Quadratic Equations Chapter 4: NCERT Exemplar Solutions for Class 10

Quadratic Equations Class 10 exemplar solutions  are created here to help students prepare for 10th board examinations. The exemplars have been prepared by subject experts with respect to the upgraded syllabus (2024-2025) and are available in PDF format and can be downloaded easily.

Chapter 4 in Class 10 Math can be consider as a main base or core chapter for Class 10 students. Besides, there will be several questions based on this chapter, and students have to solve these problems using quadratic equations. Further, while dealing with this chapter, students will get to know more about different quadratic equations and how to find solutions by either using the factorisation method or the square method. To help students crack the complete concepts of this chapter, simply acad provides you with free solutions to NCERT exemplar in an pdf format for free.

Scroll down to grab this opportunity, and free your mind with the fear of equations!

 

Quadratic Equations Class 10 exemplar solutions Question 1 to 10

Question 1.

Which of the following options is a quadratic equation?

(A) x2+2x+1=(4−x)2+3x^2 + 2x + 1 = (4 – x)^2 + 3×2+2x+1=(4−x)2+3
(B) −2×2=(5−x)(2x−25)-2x^2 = (5 – x)(2x – \frac{2}{5})−2×2=(5−x)(2x−52​)
(C) (k+1)x2+32x=7(k + 1)x^2 + \frac{3}{2}x = 7(k+1)x2+23​x=7, here k=−1k = -1k=−1
(D) x3−x2=(x−1)3x^3 – x^2 = (x – 1)^3×3−x2=(x−1)3

Answer 1: (D) x3−x2=(x−1)3x^3 – x^2 = (x – 1)^3×3−x2=(x−1)3

Explanation:
The standard form of a quadratic equation is given by ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0, where a≠0a \neq 0a=0.

(A) Expanding:
x2+2x+1=16−8x+x2+3x^2 + 2x + 1 = 16 – 8x + x^2 + 3×2+2x+1=16−8x+x2+3
Simplifying:
10x−18=010x – 18 = 010x−18=0
This is not a quadratic equation.

(B) Expanding:
−2×2=10x−2×2−2+25x-2x^2 = 10x – 2x^2 – 2 + \frac{2}{5}x−2×2=10x−2×2−2+52​x
Simplifying:
52x−10=052x – 10 = 052x−10=0
This is not a quadratic equation.

(C) Substituting k=−1k = -1k=−1:
(−1+1)x2+32x=7(-1 + 1)x^2 + \frac{3}{2}x = 7(−1+1)x2+23​x=7
Simplifying:
3x−14=03x – 14 = 03x−14=0
This is not a quadratic equation.

(D) Expanding:
x3−x2=x3−3×2+3x−1x^3 – x^2 = x^3 – 3x^2 + 3x – 1×3−x2=x3−3×2+3x−1
Simplifying:
2×2−3x+1=02x^2 – 3x + 1 = 02×2−3x+1=0
This is a quadratic equation.

Question 2.

Which among the following options is not a quadratic equation?

(A) 2(x−1)2=4×2−2x+12(x – 1)^2 = 4x^2 – 2x + 12(x−1)2=4×2−2x+1
(B) 2x−x2=x2+52x – x^2 = x^2 + 52x−x2=x2+5
(C) (2x+3)2+x2=3×2−5x(\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 – 5x(2​x+3​)2+x2=3×2−5x
(D) (x2+2x)2=x4+3+4×3(x^2 + 2x)^2 = x^4 + 3 + 4x^3(x2+2x)2=x4+3+4×3

Answer 2: (D) (x2+2x)2=x4+3+4×3(x^2 + 2x)^2 = x^4 + 3 + 4x^3(x2+2x)2=x4+3+4×3

Explanation:
A quadratic equation is represented in the form ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0, where a≠0a \neq 0a=0.

