Quadratic equations are a key part of algebra and frequently appear in CBSE Class 10 board exams. Chapter 4: Quadratic Equations focuses on forming quadratic equations from real-life situations, solving them using factorisation and the quadratic formula, and understanding the nature of roots using the discriminant.
These NCERT Solutions for Class 10 Maths Chapter 4 include important CBSE board questions asked between 2018 and 2025. All solutions are explained step by step in simple language to help students build concept clarity and score well in exams.
NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations
Q.
Find the roots of the following quadratic equationsby factorisation: (i) x2−3x−10=0 (ii) 2x2+x−6=0(iii) 2x2+7x+52=0 (iv) 2x2−x+81=0(v) 100x2−20x+1=0
Q.
In a flight of 600 km , an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km/hr and time of flight increased by 30 minutes. Find the original duration of flight.
[CBSE - 2020]
Q.
The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. Find the lengths of other two sides of the triangle.
[CBSE - 2025]
Q.
Two water taps together can fill a tank in
187 hours. The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.
[CBSE - 2019]
Q.
While designing the school year book, a teacher asked the student that the length and width of a particular photo is increased by x units each to double the area of the photo. The original photo is 18 cm long and 12 cm wide.
Based on the above information, answer the following questions:
(I) Write an algebraic equation depicting the above information.
(II) Write the corresponding quadratic equation in standard form.
(III) Can any rational value of x make the new area equal to
220 cm2?
[CBSE - 2023]
Q.
While designing the school year book, a teacher asked the student that the length and width of a particular photo is increased by x units each to double the area of the photo. The original photo is 18 cm long and 12 cm wide.Based on the above information, answer the following questions:(I) Write an algebraic equation depicting the above information.(II) Write the corresponding quadratic equation in standard form.(III) What should be the new dimensions of the enlarged photo?
[CBSE - 2023]
Q.
A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.
(i) Assuming the original length of each side of a tile be x units, make a quadratic equation from the above information.
(ii) Write the corresponding quadratic equation in standard form.
(iii) Solve the quadratic equation for x, using quadratic formula.
[CBSE - 2024]
Q.
A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.
(i) Assuming the original length of each side of a tile be x units, make a quadratic equation from the above information.
(ii) Write the corresponding quadratic equation in standard form.
(iii) Find the value of x, the length of side of a tile by factorisation.
[CBSE - 2024]
Q.
A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
[CBSE - 2018]
Q.
If x = 3 is one root of the quadratic equation
x2−2kx−6=0, then find the value of k.
[CBSE - 2018]
Q.
Solve the problems given in Example 1.
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.
Q.
Find the value of k for which the roots of the equation
3x2−10x+k=0 are reciprocal of each other.
[CBSE - 2019]
Q.
For what values of k, the roots of the equation
x2+4x+k=0 are real?
[CBSE - 2019]
Q.
The least positive value of k, for which the quadratic equation
2x2 + kx – 4=0 has rational roots, is
[CBSE - 2023]
Q.
If the roots of equation
ax2+bx+c=0, a=0 are real and equal, then which of the following relation is true?
[CBSE - 2024]
Q.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx +3 = 0
(ii) kx(x - 2) + 6 = 0
Q.
Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Q.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Q.
Find two numbers whose sum is 27 and product is 182.
Q.
A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.
[CBSE - 2025]
Class 10 Maths Chapter 4 Questions & Answers – Quadratic Equations
Important Board Questions (CBSE 2018–2025)
Q1. (Medium) [CBSE – 2025]
A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.
Solution:
Let the speed be x km/h.
Time taken = 480/x hours.
New speed = (x − 8) km/h, new time = 480/(x − 8) hours.
Given: 480/(x − 8) = 480/x + 3
480x = 480(x − 8) + 3x(x − 8)
480x = 480x − 3840 + 3x2 − 24x
3x2 − 24x − 3840 = 0
x2 − 8x − 1280 = 0
x = [8 ± √(64 + 5120)]/2 = [8 ± √5184]/2 = [8 ± 72]/2
x = 40 (positive value)
Answer: Speed of the train = 40 km/h.
