ICSE Class 10 Linear Inequalities Notes | OP Malhotra Maths (2026-27)

What Is a Linear Inequality?

A linear inequality is a mathematical statement that compares two expressions using inequality signs instead of an equals sign. The four signs you’ll work with are:
> (greater than)
< (less than)
≥ (greater than or equal to)
≤ (less than or equal to)

For example, 2x + 3 > 7 is a linear inequality in one variable.
Unlike 2x + 3 = 7, which has exactly one solution (x = 2), the inequality 2x + 3 > 7 has infinitely many solutions — any value of x greater than 2 satisfies it.

The Golden Rules of Solving Inequalities

Solving an inequality looks almost identical to solving an equation, except for one critical difference. These rules govern everything:

1. Adding or subtracting the same number from both sides does not change the direction of the inequality.
Example, If x + 3 > 5, then x +3 – 3> 2 – 3 ⇒ x > 2

2. Multiplying or dividing both sides by a positive number does not change the direction.
Example, If 2x > 6, then 2x/2 > 6/2 ⇒ x > 3

3. Multiplying or dividing both sides by a negative number REVERSES the inequality sign. This is the rule students forget most often.
Example, If −2x > 6, dividing both sides by −2 gives x < −3 (the sign flips from > to <).

4. Taking the reciprocal of both sides also reverses the inequality, provided both sides have the same sign.

Replacement Set and Solution Set

1. Replacement set is the set from which the values of the variable involved
in the inequation are chosen.

2. Solution set is the subset of the replacement set, whose elements satisfy
the given inequation.

Example: Let the given inequation be x ≤ 4, if:

1. 1. the replacement set = N, the set of natural numbers;
      the solution set = {1, 2, 3, 4}
   2. the replacement set = W, the set of whole numbers;
      the solution set = {0, 1, 2, 3, 4}
   3. the replacement set = Z or I, the set of integers;
      the solution set = {…, -2, -1, 0, 1, 2, 3, 4}
   4. the replacement set = R, the set of real numbers;
      the solution set = { x : x ∈ R and x ≤ 4 }

3. The solution set depends on the replacement set.

Worked Example 1: Basic Inequality

Solve: 3x − 5 ≤ 13, where x ∈ Z

Sol: Adding 5 to both sides:
3x ≤ 18

Dividing both sides by 3 (positive, so sign stays the same):
x ≤ 6

Since x belongs to integers,

Here , the replacement set is Z ,
and the solution set is {…, 2, 3, 4, 5, 6}

Worked Example 2: The Sign-Flip Trap

Solve: 5 − 2x > 1, where x ∈ R

Sol: Subtracting 5 from both sides:
−2x > −4

Dividing both sides by −2 (negative — sign flips):
x < 2

Worked Example 3: Combined (Double) Inequalities

Solve: −3 < 2x − 1 ≤ 5, where x ∈ R

Sol:
Add 1 to all three parts:
⇒ −3 + 1 < 2x – 1 + 1 ≤ 5 + 1
⇒ −2 < 2x ≤ 6

Divide all three parts by 2:
⇒ −2/2 < 2x/2 ≤ 6/2
⇒ −1 < x ≤ 3

Here,
the replacement set is R,
and the solution set is x∈(-1,3]

Representing Solutions on a Number Line

• The solution set of an inequation can be represented on a real number line.
• A hollow/open circle (◦) is used when the inequality is strict (>or<) – meaning that point is excluded.
• A filled/dark circle (•) is used at a number when the inequality includes “equal to” (≥or≤) – meaning that point is included.
• A ray or line segment that extends in the direction of all valid solutions.

Some examples on Number Line ,

Example 1: x > 3 , where x ∈ R

Sol:

Example 2: x ≤ -2 , where x∈R

Sol:

Example 3: -1 < x ≤ 4 , where x∈R

Sol:

Example 4: 2x-3 ≥ 1 , where x∈R

Sol: 2x-3 ≥ 1

Adding 3 to both sides ,
⇒ 2x – 3 + 3 ≥ 1 + 3
⇒ 2x ≥ 4

Dividing by 2 on both sides ,
⇒ 2x/2 ≥ 4/2
⇒ x ≥ 2

Common Mistakes 

• Forgetting that the answer must come only from the replacement set, not from all real numbers or integers in general.
• Skipping the sign-flip rule when dividing by a negative number, leading to a wrong solution set.
• Confusing roster form with set-builder form when writing the final answer — check what the question asks for.
• Including boundary values incorrectly when the inequality is strict (< or >) versus non-strict (≤ or ≥).

Practice Questions on Linear Inequalities : Exercise-4

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Self Evaluation on Quadratic Equations Class 10 ICSE OP Malhotra Maths 2026-27

Self Evaluation on Quadratic Equations

Self Evaluation on Quadratic Equations Class 10 ICSE OP Malhotra Maths 2026-27

Que-1: Solve for x and give your answer correct to 2 decimal places : x² – 10x + 6 = 0.

Sol: x² – 10x + 6 = 0
Here a = 1, b = – 10, c = 6

∴ D = b² – 4ac = (- 10)² – 4 x 1 x 6 = 100 – 24 = 76

∴ x = {−b±√(b²−4ac)}/2a={−(−10)±√76}/2×1
= {10±√(4×19)}/2 = {10±2√19}/2
= 5 ± √19
= 5 ± 4.358

∴ x₁ = 5 + 4.358 = 9.358 = 9.36
x₂ = 5 – 4.358 = 0.642 = 0.64
∴ x = 9.36, 0.64

Que-2: Solve using the quadratic formula x² – 4 + 1 = 0

Sol: x² – 4x + 1 = 0
Here a = 1, b = – 4, c = 1

D = b² – 4ac = (- 4)² – 4 x 1 x 1 = 16 – 4 = 12

∴ x = {−b±√(b²−4ac)}/2a ={−(-4)±√(12)}/2×1
= 4±√4×3/2 = 4±2√3/2
= 2 ± √3 (Dividing by 2)
= 2 ± 1.732

∴ x₁ = 2 + 1.732 = 3.732
x₂ = 2 – 1.732 = 0.268

∴ x = 3.732, 0.268

Que-3: Solve the equation 3x² – x – 7 = 0 and give your answer correct to two decimal places.

Sol: 3x² − x − 7 = 0

Here a = 3, b = −1, c = −7

D = b² − 4ac = (−1)² − 4 × 3 × (−7) m= 1 + 84 = 85

∴ x = (−b ± √(b² − 4ac)) / (2 × 3)
= (−(−1) ± √85) / 6
= (1 ± 9.22) / 6

x₁ = (1 + 9.22) / 6 = 10.22 / 6 = 1.70
x₂ = (1 − 9.22) / 6 = −8.22 / 6 = −1.37

∴ x = 1.70, −1.37

Que-4: Solve the following equation and give your answer up to two decimal places : x² – 5x – 10 = 0

Sol: x² − 5x − 10 = 0
Here a = 1, b = −5, c = −10

D = b² − 4ac = (−5)² − 4 × 1 × (−10) = 25 + 40 = 65

x = (−b ± √(b² − 4ac)) / (2 × 1)
= (−(−5) ± √65) / (2 × 1)
= (5 ± √65) / 2
= (5 ± 8.06) / 2

x₁ = (5 + 8.06) / 2 = 13.06 / 2 = 6.53
x₂ = (5 − 8.06) / 2 = −3.06 / 2 = −1.53

∴ x = 6.53, −1.53

Que-5: Solve the equation 2x – 1/x = 7. Write your answer correct to two decimal places.

Sol: 2x − ¹/_x = 7
2x² − 1 = 7x
2x² − 7x − 1 = 0

Here a = 2, b = −7, c = −1

D = b² − 4ac = (−7)² − 4 × 2 × (−1)
= 49 + 8 = 57

x = (−b ± √(b² − 4ac)) / (2 × 2)
= (−(−7) ± √57) / (2 × 2)
= (7 ± √57) / 4
= (7 ± 7.55) / 4

x₁ = (7 + 7.55) / 4 = 14.55 / 4 = 3.64
x₂ = (7 − 7.55) / 4 = −0.55 / 4 = −0.14

∴ x = 3.64, −0.14

Que-6: The bill for a number of people for overnight stay is ₹ 4800. If there were 4 more, the bill each person had to pay would have reduced by ₹ 200. Find the number of people staying overnight.

Sol: Total amount of the bill = ₹4800

Let number of persons originally = x
Then share of each person = ₹4800/x

If there were 4 more persons, then number of persons = x + 4

Then share of each person = ₹4800/(x + 4)

According to the condition,
(4800/x) − (4800/(x + 4)) = 200
(4800(x + 4) − 4800x) / x(x + 4) = 200

19200 / x(x + 4) = 200
19200 = 200x(x + 4)
96 = x(x + 4)

x² + 4x − 96 = 0
x² + 12x − 8x − 96 = 0
x(x + 12) − 8(x + 12) = 0
(x + 12)(x − 8) = 0

x + 12 = 0 or x − 8 = 0
x = −12 or x = 8

Since number of persons cannot be negative, x = 8

∴ Number of persons = 8

Que-7: An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for :
(i) The onward journey
(ii) The return journey.
If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.