(A) Expanding:
2(x2−2x+1)=4×2−2x+12(x^2 – 2x + 1) = 4x^2 – 2x + 12(x2−2x+1)=4×2−2x+1
Simplifying:
2×2+2x−1=02x^2 + 2x – 1 = 02×2+2x−1=0
This is a quadratic equation.

(B) Simplifying:
2×2−2x+5=02x^2 – 2x + 5 = 02×2−2x+5=0
This is a quadratic equation.

(C) Expanding:
2×2+26x+3=3×2−5x2x^2 + 2\sqrt{6}x + 3 = 3x^2 – 5x2x2+26​x+3=3×2−5x
Simplifying:
x2−(5+26)x−3=0x^2 – (5 + 2\sqrt{6})x – 3 = 0x2−(5+26​)x−3=0
This is a quadratic equation.

(D) Expanding:
x4+4×3+4×2=x4+3+4x2x^4 + 4x^3 + 4x^2 = x^4 + 3 + 4x^2×4+4×3+4×2=x4+3+4×2
Simplifying:
4×3−3=04x^3 – 3 = 04×3−3=0
This is a cubic equation, not a quadratic equation.

Question 3.

Which of the following equations contains 2 as a root?

(A) x2−4x+5=0x^2 – 4x + 5 = 0x2−4x+5=0
(B) x2+3x−12=0x^2 + 3x – 12 = 0x2+3x−12=0
(C) 2×2−7x+6=02x^2 – 7x + 6 = 02×2−7x+6=0
(D) 3×2−6x−2=03x^2 – 6x – 2 = 03×2−6x−2=0

Answer 3: (C) 2×2−7x+6=02x^2 – 7x + 6 = 02×2−7x+6=0

Explanation:
When 222 is a root, substituting x=2x = 2x=2 should satisfy the equation.

(A) Substituting x=2x = 2x=2:
(2)2−4(2)+5=1≠0(2)^2 – 4(2) + 5 = 1 \neq 0(2)2−4(2)+5=1=0
Thus, x=2x = 2x=2 is not a root.

(B) Substituting x=2x = 2x=2:
(2)2+3(2)−12=−2≠0(2)^2 + 3(2) – 12 = -2 \neq 0(2)2+3(2)−12=−2=0
Thus, x=2x = 2x=2 is not a root.

(C) Substituting x=2x = 2x=2:
2(2)2−7(2)+6=02(2)^2 – 7(2) + 6 = 02(2)2−7(2)+6=0
Thus, x=2x = 2x=2 is a root.

(D) Substituting x=2x = 2x=2:
3(2)2−6(2)−2=−2≠03(2)^2 – 6(2) – 2 = -2 \neq 03(2)2−6(2)−2=−2=0
Thus, x=2x = 2x=2 is not a root.

Question 4.

If 12\frac{1}{2}21​ is a root of the equation x2+kx−54=0x^2 + kx – \frac{5}{4} = 0x2+kx−45​=0, the value of kkk is:

(A) 2
(B) -2
(C) 14\frac{1}{4}41​
(D) 12\frac{1}{2}21​

Answer 4: (A) 2

Explanation:
Substituting x=12x = \frac{1}{2}x=21​ into the equation:

(12)2+k(12)−54=0\left(\frac{1}{2}\right)^2 + k\left(\frac{1}{2}\right) – \frac{5}{4} = 0(21​)2+k(21​)−45​=0

14+k2−54=0\frac{1}{4} + \frac{k}{2} – \frac{5}{4} = 041​+2k​−45​=0

k2=54−14=44\frac{k}{2} = \frac{5}{4} – \frac{1}{4} = \frac{4}{4}2k​=45​−41​=44​

k2=1\frac{k}{2} = 12k​=1

k=2k = 2k=2

Question 5.

Which of the following equations has the sum of its roots as 3?

