35. A purse has ₹1, ₹2 and ₹5 coins in the ratio 3 : 4 : 2. The total amount is ₹84. How many ₹5 coins are there?

(a) 4   (b) 8 ✔   (c) 12   (d) 10

Solution:

  1. Value of one set = 3×1 + 4×2 + 2×5 = 21.
  2. k = 84 ÷ 21 = 4.
  3. ₹5 coins = 2 × 4 = 8.

Answer: (b) 8

Q36. A bag contains ₹1,320 in ₹2, ₹5 and ₹10 coins. The number of coins is in the ratio 4 : 6 : 5. How many coins are there in all?

(a) 180   (b) 225 ✔   (c) 240   (d) 150

Solution:

  1. Value of one set = 4×2 + 6×5 + 5×10 = 88.
  2. k = 1320 ÷ 88 = 15.
  3. Total coins = (4 + 6 + 5) × 15 = 225.

Answer: (b) 225

Q37. A box has 270 coins of ₹5, ₹10 and ₹20 in the ratio 3 : 4 : 2. What is the total value of the coins?

(a) ₹2,700   (b) ₹3,150   (c) ₹2,850 ✔   (d) ₹2,400

Solution:

  1. k = 270 ÷ 9 = 30, so 90, 120 and 60 coins.
  2. Value = 90×5 + 120×10 + 60×20 = 450 + 1200 + 1200.
  3. Total = ₹2,850.

Answer: (c) ₹2,850

Q38. A cashier pays ₹2,200 in ₹5, ₹10 and ₹20 coins whose numbers are in the ratio 2 : 4 : 3. What is the total worth of the ₹5 coins?

(a) ₹100   (b) ₹400   (c) ₹200 ✔   (d) ₹250

Solution:

  1. Value of one set = 2×5 + 4×10 + 3×20 = 110.
  2. k = 2200 ÷ 110 = 20, so there are 40 coins of ₹5.
  3. Worth = 40 × 5 = ₹200.

Answer: (c) ₹200

Q39. Sunil buys an item for ₹150 and pays with a ₹500 note. The shopkeeper returns the balance in ₹2, ₹5 and ₹10 coins in the ratio 5 : 2 : 3. How many ₹5 coins does he get?

(a) 14 ✔   (b) 7   (c) 21   (d) 10

Solution:

  1. Balance = 500 − 150 = ₹350.
  2. Value of one set = 5×2 + 2×5 + 3×10 = 50, so k = 7.
  3. ₹5 coins = 2 × 7 = 14.

Answer: (a) 14

Model 10: Income, Expenditure and Savings

Income = Expenditure + Savings. Questions give two of these as ratios and one fixed value. When savings (or expenditure) are equal for two people, write incomes as ak, bk and expenses as cm, dm, then use the equal part.

💡 Trick: Equal savings: (ak − S)/(bk − S) = c/d. Equal expenditure: savings ratio is used the same way. Always check your answer by putting it back into both ratios.

Q40. The incomes of A and B are in the ratio 5 : 7 and their expenses are in the ratio 3 : 5. If each saves ₹6,000, find A’s income.

(a) ₹21,000   (b) ₹15,000 ✔   (c) ₹18,000   (d) ₹12,000

Solution:

  1. (5x − 6000)/(7x − 6000) = 3/5
  2. 25x − 30000 = 21x − 18000 → 4x = 12000 → x = 3000.
  3. A’s income = 5 × 3000 = ₹15,000. (Expenses 9,000 and 15,000 = 3 : 5 ✔)

Answer: (b) ₹15,000

Q41. The incomes of A and B are in the ratio 3 : 5 and their savings are in the ratio 5 : 9. If each spends ₹2,400, what is the difference between their incomes?

(a) ₹9,600 ✔   (b) ₹4,800   (c) ₹7,200   (d) ₹12,000

Solution:

  1. 9(3x − 2400) = 5(5x − 2400) → 27x − 21600 = 25x − 12000 → x = 4800.
  2. Incomes: 14,400 and 24,000.
  3. Difference = ₹9,600.

Answer: (a) ₹9,600

Q42. The expenditure and savings of a person are in the ratio 3 : 2. His income goes up by 20% and his expenditure by 10%. By what percent do his savings go up?

