CBSE Class 10 · Maths
CBSE Class 10 Maths Quadratic Equations Competency Based Questions
Help your child master Quadratic Equations — one of the most important and scoring chapters in CBSE Class 10 Maths — with these carefully prepared competency based questions. Each question, including MCQs, case studies, and assertion-reason types, comes with a detailed step-by-step solution. Prepared by Angle Belearn’s CBSE specialists to build real exam confidence.
CBSE Class 10 Maths Quadratic Equations — Questions with Solutions
The roots of the equation $x^{2}+3x-10=0$ are:
Solution
Option (A) is correct.
We have, $$\begin{aligned} & x^{2}+3x-10 = 0 \\ & x^{2}+5x-2x-10 = 0 \\ & x(x+5)-2(x+5) = 0 \\ & (x-2)(x+5) = 0 \\ & x = 2,-5 \end{aligned}$$
We have, $$\begin{aligned} & x^{2}+3x-10 = 0 \\ & x^{2}+5x-2x-10 = 0 \\ & x(x+5)-2(x+5) = 0 \\ & (x-2)(x+5) = 0 \\ & x = 2,-5 \end{aligned}$$
If the quadratic equation $ax^{2}+bx+c=0$ has two real and equal roots, then ‘$c$’ is equal to:
Solution
Option (D) is correct.
For equation having real and equal roots, $$\begin{aligned} D &= b^{2}-4ac = 0 \\ b^{2} &= 4ac \\ \frac{b^{2}}{4a} &= c \end{aligned}$$
For equation having real and equal roots, $$\begin{aligned} D &= b^{2}-4ac = 0 \\ b^{2} &= 4ac \\ \frac{b^{2}}{4a} &= c \end{aligned}$$
What is/are the roots of $3x^{2}=6x$?
Solution
Option (D) is correct.
We have, $$\begin{aligned} 3x^{2} &= 6x \\ 3x^{2}-6x &= 0 \\ 3x(x-2) &= 0 \\ x &= 0,\ 2 \end{aligned}$$ So, roots are 0 and 2.
We have, $$\begin{aligned} 3x^{2} &= 6x \\ 3x^{2}-6x &= 0 \\ 3x(x-2) &= 0 \\ x &= 0,\ 2 \end{aligned}$$ So, roots are 0 and 2.
If the quadratic equation $x^{2}-8x+k=0$ has real roots, then:
Solution
Option (B) is correct.
Comparing with $ax^{2}+bx+c=0$, we get $a=1,\ b=-8,\ c=k$.
Discriminant $D = b^{2}-4ac$. For real roots, $D \geq 0$: $$b^{2}-4ac = (-8)^{2}-4(1)(k) \geq 0$$ $$64-4k \geq 0 \implies k \leq 16$$
Comparing with $ax^{2}+bx+c=0$, we get $a=1,\ b=-8,\ c=k$.
Discriminant $D = b^{2}-4ac$. For real roots, $D \geq 0$: $$b^{2}-4ac = (-8)^{2}-4(1)(k) \geq 0$$ $$64-4k \geq 0 \implies k \leq 16$$
The nature of roots of the quadratic equation $9x^{2}-6x-2=0$ is:
Solution
Option (C) is correct.
For $9x^{2}-6x-2=0$: $a=9,\ b=-6,\ c=-2$.
$$b^{2}-4ac = (-6)^{2}-4 \times 9 \times (-2) = 36+72 = 108$$ Since $b^{2}-4ac > 0$, the roots are 2 distinct real roots. As the equation is of degree 2, there are exactly two roots.
For $9x^{2}-6x-2=0$: $a=9,\ b=-6,\ c=-2$.
$$b^{2}-4ac = (-6)^{2}-4 \times 9 \times (-2) = 36+72 = 108$$ Since $b^{2}-4ac > 0$, the roots are 2 distinct real roots. As the equation is of degree 2, there are exactly two roots.
Let $p$ be a prime number. The quadratic equation having its roots as factors of $p$ is:
Solution
Option (B) is correct.
Factors of $p = p \times 1$, so roots are $p$ and $1$.
The quadratic equation is: $$x^{2}-(\text{sum of roots})x+\text{product of roots}=0$$ $$\Rightarrow x^{2}-(p+1)x+p=0$$
Factors of $p = p \times 1$, so roots are $p$ and $1$.