(A) 2×2−3x+6=02x^2 – 3x + 6 = 02×2−3x+6=0
(B) −x2+3x−3=0-x^2 + 3x – 3 = 0−x2+3x−3=0
(C) 2×2−32x+1=0\sqrt{2}x^2 – \frac{3}{\sqrt{2}}x + 1 = 02​x2−2​3​x+1=0
(D) 3×2−3x+3=03x^2 – 3x + 3 = 03×2−3x+3=0

Answer 5: (B) −x2+3x−3=0-x^2 + 3x – 3 = 0−x2+3x−3=0

Explanation:
The sum of the roots for the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 is given by −ba-\frac{b}{a}−ab​.

(A) 2×2−3x+6=02x^2 – 3x + 6 = 02×2−3x+6=0
Sum of the roots: 32≠3\frac{3}{2} \neq 323​=3

(B) −x2+3x−3=0-x^2 + 3x – 3 = 0−x2+3x−3=0
Sum of the roots: −3−1=3-\frac{3}{-1} = 3−−13​=3

(C) 2×2−32x+1=0\sqrt{2}x^2 – \frac{3}{\sqrt{2}}x + 1 = 02​x2−2​3​x+1=0
Sum of the roots: 32≠3\frac{3}{2} \neq 323​=3

(D) 3×2−3x+3=03x^2 – 3x + 3 = 03×2−3x+3=0
Sum of the roots: 1≠31 \neq 31=3

Question 6.

Values of kkk for which the quadratic equation 2×2−kx+k=02x^2 – kx + k = 02×2−kx+k=0 has equal roots are:

(A) 0 only
(B) 4
(C) 8 only
(D) 0, 8

Answer 6: (D) 0, 8

Explanation:
Given equation: 2×2−kx+k=02x^2 – kx + k = 02×2−kx+k=0

For equal roots, the discriminant should be zero:
k2−8k=0k^2 – 8k = 0k2−8k=0
k(k−8)=0k(k – 8) = 0k(k−8)=0
Hence, k=0k = 0k=0 or k=8k = 8k=8.

Question 7.

Which constant should be added and subtracted to solve the quadratic equation 9×2+34x−2=09x^2 + \frac{3}{4}x – \sqrt{2} = 09×2+43​x−2​=0 by the method of completing the square?

(A) 18\frac{1}{8}81​
(B) 164\frac{1}{64}641​
(C) 14\frac{1}{4}41​
(D) 964\frac{9}{64}649​

Answer 7: (B) 164\frac{1}{64}641​

Explanation:
Given equation: 9×2+34x−2=09x^2 + \frac{3}{4}x – \sqrt{2} = 09×2+43​x−2​=0

First, convert it to the form (y+a)2=constant(y + a)^2 = \text{constant}(y+a)2=constant:
y2+14y−29=0y^2 + \frac{1}{4}y – \frac{\sqrt{2}}{9} = 0y2+41​y−92​​=0
Adding and subtracting (18)2\left(\frac{1}{8}\right)^2(81​)2, we get:
y2+14y+164=1+64264y^2 + \frac{1}{4}y + \frac{1}{64} = \frac{1 + 64\sqrt{2}}{64}y2+41​y+641​=641+642​​

Question 8.

The quadratic equation 2×2−5x+1=02x^2 – \sqrt{5}x + 1 = 02×2−5​x+1=0 has:

(A) Two distinct real roots
(B) Two equal real roots
(C) No real roots
(D) More than 2 real roots

Answer 8: (C) No real roots

Explanation:
Given equation: 2×2−5x+1=02x^2 – \sqrt{5}x + 1 = 02×2−5​x+1=0

Discriminant D=b2−4acD = b^2 – 4acD=b2−4ac
D=(−5)2−4×2×1=5−8=−3D = (-\sqrt{5})^2 – 4 \times 2 \times 1 = 5 – 8 = -3D=(−5​)2−4×2×1=5−8=−3

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

Question 9.

Which among the following equations has two distinct real roots?