Sol:  Distance travelled by an aeroplane = 400 km
Speed of aeroplane = x km/hr

(i) ∴ Time taken = 400/x hours
On return speed by increasing 40 km/hr,
The speed will be (x + 40) km/hr

(ii) ∴ Time taken = 400/(x + 40)

(iii) According to the condition,
400/x − 400/(x + 40) = 30/60
(400(x + 40) − 400x) / x(x + 40) = 1/2
16000 / (x² + 40x) = 1/2
x² + 40x = 32000
x² + 40x − 32000 = 0
x² + 200x − 160x − 32000 = 0
x(x + 200) − 160(x + 200) = 0
(x + 200)(x − 160) = 0

Either x + 200 = 0, then x = −200 but it is not possible being negative

or x − 160 = 0, then x = 160

∴ Speed = 160 km/hr

Que-8: In an auditorium, seats were arranged in rows and columns. The number of rows were equal to the number of seats in each row. When the number of rows were doubled and the number of seats in each row were reduced by 10, the total number of seats increased by 300. Find :
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after rearrangement.

Sol: In first case
Let number of rows = x

∴ Number of seats in each row = x
∴ Number of seats = x²

In second case,
Number of rows = 2x
and number of columns = x − 10

∴ Total number of seats = 2x(x − 10)

According to the condition,
2x(x − 10) = x² + 300
2x² − 20x = x² + 300
x² − 20x − 300 = 0
x² − 30x + 10x − 300 = 0
x(x − 30) + 10(x − 30) = 0
(x − 30)(x + 10) = 0

Either x = 30, or x = −10

∵ x = −10 is not possible being negative
∴ Number of rows in the original arrangement = 30

(ii) Number of seats after re-arrangement = 2x(x − 10)
= 60 × 20 = 1200

Que-9: P and Q are centres of circles of radius 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of a circle of radius x cm, which touches the above circles externally. Given that ∠PRQ = 90°, write an equation in x and solve it.

Sol: P and Q are the centres of two circles with radii 9 cm and 2 cm respectively.

PQ = 17 cm

Let x be the radius of the circle with centre R.

Since the circles touch externally,
PR = PL + LR = 9 + x cm
and QR = QM + MR = 2 + x cm

In right-angled ΔPQR,
PR² + QR² = PQ²
(9 + x)² + (2 + x)² = 17²
81 + 18x + x² + 4 + 4x + x² = 289
2x² + 22x + 85 = 289
2x² + 22x − 204 = 0
x² + 11x − 102 = 0
x² + 17x − 6x − 102 = 0
x(x + 17) − 6(x + 17) = 0
(x + 17)(x − 6) = 0

Either x = 6 or x = −17
∵ x = −17 is not possible being negative

∴ x = 6 cm

Que-10: By increasing the speed of car by 10 km/ hr, the time of journey for a distance of 72 km is reduced by 36 min. Find the original speed of the car.

Sol: Distance of journey = 72 km
Let original speed of car = x km/hr
and increased speed of car = (x + 10) km/hr

New time taken = 72/(x + 10) hrs
and time taken in first case = 72/x hrs

According to the condition,
72/x − 72/(x + 10) = 36/60
720 / x(x + 10) = 3/5
3600 = 3x(x + 10)
x² + 10x − 1200 = 0
x² + 40x − 30x − 1200 = 0
(x + 40)(x − 30) = 0

Either x = −40 or x = 30
∵ x = −40 is not possible being negative

∴ Original speed of the car = 30 km/hr

Que-11: A shopkeeper buys a certain number of books for ₹ 720. If the cost per book was ₹ 5 less, the number of books that could be bought for ₹ 720 would be 2 more. Taking the original cost of each book to be ₹ x, write an equation in x and solve it.

Sol: Price of books = ₹720

Let cost of one book = ₹x
∴ Number of books purchased = 720/x

In second case, the price of each book = (x − 5)
∴ Number of books = 720/(x − 5)

According to the condition,
720/(x − 5) − 720/x = 2
(720x − 720(x − 5)) / x(x − 5) = 2
3600 / x(x − 5) = 2
3600 = 2x(x − 5)

2x² − 10x − 3600 = 0
x² − 5x − 1800 = 0
x² − 45x + 40x − 1800 = 0
x(x − 45) + 40(x − 45) = 0
(x − 45)(x + 40) = 0

Either x = 45 or x = −40

Since, x = −40 is not possible being negative
∴ x = 45

∴ Number of books originally purchased = 720/45 = 16

Que-12: Solve the following quadratic equation for x and give your answer correct to 2 decimal places : x² – 3x – 9 = 0

Sol: x² − 3x − 9 = 0
Here a = 1, b = −3, c = −9

D = b² − 4ac = (−3)² − 4 × 1 × (−9) = 9 + 36 = 45

x = (−b ± √(b² − 4ac)) / (2 × 1)
= (−(−3) ± √45) / 2
= (3 ± 3√5) / 2
= (3 ± 6.71) / 2

x₁ = (3 + 6.71) / 2 = 9.71 / 2 = 4.86
x₂ = (3 − 6.71) / 2 = −3.71 / 2 = −1.86

∴ x = 4.86, −1.86

Que-13: Five years ago, a woman’s age was the square of her son’s age. Ten years hence her age will be twice that of her son’s age. Find :
(i) The age of her son five years ago.
(ii) The present age of the woman.

Sol: 5 years ago,
Let the age of son = x years

Then age of his mother = x² years
Present age of son = (x + 5) years
and age of mother = (x² + 5) years

10 years hence,
Age of son = x + 5 + 10 = x + 15
and age of mother = x² + 5 + 10 = x² + 15
According to the condition,
x² + 15 = 2 (x + 15) ⇒ x² + 15 = 2x + 30
⇒ x² + 15 – 2x – 30 = 0 ⇒ x² – 2x – 15 = 0
⇒ x² – 5x + 3x – 15 = 0
⇒ x (x – 5) + 3 (x – 5) = 0
⇒ (x – 5) (x + 3) = 0

Either x – 5 = 0, then x = 5
or x + 3 = 0, then x = – 3
but it is not possible being negative

(i) ∴ Age of son 5 years ago = 5 years
(ii) Present age of woman = x² + 5 = (5)² + 5 = 25 + 5 = 30 years

Que-14: Solve the following quadratic equation for x and give your answer correct to two decimal places : 5x (x + 2) = 3

Sol: 5x(x + 2) = 3
5x² + 10x = 3
5x² + 10x − 3 = 0

Here a = 5, b = 10, c = −3

D = b² − 4ac = 10² − 4 × 5 × (−3) = 100 + 60 = 160

x = (−b ± √D) / 2a
= (−10 ± √160) / (2 × 5)
= (−10 ± 12.65) / 10

x₁ = (−10 + 12.65) / 10 = 2.65 / 10 = 0.265 = 0.26
x₂ = (−10 − 12.65) / 10 = −22.65 / 10 = −2.265 = −2.26

∴ x = 0.26, −2.26

Que-15: Some students planned a picnic. The budget for the food was ₹ 480. As eight of them failed to join the party the cost of the food for each member increased by ₹ 10. Find how many students went for the picnic?

Sol:  Budget for food = ₹480

Let number of students who went to picnic = x
∴ Each share = ₹480/x

Number of students who did not go = 8
Remaining students = x − 8
and then each share = ₹480/(x − 8)

Now according to the condition,
480/(x − 8) − 480/x = 10
(480x − 480(x − 8)) / x(x − 8) = 10
3840 / (x² − 8x) = 10

10x² − 80x = 3840
10x² − 80x − 3840 = 0

x² − 8x − 384 = 0
x² − 24x + 16x − 384 = 0
x(x − 24) + 16(x − 24) = 0

(x − 24)(x + 16) = 0

Either x − 24 = 0, then x = 24
or x + 16 = 0, then x = −16 which is not possible being negative

∴ Number of students who went for the picnic = 24

Que-16: Solve the following quadratic equation and give the answer correct to two significant figures 4x² – 7x + 2 = 0.

Sol: 4x² − 7x + 2 = 0
x = (7 ± √(49 − 32)) / 8
x = (7 ± √17) / 8
x = (7 ± 4.12) / 8

Taking (+) sign,
x = 11.12 / 8 = 1.4

Taking (−) sign,
x = 2.88 / 8 = 0.36

∴ x = 1.4, 0.36

Que-17: The speed of an express train is x km/h and the speed of an ordinary train is 12 km/h less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240 km, find the speed of the express train.

Sol: 240/(x – 12) − 240/x = 1
240[(1/(x – 12)) − (1/x)] = 1
240[(x – (x – 12))/x(x – 12)] = 1
240[12/(x(x – 12))] = 1

2880/(x² – 12x) = 1

x² − 12x = 2880
x² − 12x − 2880 = 0
x² − 60x + 48x − 2880 = 0
x(x − 60) + 48(x − 60) = 0

(x − 60)(x + 48) = 0

Either x − 60 = 0, then x = 60
or x + 48 = 0, then x = −48 which is not possible being negative

∴ Speed of the express train = 60 km/h

Que-18: Without solving the following quadratic equation, find the value of ‘p’ for which the roots are equal. px² – 4x + 3 = 0.