Q2. (Medium) [CBSE – 2025]
The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. Find the lengths of other two sides of the triangle.
Solution:
Let the other two sides be a and b.
Perimeter: a + b + 25 = 60 ⇒ a + b = 35 …(1)
Right triangle: a2 + b2 = 252 = 625 …(2)
(a + b)2 = a2 + b2 + 2ab
352 = 625 + 2ab ⇒ 1225 = 625 + 2ab ⇒ 2ab = 600 ⇒ ab = 300
So a and b are roots of t2 − 35t + 300 = 0
t2 − 35t + 300 = (t − 15)(t − 20) = 0
Answer: The other sides are 15 cm and 20 cm.
Q3. (Medium) [CBSE – 2024]
A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.
(i) Assuming the original side length is x units, make a quadratic equation.
(ii) Write it in standard form.
(iii) Solve using quadratic formula.
Solution:
Floor area is same in both cases.
Original tile area = x2
Total area = 200x2
New tile side = (x + 1), new tile area = (x + 1)2
Total area = 128(x + 1)2
So, 200x2 = 128(x + 1)2 …(i)
200x2 = 128(x2 + 2x + 1)
200x2 = 128x2 + 256x + 128
72x2 − 256x − 128 = 0
Divide by 8: 9x2 − 32x − 16 = 0 …(ii)
Using quadratic formula:
x = [32 ± √(322 − 4·9·(−16))]/(2·9)
x = [32 ± √(1024 + 576)]/18 = [32 ± √1600]/18 = [32 ± 40]/18
x = 4 (positive) or x = −4/9 (reject)
Answer: x = 4 units.
Q4. (Medium) [CBSE – 2024]
Same question as above, but find x by factorisation.
Solution:
From Q3, standard form: 9x2 − 32x − 16 = 0
9x2 − 36x + 4x − 16 = 0
9x(x − 4) + 4(x − 4) = 0
(x − 4)(9x + 4) = 0
x = 4 or x = −4/9 (reject)
Answer: x = 4 units.
Q5. (Easy) [CBSE – 2024]
If the roots of ax2 + bx + c = 0 (a ≠ 0) are real and equal, then which relation is true?
Solution:
For real and equal roots, discriminant = 0.
So, b2 − 4ac = 0.
Q6. (Medium) [CBSE – 2023]
A photo is 18 cm long and 12 cm wide. Its length and width are increased by x units each to double the area.
(I) Write an algebraic equation.
(II) Write quadratic equation in standard form.
(III) Can any rational x make the new area 220 cm2?
Solution:
Original area = 18 × 12 = 216 cm2
Double area = 432 cm2
(I) (18 + x)(12 + x) = 432
(II) x2 + 30x + 216 = 432
x2 + 30x − 216 = 0
(III) If new area is 220 cm2, then (18 + x)(12 + x) = 220
x2 + 30x + 216 = 220 ⇒ x2 + 30x − 4 = 0
Discriminant = 302 − 4(1)(−4) = 916, which is not a perfect square.
So, x is not rational.
Answer: No, no rational value of x can make the area 220 cm2.
Q7. (Medium) [CBSE – 2023]
Same photo question: find the new dimensions of the enlarged photo.
Solution:
From Q6: x2 + 30x − 216 = 0
x2 + 36x − 6x − 216 = 0
(x + 36)(x − 6) = 0
x = 6 (positive)
New length = 18 + 6 = 24 cm
New width = 12 + 6 = 18 cm
Answer: New dimensions are 24 cm × 18 cm.
Q8. (Medium) [CBSE – 2023]
The least positive value of k for which 2x2 + kx − 4 = 0 has rational roots is:
Solution:
For rational roots, discriminant must be a perfect square.
D = k2 − 4(2)(−4) = k2 + 32
We need the least positive k such that k2 + 32 is a perfect square.