(A) 2×2−32x+94=02x^2 – 3\sqrt{2}x + \frac{9}{4} = 02×2−32​x+49​=0
(B) x2+x−5=0x^2 + x – 5 = 0x2+x−5=0
(C) x2+3x+22=0x^2 + 3x + 2\sqrt{2} = 0x2+3x+22​=0
(D) 5×2−3x+1=05x^2 – 3x + 1 = 05×2−3x+1=0

Answer 9: (B) x2+x−5=0x^2 + x – 5 = 0x2+x−5=0

Explanation:
For two distinct real roots, the discriminant D=b2−4ac>0D = b^2 – 4ac > 0D=b2−4ac>0.

(A) D=0D = 0D=0
(B) D=21>0D = 21 > 0D=21>0
(C) D<0D < 0D<0
(D) D<0D < 0D<0

Question 10.

Which among the following equations has no real roots?

(A) x2−4x+32=0x^2 – 4x + 3\sqrt{2} = 0x2−4x+32​=0
(B) x2+4x−32=0x^2 + 4x – 3\sqrt{2} = 0x2+4x−32​=0
(C) x2−4x−32=0x^2 – 4x – 3\sqrt{2} = 0x2−4x−32​=0
(D) 3×2+43x+4=03x^2 + 4\sqrt{3}x + 4 = 03×2+43​x+4=0

Answer 10: (A) x2−4x+32=0x^2 – 4x + 3\sqrt{2} = 0x2−4x+32​=0

Explanation:
For no real roots, the discriminant D=b2−4ac<0D = b^2 – 4ac < 0D=b2−4ac<0.

(A) D<0D < 0D<0
(B) D>0D > 0D>0
(C) D>0D > 0D>0
(D) D=0D = 0D=0

Quadratic Equations Class 10 exemplar solutions Question 11 to 20

Question 11.

(x2+1)2−x2=0(x^2 + 1)^2 – x^2 = 0(x2+1)2−x2=0 has:

(A) Four real roots
(B) Two real roots
(C) No real roots
(D) One real root

Answer 11: (C) No real roots

Explanation:
Given equation:
x4+x2+1=0x^4 + x^2 + 1 = 0x4+x2+1=0
Let y=x2y = x^2y=x2:
y2+y+1=0y^2 + y + 1 = 0y2+y+1=0
Discriminant D=1−4=−3<0D = 1 – 4 = -3 < 0D=1−4=−3<0

Thus, there are no real roots.

Question 12.

State if the following quadratic equations have two distinct real roots. Give explanations from the answer.

1. x2−3x+4=0x^2 – 3x + 4 = 0x2−3x+4=0
2. 2×2+x−1=02x^2 + x – 1 = 02×2+x−1=0
3. 2×2−6x+92=02x^2 – 6x + \frac{9}{2} = 02×2−6x+29​=0
4. 3×2−4x+1=03x^2 – 4x + 1 = 03×2−4x+1=0
5. (x+4)2−8x=0(x + 4)^2 – 8x = 0(x+4)2−8x=0
6. (x−2)2−2(x+1)=0(x – \sqrt{2})^2 – 2(x + 1) = 0(x−2​)2−2(x+1)=0
7. 2×2−32x+12=0\sqrt{2}x^2 – \frac{3}{\sqrt{2}}x + \frac{1}{\sqrt{2}} = 02​x2−2​3​x+2​1​=0
8. x(1−x)−2=0x(1 – x) – 2 = 0x(1−x)−2=0
9. (x−1)(x+2)+2=0(x – 1)(x + 2) + 2 = 0(x−1)(x+2)+2=0
10. (x+1)(x−2)+x=0(x + 1)(x – 2) + x = 0(x+1)(x−2)+x=0

Answer 12:

1. No real roots, as D<0D < 0D<0.
2. Two distinct real roots, as D>0D > 0D>0.
3. Real and equal roots, as D=0D = 0D=0.
4. Two distinct real roots, as D>0D > 0D>0.
5. No real roots, as D<0D < 0D<0.
6. Two distinct real roots, as D>0D > 0D>0.
7. Two distinct real roots, as D>0D > 0D>0.
8. No real roots, as D<0D < 0D<0.
9. Two distinct real roots, as D>0D > 0D>0.
10. Two distinct real roots, as D>0D > 0D>0.