The quadratic equation is: $$x^{2}-(\text{sum of roots})x+\text{product of roots}=0$$ $$\Rightarrow x^{2}-(p+1)x+p=0$$
Which of the following is not a quadratic equation?
Solution
Option (C) is correct.
Expanding option (C): $$\begin{aligned} 2x^{2}+3+2\sqrt{6}x+x^{2} &= 3x^{2}-5x \\ 3x^{2}+2\sqrt{6}x+3 &= 3x^{2}-5x \\ x(5+2\sqrt{6})+3 &= 0 \end{aligned}$$ This is not of the form $ax^{2}+bx+c=0$, so it is not a quadratic equation.
Expanding option (C): $$\begin{aligned} 2x^{2}+3+2\sqrt{6}x+x^{2} &= 3x^{2}-5x \\ 3x^{2}+2\sqrt{6}x+3 &= 3x^{2}-5x \\ x(5+2\sqrt{6})+3 &= 0 \end{aligned}$$ This is not of the form $ax^{2}+bx+c=0$, so it is not a quadratic equation.
Which of the following equations has 2 as its root?
Solution
Option (C) is correct.
Put $x=2$ in $2x^{2}-7x+6=0$: $$2(2)^{2}-7(2)+6 = 8-14+6 = 0$$ So, $x=2$ is a root of $2x^{2}-7x+6=0$.
Put $x=2$ in $2x^{2}-7x+6=0$: $$2(2)^{2}-7(2)+6 = 8-14+6 = 0$$ So, $x=2$ is a root of $2x^{2}-7x+6=0$.
Which of the following equations has the sum of its roots as 3?
Solution
Option (B) is correct.
For $-x^{2}+3x-3=0$, comparing with $ax^{2}+bx+c=0$: $a=-1,\ b=3,\ c=-3$. $$\text{Sum of roots} = \frac{-b}{a} = \frac{-3}{-1} = 3$$
For $-x^{2}+3x-3=0$, comparing with $ax^{2}+bx+c=0$: $a=-1,\ b=3,\ c=-3$. $$\text{Sum of roots} = \frac{-b}{a} = \frac{-3}{-1} = 3$$
The roots of the equation $x^{2}+7x+10=0$ are:
Solution
Option (A) is correct.
Given $x^{2}+7x+10=0$. Comparing with $ax^{2}+bx+c=0$: $a=1,\ b=7,\ c=10$. $$x = \frac{-7 \pm \sqrt{49-40}}{2} = \frac{-7 \pm 3}{2}$$ $$x = \frac{-7+3}{2} = -2 \quad \text{or} \quad x = \frac{-7-3}{2} = -5$$
Given $x^{2}+7x+10=0$. Comparing with $ax^{2}+bx+c=0$: $a=1,\ b=7,\ c=10$. $$x = \frac{-7 \pm \sqrt{49-40}}{2} = \frac{-7 \pm 3}{2}$$ $$x = \frac{-7+3}{2} = -2 \quad \text{or} \quad x = \frac{-7-3}{2} = -5$$
Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of $x$ km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.
(i) What will be the distance covered by Ajay’s car in two hours?
(A) $2(x+5)$ km
(B) $(x-5)$ km
(C) $2(x+10)$ km
(D) $(2x+5)$ km
(ii) Which of the following quadratic equations describes the speed of Raj’s car?
(A) $x^{2}-5x-500=0$
(B) $x^{2}+4x-400=0$
(C) $x^{2}+5x-500=0$
(D) $x^{2}-4x+400=0$
(iii) What is the speed of Raj’s car?
(A) 20 km/hour
(B) 15 km/hour
(C) 25 km/hour
(D) 10 km/hour
(iv) How much time is taken by Ajay to travel 400 km?
(A) 20 h
(B) 40 h
(C) 25 h
(D) 16 h
(v) The speed of Ajay’s car is:
(A) 25 km/hour
(B) 50 km/hour
(C) 75 km/hour
(D) None of these
(i) What will be the distance covered by Ajay’s car in two hours?
(A) $2(x+5)$ km
(B) $(x-5)$ km
(C) $2(x+10)$ km
(D) $(2x+5)$ km
(ii) Which of the following quadratic equations describes the speed of Raj’s car?
(A) $x^{2}-5x-500=0$
(B) $x^{2}+4x-400=0$
(C) $x^{2}+5x-500=0$
(D) $x^{2}-4x+400=0$
(iii) What is the speed of Raj’s car?
