ICSE Class 10 Notes on Ratio and Proportion OP Malhotra Maths (2026-27).  We Provide Notes with Step by Step Answer of Exe-6 with Self Evaluation and Revision of S Chand OP Malhotra Maths . Visit official Website CISCE  for detail information about ICSE Board Class-10.

ICSE Class 10 Notes on Ratio and Proportion OP Malhotra Maths (2026-27)

1. Proportion

In mathematics, a proportion is a statement showing that two ratios or fractions are equal.

Four quantities a, b, c and d are said to be in proportion if: a : b :: c : d
⇒ a/b = c/d

• a and d are called Extremes.
• b and c are called Means.

Fundamental Property:
Product of Extremes = Product of Means
i.e, ad = bc

Fourth Proportional
If a : b = c : x then x is called the Fourth Proportional.
x = (b × c) / a
Shortcut: Fourth Proportional = (2nd × 3rd) ÷ 1st

Third Proportional
If a : b = b : x then x is called the Third Proportional.
x = b² / a
Shortcut: Square of second number ÷ first number

Continued Proportion
Three quantities are in continued proportion if: a : b = b : c
b² = ac
Square of Mean = Product of Extremes
Example: 3 : 6 = 6 : 12

Mean Proportional
If a : x = x : b then x is called the Mean Proportional.
x = √(ab)
(Mean Proportional)² = Product of Extremes

Some important type of questions 

Q1. What number should be added to each of the four numbers 10,18,22 and 38 to make them proportional ?

Sol: Let the number to be added be x . Then ,

10+x , 18+x , 22+x and 38+ x are in proportion
(10+x):(18+x) = (22+x):(38+x) ⇒ (10+x)/(18+x) = (22+x)/(38+x)

⇒(10+x)(38+x) = (18+x)(22+x)
⇒ 380+10x+38x+x² = 396+18x+22x+x²
⇒ 380+48x = 396+40x
⇒ 8x = 16
⇒ x = 2

∴ The required number is 2 .

Q2. Find the third proportional to 5,10 .

Sol: Let x be the third proportional to 5 and 10 , then

5:10 = 10:x ⇒ 5/10 = 10/x
⇒ 5x = 100
⇒ x = 20

∴ The third proportional to 5,10 is 20.

Practice Questions on this topic : Exercise-6(a)

Important Results of Proportion

• If a:b::c:d (a/b = c/d):
• Invertendo: If a:b :: c:d then b:a :: d:c , that is a/b = c/d ⇒ b/a = d/c
• Alternendo: If a:b :: c:d then a:c :: b:d , that is a/b = c/d ⇒ a/c = b/d
• Componendo: If a:b :: c:d then (a+b):b :: (c+d):d , that is a/b = c/d ⇒ (a+b)/b = (c+d)/d
• Dividendo: If a:b :: c:d then (a-b):b :: (c-d):d , that is a/b = c/d ⇒ (a−b)/b = (c−d)/d
• Componendo-Dividendo: If a:b :: c:d then (a+b):(a-b) :: (c+d):(c-d) , that is a/b = c/d⇒ (a+b)/(a−b) = (c+d)/(c−d) ← most tested property
• Convertendo: If a:b :: c:d then a:(a-b) :: c:(c-d) , that is a/b = c/d ⇒ a/(a−b) = c/(c−d)

Some Important Type of Questions

Q1. If a:b = c:d , show that 2a+3b/2c+3d = 2a-3b/2c-3d

Sol: a/b = c/d (given)
⇒ 2a/3b = 2c/3d (Multiplying by 2/3 on both sides)

(2a+3b)/(2a-3b) = (2c+3d)/(2a-3d) (By componendo and dividendo)

(2a+3b)/(2c+3d) = (2a-3b)/(2c-3d)

Q2. Solve [2x+√(4x²+1)]/[2x-√(4x²-1)] = 4 for x , using the properties of proportion.

Sol: (2x + √(4x² − 1)) / (2x − √(4x² − 1)) = 4
⇒ [(2x + √(4x² − 1)) + (2x − √(4x² − 1))] /[(2x + √(4x² − 1)) − (2x − √(4x² − 1))] = (4 + 1) / (4 − 1) (By Componendo and Dividendo)

⇒ 4x / 2√(4x² − 1) = 5/3
⇒ 2x / √(4x² − 1) = 5/3

Squaring both sides,
⇒ [2x / √(4x² − 1)]² = (5/3)²
⇒ 4x² / (4x² − 1) = 25/9
⇒ 9(4x²) = 25(4x² − 1)
⇒ 36x² = 100x² − 25
⇒ 64x² = 25
⇒ x² = 25/64
⇒ x = ±5/8

Practice Questions on this topic : Exercise-6(b)

k-Method

When a series of fractions are equal , each of them is equal to the sum of all the numerators divided by the sum of all the denominators.

for example ,
if a/b = c/d = e/f = ……, then each ratio is equal to
(a+c+e+…)/(b+d+f+…) = sum of antecedents/sum of consequents

We can prove this by using k-method. as,
So if , a/b = c/d = e/f

Let a/b = c/d = e/f = k , Then a=bk , c=dk , e=fk

Now , a+c+e/b+d+f = bk+dk+fk/b+d+f = k(b+d+f)/(b+d+f) = k

Hence, a/b = c/d = e/f = a+c+e/b+d+f

Hence Proved .

So , this was k-Method.

Application of k-Method to Continued Proportion

Q1. If a,b,c are in continued proportion , prove that (a+b):(b+c) = a²(b-c) : b²(a-b)

Sol: Given,
(a + b)/(b + c) = a²(b − c)/b²(a − b)

Since a, b, c are in continued proportion,
a/b = b/c

Let, a/b = b/c = k

Then,
a = bk and b = ck

Hence,
a = bk = (ck)k = ck²

LHS = (a + b)/(b + c)
LHS = (ck² + ck)/(ck + c)
LHS = ck(k + 1)/c(k + 1)
LHS = [ck(k + 1)]/[c(k + 1)]
LHS = k ………. (I)

RHS = a²(b − c)/b²(a − b)
RHS = (ck²)²(ck − c)/(ck)²(ck² − ck)
RHS = c²k⁴(c(k − 1))/c²k²(ck(k − 1))
RHS = c³k⁴(k − 1)/c³k³(k − 1)
RHS = k ………. (II)

From (I) and (II),
LHS = RHS

Q.2 If a,b,c,d are in continued proportion prove that a:d = (a-b)³:(b-c)³

Sol: a, b, c, d are in continued proportion

a/b = b/c = c/d = k (say)

c = dk,
b = ck = dk²,
a = bk = dk³

a : d = dk³ : d = k³
∴ a − d = d(k³ − 1)
(a − b)³ = (dk³ − dk²)³

L.H.S.
a/d = a ÷ d
= dk³ / d
= k³

R.H.S.
(a − b)³ / (b − c)³
= (dk³ − dk²)³ / (dk² − dk)³
= [dk²(k − 1)]³ / [dk(k − 1)]³
= d³k⁶(k − 1)³ / d³k³(k − 1)³
= k³

∴ L.H.S. = R.H.S.

Practice Questions on this topic : Exercise-6(c)

Self Evaluation and Revision: Coming Soon