Question 13.

Say whether the following statements are true or false. Explain your answers.

1. Every quadratic equation contains exactly one root.
2. Every quadratic equation contains at least one real root.
3. Every quadratic equation contains at least two roots.
4. Every quadratic equation contains at most two roots.
5. If the coefficient for x2x^2×2 and the constant term for the quadratic equation have opposite signs, then the quadratic equation contains real roots.
6. If the coefficient for x2x^2×2 and the constant term have the same sign, as well as when the coefficient of xxx term is zero, then the quadratic equation has no real roots.

Answer 13:

1. False. A quadratic equation can have no real roots, one real root (repeated), or two real roots.
2. False. A quadratic equation can have no real roots (when the discriminant is negative).
3. False. A quadratic equation always has two roots, but they might be complex or identical.
4. True. Every quadratic equation has at most two roots.
5. True. If the signs of the coefficient of x2x^2×2 and the constant term are opposite, the equation generally has real roots because the discriminant is positive.
6. True. If the signs are the same and the xxx coefficient is zero, the equation may have no real roots (negative discriminant).

Question 14.

The quadratic equation having an integral coefficient has integral roots. Give explanations for your answer.

Answer 14:
No, the quadratic equation with integral coefficients may or may not contain integral roots.

Explanation:
For the equation 8×2−2x−1=08x^2 – 2x – 1 = 08×2−2x−1=0, the roots are 12\frac{1}{2}21​ and −14-\frac{1}{4}−41​, which are not integers. Therefore, an equation with integral coefficients can have non-integral roots.

Question 15.

Find out the roots for the quadratic equations given below by using the quadratic formula in each of them:

1. 2×2−3x−5=02x^2 – 3x – 5 = 02×2−3x−5=0
2. 5×2+13x+8=05x^2 + 13x + 8 = 05×2+13x+8=0
3. −3×2+5x+12=0-3x^2 + 5x + 12 = 0−3×2+5x+12=0
4. −x2+7x−10=0-x^2 + 7x – 10 = 0−x2+7x−10=0
5. x2+22x−6=0x^2 + 2\sqrt{2}x – 6 = 0x2+22​x−6=0
6. x2−35x+10=0x^2 – 3\sqrt{5}x + 10 = 0x2−35​x+10=0
7. (12)x2−11x+1=0\left(\frac{1}{2}\right)x^2 – \sqrt{11}x + 1 = 0(21​)x2−11​x+1=0

Answer 15:

1. Quadratic formula:
   x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac​​
   Roots: x=−1x = -1x=−1, x=52x = \frac{5}{2}x=25​
2. Roots: x=−1x = -1x=−1, x=−85x = \frac{-8}{5}x=5−8​
3. Roots: x=3x = 3x=3, x=−43x = \frac{-4}{3}x=3−4​
4. Roots: x=5x = 5x=5, x=2x = 2x=2
5. Roots: x=2x = \sqrt{2}x=2​, x=−32x = -3\sqrt{2}x=−32​
6. Roots: x=25x = 2\sqrt{5}x=25​, x=5x = \sqrt{5}x=5​
7. Roots: x=3+11x = 3 + \sqrt{11}x=3+11​, x=3−11x = 3 – \sqrt{11}x=3−11​

Question 16.

