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).

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

Key Algebraic Properties (Proportion Rules)

If a / b = c / d:

• 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

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

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

1. Calculate Cross Product Difference 1: |(a × d) − (b × c)| units.
2. Calculate Cross Product Difference 2: |(c × T₂) − (d × T₁)|.
3. Equate: Unit Value = |(c × T₂) − (d × T₁)| / |(a × d) − (b × c)|
4. Multiply the unit value back into the OLD ratio to find the initial values.

C. Coin & Denomination Problems

When given the ratio of the number of coins and the total amount in rupees:

1. Convert denominations into a uniform unit (e.g., ₹1 = ₹1, 50p = ₹0.50, 25p = ₹0.25).
2. Value Ratio = (Coin Ratio) × (Denomination Value).
3. Sum the value ratio units and equate to the total given amount to find the multiplier x.
4. 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)

Some Solved Problems on Ratio and Proportion

1. 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.

(A) 42

(B) 56

(C) 70

(D) 84

Answer: (B) 56

Solution: Let numbers be 3x and 4x. Sum =7x.

3x+84x+8=45⟹53x+8=44x+8

15x+40=16x+32⟹x=8

Sum =7×8=56

2. If A:B=2:3, B:C=4:5, and C:D=6:7, find the ratio A:D.

(A) 16:35

(B) 8:35

(C) 24:35

(D) 12:35

Answer: (A) 16:35

Solution: AD=ABBCCD=234567=2×4×21×5×7=1635

3. 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.

(A) 360

(B) 720

(C) 800

(D) 760

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

4. 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.

(A) ₹4,000

(B) ₹5,000

(C) ₹3,200

(D) ₹4,500

Answer: (A) ₹4,000

Solution: Let incomes be 4x and 5x. Expenditures are 4x-500 and 5x-500.

4x-5005x-500=79⟹94x-500=75x-500

36x-4500=35x-3500⟹x=1000

A’s income =4×1000=₹4,000

5. What number should be added to each of 6,14,18, and 38 so that the resulting numbers are in proportion?

(A) 1

(B) 2

(C) 3

(D) 4

Answer: (B) 2

Solution: Let x be added: 6+x14+x=18+x38+x

6+x38+x=14+x18+x

228+44x+x2=252+32x+x2

12x=24⟹x=2

6. Find the mean proportional between 3+2 and 12-32.

(A) 7

(B) 27

(C) 7

(D) 14

Answer: (B) 27

Solution: Simplify second term: 12-32=12-42=43-2

Mean proportional =3+2×43-2

=4×32-22=4×9-2=28=27

7. 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.

(A) 20

(B) 30

(C) 48

(D) 35

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

8. 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?

(A) 2:3:4

(B) 6:7:8

(C) 6:8:9

(D) 3:4:5

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

9. 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?

(A) 20 litres

(B) 40 litres

(C) 60 litres

(D) 80 litres

Answer: (C) 60 litres

Solution: Milk =23×60=40 litres, Water =20 litres.

Let x litres of water be added.

4020+x=12⟹80=20+x⟹x=60 litres

10. 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?

(A) 2:1

(B) 11:6

(C) 5:2

(D) 3:1

Answer: (A) 2:1

Solution: Let ages be 7x and 3x.

7x×3x=756⟹21×2=756⟹x2=36⟹x=6

Present ages: Father =42, Son =18.

Ages after 6 years: Father =48, Son =24.

Ratio =48:24=2:1

11. The mean proportional of two numbers is 18 and their third proportional is 144. Find the larger of the two numbers.

(A) 9

(B) 12

(C) 36

(D) 48

Answer: (C) 36

Solution: Let numbers be a and b.

Mean proportional: ab=18⟹ab=324

Third proportional: b2a=144⟹a=b2144

Substitute a: b3144=324⟹b3=324×144=182122=18322×3618=46656=363

b=36⟹a=32436=9

Larger number =36

12. 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.

(A) ₹1,200

(B) ₹900

(C) ₹1,440

(D) ₹1,000

Answer: (A) ₹1,200

Solution: If A:B=3:4 and B:C=5:6, combined ratio =15:20:24.

B’s share =2059×3540=₹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-95x-9=1223⟹233x-9=125x-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: 2a+b+c=21k⟹214=21k⟹28=21k⟹k=43.

a+b=6×43=8.

Since a+b+c=14, c=14-a+b=14-8=6.