Sol: px² − 4x + 3 = 0 …(i)

Compare (i) with ax² + bx + c = 0
Here a = p, b = −4, c = 3

∴ D = b² − 4ac = (−4)² − 4 × p × 3 = 16 − 12p

As roots are equal,
∴ D = 0

16 − 12p = 0
16 = 12p

p = 16/12

∴ p = 4/3

Que-19: Solve the following equation : x – 18/x = 6. Give your answer correct to two significant figures.

Sol: x − 18/x = 6

x² − 6x − 18 = 0
a = 1, b = −6, c = −18

x = (−b ± √(b² − 4ac)) / 2a = (6 ± √(36 + 72)) / 2
= (6 ± √108) / 2
= (6 ± 6√3) / 2
= 3(1 + 1.73) or 3(1 − 1.73)
= 3 × 2.73 or 3 × (−0.73)

= 8.19 or −2.19

∴ x = 8.19, −2.19

Que-20: ₹ 480 is divided equally among ‘x’ children. If the number of children were 20 more then each would have got ₹ 12 less. Find ‘x’.

Sol: Share of each child = ₹480/x

According to the question :
480/(x + 20) = 480/x − 12
480/(x + 20) = (480 − 12x)/x
480/(x + 20) = 12(40 − x)/x
480x = 12(x + 20)(40 − x)

(x + 20)(40 − x) = 40x

40x − x² + 800 − 20x = 40x
x² + 20x − 800 = 0
x² + 40x − 20x − 800 = 0
x(x + 40) − 20(x + 40) = 0

(x + 40)(x − 20) = 0

x = −40 or x = 20

Negative value of x is not possible.
∴ Number of children = 20

Que-21: Without solving the following quadratic equation, find the value of ‘m’ for which the given equation has real and equal roots.
x² + 2 (m – 1) x + (m + 5) = 0.

Sol: Here, a = 1, b = 2(m − 1), c = m + 5

Discriminant, D = b² − 4ac
= 4(m − 1)² − 4(1)(m + 5)
= 4(m² + 1 − 2m) − 4(m + 5)
= 4m² + 4 − 8m − 4m − 20
= 4m² − 12m − 16

For real and equal roots, D = 0
4m² − 12m − 16 = 0
m² − 3m − 4 = 0
m² − 4m + m − 4 = 0
m(m − 4) + 1(m − 4) = 0

(m − 4)(m + 1) = 0

∴ m = 4 or m = −1

Que-22: A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/ h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.

Sol: Let the original speed of the car be x km/h.
Time taken to cover 400 km = 400/x h …(i)

New speed = (x + 12) km/h
New time taken to cover 400 km = 400/(x + 12) h …(ii)

Time taken for journey would have been 1 hour 40 minutes less.

1 hour 40 minutes = 1 + 40/60 = 5/3 hours

∴ From (i) and (ii),

(400/x) − 400/(x + 12) = 5/3
(400(x + 12) − 400x)/x(x + 12) = 5/3

⇒400(x + 12 − x)/(x² + 12x) = 5/3

1200x + 14400 = 5x² + 60x
5x² − 1140x − 14400 = 0
x² − 228x − 2880 = 0
x² + 60x − 48x − 2880 = 0
x(x + 60) − 48(x + 60) = 0

(x + 60)(x − 48) = 0
x = 48 or x = −60

x = 48 (Rejecting x = −60, being speed)

Hence, original speed of the car = 48 km/h.

Que-23: (i) Solve the following equation and calculate the answer correct to two decimal places: x² – 5x – 10 = 0
(ii) Without solving the following quadratic equation, find the value of ‘p’ for which the given equation has real and equal roots x2 + (p – 3)x + p = 0.

Sol: (i) x² − 5x − 10 = 0
Here, a = 1, b = −5, c = −10

x = (−b ± √(b² − 4ac)) / 2a = (5 ± √65) / 2
= (5 ± 8.06) / 2

x₁ = (5 + 8.06) / 2 = 13.06 / 2 = 6.53
x₂ = (5 − 8.06) / 2 = −3.06 / 2 = −1.53

∴ x = 6.53, −1.53

(ii) x² + (p − 3)x + p = 0

Equation has real and equal roots.

∴ b² − 4ac = 0
(p − 3)² − 4(1)(p) = 0

p² + 9 − 6p − 4p = 0
p² − 10p + 9 = 0
p² − 9p − p + 9 = 0

p(p − 9) − 1(p − 9) = 0
(p − 1)(p − 9) = 0

∴ p = 1, 9

Que-24: A shopkeeper purchases a certain number of books for ₹ 960. If the cost per book was ₹ 8 less, the number of books that could be purchased for ₹ 960 would be 4 more. Write an equation, taking the original cost of each book to be ₹ x, and solve it to find the original cost of the books.

Sol: Let original cost = ₹ x

No. of books bought = 960/x

New cost of books = ₹ (x − 8)
∴ No. of books bought = 960/(x − 8)

∴ According to condition,
960/(x − 8) − 960/x = 4
960[(1/(x − 8)) − (1/x)] = 4
(x − (x − 8)) / x(x − 8) = 4/960
(x − x + 8) / (x² − 8x) = 4/960
8/(x² − 8x) = 1/240

x² − 8x = 8 × 240
x² − 8x − 1920 = 0

Now, x = (−b ± √(b² − 4ac)) / 2a
x = (−(−8) ± √((−8)² − 4(1)(−1920))) / 2
= (8 ± √(64 + 7680)) / 2
= (8 ± √7744) / 2
= (8 ± 88) / 2

= 96/2 , −80/2

= 48, −40 (rejecting)
∴ No. of books = 48

Que-25: Solve for x using the quadratic formula. Write your answer correct to two significant figures. (x – 1)² – 3x + 4 = 0.

Sol: (x − 1)² − 3x + 4 = 0
x² + 1 − 2x − 3x + 4 = 0
x² − 5x + 5 = 0

Now, x = (−b ± √(b² − 4ac)) / 2a
Here, a = 1, b = −5, c = 5

x = (−(−5) ± √((−5)² − 4(1)(5))) / 2
= (5 ± √(25 − 20)) / 2
= (5 ± √5) / 2
= (5 ± 2.236) / 2
= 7.236/2   or   2.764/2
= 3.618, 1.382

∴ x = 3.62, 1.38

Que-26: A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number.

Sol: Let 2-digit number = xy = 10x + y
Reversed digits = yx = 10y + x

Acc. to question xy = 6
y = 6/x
and 10x + y + 9 = 10y + x

10x + 6/x + 9 = 10 x 6/x + x
10x² + 6 + 9x = 60 + x²
10x² – x² + 9x + 6 – 60 = 0
9x² + 9x – 54 = 0 ⇒ x² + x – 6 = 0 ⇒ x² + 3x – 2x – 6 = 0
x(x + 3) – 2(x + 3) = 0

⇒ (x – 2) (x + 3) = 0
⇒ x = 2 or – 3 (rejecting -3)

putting the value of x in (i)
y = 6/2 = 3
∴ 2-digit = 10x + y = 10 x 2 + 3 = 23

Que-27: Find the value of ‘k’ for which x = 3 is a solution of the quadratic equation, (k + 2)x² – kx + 6 = 0.
Thus find the other root of the equation.

Sol: (k + 2)x² – kx + 6 = 0 … (i)

Substitute x = 3 in equation (1)
(k + 2) (3)² – k(3) + 6 = 0
⇒ 9(k + 2) – 3k + 6 = 0
⇒ 9k + 18 – 3k + 6 = 0
⇒ 6k + 24 = 0
⇒ 6k = – 24
⇒ k = −24/6 = 4

∴ k = – 4

Now, substituting = – 4 in equation (i), we get,
(- 4 + 2)x² – (- 4)x + 6 = 0
⇒ – 2x² + 4x + 6 = 0
⇒ x² – 2x – 3 = 0 (Dividingby2)
⇒ x² – 3x + x – 3 = 0
⇒ x(x – 3) + 1(x – 3) = 0
⇒ (x + 1) (x – 3) = 0

So, the roots are x = -1 and x = 3
Thus, the other root of the equation is x = – 1

Que-28: Sum of two natural numbers is 8 and the difference of their reciprocal is 2/15. Find the numbers.

Sol: Let x and y be two numbers Given that
x + y = 8 … (i)
and 1/x – 1/y = /215 … (ii)

From equation (i), we have, y = 8 – x

Substituting the value of y in equation (ii), we have,
1/x−1/8−x=2/15
⇒ 8−x−x/x(8−x)=2/15
⇒ 8−2x/x(8−x)=2/15
⇒ 4−x/x(8−x)=1/15
⇒ 15(4 – x) = x(8 – x)

⇒ 60 – 15x = 8x – x²
⇒ x² – 15x – 8x + 60 = 0
⇒ x² – 23x + 60 = 0
⇒ x² – 20x – 3x + 60 = 0
⇒ x(x – 20) – 3(x – 20) = 0
⇒ (x – 3) (x – 20) = 0
⇒ (x – 3) = 0 or (x – 20) = 0
⇒ x = 3 or x = 20

Since sum of two natural numbers is 8 : x cannot be equal to 20
Thus x = 3
From equation (1), y = 8 – x = 8 -3 = 5
Thus the values of x and y are 3 and 5 respectively.

Que-29: Solve the quadratic equation x² – 3(x + 3) = 0; Give your answer correct two significant figures.