Check if the following are quadratic equations:

1. (x+1)2=2(x−3)(x + 1)^2 = 2(x – 3)(x+1)2=2(x−3)
2. x2−2x=−2(3−x)x^2 – 2x = -2(3 – x)x2−2x=−2(3−x)
3. (x−2)(x+1)=(x−1)(x+3)(x – 2)(x + 1) = (x – 1)(x + 3)(x−2)(x+1)=(x−1)(x+3)
4. (x−3)(2x+1)=x(x+5)(x – 3)(2x + 1) = x(x + 5)(x−3)(2x+1)=x(x+5)
5. (2x−1)(x−3)=(x+5)(x−1)(2x – 1)(x – 3) = (x + 5)(x – 1)(2x−1)(x−3)=(x+5)(x−1)
6. x2+3x+1=(x−2)2x^2 + 3x + 1 = (x – 2)^2×2+3x+1=(x−2)2
7. (x+2)3=2x(x2−1)(x + 2)^3 = 2x(x^2 – 1)(x+2)3=2x(x2−1)
8. x3−4×2−x+1=(x−2)3x^3 – 4x^2 – x + 1 = (x – 2)^3×3−4×2−x+1=(x−2)3

Answer 16:

1. Quadratic equation.
2. Quadratic equation.
3. Not a quadratic equation.
4. Quadratic equation.
5. Quadratic equation.
6. Not a quadratic equation.
7. Not a quadratic equation.
8. Quadratic equation.

Question 17.

Find out the roots of quadratic equations by factorization:

1. 2×2+7x+52=0\sqrt{2} x^2 + 7x + 5\sqrt{2} = 02​x2+7x+52​=0
2. 100×2−20x+1=0100x^2 – 20x + 1 = 0100×2−20x+1=0

Answer 17:

1. Roots: x=−52x = \frac{-5}{\sqrt{2}}x=2​−5​, x=−2x = -\sqrt{2}x=−2​
2. Roots: x=110x = \frac{1}{10}x=101​, x=110x = \frac{1}{10}x=101​

Question 18.

Find out the two consecutive positive integers, the sum of whose squares is 365.

Answer 18:
Assume the two consecutive positive integers as xxx and x+1x + 1x+1.

x2+(x+1)2=365x^2 + (x + 1)^2 = 365×2+(x+1)2=365
2×2+2x−364=02x^2 + 2x – 364 = 02×2+2x−364=0
Roots: x=13x = 13x=13, x=14x = 14x=14
Thus, the integers are 13 and 14.

Question 19.

Solve for the quadratic equation 2×2−7x+3=02x^2 – 7x + 3 = 02×2−7x+3=0 by using the quadratic formula.

Answer 19:
x=7±49−244x = \frac{7 \pm \sqrt{49 – 24}}{4}x=47±49−24​​
Roots: x=3x = 3x=3, x=12x = \frac{1}{2}x=21​

Question 20.

Find out the values for kkk for each of the following quadratic equations, such that they have two equal roots.

1. 2×2+kx+3=02x^2 + kx + 3 = 02×2+kx+3=0
2. kx(x−2)+6=0kx(x – 2) + 6 = 0kx(x−2)+6=0

Answer 20:

1. k=±26k = \pm 2\sqrt{6}k=±26​
2. k=6k = 6k=6

Quadratic Equations Class 10 exemplar solutions Question 21 to 30

Question 21.

Find out the natural number whose square diminished by 84 is equal to the thrice for 8 more than the given number.

Answer 21:
Let the natural number be xxx.

x2−84=3(x+8)x^2 – 84 = 3(x + 8)x2−84=3(x+8)
x=12x = 12x=12

Question 22.

A natural number, when increased by 12, is the same as 160 times its reciprocal. Find out the number.

Answer 22:
Let the number be xxx.

x+12=160xx + 12 = \frac{160}{x}x+12=x160​
x=8x = 8x=8

Question 23.

If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?

Answer 23:
Let Zeba’s age be xxx.

(x−5)2=5x+11(x – 5)^2 = 5x + 11(x−5)2=5x+11
x=14x = 14x=14

Question 24.

Show the following situations in the form of quadratic equations:

1. The area of a rectangular plot is 528 m². The length of the plot (in meters) is one more than twice the breadth. Find the length and breadth of the plot.
2. The product of two consecutive positive integers is 306. Find the integers.
3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. Find Rohan’s present age.
4. A train travels a distance of 480 km at a constant speed. If the speed had been 8 km/h less, what would be the speed of the train?