(a) 30%   (b) 25%   (c) 35% ✔   (d) 40%

Solution:

  1. Take income = 100: expenditure 60, savings 40.
  2. New income 120, new expenditure 66, new savings 54.
  3. Increase = 14/40 × 100 = 35%.

Answer: (c) 35%

Q43. The incomes of Kiran and Naresh are in the ratio 4 : 7 and their expenses are in the ratio 6 : 11. Kiran saves ₹500 out of an income of ₹3,200. How much does Naresh save?

(a) ₹700   (b) ₹650 ✔   (c) ₹600   (d) ₹550

Solution:

  1. 4 parts = 3200, so Naresh’s income = 7 × 800 = ₹5,600.
  2. Kiran’s expense = 3200 − 500 = 2700 = 6 parts, so 1 part = 450.
  3. Naresh’s expense = 11 × 450 = 4,950; savings = 5600 − 4950 = ₹650.

Answer: (b) ₹650

Q44. A, B and C spend 75%, 80% and 60% of their incomes respectively. Their savings are in the ratio 5 : 6 : 12. If C earns ₹4,000 more than A, find B’s income.

(a) ₹8,000   (b) ₹12,000 ✔   (c) ₹10,000   (d) ₹16,000

Solution:

  1. Savings: 25% of A : 20% of B : 40% of C = 5 : 6 : 12.
  2. So A = 5/0.25 = 20, B = 6/0.2 = 30, C = 12/0.4 = 30 parts.
  3. C − A = 10 parts = 4000, so 1 part = 400. B = 30 × 400 = ₹12,000.

Answer: (b) ₹12,000

Q45. Last year the incomes of A and B were in the ratio 5 : 4. A’s last-year to this-year income ratio is 3 : 4 and B’s is 2 : 3. If their total income this year is ₹7,60,000, find A’s income this year.

(a) ₹3,60,000   (b) ₹4,00,000 ✔   (c) ₹3,00,000   (d) ₹4,20,000

Solution:

  1. Take last year A = 15m, B = 12m (5 : 4, with 15 divisible by 3 and 12 by 2).
  2. This year A = 15m × 4/3 = 20m, B = 12m × 3/2 = 18m. Total = 38m = 7,60,000, so m = 20,000.
  3. A this year = 20 × 20000 = ₹4,00,000.

Answer: (b) ₹4,00,000

Model 11: Compound, Duplicate, Triplicate, Sub-duplicate and Inverse Ratio

Compound ratio of a : b and c : d is ac : bd. Duplicate = a² : b², triplicate = a³ : b³, sub-duplicate = √a : √b, sub-triplicate = ∛a : ∛b, inverse (reciprocal) = b : a, i.e. 1/a : 1/b.

💡 Trick: With decimals or fractions, first turn the ratio into whole numbers (multiply both sides by the same number), then apply the rule.

Q46. Find the compound ratio of 2 : 3, 4 : 5 and 9 : 16.

(a) 3 : 10 ✔   (b) 10 : 3   (c) 8 : 15   (d) 2 : 5

Solution:

  1. Multiply the first terms and the second terms: (2×4×9) : (3×5×16) = 72 : 240.
  2. Simplify: 3 : 10.

Answer: (a) 3 : 10

Q47. Find the duplicate ratio of 13 : 15.

(a) 26 : 30   (b) 169 : 225 ✔   (c) 225 : 169   (d) 2197 : 3375

Solution:

  1. Duplicate = 13² : 15² = 169 : 225.

Answer: (b) 169 : 225

Q48. Find the sub-duplicate ratio of 81 : 196.

(a) 9 : 14 ✔   (b) 14 : 9   (c) 81 : 196   (d) 3 : 7

Solution:

  1. √81 : √196 = 9 : 14.

Answer: (a) 9 : 14

Q49. Find the triplicate ratio of 0.2 : 0.7.

(a) 4 : 49   (b) 8 : 343 ✔   (c) 2 : 7   (d) 343 : 8

Solution:

  1. 0.2 : 0.7 = 2 : 7.
  2. Triplicate = 2³ : 7³ = 8 : 343.

Answer: (b) 8 : 343

Q50. Find the sub-triplicate ratio of 216 : 343.

(a) 7 : 6   (b) 36 : 49   (c) 6 : 7 ✔   (d) 2 : 3

Solution:

  1. ∛216 : ∛343 = 6 : 7.

Answer: (c) 6 : 7

Q51. Find the inverse ratio of 1/27 : 1/81.