For k = 2: D = 4 + 32 = 36 = 62 ✔
Answer: Least positive k = 2.
Q9. (Medium) [CBSE – 2019]
Two water taps together can fill a tank in 1 7/8 hours. The tap with longer diameter takes 2 hours less than the smaller one to fill the tank separately. Find the time taken by each tap.
Solution:
1 7/8 hours = 15/8 hours.
Combined rate = 1 ÷ (15/8) = 8/15 tank per hour.
Let smaller tap take t hours, larger tap take (t − 2) hours.
1/t + 1/(t − 2) = 8/15
(2t − 2) / [t(t − 2)] = 8/15
15(2t − 2) = 8t(t − 2)
30t − 30 = 8t2 − 16t
8t2 − 46t + 30 = 0
4t2 − 23t + 15 = 0
(4t − 3)(t − 5) = 0
t = 5 (valid), t = 3/4 (reject as t > 2)
Smaller tap = 5 hours
Larger tap = 5 − 2 = 3 hours
Q10. (Medium) [CBSE – 2018]
A plane left 30 minutes late. To reach the destination 1500 km away in time, it increased its speed by 100 km/h. Find its usual speed.
Solution:
Let usual speed be x km/h.
Usual time = 1500/x hours.
New speed = x + 100, new time = 1500/(x + 100).
Since it left 1/2 hour late, it must save 1/2 hour in flying time:
1500/(x + 100) = 1500/x − 1/2
Solving gives x = 500 (positive)
Answer: Usual speed = 500 km/h.
Q11. (Medium) [CBSE – 2020]
In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed was reduced by 200 km/h and the time increased by 30 minutes. Find the original duration of flight.
Solution:
Let original speed be x km/h.
Original time = 600/x hours.
Reduced speed = x − 200, new time = 600/(x − 200).
Given time increased by 1/2 hour:
600/(x − 200) = 600/x + 1/2
Solving gives x = 600 km/h.
Original duration = 600/600 = 1 hour.
Q12. (Easy) [CBSE – 2018]
If x = 3 is one root of x2 − 2kx − 6 = 0, find k.
Solution:
Put x = 3:
9 − 2k(3) − 6 = 0 ⇒ 3 − 6k = 0 ⇒ k = 1/2
Q13. (Medium) [CBSE – 2019]
Find k for which the roots of 3x2 − 10x + k = 0 are reciprocal of each other.
Solution:
If roots are reciprocal, product of roots = 1.
For ax2 + bx + c = 0, product = c/a.
Here, product = k/3 = 1 ⇒ k = 3.
Q14. (Easy) [CBSE – 2019]
For what values of k, the roots of x2 + 4x + k = 0 are real?
Solution:
For real roots, discriminant ≥ 0.
D = 42 − 4(1)(k) = 16 − 4k ≥ 0 ⇒ k ≤ 4
Q15. (Medium)
Find the values of k so that the quadratic equations have two equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx(x − 2) + 6 = 0
Solution:
(i) Equal roots ⇒ D = 0
D = k2 − 4·2·3 = k2 − 24 = 0
k = ±√24 = ±2√6
(ii) kx(x − 2) + 6 = 0 ⇒ kx2 − 2kx + 6 = 0
Here a = k, b = −2k, c = 6
D = (−2k)2 − 4(k)(6) = 4k2 − 24k = 4k(k − 6)
For equal roots: D = 0 ⇒ 4k(k − 6) = 0
k = 0 or k = 6, but k = 0 makes it not a quadratic.
So k = 6.
Q16. (Medium)
Find two numbers whose sum is 27 and product is 182.
Solution:
Let numbers be x and y.
x + y = 27 and xy = 182.
So t2 − 27t + 182 = 0
t2 − 27t + 182 = (t − 13)(t − 14) = 0
Answer: The numbers are 13 and 14.
Q17. (Medium)
Solve the problems given in Example 1.