Answer 24:

1. 2×2+x−528=02x^2 + x – 528 = 02×2+x−528=0
2. x2+x−306=0x^2 + x – 306 = 0x2+x−306=0
3. x2+32x−273=0x^2 + 32x – 273 = 0x2+32x−273=0
4. x2−8x−1280=0x^2 – 8x – 1280 = 0x2−8x−1280=0

Question 25.

Find out two numbers whose sum is 27 and the product is 182.

Answer 25:
Let the numbers be xxx and 27−x27 – x27−x.

x(27−x)=182x(27 – x) = 182x(27−x)=182
Numbers: 13 and 14.

Question 26.

Find out the two consecutive positive integers, sum of whose squares is 365.

Answer 26:
Let the two consecutive positive integers be xxx and x+1x + 1x+1.

x2+(x+1)2=365x^2 + (x + 1)^2 = 365×2+(x+1)2=365
Integers: 13 and 14.

Question 27.

The altitude for a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find out the other two sides.

Answer 27:
Let the base be xxx cm.

x2+(x−7)2=132x^2 + (x – 7)^2 = 13^2×2+(x−7)2=132
Sides: 12 cm (base), 5 cm (altitude).

Question 28.

The cottage industry produces a certain number of pottery articles in a day. It was observed on the particular day that the cost for production of each article (in rupees) was 3 more than twice the number of articles produced on that particular day. If the total cost of production on that day was Rs.90, find out the number of articles produced and the cost for each article.

Answer 28:
Let the number of articles produced be xxx.

x(2x+3)=90x(2x + 3) = 90x(2x+3)=90
Number of articles: 6, Cost: Rs 15.

 

Question 29.

Find out the roots of the following quadratic equations, if they exist by completing the square method:

1. 2×2−7x+3=02x^2 – 7x + 3 = 02×2−7x+3=0
2. 2×2+x−4=02x^2 + x – 4 = 02×2+x−4=0
3. 4×2+43x+3=04x^2 + 4\sqrt{3}x + 3 = 04×2+43​x+3=0
4. 2×2+x+4=02x^2 + x + 4 = 02×2+x+4=0

Answer 29:

1. Roots: x=3x = 3x=3, x=12x = \frac{1}{2}x=21​
2. Roots: x=33−14x = \frac{\sqrt{33} – 1}{4}x=433​−1​, x=−33−14x = \frac{-\sqrt{33} – 1}{4}x=4−33​−1​
3. Roots: x=−32x = -\frac{\sqrt{3}}{2}x=−23​​
4. No real roots.

Question 30.

Find out the roots for the quadratic equations given by applying the quadratic formula.

Answer 30:

1. 2×2−7x+3=02x^2 – 7x + 3 = 02×2−7x+3=0
   Roots: x=3x = 3x=3, x=12x = \frac{1}{2}x=21​
2. 2×2+x−4=02x^2 + x – 4 = 02×2+x−4=0
   Roots: x=33−14x = \frac{\sqrt{33} – 1}{4}x=433​−1​, x=−33−14x = \frac{-\sqrt{33} – 1}{4}x=4−33​−1​
3. 4×2+43x+3=04x^2 + 4\sqrt{3}x + 3 = 04×2+43​x+3=0
   Roots: x=−32x = -\frac{\sqrt{3}}{2}x=−23​​
4. 2×2+x+4=02x^2 + x + 4 = 02×2+x+4=0
   No real roots.

Quadratic Equations Class 10 exemplar solutions Question 31 to 41

Question 31.

Find the roots for the given equations:

1. x−1x=3x – \frac{1}{x} = 3x−x1​=3, x≠0x \neq 0x=0
2. 1x+4−1x−7=1130\frac{1}{x+4} – \frac{1}{x-7} = \frac{11}{30}x+41​−x−71​=3011​, x=−4,7x = -4, 7x=−4,7

Answer 31:

1. Roots: x=3+132x = \frac{3 + \sqrt{13}}{2}x=23+13​​, x=3−132x = \frac{3 – \sqrt{13}}{2}x=23−13​​
2. Roots: x=1x = 1x=1, x=2x = 2x=2

Question 32.