(a) 3 : 1   (b) 1 : 3 ✔   (c) 27 : 81   (d) 81 : 1

Solution:

  1. 1/27 : 1/81 = 81 : 27 = 3 : 1 (multiply by 81×27 and cancel).
  2. Inverse = 1 : 3.

Answer: (b) 1 : 3

Model 12: Proportion with Three Given Numbers (Fourth and First Proportional)

a : b :: c : d means a/b = c/d, so a × d = b × c (product of extremes = product of means). The fourth proportional of a, b, c is d = (b × c)/a. If the missing number is the first one, x = (a × b)/c from x : a :: b : c.

💡 Trick: Fourth proportional = (2nd × 3rd) ÷ 1st. Just remember “extremes product = means product” and you can find any missing place.

Q52. Find the fourth proportional to 6, 15 and 22.

(a) 50   (b) 60   (c) 55 ✔   (d) 44

Solution:

  1. 6 : 15 :: 22 : x → 6x = 15 × 22 = 330.
  2. x = 55.

Answer: (c) 55

Q53. Find the fourth proportional to 3, 7 and 18.

(a) 42 ✔   (b) 36   (c) 49   (d) 54

Solution:

  1. x = (7 × 18)/3 = 126/3 = 42.

Answer: (a) 42

Q54. What is the ratio of the fourth proportional of 3, 7, 9 to the fourth proportional of 4, 10, 12?

(a) 3 : 4   (b) 7 : 10 ✔   (c) 10 : 7   (d) 21 : 40

Solution:

  1. First: (7 × 9)/3 = 21. Second: (10 × 12)/4 = 30.
  2. Ratio = 21 : 30 = 7 : 10.

Answer: (b) 7 : 10

Q55. Find x if x : 9 :: 12 : 36 (x is called the first proportional).

(a) 4   (b) 27   (c) 3 ✔   (d) 6

Solution:

  1. Product of extremes = product of means: 36x = 9 × 12 = 108.
  2. x = 3.

Answer: (c) 3

Q56. The fourth proportional to 4, a and 9a is 36. Find a (a > 0).

(a) 2   (b) 4 ✔   (c) 6   (d) 3

Solution:

  1. 4 : a :: 9a : 36 → 4 × 36 = 9a × a.
  2. 9a² = 144, a² = 16, a = 4.

Answer: (b) 4

Model 13: Mean Proportional and Third Proportional (Two Given Numbers)

If a : b :: b : c (continued proportion), then b² = ac. b is the mean proportional = √(ac), and c is the third proportional = b²/a.

💡 Trick: Mean proportional = √(a × b). Third proportional of a and b = b² ÷ a. For decimals, square-root numerator and denominator separately.

Q57. Find the mean proportional between 3 and 48.

(a) 24   (b) 16   (c) 12 ✔   (d) 9

Solution:

  1. √(3 × 48) = √144 = 12.

Answer: (c) 12

Q58. Find the mean proportional between 0.25 and 6.25.

(a) 1.25 ✔   (b) 2.5   (c) 12.5   (d) 1.5

Solution:

  1. √(0.25 × 6.25) = 0.5 × 2.5 = 1.25.

Answer: (a) 1.25

Q59. Find the third proportional to 16 and 28.

(a) 40   (b) 49 ✔   (c) 56   (d) 42

Solution:

  1. 16 : 28 :: 28 : x → x = 28²/16 = 784/16 = 49.

Answer: (b) 49

Q60. A is the mean proportional of 18 and 8, and B is the mean proportional of 50 and 2. Find 2A + 3B.

(a) 54 ✔   (b) 48   (c) 60   (d) 44

Solution:

  1. A = √(18 × 8) = √144 = 12. B = √(50 × 2) = √100 = 10.
  2. 2A + 3B = 24 + 30 = 54.

Answer: (a) 54

Q61. The third proportional of 4 and x is 25, and the third proportional of 3 and y is 12 (x, y > 0). Find x : y.

(a) 3 : 5   (b) 5 : 3 ✔   (c) 2 : 1   (d) 25 : 12

Solution:

  1. x²/4 = 25 → x² = 100 → x = 10.
  2. y²/3 = 12 → y² = 36 → y = 6.
  3. x : y = 10 : 6 = 5 : 3.

Answer: (b) 5 : 3

Q62. If p is the third proportional to 9 and 15, and q is the fourth proportional to 4, 7 and 20, find 2p + q.

