NCERT Exemplar Class 10 Maths Solutions Chapter 4 Quadratic Equations

CBSETuts.com provides you Free PDF download of NCERT Exemplar Class 10 Maths chapter 4 Quadratic Equations solved by expert teachers as per NCERT (CBSE) book guidelines. All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

You can also Download NCERT Solutions for class 10 Maths in Hindi to help you to revise complete Syllabus and score more marks in your examinations.

NCERT Exemplar Problems Class 10 Maths Solutions Chapter 4 Quadratic Equations

Exercise 4.1 Multiple Choice Questions (MCQs)

Question 1:
Which of the following is a quadratic equation?
(a) x² + 2x + 1 = (4 – x)² + 3
(b) – 2x² = (5 – x) (2x – \(\frac { 2 }{ 5 } \)
(c) (k + 1)x² + – \(\frac { 3 }{ 2 } \)x = 7, where k = -1
(d) x³ – x² =(x -1)^3
Solution:

Question 2:
Which of the following is not a quadratic equation?
(a) 2 (x -1)² = 4x² – 2x +1
(b) 2x – x² = x² + 5
(c) (-√2X +√3)² = 3x² – 5x
(d) (x² + 2x)² = x⁴ + 3 + 4x^2
Solution:

Question 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
Solution:

Question 4:
If \(\frac { 1 }{ 2 } \) is a root of the equation x² + kx –\(\frac { 5 }{ 4 } \) = 0, then the value of k is
(a) 2         (b) -2                          (c) \(\frac { 1 }{ 4 } \)                             (d)\(\frac { 1 }{ 2 } \)
Solution:

Question 5:
Which of the following equations has the sum of its roots as 3?
(a) 2x² – 3x + 6 = 0
(b) -x² + 3x – 3 = 0
(c) √2x² – \(\frac { 3 }{ \sqrt { 2 }}\)x + 1 = 0
(d) 3x² – 3x + 3 = 0
Solution:

Question 6:
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
Solution:
(d)
Given equation is 2x² – kx + k- 0
On comparing with ax² + bx + c = 0, we get
a = 2, b= – k and c = k
For equal roots, the discriminant must be zero.

Hence, the required values of k are 0 and 8.

Question 7:
Which constant must be added and subtracted to solve the quadratic
equation 9x² +\(\frac { 3 }{ 4 } \) x – √2= 0 by the method of completing the square?
(a) \(\frac { 1 }{ 8 } \)                  (b) \(\frac { 1 }{ 64 } \)                           (c) \(\frac { 1 }{ 4 } \)                 (d) \(\frac { 9 }{ 64 } \)
Solution:

Question 8:
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
Solution:

Question 9:
Which of the following equations has two distinct real roots?
(a)2x²-3√2x +\(\frac { 9 }{ 4 } \) =0
(b) x² + x – 5 =0
(c) x² + 3x + 2√2 =0
(d)5x²-3x + 1=0
Solution:

Question 10:
Which of the following equations has no real roots?
(a) x² – 4x + 3√2 =0
(b)x²+4x-3√2=0
(c) x² – 4x – 3√2 = 0
(d) 3x² + 4√3x + 4=0
Solution:

Question 11:
(x² +1)² – x² = 0 has
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real root
Solution:

Exercise 4.2 Very Short Answer Type Questions

Question 1:
State whether the following quadratic equations have two distinct real roots. Justify your answer.

Solution:

Question 2:
Write whether the following statements are true or false. Justify your answers.
(i) Every quadratic equation has exactly one root.
(ii) Every quadratic equation has atleast one real root.
(iii) Every quadratic equation has atleast two roots.
(iv) Every quadratic equation has atmost two roots.
(v) If the coefficient of x² and the constant term of a quadratic equation have opposite
signs, then the quadratic equation has real roots.
(vi) If the coefficient of x² and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.
Solution:
(i)   False, since a quadratic equation has two and only two roots.
(ii)  False, for example x² + 4 = O has no real root.
(iii) False, since a quadratic equation has two and only two roots.
(vi)  True, because every quadratic polynomial has atmost two roots.
(v)   True, since in this case discriminant is always positive, so it has always real roots, e., ac < 0 and so, b² – 4ac > 0.
(vi) True, since in this case discriminant is always negative, so it has no real roots e., if b = 0, then b² – 4ac ⇒ – 4ac < 0and ac>0.

polynomial root calculator, Factoring Polynomials. Factoring a polynomial is the opposite process of multiplying polynomials.

Question 3:
A quadratic equation with integral coefficient has integral roots. Justify your answer.
Solution:
No, consider the quadratic equation 2x² + x-6=0 with integral coefficient. The roots of
the given quadratic equation are -2 and \(\frac { 3 }{ 2 } \) which are not integers.

Question 4:
Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.
Solution:
Yes, consider the quadratic equation 2x² + x – 4 = Owith rational coefficient. The roots ofthe given quadratic equation are \(\frac { -1+\sqrt { 33 } }{4}\) and \(\frac { -1-\sqrt { 33 } }{ 4 }\) are irrational.