A train traveling at a constant speed of 360 km will have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find out the original speed of the train.

Answer 32:
Let the original speed be xxx km/h.

360x−360x+5=45\frac{360}{x} – \frac{360}{x + 5} = \frac{4}{5}x360​−x+5360​=54​
Speed: 45 km/h.

Question 33.

(i) John and Jivanti together have 45 marbles. If both of them lose 5 marbles each, and the product for the number of each marble is now 124, then find out how many marbles they had to begin with.

(ii) The cottage industry produces a certain number of toys in a day. The cost of production for 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. Find out the number of toys produced on that particular day.

Answer 33:

(i) Number of marbles: John = 36, Jivanti = 9 or John = 9, Jivanti = 36.
(ii) Number of toys: 25 or 30.

Question 34.

The sum of the reciprocals of Rehman’s ages in years 3 years ago and 5 years from now is 13\frac{1}{3}31​. Find out his present age.

Answer 34:
Let Rehman’s present age be xxx years.

1x−3+1x+5=13\frac{1}{x – 3} + \frac{1}{x + 5} = \frac{1}{3}x−31​+x+51​=31​
Age: 7 years.

Question 35.

In a class test, the sum of Shefali’s marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find out her marks in the two subjects.

Answer 35:
Marks: Mathematics = 12, English = 18 or Mathematics = 13, English = 17.

Question 36.

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find out the sides of the field.

Answer 36:
Shorter side = 90 m, Longer side = 120 m.

Question 37.

The difference of squares for two numbers is 180. The square of the smaller number is 8 times the larger number. Find out the two numbers.

Answer 37:
Larger number = 18, Smaller number = 12 or 18 and -12.

Question 38.

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 out the speed of the train.

Answer 38:
Speed: 40 km/h.

Question 39.

Two water taps together could fill a tank in 9389\frac{3}{8}983​ hours. The tap for a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find out the time in which each tap can separately fill the tank.

Answer 39:
Smaller pipe: 25 hours, Larger pipe: 15 hours.

Question 40.

An express train takes 1 hour less than a passenger train to travel 132 km from Mysore to Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find out the average speed of the two trains.

Answer 40:
Passenger train speed: 33 km/h, Express train speed: 44 km/h.

Question 41.

The sum of the areas of two squares is 468 m². If the difference for their perimeters is 24 m, find out the sides of the two squares.

Answer 41:
Sides: 12 m and 18 m.

Frequently Asked Questions on NCERT Exemplar Solutions for Class 10 Maths Chapter 4

Q1: How many problems are in NCERT Exemplar Solutions for Class 10 Maths Chapter 4?

Exercise 4.1 has 5 questions, Exercise 4.2 has 3 questions, Exercise 4.3 has 1 question with many sub-questions, and Exercise 4.4 has 4 questions. Each exercise includes long answers, short answers, and MCQs, covering key exam topics. These solutions are designed by subject experts at SimplyAcad.

Q2: What topics are covered in NCERT Exemplar Solutions for Class 10 Maths Chapter 4?

This chapter includes:

1. Representing situations as quadratic equations.
2. Solving quadratic equations using factorization.
3. Solving quadratic equations by completing the square.
4. Determining the nature of roots.

Q3: What are the roots of quadratic equations in Chapter 4?

The roots of a quadratic equation are the values of the variable that satisfy the equation. In simpler terms, if x = α is a root, then f(α) = 0. The real roots are where the graph of y = f(x) crosses the x-axis. Here are some key points:

1. If c = 0, one root is zero, and the other is -b/a.
2. If b = c = 0, both roots are zero.
3. If a = c, the roots are reciprocals of each other.