A rectangular plot has a perimeter of 46 m and an area of 120 m2. Find its length and breadth.

Length plus breadth is 23 m, so take sides x and 23 - x. Then x(23 - x) = 120 gives x2 - 23x + 120 = 0, so (x - 15)(x - 8) = 0. The plot is 15 m by 8 m.

Solving Problems (Based on Quadratic Equations)

Q: The product of two consecutive positive even integers is 168. Find them.

x(x + 2) = 168 gives x2 + 2x - 168 = 0, so (x + 14)(x - 12) = 0. Rejecting -14, the integers are 12 and 14.

Q: A polygon with n sides has (n(n - 3))/(2) diagonals. How many sides has a polygon with 54 diagonals?

n(n - 3) = 108 gives n2 - 3n - 108 = 0, so (n - 12)(n + 9) = 0. A number of sides cannot be negative, so n = 12.

Q: The product of Neha's age two years ago and her age three years from now is 84. Find her present age.

(x - 2)(x + 3) = 84 gives x2 + x - 90 = 0, so (x + 10)(x - 9) = 0. Rejecting -10, Neha is 9 years old (check: 7 × 12 = 84).

Q: The hypotenuse of a right-angled triangle is 17 cm, and one side is 7 cm longer than the other. Find the two sides.

x2 + (x + 7)2 = 289 gives x2 + 7x - 120 = 0, so (x + 15)(x - 8) = 0. The sides are 8 cm and 15 cm.

Q: A cyclist rides 48 km. Riding 4 km h-1 faster would save 1 hour. Find the usual speed.

(48)/(x) - (48)/(x + 4) = 1 gives 192 = x2 + 4x, so x2 + 4x - 192 = 0 and (x + 16)(x - 12) = 0. The usual speed is 12 km h-1.

ICSE Class 10 · Mathematics · Chapter 6

What this chapter covers

From a word problem to a quadratic equation
Problems on numbers
Problems on ages
Problems on area, perimeter and right triangles
Problems on speed, distance and time
Problems on time and work
Problems on money and quantities
Checking roots and writing the answer

Key points

A word problem based on quadratic equations is solved by framing an equation from its conditions, reducing it to ax² + bx + c = 0, and solving.
Use one letter for one unknown and write every other quantity in terms of it.
Solve by factorisation when the factors are easy; otherwise use x = (-b ± √(b² - 4ac))/(2a).
Reject a root that is negative for a length, speed, time, age, price or count, fractional for a count, or against a condition of the problem; state the reason.
Consecutive integers are x and x + 1; consecutive even or odd integers are x and x + 2.
A two-digit number with digits t and u is 10t + u; reversed, it is 10u + t.
For a path of width w round a rectangle, each outer side grows by 2w.
Time = distance ÷ speed; convert minutes to hours when speed is in km h^-1.
Downstream speed is u + v and upstream speed is u - v.
In work problems, the rates add: (1)/(x) + (1)/(y) = (1)/(T).
Number of items = total amount ÷ price of one item.
Always answer the question asked, with units, and check the answer against the original words.

Questions answered

Define an admissible root of a framed quadratic equation.

An admissible root is a root of the framed equation that also satisfies every condition stated or implied in the problem, such as being positive for a length or a whole number for a count.

The product of two consecutive positive even integers is 168. Find them.

x(x + 2) = 168 gives x² + 2x - 168 = 0, so (x + 14)(x - 12) = 0. Rejecting -14, the integers are 12 and 14.

A polygon with n sides has (n(n - 3))/(2) diagonals. How many sides has a polygon with 54 diagonals?

n(n - 3) = 108 gives n² - 3n - 108 = 0, so (n - 12)(n + 9) = 0. A number of sides cannot be negative, so n = 12.

A two-digit number has a product of digits equal to 24. When 18 is added to the number, its digits are reversed. Find the number.

With tens digit t and units digit u, 10t + u + 18 = 10u + t gives u = t + 2. Then t(t + 2) = 24, so t² + 2t - 24 = 0 and (t + 6)(t - 4) = 0. Rejecting -6, t = 4 and u = 6; the number is 46.

The product of Neha's age two years ago and her age three years from now is 84. Find her present age.

(x - 2)(x + 3) = 84 gives x² + x - 90 = 0, so (x + 10)(x - 9) = 0. Rejecting -10, Neha is 9 years old (check: 7 × 12 = 84).

A rectangular plot has a perimeter of 46 m and an area of 120 m². Find its length and breadth.

Length plus breadth is 23 m, so take sides x and 23 - x. Then x(23 - x) = 120 gives x² - 23x + 120 = 0, so (x - 15)(x - 8) = 0. The plot is 15 m by 8 m.

The hypotenuse of a right-angled triangle is 17 cm, and one side is 7 cm longer than the other. Find the two sides.

x² + (x + 7)² = 289 gives x² + 7x - 120 = 0, so (x + 15)(x - 8) = 0. The sides are 8 cm and 15 cm.

A cyclist rides 48 km. Riding 4 km h^-1 faster would save 1 hour. Find the usual speed.

(48)/(x) - (48)/(x + 4) = 1 gives 192 = x² + 4x, so x² + 4x - 192 = 0 and (x + 16)(x - 12) = 0. The usual speed is 12 km h^-1.