(A) 20 km/hour
(B) 15 km/hour
(C) 25 km/hour
(D) 10 km/hour
(iv) How much time is taken by Ajay to travel 400 km?
(A) 20 h
(B) 40 h
(C) 25 h
(D) 16 h
(v) The speed of Ajay’s car is:
(A) 25 km/hour
(B) 50 km/hour
(C) 75 km/hour
(D) None of these
Answer
(i) Option (A) is correct.
Speed of Raj’s car $= x$ km/h; Speed of Ajay’s car $= (x+5)$ km/h.
Distance covered by Ajay in 2 hours $= 2(x+5)$ km.
(ii) Option (C) is correct.
Time taken by Raj $= \dfrac{400}{x}$ h; Time taken by Ajay $= \dfrac{400}{x+5}$ h.
Since Raj took 4 hours more: $$\frac{400}{x}-\frac{400}{x+5} = 4 \implies \frac{100}{x}-\frac{100}{x+5} = 1$$ $$100(x+5)-100x = x(x+5) \implies x^{2}+5x-500=0$$
(iii) Option (A) is correct.
Solving $x^{2}+5x-500=0$: $$(x+25)(x-20)=0 \implies x=20 \text{ (taking positive value)}$$ Speed of Raj’s car is 20 km/hour.
(iv) Option (D) is correct.
Time taken by Ajay $= \dfrac{400}{x+5} = \dfrac{400}{20+5} = 16$ h.
(v) Option (A) is correct.
Speed of Ajay’s car $= x+5 = 20+5 = 25$ km/hour.
Speed of Raj’s car $= x$ km/h; Speed of Ajay’s car $= (x+5)$ km/h.
Distance covered by Ajay in 2 hours $= 2(x+5)$ km.
(ii) Option (C) is correct.
Time taken by Raj $= \dfrac{400}{x}$ h; Time taken by Ajay $= \dfrac{400}{x+5}$ h.
Since Raj took 4 hours more: $$\frac{400}{x}-\frac{400}{x+5} = 4 \implies \frac{100}{x}-\frac{100}{x+5} = 1$$ $$100(x+5)-100x = x(x+5) \implies x^{2}+5x-500=0$$
(iii) Option (A) is correct.
Solving $x^{2}+5x-500=0$: $$(x+25)(x-20)=0 \implies x=20 \text{ (taking positive value)}$$ Speed of Raj’s car is 20 km/hour.
(iv) Option (D) is correct.
Time taken by Ajay $= \dfrac{400}{x+5} = \dfrac{400}{20+5} = 16$ h.
(v) Option (A) is correct.
Speed of Ajay’s car $= x+5 = 20+5 = 25$ km/hour.
Digital images consist of pixels. A pixel can be considered as the smallest unit on a display screen in a mobile or a computer. The number of pixels, their size and colours depend on the display screen and its graphic card. Display screens are rectangular in shape and their size is defined as the length of the diagonal.
Amit is designing a web page for a display on a screen whose size is 1000 pixels. The width of the screen is 800 pixels.
(i) Which of the following equations can be used to calculate the height $(h)$ of the screen?
(A) $h^{2}+200 \times 1800=0$
(B) $h^{2}-200 \times 1800=0$
(C) $h^{2}-200=0$
(D) $h^{2}-1800=0$
(ii) If a screen of size 13 inches has a width 7 inches more than height, then which of the following expression applies?
(A) $x^{2}-(x+7)^{2}=13^{2}$
(B) $x^{2}+13^{2}=(x+7)^{2}$
(C) $x^{2}+(x+7)^{2}=13^{2}$
(D) $x^{2}-(x+7)^{2}=13^{2}$
(iii) Using the problem in (ii), the height and width are respectively (in inches):
(A) 6, 13
(B) 5, 12
(C) 4, 11
(D) 9, 16
(iv) If the quadratic equation is $h^{2}-200 \times 1800=0$, then the positive value of $h$ is:
(A) $\pm 600$
(B) $\pm 900$
(C) 900
(D) 600
(v) Which of the following is not a method of solving a quadratic equation?
(A) Factorisation
(B) Completing the square
(C) Using quadratic formula
(D) Identifying the nature of the root.
Amit is designing a web page for a display on a screen whose size is 1000 pixels. The width of the screen is 800 pixels.
(i) Which of the following equations can be used to calculate the height $(h)$ of the screen?
