Ratio & Proportion
Concepts & Definitions
Ratio & Proportion: Complete Concepts, Shortcuts & Solved Problems
1. Concepts & Definitions
- Ratio (a : b): A comparison of two quantities of the same kind by division. In a : b, a is called the antecedent (numerator) and b is called the consequent (denominator). Multiplying or dividing both terms by the same non-zero constant does not change the ratio.
- Proportion (a : b :: c : d): An equation stating that two ratios are equal (a / b = c / d).
- Product Rule: Product of extremes (a × d) = Product of means (b × c).
2. Standard Types of Ratios & Proportions
| Term | Formula / Relation | Example / Logic |
|---|---|---|
| Duplicate Ratio | a² : b² | Ratio of squares |
| Triplicate Ratio | a³ : b³ | Ratio of cubes |
| Sub-duplicate Ratio | √a : √b | Square root of terms |
| Sub-triplicate Ratio | ∛a : ∛b | Cube root of terms |
| Compound Ratio | (a₁×a₂×…) : (b₁×b₂×…) | Product of corresponding terms |
| Mean Proportional | x = √(a×b) | For a : x :: x : b |
| Third Proportional | c = (b²) / a | For a : b :: b : c |
| Fourth Proportional | d = (b×c) / a | For a : b :: c : d |
3. Key Algebraic Properties (Proportion Rules)
If a / b = c / d, then:
- Componendo: (a+b) / b = (c+d) / d
- Dividendo: (a−b) / b = (c−d) / d
- Componendo & Dividendo: (a+b) / (a−b) = (c+d) / (c−d)
- Invertendo: b / a = d / c
- Alternendo: a / c = b / d
4. Essential Shortcuts & Solution Methods
A. Combining Chain Ratios (LCM & “Hide and Seek” Methods)
- Combining Two Ratios: If A : B = a : b and B : C = c : d, then A : B : C = (a × c) : (b × c) : (b × d). (Alternatively, make the common term B equal by taking the LCM of its values).
- Equations to Ratio (“Hide-and-Seek” Method): If 3A = 4B = 5C, find A : B : C:
- Hide A: Multiply remaining coefficients (4 × 5 = 20)
- Hide B: Multiply remaining coefficients (3 × 5 = 15)
- Hide C: Multiply remaining coefficients (3 × 4 = 12)
- Result: A : B : C = 20 : 15 : 12
B. The Cross-Multiplication Shortcut (For Ages & Income-Expenditure Changes)
Useful when an initial ratio a : b changes to a new ratio c : d after adding/subtracting values or time gaps (T₁, T₂).
- Old Ratio: a : b
- New Ratio: c : d
- Change / Time Gap: T₁ : T₂
- Calculate Cross Product Difference 1: |(a × d) − (b × c)| units.
- Calculate Cross Product Difference 2: |(c × T₂) − (d × T₁)|.
- Equate: Unit Value = |(c × T₂) − (d × T₁)| / |(a × d) − (b × c)|
- Multiply the unit value back into the OLD ratio to find the initial values.
C. Coin & Denomination Problems
- Convert denominations into a uniform unit (e.g., ₹1 = ₹1, 50p = ₹0.50, 25p = ₹0.25).
- Value Ratio = (Coin Ratio) × (Denomination Value).
- Sum the value ratio units and equate to the total given amount to find the multiplier x.
- Number of specific coins = (Coin Ratio Term) × x.
D. Mixture & Alligation Rule
To find the ratio in which two ingredients at different prices/concentrations must be mixed to obtain a mean price/concentration:
Ratio of Cheaper : Dearer = (Dearer Price − Mean Price) : (Mean Price − Cheaper Price)
E. Partnership Problems
Profit sharing is directly proportional to the product of Investment and Time Period:
Profit Share Ratio = (I₁ × T₁) : (I₂ × T₂) : (I₃ × T₃)
5. Solved Problems on Ratio and Proportion
Q1. Two numbers are in the ratio 3:4. If 8 is added to each number, the ratio becomes 4:5. Find the sum of the two numbers.
Answer: (B) 56
Solution: Let numbers be 3x and 4x. Sum = 7x.
(3x+8)/(4x+8) = 4/5 =; 5(3x+8) = 4(4x+8)
15x + 40 = 16x + 32 =; x = 8
Sum = 7 × 8 = 56
Q2. If A:B = 2:3, B:C = 4:5, and C:D = 6:7, find the ratio A:D.
Answer: (A) 16:35
Solution: A/D = (A/B) × (B/C) × (C/D) = (2/3) × (4/5) × (6/7) = (2 × 4 × 2) / (1 × 5 × 7) = 16:35
Q3. A bag contains coins of 50p, 25p, and 10p in the ratio 5:9:4, amounting to ₹206. Find the total number of coins in the bag.
Answer: (B) 720
Solution: Let the number of 50p, 25p, and 10p coins be 5x, 9x, and 4x.
Total Value in Rupees: 5x × 0.50 + 9x × 0.25 + 4x × 0.10 = 206
2.5x + 2.25x + 0.40x = 206 =; 5.15x = 206 =; x = 40
Total coins = 5x + 9x + 4x = 18x = 18 × 40 = 720
Q4. The ratio of monthly incomes of two persons A and B is 4:5 and the ratio of their expenditures is 7:9. If each saves ₹500 per month, find the monthly income of A.