Question 5:
Does there exist a quadratic equation whose coefficient are all distinct irrationals but both the roots are rationals? why?
Solution:
Yes, consider the quadratic equation with all distinct irrationals coefficients i.e., √3x² – 7 √3x + 12√3 = 0. The roots of this quadratic equation are 3 and 4, which are rationals.

Question 6:
Is 0.2 a root of the equation x² – 0.4 = 0? Justify your answer.
Solution:
No, since 0.2 does not satisfy the quadratic equation i.e., ( 0.2)² – 0.4 = 0.04 – 0.4 ≠0.

Question 7:
If b = 0, c < 0, is it true that the roots of x² + bx + c = 0 are numerically equal and opposite in sign? Justify your answer.
Solution:
Given that, b = 0andc < 0and quadratic equation,

So, the roots of x² + bx+c = O are numerically equal and opposite in sign.

Exercise 4.3 Short Answer Type Questions

Question 1:

Find the roots of the quadratic equations by using the quadratic formula irt each of the following
Solution:

Question 2:
Find the roots of the following quadratic equations by the factorisation method.

Solution:

Exercise 4.4 Long Answer Type Questions

Question 1:
Find whether the following equations have real roots. If real roots exist, find them
(i) 8x² +2x -3=0
(ii) – 2x² + 3x + 2 = 0
(iii) 5x² – 2x- 10 = 0
(iv) \(\frac { 1 }{ 2x-3 } +\frac { 1 }{ x-5 }\)= 1,x≠\(\frac { 3 }{ 2 } \),5
(v) x² + 5 √5x – 70 = 0
Solution:

Question 2:
Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.
Solution:
Let n be a required natural number.
Square of a natural number diminished by 84 = n² – 84 and thrice of 8 more than the natural number = 3 (n + 8)
Now, by given condition,

Question 3:
A natural number, when increased by 12, equals 160 times its reciprocal.Find the number.
Solution:
Let the natural number be x.
According to the question,

Now, x + 20 = 0 ⇒ x = – 20 which is not possible because natural number is always greater than zero and x-8 = 0⇒x = 8.
Hence, the required natural number is 8.

Question 4:
A train, travelling at a uniform speed for 360 km, would have taken 48 min less to travel the same distance, if its speed were 5 km/h more. Find the original speed of the train.
Solution:
Let the original speed of the train = x km/h
Then, the increased speed of the train = (x + 5) km/h                   [by given condition]
and distance = 360 km
According to the question,

Now, x+50=0⇒x=-50
which is not possible because speed cannot be negative and x – 45 = 0⇒x = 45. Hence, the original speed of the train = 45 km/h

Question 5:
If Zeba were younger by 5 yr 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?
Solution:
Let the actual age of Zeba = x yr,
Herage when she was 5 yr younger = (x – 5)yr.
Now, by given condition,
Square of her age = 11 more than five times her actual age

[here, x ≠ 1 because her age is x – 5. So, x – 5 = 1 – 5 = – 4 i.e., age cannot be negative]  Hence, required Zeba’s age now is 14 yr.

Question 6:
At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age. Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.
Solution:
Let Nisha’s present age be x yr.
Then, Asha’s present age = x² + 2                                                               [by given condition]
Now, when Nisha grows to her mother’s present age i.e., after [(x² +2)- x] yr. Then, Asha’s age also increased by [(x² + 2) – x]yr.
Again by given condition,
Age of Asha = One years less than 10 times the present age of Nisha

Hence, required age of Nisha = 5yr
and required age of Asha = x² + 2 = (5)² + 2 = 25 + 2 = 27 yr

Question 7:
In the centre of a rectangular lawn of dimensions 50m x 40m, a rectangular pond has to be constructed, so that the area of the grass surrounding the pond would be 1184 m²[see figure]. Find the length and breadth of the pond.

Solution:
Given that a rectangular pond has to be constructed in the centre of a rectangular lawn of dimensions 50 mx 40 m So, the distance between pond and lawn would be same around the pond. Say x m.

Now, length of rectangular lawn (l₁) = 50 m and breadth of rectangular lawn (b₁) = 40 m Length of rectangular pond (l₂)= 50 – (x + x) = 50 – 2x
and breadth of rectangular pond (b₂) = 40 – (x+ x)= 40-2x
Also, area of the grass surrounding the pond =1184 m²
Area of rectangular lawn – Area of rectangular pond

∴                                                                  x=8
[At x = 37, length and breadth of pond are -24 and -34, respectively but length and
breadth cannot be negative. So, x = 37 cannot be possible]
∴                                         Length of pond = 50- 2x= 50- 2 (8)= 50- 16= 34 m
and                             breadth of pond = 40 – 2x = 40 – 2(8) = 40 – 16 = 24 m                       ‘
Hence, required length and breadth of pond are 34 m and 24 m, respectively.

Question 8:
At t min past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 min less than  \(\frac { { t }^{ 2 } }{ 4 } \)min. Find t.
Solution:
We know that, the time between 2 pm to 3 pm = 1 h = 60 min
Given that, at f min past 2pm, the time needed by the min hand of a clock to show 3 pm was found to be 3 min less than \(\frac { { t }^{ 2 } }{ 4 } \) min i,e

Hence, the required value off is 14 min.

NCERT Exemplar ProblemsNCERT Exemplar MathsNCERT Exemplar Science