(A) $h^{2}+200 \times 1800=0$
(B) $h^{2}-200 \times 1800=0$
(C) $h^{2}-200=0$
(D) $h^{2}-1800=0$
(ii) If a screen of size 13 inches has a width 7 inches more than height, then which of the following expression applies?
(A) $x^{2}-(x+7)^{2}=13^{2}$
(B) $x^{2}+13^{2}=(x+7)^{2}$
(C) $x^{2}+(x+7)^{2}=13^{2}$
(D) $x^{2}-(x+7)^{2}=13^{2}$
(iii) Using the problem in (ii), the height and width are respectively (in inches):
(A) 6, 13
(B) 5, 12
(C) 4, 11
(D) 9, 16
(iv) If the quadratic equation is $h^{2}-200 \times 1800=0$, then the positive value of $h$ is:
(A) $\pm 600$
(B) $\pm 900$
(C) 900
(D) 600
(v) Which of the following is not a method of solving a quadratic equation?
(A) Factorisation
(B) Completing the square
(C) Using quadratic formula
(D) Identifying the nature of the root.
Answer
(i) Option (B) is correct.
Using Pythagoras theorem: $h^{2} = (1000)^{2}-(800)^{2} = (1800)(200)$
So $h^{2}-200 \times 1800 = 0$.
(ii) Option (C) is correct.
Let height $= x$. Then width $= x+7$ and diagonal $= 13$.
By Pythagoras: $x^{2}+(x+7)^{2}=13^{2}$.
(iii) Option (B) is correct.
$x^{2}+(x+7)^{2}=169 \implies 2x^{2}+14x-120=0 \implies x^{2}+7x-60=0$
$(x+12)(x-5)=0 \implies x=5$ (height cannot be negative).
Height = 5 inches, Width = 12 inches.
(iv) Option (D) is correct.
$h^{2}=200 \times 1800 = 360000 \implies h = \pm 600$.
Positive value of $h$ is 600.
(v) Option (D) is correct.
Identifying the nature of roots uses the discriminant but is not a method of solving a quadratic equation. The three methods are: factorisation, completing the square, and the quadratic formula.
Using Pythagoras theorem: $h^{2} = (1000)^{2}-(800)^{2} = (1800)(200)$
So $h^{2}-200 \times 1800 = 0$.
(ii) Option (C) is correct.
Let height $= x$. Then width $= x+7$ and diagonal $= 13$.
By Pythagoras: $x^{2}+(x+7)^{2}=13^{2}$.
(iii) Option (B) is correct.
$x^{2}+(x+7)^{2}=169 \implies 2x^{2}+14x-120=0 \implies x^{2}+7x-60=0$
$(x+12)(x-5)=0 \implies x=5$ (height cannot be negative).
Height = 5 inches, Width = 12 inches.
(iv) Option (D) is correct.
$h^{2}=200 \times 1800 = 360000 \implies h = \pm 600$.
Positive value of $h$ is 600.
(v) Option (D) is correct.
Identifying the nature of roots uses the discriminant but is not a method of solving a quadratic equation. The three methods are: factorisation, completing the square, and the quadratic formula.
There is a triangular playground as shown in the figure below.
As shown in the figure of the right-angled triangle playground, the length of the sides are $5x$ cm and $(3x-1)$ cm and the area of the triangle is $60$ cm².
(i) Find the value of $x$. OR Find the length of AC.
(ii) Find the length of AB.
(iii) Find the perimeter of $\triangle ABC$.
As shown in the figure of the right-angled triangle playground, the length of the sides are $5x$ cm and $(3x-1)$ cm and the area of the triangle is $60$ cm².
(i) Find the value of $x$. OR Find the length of AC.
(ii) Find the length of AB.
(iii) Find the perimeter of $\triangle ABC$.
Answer
(i) Area of triangle = 60 cm²
$$\frac{1}{2} \times AB \times BC = 60 \implies (5x)(3x-1) = 120$$
$$15x^{2}-5x-120=0 \implies 3x^{2}-x-24=0$$
$$3x^{2}-9x+8x-24=0 \implies 3x(x-3)+8(x-3)=0$$
$$(x-3)(3x+8)=0 \implies x=3 \text{ (length cannot be negative)}$$ OR $AB=15$ cm, $BC = (3 \times 3-1) = 8$ cm.
By Pythagoras: $AC^{2}=15^{2}+8^{2}=225+64=289 \implies AC=17$ cm.