(a) 75   (b) 85 ✔   (c) 90   (d) 80

Solution:

  1. p = 15²/9 = 225/9 = 25.
  2. q = (7 × 20)/4 = 35.
  3. 2p + q = 50 + 35 = 85.

Answer: (b) 85

Q63. Find the mean proportional between 2a³ and 18/a (a > 0).

(a) 6a²   (b) 36a   (c) 6a ✔   (d) 3a

Solution:

  1. √(2a³ × 18/a) = √(36a²) = 6a.

Answer: (c) 6a

Model 14: Number Added or Subtracted to Make Four Numbers Proportional

If x is added to a, b, c, d so that (a + x) : (b + x) :: (c + x) : (d + x), then (a + x)(d + x) = (b + x)(c + x). Expanding gives x = (ad − bc) ÷ [(b + c) − (a + d)]. For subtraction, x = (ad − bc) ÷ [(a + d) − (b + c)].

💡 Trick: In the exam the fastest way is often to try the options: add (or subtract) the option value and check whether 1st × 4th = 2nd × 3rd.

Q64. Which number should be added to each of 5, 13, 8 and 20 so that they become proportional?

(a) 2   (b) 1 ✔   (c) 3   (d) 4

Solution:

  1. x = (5×20 − 13×8) ÷ [(13 + 8) − (5 + 20)] = (100 − 104) ÷ (−4) = 1.
  2. Check: 6, 14, 9, 21 → 6 × 21 = 126 = 14 × 9 ✔

Answer: (b) 1

Q65. What number should be subtracted from each of 17, 39, 23 and 57 so that the remainders are proportional?

(a) 5   (b) 6 ✔   (c) 7   (d) 4

Solution:

  1. x = (17×57 − 39×23) ÷ [(17 + 57) − (39 + 23)] = (969 − 897) ÷ 12 = 6.
  2. Check: 11, 33, 17, 51 → 561 = 561 ✔

Answer: (b) 6

Q66. When x is added to each of 4, 7, 16 and 25, the numbers are in proportion. What is the mean proportional between (x + 2) and (x + 14)?

(a) 6   (b) 8 ✔   (c) 12   (d) 9

Solution:

  1. x = (4×25 − 7×16) ÷ [(7 + 16) − (4 + 25)] = (100 − 112) ÷ (−6) = 2.
  2. x + 2 = 4 and x + 14 = 16.
  3. Mean proportional = √(4 × 16) = 8.

Answer: (b) 8

Q67. If x is subtracted from each of 31, 45, 57 and 85, the numbers obtained are in proportion. Find (5x + 3) : (3x + 12).

(a) 28 : 27 ✔   (b) 27 : 28   (c) 1 : 1   (d) 4 : 3

Solution:

  1. x = (31×85 − 45×57) ÷ [(31 + 85) − (45 + 57)] = (2635 − 2565) ÷ 14 = 5.
  2. Check: 26, 40, 52, 80 → 2080 = 2080 ✔
  3. (5×5 + 3) : (3×5 + 12) = 28 : 27.

Answer: (a) 28 : 27

Model 15: Direct and Inverse Variation (incl. Price ∝ Square of Weight)

A varies directly as B means A = kB (A ∝ B). A varies inversely as B means A × B = k. Find k from the given pair, then use it. If price ∝ (weight)², a broken stone loses value because the squares of the parts add up to less than the square of the whole.

💡 Trick: Broken stone: if pieces are in ratio a : b, original value ∝ (a + b)² and new value ∝ a² + b². Loss = (a + b)² − (a² + b²) = 2ab parts.

Q68. A varies directly as (B + 12). A = 90 when B = 18. Find A when B = 48.

(a) 150   (b) 180 ✔   (c) 160   (d) 200

Solution:

  1. A = k(B + 12) → 90 = k × 30 → k = 3.
  2. When B = 48: A = 3 × 60 = 180.

Answer: (b) 180

Q69. x varies directly as y and inversely as z. x = 6 when y = 4 and z = 10. Find x when y = 9 and z = 5.

(a) 18   (b) 24   (c) 27 ✔   (d) 30

Solution:

  1. x = k·y/z → 6 = k × 4/10 → k = 15.
  2. x = 15 × 9/5 = 27.

Answer: (c) 27

Q70. P varies inversely as Q. P = 12 when Q = 15. Find P when Q = 20.