Sol: x² − 3(x + 3) = 0
x² − 3x − 9 = 0
a = 1, b = −3, c = −9

x = (−b ± √(b² − 4ac)) / 2a

x = (−(−3) ± √((−3)² − 4(1)(−9))) / 2(1)
x = (3 ± √(9 + 36)) / 2
x = (3 ± √45) / 2
x = (3 ± √(9 × 5)) / 2

x = (3 + 3√5) / 2   or   x = (3 − 3√5) / 2
x = (3 + 3(2.236)) / 2   or   x = (3 − 3(2.236)) / 2
x = (3 + 6.708) / 2   or   x = (3 − 6.708) / 2

x = 9.708 / 2   or   x = −3.708 / 2

x = 4.854   or   −1.854
x = 4.85   or   x = −1.85

x = 4.9   or   x = −1.9

Que-30: A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to covers the total distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate ‘x’.

Sol: Time taken by the bus with moving at speed x km/h = 240/x
Time taken by the bus with moving at speed (x − 10) km/h = 240/(x − 10)

According to the given condition,
2 = 240/(x − 10) − 240/x
2 = 240 (1/(x − 10) − 1/x)
1/120 = 1/(x − 10) − 1/x
1/120 = (x − x + 10) / x(x − 10)
1/120 = 10 / x(x − 10)
x(x − 10) = 10 × 120

x² − 10x = 1200
x² − 10x − 1200 = 0
x² − 40x + 30x − 1200 = 0

x(x − 40) + 30(x − 40) = 0
(x − 40)(x + 30) = 0

x − 40 = 0   or   x + 30 = 0
x = 40   or   x = −30

Since, the speed cannot be negative, the uniform speed is 40 km/h.

Very Short Answer Type Questions (VSA)

Que-1: If (x + 5)(x − 3) = 0, then x = __

Sol: Using the zero product property:

x + 5 = 0    or    x − 3 = 0
x = −5    or    x = 3

Answer: x = −5, 3

Que-2: Solve : 5y² + y = 0

Sol: 5y² + y = 0

Take common factor y:
y(5y + 1) = 0
y = 0    or    5y + 1 = 0

y = −1/5

Answer: y = 0, −1/5

Que-3: Solve : x² − 9 = 0

Sol: x² − 9 = 0
x² – 3² = 0
(x-3)(x+3) = 0
x = ±3

Answer: x = 3, −3

Que-4: Solve by using the quadratic formula: −x² + 7x − 10 = 0

Sol: Given:
a = −1, b = 7, c = −10

Quadratic Formula:
x = (−b ± √(b² − 4ac)) / 2a

Substituting values:
x = (−7 ± √(7² − 4(−1)(−10))) / 2(−1)
x = (−7 ± √(49 − 40)) / −2
x = (−7 ± 3) / −2

x = −4 / −2 = 2
x = −10 / −2 = 5

Answer: x = 2, 5

Que-5: Solve : x + 9/x = 6

Sol: Multiply both sides by x:
x² + 9 = 6x
x² − 6x + 9 = 0
(x − 3)² = 0

x = 3

Answer: x = 3

Que-6: What are the values of a, b and c for the equation 4x² = 8x − 3 in the quadratic formula?

Sol: Write in standard form:
4x² − 8x + 3 = 0

Comparing with ax² + bx + c = 0:
a = 4, b = −8, c = 3

Answer: a = 4, b = −8, c = 3

Que-7: The discriminant for the quadratic equation ax² + bx + c = 0, a ≠ 0 is __

Sol: D = b² − 4ac

Answer: b² − 4ac

Que-8: Comment upon the roots of a quadratic equation ax² + bx + c = 0 if the discriminant is positive and is not a perfect square.

Sol: If D = b² − 4ac > 0 and D is not a perfect square, then the roots are:

• Real
• Distinct (unequal)
• Irrational

Answer: The roots are real, distinct and irrational.

Que-9: Find the value of k so that the equation x² − 4x + k = 0 has real and equal roots.

Sol: For real and equal roots:
D = 0
b²-4ac = 0
(−4)² − 4(1)(k) = 0
16 − 4k = 0
4k = 16
k = 4

Answer: k = 4

Que-10: Find the discriminant of the quadratic equation 3√3x² + 10x + √3 = 0

Sol: Given:
a = 3√3, b = 10, c = √3

D = b² − 4ac
D = 10² − 4(3√3)(√3)
D = 100 − 36

D = 64

Answer: D = 64

Multiple Choice Questions (MCQs)

1. If the quadratic equation mx² + 2x + m = 0 has two equal roots,
then find the value of m

(a) +1
(b) 0, 2
(c) 0, 1
(d) −1, 0

Sol: (a) +1 
Given:
mx² + 2x + m = 0

For equal roots, the discriminant must be zero.
D = b² − 4ac

Here, a = m, b = 2, c = m

D = (2)² − 4(m)(m)
= 4 − 4m²

For equal roots:
4 − 4m² = 0
1 − m² = 0
m² = 1
m = ±1

Answer: m = ±1

2. The roots of the quadratic equation
2x² − x − 6 = 0 are

(a) −2, 3/2
(b) 2, -3/2
(c) −2, -3/2
(d) 2, 3/2

Sol: (b) 2, −3/2
Given:
2x² − x − 6 = 0

Using the quadratic formula:
x = [−b ± √(b² − 4ac)] / 2a

Here, a = 2, b = −1, c = −6
x = [−(−1) ± √{(−1)² − 4(2)(−6)}] / 2(2)
= [1 ± √(1 + 48)] / 4
= [1 ± √49] / 4
= (1 ± 7) / 4

x = (1 + 7) / 4 = 2
x = (1 − 7) / 4 = −3/2

Roots: 2, −3/2

Answer: (b) 2, −3/2

3. Which of the following equations has 2 as a root.

(a) x² − 4x + 5 = 0
(b) x² + 3x − 12 = 0
(c) 2x² − 7x + 6 = 0
(d) 3x² − 6x − 2 = 0

Sol: (c) 2x² − 7x + 6 = 0
To check whether 2 is a root, substitute x = 2 in each equation.

(a) x² − 4x + 5 = 0
= (2)² − 4(2) + 5
= 4 − 8 + 5
= 1 ≠ 0

So, 2 is not a root.

(b) x² + 3x − 12 = 0
= (2)² + 3(2) − 12
= 4 + 6 − 12
= −2 ≠ 0

So, 2 is not a root.

(c) 2x² − 7x + 6 = 0
= 2(2)² − 7(2) + 6
= 8 − 14 + 6
= 0

So, 2 is a root.

(d) 3x² − 6x − 2 = 0
= 3(2)² − 6(2) − 2
= 12 − 12 − 2
= −2 ≠ 0

So, 2 is not a root.

Answer: (c) 2x² − 7x + 6 = 0

4. If ¹⁄₂ is a root of the equation x² + kx − ⁵⁄₄ = 0, then the value of k is

(a) 2
(b) −2
(c) ¹⁄₄
(d) ¹⁄₂

Sol: (a) 2
Given:
x² + kx − 5/4 = 0

Since 1/2 is a root, substitute x = 1/2 in the equation.

(1/2)² + k(1/2) − 5/4 = 0
1/4 + k/2 − 5/4 = 0
k/2 − 1 = 0
k/2 = 1
k = 2

Answer: (a) 2

5. The quadratic equation 2x² − √5x + 1 = 0 has

(a) two distinct real roots
(b) two equal real roots
(c) no real roots
(d) more than 2 real roots

Sol: (c) no real roots
Given:
2x² − √5x + 1 = 0

Here, a = 2, b = −√5, c = 1

Discriminant, D = b² − 4ac
D = (−√5)² − 4(2)(1)
= 5 − 8
= −3

Since D < 0, the equation has no real roots.

Answer: (c) no real roots

6. The sum of a number as its square is 20, then, the number is

(a) −5, or 4
(b) 2 or 3
(c) −5 only
(d) 5 or −4

Sol: (a) −5, or 4
Let the number be x.

According to the question,
x + x² = 20
x² + x − 20 = 0

Factoring,
(x + 5)(x − 4) = 0
x + 5 = 0   or   x − 4 = 0
x = −5   or   x = 4

Answer: (a) −5, or 4

7. The roots of the quadratic equation x² − 2√3x − 22 = 0 are

(a) non-real
(b) real, rational and equal
(c) real, irrational and unequal
(d) real, rational and unequal

Sol: (c) real, irrational and unequal
Given:
x² − 2√3x − 22 = 0

Here, a = 1, b = −2√3, c = −22

Discriminant, D = b² − 4ac
D = (−2√3)² − 4(1)(−22)
= 12 + 88
= 100

Since D > 0, the roots are real and unequal.

Also, √D = √100 = 10, which is rational.

x = [−b ± √D] / 2a
= [2√3 ± 10] / 2
= √3 ± 5

Since √3 is irrational, both roots are irrational.