Answer: (A) ₹4,000
Solution: Let incomes be 4x and 5x. Expenditures are (4x-500) and (5x-500).
(4x-500)/(5x-500) = 7/9 &implies; 9(4x-500) = 7(5x-500)
36x – 4500 = 35x – 3500 =; x = 1000
A’s income = 4 × 1000 = ₹4,000
Q5. What number should be added to each of 6, 14, 18, and 38 so that the resulting numbers are in proportion?
Answer: (B) 2
Solution: Let x be added: (6+x)/(14+x) = (18+x)/(38+x)
(6+x)(38+x) = (14+x)(18+x)
228 + 44x + x² = 252 + 32x + x²
12x = 24 =; x = 2
Q6. Find the mean proportional between (3+√2) and (12-√32).
Answer: (B) 2√7
Solution: Simplify second term: 12 – √32 = 12 – 4√2 = 4(3-√2)
Mean proportional = √((3+√2) × 4(3-√2))
= √(4 × (3² – (√2)²)) = √(4 × (9-2)) = √28 = 2√7
Q7. The sum of three numbers is 98. If the ratio of the first to the second is 2:3 and that of the second to the third is 5:8, find the second number.
Answer: (B) 30
Solution:
A : B = 2 : 3 = 10 : 15
B : C = 5 : 8 = 15 : 24
A : B : C = 10 : 15 : 24
Sum = 10x + 15x + 24x = 49x = 98 =; x = 2
Second number (B) = 15 × 2 = 30
Q8. Seats for Mathematics, Physics, and Biology in a school are in the ratio 5:7:8. There is a proposal to increase these seats by 40%, 50%, and 75% respectively. What will be the ratio of increased seats?
Answer: (A) 2:3:4
Solution: Original seats = 5x, 7x, 8x.
New seats: Math = 5x × 1.40 = 7x, Physics = 7x × 1.50 = 10.5x, Bio = 8x × 1.75 = 14x
New Ratio = 7 : 10.5 : 14 = 14 : 21 : 28 = 2 : 3 : 4
Q9. In a mixture of 60 litres, the ratio of milk and water is 2:1. If this ratio is to be made 1:2, how many litres of water should be added?
Answer: (C) 60 litres
Solution: Milk = (2/3) × 60 = 40 litres, Water = 20 litres.
Let x litres of water be added.
40 / (20+x) = 1/2 =; 80 = 20 + x =; x = 60 litres
Q10. The ratio of the age of a father to his son is 7:3. If the product of their ages is 756, what will be the ratio of their ages after 6 years?
Answer: (A) 2:1
Solution: Let ages be 7x and 3x.
7x × 3x = 756 =; 21x² = 756 =; x² = 36 =; x = 6
Present ages: Father = 42, Son = 18.
Ages after 6 years: Father = 48, Son = 24.
Ratio = 48 : 24 = 2 : 1
Q11. The mean proportional of two numbers is 18 and their third proportional is 144. Find the larger of the two numbers.
Answer: (C) 36
Solution: Let numbers be a and b.
Mean proportional: √(ab) = 18 =; ab = 324
Third proportional: b² / a = 144 =; a = b² / 144
Substitute a: b³ / 144 = 324 =s; b³ = 324 × 144 = 46656 = 36³
b = 36 =; a = 324 / 36 = 9
Larger number = 36
Q12. A sum of ₹3,700 is divided among A, B, and C such that A’s share : B’s share = 3:4 and B’s share : C’s share = 5:6. Find B’s share.
Answer: (A) ₹1,200
Solution: If A:B = 3:4 and B:C = 5:6, combined ratio = 15 : 20 : 24.
B’s share = (20 / 59) × 3540 = ₹1,200
13. The ratio of two numbers is 3: 5. If 9 is subtracted from both, their ratio becomes 12: 23. Find the smaller number.
(A) 27
(B) 33
(C) 45
(D) 55
Answer: (B) 33
Solution: Let numbers be 3x and 5x.
(3x – 9) / (5x – 9) = 12 / 23 ⇒ 23(3x – 9) = 12(5x – 9)
=
69x – 207 = 60x – 108 ⇒ 9x = 99 ⇒ x = 11
Smaller number = 3 × 11 = 33
14. An employer reduces the number of his employees in the ratio 9: 8 and increases their wages in the ratio 14: 15. In what ratio is the total wage bill changed?
(A) 20: 21
(B) 21: 20
(C) 27: 25
(D) 25: 27
Answer: (B) 21: 20
Solution: Ratio of total wage bill = (9 × 14): (8 × 15) = 126: 120
Simplify by 21: 20
15. If (a + b): (b + c): (c + a) = 6: 7: 8 and a + b + c = 14, find the value of c.
(A) 6
(B) 7
(C) 8
(D) 14
Answer: (A) 6
Solution: Let (a + b) = 6k, (b + c) = 7k, (c + a) = 8k.
Adding all three: 2(a + b + c) = 21k ⇒ 2(14) = 21k ⇒ 28 = 21k ⇒ k = 4 / 3.
a + b = 6 × (4 / 3) = 8.
Since a + b + c = 14, c = 14 − (a + b) = 14 − 8 = 6.