(a) 16   (b) 10   (c) 9 ✔   (d) 8

Solution:

  1. P × Q = 12 × 15 = 180.
  2. P = 180 ÷ 20 = 9.

Answer: (c) 9

Q71. The price of a diamond varies as the square of its weight. A diamond worth ₹1,44,000 breaks into two pieces whose weights are in the ratio 3 : 5. Find the loss.

(a) ₹54,000   (b) ₹72,000   (c) ₹67,500 ✔   (d) ₹60,000

Solution:

  1. Whole = (3 + 5)² = 64 parts = 1,44,000, so 1 part = 2,250.
  2. Pieces = 3² + 5² = 34 parts. Loss = 64 − 34 = 30 parts (= 2 × 3 × 5).
  3. Loss = 30 × 2250 = ₹67,500.

Answer: (c) ₹67,500

Q72. The value of a stone is proportional to the square of its weight. A stone worth ₹2,52,000 breaks into three pieces with weights in the ratio 1 : 2 : 3. What is the total value of the pieces?

(a) ₹84,000   (b) ₹1,26,000   (c) ₹98,000 ✔   (d) ₹1,12,000

Solution:

  1. Whole = (1 + 2 + 3)² = 36 parts = 2,52,000, so 1 part = 7,000.
  2. Pieces = 1 + 4 + 9 = 14 parts.
  3. Value = 14 × 7000 = ₹98,000.

Answer: (c) ₹98,000

Model 16: Mixed and Tricky Ratio Questions (Exams, Mixtures, Wages, Bills)

These look different but use the same tools: give each group a variable, write the given ratios as equations, and solve. Draw a small table (group vs condition) before writing anything.

💡 Trick: For two mixtures mixed in equal weights, first make both totals equal (LCM of the totals), then add item by item.

Q73. In an exam, the ratio of passed to failed students is 7 : 2. If 18 of the students who passed had failed instead, the ratio would have been 5 : 4. How many students wrote the exam?

(a) 72   (b) 81 ✔   (c) 90   (d) 63

Solution:

  1. (7x − 18)/(2x + 18) = 5/4 → 28x − 72 = 10x + 90 → 18x = 162 → x = 9.
  2. Total = 9x = 81. (Check: 45 : 36 = 5 : 4 ✔)

Answer: (b) 81

Q74. On a planet the ratio of land to water on the whole surface is 3 : 7. In its northern half, land to water is 2 : 3. Find the ratio of land to water in the southern half.

(a) 1 : 4 ✔   (b) 2 : 3   (c) 1 : 3   (d) 3 : 7

Solution:

  1. Take the whole surface = 10 units, so each half = 5 units.
  2. North: land 2, water 3. Whole: land 3, water 7.
  3. South: land 3 − 2 = 1, water 7 − 3 = 4 → 1 : 4.

Answer: (a) 1 : 4

Q75. Barrel 1 has three teas A, D, N in the ratio 3 : 2 : 1. Barrel 2 has D, N, K in the ratio 4 : 3 : 5. Equal weights from both barrels are mixed. Find the ratio of D to N in the mixture.

(a) 6 : 4   (b) 8 : 5 ✔   (c) 2 : 1   (d) 4 : 3

Solution:

  1. Barrel 1 total = 6, Barrel 2 total = 12. Make them equal: Barrel 1 ×2 → A 6, D 4, N 2.
  2. Barrel 2: D 4, N 3, K 5.
  3. Mixture: D = 4 + 4 = 8, N = 2 + 3 = 5 → 8 : 5.

Answer: (b) 8 : 5

Q76. A man ordered 5 pairs of black socks and some pairs of brown socks. A black pair costs twice a brown pair. The clerk swapped the numbers of black and brown pairs, which raised the bill by 50%. Find the ratio of black to brown pairs in the original order.

(a) 1 : 2   (b) 2 : 5   (c) 1 : 4 ✔   (d) 1 : 3

Solution:

  1. Let brown pairs = b and brown price = p, so black price = 2p.
  2. Original bill = 10p + bp; swapped bill = 2bp + 5p.
  3. 2b + 5 = 1.5(10 + b) → 0.5b = 10 → b = 20. Ratio = 5 : 20 = 1 : 4.