Answer: (c) real, irrational and unequal

8. The value of a for which the equation 2x² + 2√6x + a = 0 has equal roots, is

(a) 3
(b) 4
(c) 2
(d) √3

Sol: (a) 3
Given:
2x² + 2√6x + a = 0

For equal roots, the discriminant must be zero.
D = b² − 4ac

Here, a = 2, b = 2√6, c = a

D = (2√6)² − 4(2)(a)
= 24 − 8a

For equal roots:
24 − 8a = 0
8a = 24
a = 3

Answer: (a) 3

9. If one root of equation x² + ax + 12 = 0 is 4 while the equation x² + ax + b = 0 has equal roots, then the value of b is

(a) 4/49
(b) 49/4
(c) 7/4
(d) 4/7

Sol: (b) 49/4
Given:
x² + ax + 12 = 0 has 4 as a root.

Substituting x = 4,
4² + 4a + 12 = 0
16 + 4a + 12 = 0
28 + 4a = 0
a = −7

Now consider the equation:
x² + ax + b = 0
x² − 7x + b = 0

Since it has equal roots, its discriminant is zero.
D = b² − 4ac
= (−7)² − 4(1)(b)
49 − 4b = 0
4b = 49
b = 49/4

Answer: (b) 49/4

10. Value(s) of k for which the quadratic equation 2x² − kx + k = 0 has equal roots is/are

(a) 0
(b) 4
(c) 8
(d) 0, 8

Sol: (d) 0, 8
Given:
2x² − kx + k = 0

For equal roots, the discriminant must be zero.
D = b² − 4ac

Here, a = 2, b = −k, c = k

D = (−k)² − 4(2)(k)
= k² − 8k

For equal roots:
k² − 8k = 0
k(k − 8) = 0
k = 0 or k = 8

Answer: (d) 0, 8

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Quadratic Equations ICSE Class 10 Maths Notes OP Malhotra 2026-27

What is  a quadratic equation ?

A quadratic equation in one variable is an equation in which the highest power of the variable is two.

The standard form of a quadratic equation is ax²+bx+c = 0 ; a,b,c∈R , a≠0
Thus, the equation 3x²+2x-1 = 0 is in standard form, here a=3 , b=3 , c=-1

The value of x satisfying the equation are called the zeroes or roots of the equation.
Thus a real number α is said to be root of the equation if aα²+bα+c=0

How to solve quadratic equation by factorising ?

Example,

Solve the equation (9/2)x = 5 + x²  by factorization:

Step 1: Clear all fractions and brackets, if necessary

9x = 2(5 + x²)

Step 2: Transpose all the terms to the left hand side to get an equation in the form
ax² + bx + c = 0

9x = 2x² + 10
⇒ 2x² − 9x + 10 = 0

Step 3: Factorise the expression on the left hand side.

2x² − 9x + 10 = 0
⇒ 2x² − 5x − 4x + 10 = 0
⇒ x(2x − 5) − 2(2x − 5) = 0
⇒ (x − 2)(2x − 5) = 0

Step 4: Put each factor equal to zero and solve

(x − 2)(2x − 5) = 0

⇒ x − 2 = 0       2x − 5 = 0

⇒ x = 2;          2x = 5

⇒ x = 2;          x = ⁵/₂

Thus, we have, x = 2 or x = ⁵/₂

Practice Questions on Factorisation :- Exercise-5(a)

Equations Reducible to Quadratic Equations

Many equations that are not in the standard form ax²+bx+c = 0 can be reduced to quadratic equation by using suitable algebraic transformation. Here you will learn reduce the non standard form to standard form of quadratic equation with some examples.

Method

1. Make a suitable substitution.
2. Reduce the equation to a quadratic form.
3. Solve for the new variable.
4. Substitute back to get the value of x.

Example 1 :- Solve: x⁴ − 5x² + 4 = 0

Sol: Let x² = t

Then t² − 5t + 4 = 0
(t − 1)(t − 4) = 0

t = 1, 4

Substituting back:

x² = 1 ⇒ x = ±1

x² = 4 ⇒ x = ±2

Ans :- x = ±1, ±2

Example 2 :- Solve: x² + 1/x² = 14

Sol: Using the identity
(x + 1/x)² = x² + 1/x² + 2

(x + 1/x)² = 14 + 2 = 16

Let x + 1/x = t

Then
t² = 16

t = ±4

When t = 4
x + 1/x = 4

Multiplying by x,
x² − 4x + 1 = 0

x = 2 ± √3

When t = -4
x + 1/x = -4

Multiplying by x,
x² + 4x + 1 = 0

x = -2 ± √3

Ans :- x = 2 + √3, 2 − √3, -2 + √3, -2 − √3

Example 3 :- Solve: x² + 1/x² = 7

Sol: Using the identity
(x − 1/x)² = x² + 1/x² − 2

(x − 1/x)² = 7 − 2 = 5

Let x − 1/x = t

Then
t² = 5

t = ±√5

When t = √5
x − 1/x = √5

Multiplying by x,
x² − √5x − 1 = 0

x = (√5 + 3)/2, (√5 − 3)/2

When t = -√5
x − 1/x = -√5

Multiplying by x,
x² + √5x − 1 = 0

x = (-√5 + 3)/2, (-√5 − 3)/2

Ans :- x = (√5 + 3)/2,
(√5 − 3)/2,
(-√5 + 3)/2,
(-√5 − 3)/2

Example 4 :- Solve: 4^x − 3 × 2^x+2 + 32 = 0

Sol: Let 4^x = (2²)^x = 2^2x
and 3 × 2^x+2 = 3 × 4 × 2^x = 12 × 2^x (as 2^x+2 = 2^x.2²)

Therefore,
2^2x − 12 × 2^x + 32 = 0

Substitute
2^x = t

Then
t² − 12t + 32 = 0

Solve the quadratic equation
t² − 12t + 32 = 0

(t − 4)(t − 8) = 0

t = 4 or t = 8

Substitute back
t = 2^x

Case 1: 2^x = 4 = 2²

x = 2

Case 2: 2^x = 8 = 2³

x = 3

Ans :- x = 2 or x = 3

Important Identities

(x + 1/x)² = x² + 1/x² + 2
(x − 1/x)² = x² + 1/x² − 2

Practice Questions on Equations reducible to quadratic :- Exercise-5(b)

Finding Roots of the Quadratic using Quadratic Formula

general quadratic equation – ax²+bx+c = 0

Quadratic Formula ⇒ x = {-b±√(b²-4ac)}/2a

where,
a is coefficient of x²
b is coefficient of x
c is constant

Practice Questions on Solving quadratic using quadratic formula :- Exercise-5(c)

Nature of the roots of a quadratic equation

The nature of the roots of a quadratic equation depends upon the value of discriminant b² − 4ac.

1. If b² − 4ac > 0, the roots are real and unequal.
2. If b² − 4ac = 0, the roots are real and equal.
3. If b² − 4ac < 0, the roots are imaginary (not real).

If ax² + bx + c, a ≠ 0, can be reduced to the product of two linear factors, then the roots of the quadratic equation ax² + bx + c = 0 can be found by equating each factor to zero.

Practice Questions on Nature of Roots :- Exercise-5(d)

Word Problems involving Quadratic Equations

Steps to Solve Word Problems

1. Read the question carefully.
2. Let the unknown quantity be x.
3. Form an equation according to the given conditions.
4. Simplify it into the standard form: ax² + bx + c = 0.
5. Solve the quadratic equation by factorization, completing the square, or quadratic formula.
6. Verify the answer and reject any impossible value (e.g., negative age, negative length).

Common Types of Word Problems

1. Number Problems

• Let the number be x.
• Represent consecutive numbers as x + 1, x + 2, etc.
• Form an equation using the given condition.

2. Age Problems

• Present age = x
• Age after n years = x + n
• Age n years ago = x − n

3. Geometry Problems

• Length = x
• Breadth = x ± k
• Use formulas for area or perimeter to form a quadratic equation.

4. Speed, Distance and Time Problems

• Distance = Speed × Time
• Express speed or time in terms of x and form the equation.

5. Area and Dimension Problems

• Use area formulas of rectangles, squares, etc.
• Convert the given condition into a quadratic equation.

6. Product and Sum Conditions

• If two numbers differ by a known value, let them be x and x + k.
• Use the given sum/product relationship to form the equation.

Word Problems on Quadratic Equations :- Exercise-5(e)

In this chapter, we study all the topics on Quadratic Equations and do some practice questions also. Here we solve extra practice questions on this chapter for better understanding.

Here is the link for extra practice questions on Quadratic Equations :- Self Evaluation

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OP Malhotra Notes on Circle Class-9 S.Chand ICSE Maths 2026-27

Circle

A circle is a locus of a point which moves in such way that its distance from a fixed point is always constant. The fixed point is called the centre of the circle.
The line segment joining any two points on a circle is called a chord of a circle.

Chord of Properties

1. A straight line drawn from the centre of a circle to bisect a chord which is not a diameter is at right angles to chord.
2. Perpendicular drawn to a chord from the centre of a circle bisects the chord.
3. Equal chords of a circle are equidistant from centre.
4. Chords which are equidistant from the centre are equal in lengths.
5. There is one and only circle which passes though three given points not in a straight line.
6. The perpendicular bisector of a chord of a circle always passes through its centre
7. Perpendicular bisectors of two chords of a circle intersect at its centre

Practice Questions on Chord Properties :- Exercise-13(a)

Arc Definition

An arc is a part of the circumference of a circle.
An arc less than one-half of the whole arc of a circle is called a minor arc of the circle, and an arc greater than one-half of the whole arc of a circle is called a major arc of a circle.