(i) John and Jivanti together have 45 marbles. Both lost 5 each and the product of remaining marbles is 124. Find original marbles.
(ii) A cottage industry produces toys. Cost per toy = 55 − (number of toys). Total cost = ₹750. Find number of toys.
Solution:
(i) Let John = x, Jivanti = y.
x + y = 45 …(1)
(x − 5)(y − 5) = 124 …(2)
Using y = 45 − x in (2):
(x − 5)(40 − x) = 124
−x2 + 45x − 200 = 124
x2 − 45x + 324 = 0
(x − 36)(x − 9) = 0
x = 36 or 9.
So they originally had 36 and 9 marbles.
(ii) Let number of toys produced = n.
Cost per toy = (55 − n).
Total cost: n(55 − n) = 750
−n2 + 55n − 750 = 0 ⇒ n2 − 55n + 750 = 0
(n − 25)(n − 30) = 0
Answer: n = 25 or 30 toys.
Q18. (Medium)
Find the roots of the following quadratic equations by factorisation:
(i) x2 − 3x − 10 = 0
(ii) 2x2 + x − 6 = 0
(iii) 2x2 + 7x + 5/2 = 0
(iv) 2x2 − x + 1/8 = 0
(v) 100x2 − 20x + 1 = 0
Solution:
(i) x2 − 3x − 10 = (x − 5)(x + 2) = 0
Roots: 5, −2
(ii) 2x2 + x − 6 = (2x − 3)(x + 2) = 0
Roots: 3/2, −2
(iii) 2x2 + 7x + 5/2 = 0
Multiply by 2: 4x2 + 14x + 5 = (2x + 1)(2x + 5) = 0
Roots: −1/2, −5/2
(iv) 2x2 − x + 1/8 = 0
Multiply by 8: 16x2 − 8x + 1 = (4x − 1)2 = 0
Root: 1/4 (repeated)
(v) 100x2 − 20x + 1 = (10x − 1)2 = 0
Root: 1/10 (repeated)
Q19. (Medium)
Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let sides be a and b (a > b).
a2 + b2 = 468 …(1)
Perimeter difference: 4a − 4b = 24 ⇒ a − b = 6 ⇒ a = b + 6
Put in (1): (b + 6)2 + b2 = 468
b2 + 12b + 36 + b2 = 468
2b2 + 12b − 432 = 0
b2 + 6b − 216 = 0
(b − 12)(b + 18) = 0 ⇒ b = 12 (positive)
So a = 12 + 6 = 18
Answer: Sides are 18 m and 12 m.
Q20. (Medium)
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let speed be x km/h.
Time = 360/x.
New speed = x + 5, new time = 360/(x + 5).
Given: 360/(x + 5) = 360/x − 1
360x = 360(x + 5) − x(x + 5)
360x = 360x + 1800 − x2 − 5x
x2 + 5x − 1800 = 0
(x + 45)(x − 40) = 0
x = 40 (positive)
Answer: Speed of the train = 40 km/h.
NCERT Solutions for Class 10 Maths Chapter 4 – FAQs
Q1. Why is Chapter 4 Quadratic Equations important for the Class 10 board exam?
Quadratic Equations is a high-weightage chapter in CBSE Class 10. Many board questions are asked from word problems, nature of roots (discriminant), and solving methods like factorisation and quadratic formula.
Q2. Which topics are most important in this chapter?
- Forming quadratic equations from real-life situations
- Solving by factorisation
- Solving by quadratic formula
- Discriminant and nature of roots (real/equal/imaginary)
Q3. What is the condition for real and equal roots?
For ax2 + bx + c = 0, roots are real and equal when b2 − 4ac = 0.
Q4. When do quadratic equations have no real roots?
If the discriminant is negative, i.e., b2 − 4ac < 0, then the quadratic equation has no real roots.
Q5. How should students prepare Chapter 4 for full marks?
Students should practise word problems regularly, revise discriminant conditions, and solve NCERT questions + previous year CBSE questions to improve speed and accuracy.