Answer: (c) 1 : 4

Q77. 44 men, women and children work in a factory. Their numbers are in the ratio 3 : 5 : 3, and the daily wage of a man, a woman and a child is in the ratio 4 : 2 : 1. If the total daily wage bill is ₹3,000, find the daily wage of a woman.

(a) ₹40   (b) ₹60 ✔   (c) ₹30   (d) ₹120

Solution:

  1. Numbers: 44 ÷ 11 = 4 → 12 men, 20 women, 12 children.
  2. Let wages be 4w, 2w, w: 12×4w + 20×2w + 12×w = 100w = 3000 → w = 30.
  3. Woman’s wage = 2w = ₹60.

Answer: (b) ₹60

How Many Days to Complete Ratio and Proportion?

Short answer: 7 to 10 days for a beginner studying 2–3 hours a day, and 3 to 4 days if you have done the chapter before and only need revision plus practice. The exact count depends on your calculation speed, not on how many videos you watch. The plan below covers all 16 models in 8 days and keeps days 9–10 for mixed practice and a test. If you are following our full SSC Maths plan, this chapter fits in the arithmetic block of that plan.

DayModelsWhat to coverPractice target
Day 1Concepts + Models 1–2What is a ratio, k-method, types of ratio, expression value, C&D25 Qs
Day 2Models 3–5Comparing ratios, combining a : b : c : d, equal products (3x = 5y = 4z)30 Qs
Day 3Models 6–7Pair sums, a : (b + c), ratio after % or fixed increase25 Qs
Day 4Models 8–9Adding/removing a number, boys–girls, coins and notes30 Qs
Day 5Model 10Income–expenditure–savings, all four sub-types25 Qs
Day 6Models 11–13Compound/duplicate/inverse ratio, fourth, third and mean proportional35 Qs
Day 7Models 14–15Number added to make proportion, direct & inverse variation, broken stone25 Qs
Day 8Model 16 + error notebookMixed tricky questions; redo every wrong answer from Days 1–725 Qs
Day 9Previous-year pattern practiceTopic-wise PYQs from SSC, RRB and banking papers (official papers / mocks)50 Qs
Day 10Timed test + formula revision25 questions in 20 minutes, then the formula box below25 Qs + review
Your levelTime per dayDays needed
Complete beginner2–3 hours8–10 days
Knows basics, slow in calculation2 hours6–7 days
Revising (done it before)1.5–2 hours3–4 days (Day 1–2 models fast, then Days 8–10)
Working professional1 hour12–14 days (one row of the table every 1.5 days)

Do not skip: After the chapter, do 10 ratio questions every Sunday for a month. Ratio is a speed chapter, and speed fades fast if you stop practising.

Ratio and Proportion Formulas: Quick Revision Box

ConceptFormula / Rule
k-methodA : B = a : b → A = ak, B = bk
Sharing an amount S in a : b : cShare of A = S × a/(a + b + c)
Combining a : b and b : cMake b equal; for chains multiply by the missing terms
px = qy = rzx : y : z = 1/p : 1/q : 1/r = qr : pr : pq
Pair sums(a+b) + (b+c) + (c+a) = 2(a + b + c)
Componendo–Dividendoa/b = c/d ⇒ (a+b)/(a−b) = (c+d)/(c−d)
Same number x added to a : bDifference stays the same; equate differences
CoinsValue per set = ฮฃ (ratio term × coin value); k = Total ÷ value per set
IncomeIncome = Expenditure + Savings
Duplicate / Triplicatea² : b² / a³ : b³
Sub-duplicate / Sub-triplicate√a : √b / ∛a : ∛b
Inverse ratiob : a
Compound ratio(a : b), (c : d), (e : f) → ace : bdf
Proportion rulea : b :: c : d ⇒ ad = bc
Fourth proportionald = bc / a
Third proportionalc = b² / a
Mean proportional√(ab)
Number x added to make a, b, c, d proportionalx = (ad − bc) / [(b + c) − (a + d)]
Number x subtractedx = (ad − bc) / [(a + d) − (b + c)]
Direct / inverse variationA = kB / A × B = k
Stone broken in a : b (price ∝ weight²)Loss = 2ab parts out of (a + b)²