Segment Definition

A chord of a circle divides it into two parts. Each part is called segment.
The part containing the minor arc is called the minor segment, and the part containing the major arc is called the major segment.

Sector Definition

The region bounded by an arc and two radii, joining the centre to the end points of an arc, is called a sector

Practice Questions on Angles and Proving :- Exercise-13(b)

In this chapter, we study all the topics on Circle like Chord, Arc, Segment, Sector and Chord Properties and do some practice questions also. Here we solve extra practice questions on this chapter for better understanding.

For extra practice questions on Circle :- Chapter Test

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OP Malhotra Coordinates and Graphs of Simultaneous Linear Equations Class-9 S.Chand ICSE Maths Ch-20

Linear Equations :

The equation of a straight line is the linear equation. It could be in one variable or two variables.

Linear Equation in One Variable

The equation with one variable in it is known as a Linear Equation in One Variable.
The general form is
px + q = s, where p, q and s are real numbers and p ≠ 0.
Example : x + 5 = 10
y – 3 = 19
These are called Linear Equations in One Variable because the highest degree of the variable is one.

Practice Questions on Linear Equation :- Exercise-20(a)

Graph of the Linear Equation in One Variable :

We can mark the point of the linear equation in one variable on the number line.

Solution of a Linear Equation :

• There is only one solution in the linear equation in one variable but there are infinitely many solutions in the linear equation in two variables.
• As there are two variables, the solution will be in the form of an ordered pair, i.e. (x, y).
• The pair which satisfies the equation is the solution of that particular equation.

Practice Questions on Graph of Linear Equation In One Variable :- Exercise-20(b)

How to solve a pair of simultaneous equations graphically?

In order to solve a pair of simultaneous equations graphically, we first draw the graph of the two equations simultaneously. We get two straight lines intersecting each other at a common point. This common point intersection of two lines gives the solution of the pair of simultaneous equations.

Step by Step Process: 

• Rearrange the equations: Express both equations in standard slope-intercept form (y = mx + c).
• Find coordinate points: Calculate at least two (x, y) pairs for each equation by plugging in random (x) values.
• Draw the grid: Set up an (x)-axis and (y)-axis on graph paper with an even, clear scale.
• Plot the lines: Mark the coordinate points for the first equation and connect them with a straight line, then repeat for the second equation.
• Identify the intersection: Locate the exact point where the two straight lines cross each other.
• Read the solution: Note the (x)-coordinate and (y)-coordinate of this intersection point (x, y).
• Verify your answer: Substitute these (x) and (y) values back into your original equations to ensure they work

Practice Questions on Solving Simultaneous Linear Equation :- Exercise-20(c)

Graphical representation of a linear equation in 2 variables :

• Any linear equation in the standard form ax+by+c=0 has a pair of solutions in the form (x,y), that can be represented in the coordinate plane.
• When an equation is represented graphically, it is a straight line that may or may not cut the coordinate axes.

Calculate Distance Between Two Graphs

We know that distance between two points P (x1 y1) and Q (x2, y2)
= √[(x2−x1)²+(y2−y1)²].

Practice Questions on Distance Between Two Graphs :- Exercise-20(d)

In this chapter, we study all the topics on Coordinates and Graphs of Simultaneous Linear Equations like Linear Equation In One Variable Graph, Distance Between Two Graphs, etc. and do some practice questions also. Here we solve extra practice questions on this chapter for better understanding.

For extra practice questions on Coordinates and Graphs :- Chapter Test

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OP Malhotra Notes on Trigonometrical Ratios Class-9 S.Chand ICSE Maths 2026-27

Trigonometric Ratios Definition :

It is defined as the values of all the trigonometric function based on the value of the ratio of sides in a right-angled triangle. The ratios of sides of a right-angled triangle with respect to any of its acute angles are known as the trigonometric ratios of that particular angle. Consider a right-angled triangle, right-angled

• sine: Sine of an angle is defined as the ratio of the side opposite(perpendicular side) to that angle to the hypotenuse.
• cosine: Cosine of an angle is defined as the ratio of the side adjacent to that angle to the hypotenuse.
• tangent: Tangent of an angle is defined as the ratio of the side opposite to that angle to the side adjacent to that angle.
• cosecant: Cosecant is a multiplicative inverse of sine.
• secant: Secant is a multiplicative inverse of cosine.
• cotangent: Cotangent is the multiplicative inverse of the tangent.

Practice Questions on Finding Trigonometrical ratios :- Exercise-19(a)

Finding Angles

sin θ = Perpendicular / Hypotenuse
cos θ = Base / Hypotenuse
tan θ = Perpendicular / Base
cot θ = Base / Perpendicular
sec θ = Hypotenuse / Base
cosec θ = Hypotenuse / Perpendicular

Practice Questions on Finding Angles :- Exercise-19(b)

Trigonometry Applications :

Trigonometry is one of the most important branches of mathematics. Some of the applications of trigonometry are:

• Measuring the heights of towers or big mountains
• Determining the distance of the shore from the sea
• Finding the distance between two celestial bodies
• Determining the power output of solar cell panels at different inclinations
• Representing different physical quantities such as mechanical waves, electromagnetic waves, etc.

It is evident from the above examples that trigonometry has its involvement in a major part of our day-to-day life and much more. In most of the applications listed above, something was being measured and that is what trigonometry is all about.

Practice Questions on Trigonometry Application :- Exercise-19(c)
and for problems on this topic is :- Exercise-19(d)

In this chapter, we study all the topics on Trigonometrical Ratios like Trigonometry Ratios, Finding Angles and Trigonometrical Applications and do some practice questions also. Here we solve extra practice questions on this chapter for better understanding.

For extra practice questions on Trigonometrical Ratios :- Chapter Test

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Self Evaluation & Revision on Shares and Dividends ICSE Class 10 Maths OP Malhotra 2026-27

Self Evaluation on Shares and Dividends

Self Evaluation & Revision on Shares and Dividends ICSE Class 10 Maths OP Malhotra 2026-27

Que-1: A dividend of 9% was declared on ₹ 100 shares selling at a certain price. If the rate of return is 7(1/2)%, calculate :
(i) the market value of the share
(ii) the amount to be invested to obtain an annual dividend of ₹ 630.

Sol: Market price of each share = x
Face value of each share = ₹ 100
Rate of dividend = 9%
Rate of return on investment = 7(1/2)% = 15/2%
(i) ∴ x×15/(100×2) = 9 ⇒ x = (9×100×2)/15 = ₹ 120
∴ Market value of each share = ₹ 120

(ii) Amount of dividend = ₹ 630
∴ Investment = (630×100)/(15/2) = (630×100×2)/15
= ₹ 42 x 2 x 100 = ₹ 8400

Que-2: A man invests ₹ 8800 in buying shares of face value of rupees hundred each at a premium of 10% in a company. If he earns ₹ 1200 at the end of the year as dividend, find
(i) the number of shares he has in the company ?
(ii) the dividend percentage per share.

Sol: Investment = ₹ 8800
Face value of each share = ₹ 1100
Market value at a premium of 10%
= ₹ 100 + 10 = ₹ 110
Total dividend he received = ₹ 1200
(i ) Number of shares = Investment/MV
= 8800/110 = 80

(ii) Face value of each share = ₹ 100 x 80 = ₹ 8000
∴ Rate of dividend per share = 1200/8000 x 100
= 15%

Que-3: A man wants to buy 62 shares available at ₹ 132 (par value of ₹ 100).
(i) How much should he invest ₹
(ii) If the dividend is 7.5%, what will be his annual income?
(iii) If he wants to increase income by ₹ 150, how many extra shares should he buy?

Sol: Number of shares = 62
Market value of each share = ₹ 132
Face value = ₹ 100
(i) His investment = ₹ 132 x 62 = ₹ 8184

(ii) Rate of dividend = 7.5% = 15/2 % p.a.
Annual income = ₹ 62 x 100 x 15/(2×100)
= ₹ 465

(iii) Extra income he wants = ₹ 150
Then annual income = ₹ 465 + 150 = ₹ 615
∴ Number of shares = (615×100×2)/(15×100) = 82
∴ Extra share he has to buy = 82 – 62 = 20

Que-4: A man invests ₹ 20,020 in buying shares of nominal value ₹ 26 at 10% premium. The dividend on the shares is 15% per annum. Calculate :
(i) The number of shares he buys.
(ii) The dividend he receives annually.
(iii) The rate of interest he gets on his money.

Sol: Investment = ₹ 20020
Nominal value of each share = ₹ 26
Market value at 10% premium
= ₹ 26×(100+10)/100
= ₹ 26×110/100 = ₹ 2860/100 = ₹ 28.60
Rate of dividend = 15%

(i) Number of share he bought = 20020/28.60
= 20020×100/2860 = 70

(ii) Total dividend per year = 700 x 26 x 15%
= 700×26×15/100 = ₹ 2730

(iii) Rate of interest on investment
= 2730×100/20020 = 13.636 %
= 13.46%

Que-5: A man invested ₹ 45,000 in 15% ₹ 100 shares quoted at ₹ 125. When the market value of these shares rose to ₹ 140, he sold some shares, just enough to raise ₹ 8400. Calculate :
(i) the number of shares he still holds;
(ii) the dividend due to him on these remaining shares.