Past board exam questions

ICSE 2027 sample paper · 3 marks

A rectangle measures 12 m by 10 m. The shaded region is a strip of uniform width x m along its top side and its right side (an L-shape), leaving an unshaded rectangle in the bottom-left corner. The area of the shaded region is 57 m². Form an equation in x and solve it to find the possible value of x.

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Unshaded rectangle = (12 - x)(10 - x). 12 × 10 - (12 - x)(10 - x) = 57 120 - (120 - 22x + x²) = 57 ⇒ x² - 22x + 57 = 0 (x - 19)(x - 3) = 0 ⇒ x = 19 or x = 3 x = 19 is not possible (it is more than 10 m), so x = 3 m.

ICSE 2026 sample paper · 3 marks

The sum of the squares of three consecutive even numbers is 596. Find the numbers.

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Let the numbers be x − 2, x, x + 2. (x - 2)² + x² + (x + 2)² = 596 ⇒ 3x² + 8 = 596 ⇒ x² = 196 ⇒ x = ± 14. The numbers are 12, 14 and 16 (taking natural numbers; x = −14 gives −16, −14, −12).

ICSE 2026 board paper · 4 marks

The difference of two natural numbers is 5 and sum of their reciprocals is (3)/(10). Find the two numbers.

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Let the numbers be x and x + 5. (1)/(x) + (1)/(x + 5) = (3)/(10) ⇒ 10(2x + 5) = 3x(x + 5) ⇒ 3x² - 5x - 50 = 0 3x² - 15x + 10x - 50 = 0 ⇒ (x - 5)(3x + 10) = 0 ⇒ x = 5 or x = -(10)/(3). The numbers are natural, so x = 5. The two numbers are 5 and 10. (Check: (1)/(5) + (1)/(10) = (3)/(10) ✓)

ICSE 2025 compartment paper · 3 marks

The sum of two numbers is 2 and the sum of their reciprocals is 2.25. Find the numbers.

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Let the numbers be x and 2 - x. (1)/(x) + (1)/(2 - x) = (9)/(4) ⇒ (2)/(x(2 - x)) = (9)/(4) ⇒ 9x(2 - x) = 8 ⇒ 9x² - 18x + 8 = 0 ⇒ (3x - 4)(3x - 2) = 0 ⇒ x = (4)/(3) or (2)/(3). The numbers are (4)/(3) and (2)/(3). (Check: (3)/(4) + (3)/(2) = 2.25.)

ICSE 2025 sample paper · 3 marks

A car travels a distance of 72 km at a certain average speed of x km per hour and then travels a distance of 81 km at an average speed of 6 km per hour more than its original average speed. If it takes 3 hours to complete the total journey then form a quadratic equation and solve it to find its original average speed.

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Time for 72 km = (72)/(x) h; time for 81 km = (81)/(x + 6) h. (72)/(x) + (81)/(x + 6) = 3 ⇒ (24)/(x) + (27)/(x + 6) = 1 24(x + 6) + 27x = x(x + 6) ⇒ 51x + 144 = x² + 6x x² - 45x - 144 = 0 ⇒ (x - 48)(x + 3) = 0 ⇒ x = 48 or x = -3. Speed cannot be negative, so the original average speed is 48 km/h.

ICSE 2025 board paper · 4 marks

Mr. and Mrs. Das were travelling by car from Delhi to Kasauli for a holiday. Distance between Delhi and Kasauli is approximately 350 km (via NH 152D). Due to heavy rain they had to slow down. The average speed of the car was reduced by 20 km/hr and time of the journey increased by 2 hours. Find: (a) the original speed of the car. (b) with the reduced speed, the number of hours they took to reach their destination.

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(a) Let the original speed be x km/h. (350)/(x - 20) - (350)/(x) = 2 ⇒ 350 × 20 = 2x(x - 20) ⇒ x² - 20x - 3500 = 0 ⇒ (x - 70)(x + 50) = 0. x = 70 (speed cannot be negative). Original speed = 70 km/h. (b) Reduced speed = 50 km/h; time = (350)/(50) = 7 hours.

ICSE 2025 compartment paper · 4 marks

Divide 20 into two parts such that the sum of their squares is 272. The larger of two parts is square of the other. Assuming the smaller part to be 'x', form an equation and solve it to find the two parts.

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Smaller part = x, larger part = 20 - x. Sum of squares: x² + (20 - x)² = 272 ⇒ 2x² - 40x + 400 = 272 ⇒ x² - 20x + 64 = 0 ⇒ (x - 4)(x - 16) = 0 ⇒ x = 4 or 16. Since x is the smaller part, x = 4; the larger part = 16. Check: 16 = 4² (larger is the square of the smaller) and 4² + 16² = 16 + 256 = 272. ✓ (Equivalently, x + x² = 20 ⇒ (x + 5)(x - 4) = 0 ⇒ x = 4.) The two parts are 4 and 16.

ICSE 2023 board paper · 3 marks

A man covers a distance of 100 km, travelling with a uniform speed of x km/hr. Had the speed been 5 km/hr more it would have taken 1 hour less. Find x the original speed.

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(100)/(x) - (100)/(x + 5) = 1 ⇒ (500)/(x(x + 5)) = 1 ⇒ x² + 5x - 500 = 0. (x + 25)(x - 20) = 0 ⇒ x = 20 or x = -25. Speed cannot be negative, so x = 20 km/hr.