Tips for Telugu-Medium Students

  • Learn the words once: Ratio = เฐจిเฐท్เฐชเฐค్เฐคి (nishpatti), Proportion = เฐ…เฐจుเฐชాเฐคం (anupatam), Mean proportional = เฐฎเฐง్เฐฏเฐฎ เฐ…เฐจుเฐชాเฐคం. Once the word clicks, English questions stop looking scary.
  • Read “of”, “to” and “each” slowly. “A to B is 3 : 4” and “B to A is 3 : 4” are opposite answers. Most mistakes here come from reading too fast, not from the maths.
  • Number of coins ≠ value of coins. This one trap appears again and again. Write “No.” and “Value” as two separate rows every time.
  • Practise in English even if you write in Telugu. RRB and some SSC exams offer Telugu, but CGL Tier-I is in English and Hindi only. Bilingual practice keeps both doors open.
  • Tables over theory. Learn squares to 30 and cubes to 15. Duplicate, triplicate and mean proportional questions become one-line answers.

Ratio and Proportion – Frequently Asked Questions

1. What is ratio and proportion in maths?
A ratio compares two quantities of the same kind by division, like 3 : 2. A proportion says two ratios are equal, like 3 : 2 :: 6 : 4. Ratio is one comparison; proportion is an equality of two comparisons.

2. What is the formula of ratio?
Ratio of a to b = a/b, written a : b. To share an amount S in the ratio a : b, A gets S × a/(a + b) and B gets S × b/(a + b).

3. What is the formula of proportion?
If a : b :: c : d, then a × d = b × c. The product of the extremes (a and d) equals the product of the means (b and c).

4. How do you find the fourth proportional?
For a, b, c the fourth proportional is (b × c)/a. For 4, 6 and 10 it is 60/4 = 15.

5. How do you find the mean proportional?
Mean proportional of a and b is √(a × b). For 4 and 25 it is √100 = 10.

6. What is the third proportional?
For a and b, the third proportional c satisfies a : b :: b : c, so c = b²/a. For 4 and 6 it is 36/4 = 9.

7. What is continued proportion?
Three numbers a, b, c are in continued proportion when a : b = b : c, so b² = ac. Example: 3, 6, 12.

8. What is duplicate ratio?
The duplicate ratio of a : b is a² : b². For 3 : 4 it is 9 : 16. Triplicate is a³ : b³ and sub-duplicate is √a : √b.

9. What is inverse ratio?
The inverse (reciprocal) ratio of a : b is b : a, or 1/a : 1/b. The inverse of 5 : 8 is 8 : 5.

10. What is compound ratio?
Multiply the ratios term by term. The compound ratio of 2 : 3 and 4 : 5 is 8 : 15.

11. How do you combine a : b and b : c into a : b : c?
Make the common term b equal in both. For a : b = 2 : 3 and b : c = 4 : 5, multiply to get 8 : 12 and 12 : 15, so a : b : c = 8 : 12 : 15.

12. What is componendo and dividendo?
If a/b = c/d, then (a + b)/(a − b) = (c + d)/(c − d). It helps you find x : y quickly from expressions like (3x + 2y)/(3x − 2y).

13. How do you divide money in a ratio?
Add the ratio terms, divide the amount by that total to get one part, then multiply by each person’s terms. ₹900 in 2 : 7 → 1 part = 100, shares 200 and 700.

14. What is the difference between direct and inverse proportion?
In direct proportion both quantities change in the same direction (A = kB). In inverse proportion one goes up as the other goes down (A × B = k).

15. How many days are needed to finish ratio and proportion?
About 7–10 days at 2–3 hours a day for beginners, and 3–4 days for revision. The day-wise table above breaks it down by model.

16. Is ratio and proportion important for SSC CGL?
Yes. It gives a few direct questions and also powers partnership, ages, mixture and DI questions, so it affects more marks than it looks.

17. Which chapters depend on ratio and proportion?
Partnership, ages, mixture & alligation, time & work, time & distance (speed ratios), mensuration ratio questions and data interpretation.

18. Can I solve ratio questions using options?
Yes, and often you should. For “find x : y” or “which number makes them proportional” questions, putting each option back is usually faster than full algebra.

Related Posts on Nck115

Next in this series: Partnership and Ages, both of which use the k-method you just practised.

Ratio and Proportion Questions Ratio and Proportion Formulas Ratio Proportion for SSC CGL Ratio Questions for RRB NTPC Mean Third Fourth Proportional Ratio Proportion Tricks

Note: All explanations and questions are original Nck115 content, written in the pattern of previous-year questions with changed numbers. Every answer was verified before publishing. Weightage figures are approximate and not official.

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