Sol: Total investment = ₹ 45000
Face value of each share = ₹ 100
Market value = ₹ 125
Rate of dividend = 15%
∴ Number of shares = 4500/125
He sells some shares at the rate of ₹ 140
(i) Raise his income ₹ 8400
∴ Number of shares he sells = 8400/140 = 60
Remaining shares = 360 – 60 = 300
Dividend on remaining shares = 300 x 100 x 15%
= ₹300×100×15/100 = ₹ 4500

Que-6: Mr. Tewari invested ₹ 29,040 in 15%, ₹ 100 shares quoted at a premium of 20%. Calculate :
(i) The number of shares bought by Mr. Tewari.
(ii) Mr. Tewari’s income from the investment.
(iii) The percentage return on his investment.

Sol: Investment made by Tewari = ₹ 29040
Face value of each share = ₹ 100
Market value at a premium of 20%
= ₹ 100 + 20 = ₹ 120
Rate of dividend = 15%
(i) Number of shares bought = ₹ 29040/120 = 242
(ii) Income from investment = ₹ 242 x 100 x 15%
= ₹ 242 x 100 x 15/100 = ₹ 3630

(iii) Percentage income on investment
= ₹ 3630×100/29040 = 12.5%

Que-7: Mr. Ram Gopal invested ₹ 8000 in 7% ₹ 100 shares at ₹ 80. After a year he sold these shares at ₹ 75 each and invested the proceeds (including his dividend) in 18%, ₹ 25 shares at ₹ 41. Find :
(i) his dividend for the first year.
(ii) his annual income in the second year.
(iii) the percentage increase in his return on his original investment.

Sol: Investment made by Ram Gopal = ₹ 8000
Face value of each share = ₹ 100
Market value = ₹ 80
Rate of dividend = 7%
Number of shares = ₹ 8000/80 = 100
(i) Dividend for the first year = ₹ 100 x 100 x 7%
= 100×100×7/100 = ₹ 700

(ii) M.V. of second year = ₹ 75
∴ Sale proceed = ₹ 100 x 75 = ₹ 7500
Total investment including dividend = ₹ 7500 + 700 = ₹ 8200
Rate of dividend in second year = 18%
M.V. = ₹ 41
Face value = ₹ 25
∴ Number of shares bought = 8200/41 = 200
Nominal value of 200 share = ₹ 25 x 200 = ₹ 5000
∴ Dividend = ₹ 5000 x 18%
= ₹ 5000 x 18/100 = ₹ 900

(iii) Increase in income = ₹ 900 – ₹ 700 = ₹ 200
∴ Increase percent = 200×100/8000
= 5/2 % = 2.5%

Que-8: Ajay owns 560 shares of a company. The face value of each share is ₹ 25. The company declares a dividend of 9%. Calculate :
(i) The dividend that Ajay will get.
(ii) The rate of interest on his investment, if Ajay had paid ₹ 30 for each share.

Sol: Ajay has shares of a company = 560
Face value of each share = ₹ 25
Rate of dividend = 9%
(i) Face value of 560 shares = ₹ 25 x 560
= 114000
∴ Total dividend he received
= ₹ 14000 x 9%
= ₹ 14000 x 9/100 = ₹ 1260

(ii) M.V. of each share = ₹ 30
∴ Total investment = ₹ 30 x 560 = ₹ 16800
Rate of interest on his investment = 1260×100/16800 = 7.5 %

Que-9: A company with 4000 shares of nominal value of ₹ 110 each declares an annual dividend of 15%. Calculate:
(i) The total amount of dividend paid by the company.
(ii) The annual income of Shah Rukh who holds 88 shares in the company.
(iii) If he received only 10% on his investment, find the price Shah Rukh paid for each share.

Sol: Number of shares = 4000
Nominal value of each share = ₹ 110
Rate of dividend = 15%
(i) ∴ Total amount of dividend
= ₹ 4000 x 110 x 15%
= ₹ 4000×110×15/100
= ₹ 66000

(ii) Face of 88 shares = ₹ 110 x 88 = ₹ 9680
∴ Annual income of Shah Rukh
= ₹ 9680×15/100 = ₹ 1452

(iii) Interest on investment made by Shah Rukh = 10%
∴ Price (value) of each share paid by Shah

Que-10: Amit Kumar invests ₹ 36,000 in buying ₹ 100 shares at ₹ 10 premium. The dividend is 15% per annum. Find
(i) the number of shares he buys.
(ii) his yearly dividend.
(iii) the percentage return on his investment.
Give your answer correct to the nearest whole number.

Sol: Investment = ₹ 36000
Face value = ₹ 100
Premium = ₹ 20, dividend = 15%
(i) No. of shares = 36000/120 = 300

(ii) Dividend = 15% of (100 x 300)
= 15/100 x 30000 = ₹ 4500

(iii) Per cent of return on investment
= 45000/36000 x 100 = 450/36 = 12.5% = 13%

Que-11: Vivek invests ₹ 4,500 in 8% ₹ 10 shares at ₹ 15. He sells the shares when the price rises to ₹ 30, and invests the proceeds in 12% ₹ 100 shares at ₹ 125. Calculate.
(i) the sale proceeds,
(ii) the number of ₹ 125 shares he buys,
(iii) the change in his annual income from dividend.

Sol: (i) If price of share bought is ₹ 15, then face value of share = Rs. 10
If price of share bought is ₹ 4500, then face value of share bought
= 10/15 x 4500 = ₹ 3000
Total face value of ₹ 10 shares = ₹ 3000 Income = 8%
= 8/100 x 3000 = Rs. 240
By selling ₹ 10 share money received = ₹ 30
By selling Rs. 3000 shares money
= 30/10 x 3000 = ₹ 9000

(ii) By investing ₹ 125, no. of share of ₹ 100 bought = 1
By investing ₹ 9000, no. of share of ₹ 100 bought = 1/125 x 9000 = 72
∴ No. of ₹ 125 shares bought = 72

(iii) By investing ₹ 125 in ₹ 100 share, income = ₹ 12
By investing Rs. 9000 in ₹ 100 share, income 12
= 12/125 x 9000 = ₹ 864
Increase in income = ₹ 864 – ₹ 240 = ₹ 624

Que-12: Mr. Parekh invested ₹ 52,000 on ₹ 100 shares at a discount of ₹ 20 paying 8% dividend. At the end of one year he sells the shares at a premium of ₹ 20. Find :
(i) The annual dividend.
(ii) The profit earned including his dividend.

Sol: Investment = ₹ 52000
Face value of 1 share = ₹ 100
Market value of 1 share = ₹ 100 – 20 = ₹ 80
No. of shares = 52000/80 = 650

(i) Annual dividend = 8/100 x 650 x 100 = ₹ 5200
(ii) S.P. of 1 share = ₹ 100 + 20 = ₹ 120

S.P. of 650 shares = ₹ 120 x 650 = ₹ 78000
C.P. of 650 shares = ₹ 100 x 650 = ₹ 65000
Profit = S.P. – C.P.
= ₹ 78000 – ₹ 52000 = ₹ 26000
Profit including dividend = ₹ 26000 + ₹ 5200 = ₹ 31200

Que-13: A man invests ₹ 9600 on ₹ 100 shares at ₹ 80. If the company pays him 18% dividend, find :
(i) the number of shares he buys.
(ii) his total dividend.
(iii) his percentage return on the shares.

Sol: Amount of investment = ₹ 9600
Price of one share = ₹ 80
(i) ∴ No. of shares bought = ₹ 9600/80 = 120

(ii) Face value of 120 shares = ₹ 120 x 100 = ₹ 12000
Rate of dividend = 18%
Dividend = ₹ 12,000×18/100 = ₹ 2160

(iii) By investing ₹ 9600, returned obtained = ₹ 2160
So, percentage return = 2.160×100/9600 = 22.5%

Que-14: Salman buys 50 shares of face value ₹ 100 available at ₹ 7132.
(i) What is his investment?
(ii) If the dividend is 7.5%, what will be his annual income?
(iii) If he wants to increase his annual income by ₹ 150, how many extra shares should he buy?

Sol: F.V. = ₹ 100
(i) M.V. = ₹ 132, no. of shares = 50
Investment = no. of shares x MV.
= 50 x 132 = ₹ 6600

(ii) Income per share = 7.5% of F.V.
= (75/10×100) x 100 = ₹ 7.5
∴ Annual income = 7.5 x 50 = ₹ 375

(iii) New annual income = 375 + 150 = ₹ 525
∴ No. of shares = 525/7.5 = 70
∴ No. of extra share = 70 – 50 = 20

Que-15: Salman invests a sum of money in ₹ 50 shares, paying 15% dividend quoted at 20% premium. If his annual dividend is ₹ 600, calculate:
(i) the number of shares he bought.
(ii) his total investment.
(iii) the rate of return on his investment.

Sol: Nominal value = ₹ 50
Dividend on 1 share = 15/100 x ₹ 50 = ₹ 7.50
Total Dividend to Salman = ₹ 600
(i) No. of shares Salman bought = 600/7.50
= 600×100/750 = 80

(ii) Premium on 1 share = 50 x 20/100 = ₹ 10
Market value of 1 share = 50 + 10 = ₹ 60
Total investment for 80 shares = 80 x 60 = ₹ 4800

(iii) Rate of return = Total dividend/Total investment ×100
= 600/4800 x 100 = 12.5%

Que-16: Rohit invested ₹ 9,600 on ₹ 100 shares at ₹ 20 premium paying 8% dividend. Rohit sold the shares when the price rose to ₹ 160. He invested the proceeds (excluding dividend) in 10% ₹ 50 shares at ₹ 40. Find the:
(i) original number of shares.
(ii) sale proceeds.
(iii) new number of shares.
(iv) change in the two dividends.

Sol: (i) 100 shares at ₹ 20 premium means
Nominal value of the share is ₹ 100
and its marked value = 100 + 20 = ₹ 120
Money required to buy 1 share = ₹ 120
Number of shares
= Money invested/Market value of 1 share
= 9600/120 = 80 shares

(ii) Dividend on 1 share = 8% of N.V.
= 8% of 100 = 8
Total dividend on 80 shares = 80 x 8 = ₹ 640
Each share is sold at ₹ 160
∴ The sale proceeds = 80 x ₹ 160
= ₹ 12800

(iii) New investment = ₹ 12800
Divident=10%
Net value = 50
Market value = ₹ 40
∴ Number of shares = Investment/Market value
= 12800/40
= 340 shares

(iv) Now, dividend on 1 share = 10% of N.V.
= 10% of 50 = 5
∴ Dividend on 340 shares = 1600
Change in two dividends = ₹ 1600 – ₹ 640 = ₹ 960

Que-17: Ashok invested ₹ 26,400 on 12%, ₹ 25 shares of a company. If he receives a dividend of ₹ 2,475. Find the :
(i) number of shares he bought.
(ii) market value of each share.

Sol: Investment = ₹ 26400
Rate of divident = 12%
Divident = ₹ 2475
Face value of one share = ₹ 25
Total dividend = Number of shares x Rate of dividend x Face value of one share
2475 = Number of shares x 12/100 x 25
Number of shares = 2475/3 = 825
Market value of one share Investment
= Investment/Number of shares bought
Market value of one share = 26400/825 = ₹ 32
Ashok bought 825 shares and market value of each share is ₹ 32.

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OP Malhotra Notes on Surface Area and Volume of 3D Solids Class-9 S.Chand ICSE Maths 2026-27

Cuboid :

A cuboid is a three dimensional Shape. The cuboid is made from six rectangular faces, which are placed at right angles. The total surface area of a cuboid is equal to the sum of the areas of its six rectangular faces.

Lateral Surface Area of a Cuboid :

Lateral surface area (LSA) is the area of all the sides apart from the top and bottom faces.
The lateral surface area of the cuboid
= Area of face AEHD + Area of face BFGC + Area of face ABFE + Area of face DHGC
$=2(b\times h)+2(l\times h)$
LSA (cuboid) = 2h(l + b)

Volume of a Cuboid :

The volume of a cuboid is the product of its dimensions.
Volume of a cuboid = length × breadth × height lbh
Where l is the length of the cuboid, b is the breadth, and h is the height of the cuboid.

Cube :

A cuboid whose length, breadth and height are all equal, is called a cube. It is a three-dimensional shape bounded by six equal squares. It has 12 edges and 8 vertices.

Lateral Surface area of a cube :

Lateral surface area (LSA) is the area of all the sides apart from the top and bottom faces.
>Lateral surface area of cube = $2(a\times a+a\times a)=4{\mathrm{a²}}^{}$

Volume of a Cube :

The volume of a cube = base × area × height
Since all dimensions are identical, the volume of the cube $={\mathrm{a³}}^{}$
Where a is the length of the edge of the cube.

Practice Questions on Cube and Cuboid :- Exercise-18(a)

Right Circular Cylinder :

A right circular cylinder is a closed solid that has two parallel circular bases connected by a curved surface in which the two bases are exactly over each other and the axis is at right angles to the base.

The curved Surface area of a right circular cylinder :

Take a cylinder of base radius r and height h units. The curved surface of this cylinder, if opened along the diameter (d = 2r) of the circular base will be transformed into a rectangle of length $2\pi r$ and height h units.

Sphere :

A sphere is a closed three-dimensional solid figure, where all the points on the surface of the sphere are equidistant from the common fixed point called “centre”. The equidistant is called the “radius”.

Surface area of a Sphere :

The surface area of a sphere of radius r = 4 times the area of a circle of radius r = 4(πr²)
For a sphere Curved surface area (CSA) = Total Surface area(TSA) = 4πr²

Cross-section

A cross-section is the exposed 2D surface or face obtained by making a straight cut through a 3D solid object.

Uniform Cross-section

A solid has a uniform cross-section if every slice made parallel to its base results in an identical shape and size. Cubes, cuboids, cylinders, and prisms are standard solids with a uniform cross-section.

Practice Questions on Area of Cross-section :- Exercise-18(b)

In this chapter, we study all the topics on Surface Area and Volume of 3D Solids like Cube, Cuboid, Area of cross-section and do some practice questions also. Here we solve extra practice questions on this chapter for better understanding.

For extra practice questions on Surface Area and Volume of 3D Solids :- Chapter Test

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OP Malhotra Notes on Circle Circumference and Area Class-9 S.Chand ICSE Maths

Circle Definition

A circle is a closed two-dimensional figure in which the set of all the points in the plane is equidistant from a given point called “centre”. Every line that passes through the circle forms the line of reflection symmetry

• Arc – It is basically the connected curve of a circle
• Sector – A region bounded by two radii and an arc.
• Segment- A region bounded by a chord and an arc lying between the chord’s endpoints. It is to be noted that segments do not contain the centre.

• Centre – It is the midpoint of a circle.
• Chord- A line segment whose endpoints lie on the circle
• Diameter- A line segment having both the endpoints on the circle
• Radius- A line segment connecting the centre of a circle to any point on the circle itself.
• Secant- A straight line cutting the circle at two points. It is also called as an extended chord.
• Tangent- A coplanar straight line touching the circle at a single point.

Circle Formulas for Circumference

 Practice Questions on Circle Circumference :- Exercise-17(a)

Properties of Circles

Here are some basic properties of circles.

• The outer line of a circle is at equidistant from the center.
• The diameter of the circle divides it into two equal parts
• Circles which have equal radii are congruent to each other.
• Circles which are different in size or having different radii are similar.
• The diameter of the circle is the largest chord and is double of the radius.

Circle Formulas for Area

Practice Questions on Circle Area :- Exercise-17(b)

Some More Points On Circle

Common Sectors

The Quadrant and Semicircle are two special types of Sector:

Quarter of a circle is called a Quadrant.

Half a circle is called a Semicircle.

Lines

A line that “just touches” the circle as it passes by is called a Tangent.
>Line that cuts the circle at two points is called a Secant.
>>A line segment that goes from one point to another on the circle’s circumference is called a Chord</b>.
>If it passes through the centre it is called a Diameter.
And a part of the circumference is called an

Secant of circle :

A line that intersects a circle at two points then it is called Secant of circle.

In this chapter, we study all the topics on Circle Circumference and Area like Area, Circumference, Lines, Segments and Secants and do some practice questions also. Here we solve extra practice questions on this chapter for better understanding.

For extra practice questions on Circle Circumference and Area :- Chapter Test

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Shares and Dividends ICSE Class 10 Maths Notes OP Malhotra 2026-27

Introduction

• A share is one of a finite number of portions in the capital of a company.
• The person who subscribe in shares is called a share holder.
• The profit distributed by the company among shareholders is called dividend.

Types of shares – There are two types of shares namely Common and Preferred 
The common shares are also called Equity shares.

Important Definitions –

1. The amount invested to start a company is called the capital stock.
2. Face Value(F.V) or Nominal Value(N.V) is the original value printed on the share.
3. Market Value(M.V) is the price at which the share is bought or sold in the market.

Important Notes –

1. If the market value of a share is the same as it’s nominal value , the share is said to at par.
2. If the market value of share is more than it’s face value , the share is said to at above par or at premium.
3. If the market value of share is less than it’s face value , the share is said to be below par or at a discount.
4. The dividend is always calculated on the face value and not on the market value.
5. The dividend is expressed as a percentage of the nominal value of the share.
6. The shares are generally of two types:
   (i) Preferred shares
   (ii) Common or ordinary shares

Important Formulas –

1. Investment = Number of shares × Market Value of 1 share
2. Number of shares bought = Sum invested/(M.V of 1 share)
3. Number of shares bought = Total dividend/(Dividend on 1 share)
4. Number of shares bought = Total Income/(Income on 1 share)
5. Income = Number of shares × rate of dividend × F.V
6. F.V = Face value = Nominal value = N.V
7. Return % = Income Profit %

So , this was Shares and Dividends Maths ICSE Notes Class 10 OP Malhotra .
Now you should solve some practice problems.

Questions on this topic :-

• Exe-3A
• Exe-3B

In this chapter, we study all the topics on Shares and Dividends and do some practice questions also. Here we solve extra practice questions on this chapter for better understanding.

Here is the link for extra practice questions on Shares and Dividends :- Self Evaluation

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