(ii) $AB = 5x = 5 \times 3 = 15$ cm.
(iii) $AB=15$ cm, $BC=8$ cm, $AC=17$ cm.
Perimeter $= 15+8+17 = 40$ cm.
By Pythagoras: $AC^{2}=15^{2}+8^{2}=225+64=289 \implies AC=17$ cm.
(ii) $AB = 5x = 5 \times 3 = 15$ cm.
(iii) $AB=15$ cm, $BC=8$ cm, $AC=17$ cm.
Perimeter $= 15+8+17 = 40$ cm.
Japan’s LO series Maglev is the fastest train in the world, with a speed record of 602 km/h. Suppose a fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km/h less than that of the fast train, then answer the following questions:
(i) Find the speed of the slow train.
(ii) Find the speed of the fast train.
(iii) How much time is taken by the slow train to cover 600 km? OR How much time is taken by the fast train to cover the same distance?
(i) Find the speed of the slow train.
(ii) Find the speed of the fast train.
(iii) How much time is taken by the slow train to cover 600 km? OR How much time is taken by the fast train to cover the same distance?
Answer
(i) Let speed of slow train $= x$ km/h, so speed of fast train $= (x+10)$ km/h.
$$\frac{600}{x}-\frac{600}{x+10}=3 \implies \frac{6000}{x(x+10)}=3$$
$$x^{2}+10x-2000=0 \implies (x+50)(x-40)=0 \implies x=40 \text{ km/h}$$
Speed of slow train is 40 km/h.
(ii) Speed of fast train $= 40+10 = $ 50 km/h.
(iii) Time taken by slow train $= \dfrac{600}{40} = $ 15 h.
OR Time taken by fast train $= \dfrac{600}{50} = $ 12 h.
(ii) Speed of fast train $= 40+10 = $ 50 km/h.
(iii) Time taken by slow train $= \dfrac{600}{40} = $ 15 h.
OR Time taken by fast train $= \dfrac{600}{50} = $ 12 h.
A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has only 30 m of barbed wire, he fences three sides of the rectangular garden letting the compound wall of his house act as the fourth side-fence.
(i) Represent the given problem in quadratic form.
(ii) Find the length of the vegetable garden.
(iii) If the length of the vegetable garden is 5 m, then find the breadth. OR If length is 5 m and breadth is 20 m, find the perimeter.
(i) Represent the given problem in quadratic form.
(ii) Find the length of the vegetable garden.
(iii) If the length of the vegetable garden is 5 m, then find the breadth. OR If length is 5 m and breadth is 20 m, find the perimeter.
Answer
(i) Let one side be $x$ m and the other side be $y$ m. Then $x+y+x=30 \implies y=30-2x$.
Area $= xy = 100 \implies x(30-2x)=100$ $$30x-2x^{2}=100 \implies x^{2}-15x+50=0$$
(ii) Solving $x^{2}-15x+50=0$: $$(x-10)(x-5)=0 \implies x=5 \text{ or } 10$$ Length of the vegetable garden is 5 m or 10 m.
(iii) If length $= 5$ m: $5 \times b = 100 \implies b = 20$ m.
OR Perimeter $= 2(5+20) = 50$ m.
Area $= xy = 100 \implies x(30-2x)=100$ $$30x-2x^{2}=100 \implies x^{2}-15x+50=0$$
(ii) Solving $x^{2}-15x+50=0$: $$(x-10)(x-5)=0 \implies x=5 \text{ or } 10$$ Length of the vegetable garden is 5 m or 10 m.
(iii) If length $= 5$ m: $5 \times b = 100 \implies b = 20$ m.
OR Perimeter $= 2(5+20) = 50$ m.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If $5+\sqrt{7}$ is a root of a quadratic equation with rational coefficients, then its other root is $5-\sqrt{7}$.
Reason (R): Surd roots of a quadratic equation with rational coefficients occur in conjugate pairs.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If $5+\sqrt{7}$ is a root of a quadratic equation with rational coefficients, then its other root is $5-\sqrt{7}$.
Reason (R): Surd roots of a quadratic equation with rational coefficients occur in conjugate pairs.
Answer
Option (A) is correct.
In quadratic equations with rational coefficients, irrational roots occur in conjugate pairs. If one root $= 5+\sqrt{7}$, then the second root $= 5-\sqrt{7}$. Hence, both Assertion and Reason are true, and the Reason is the correct explanation for the Assertion.
In quadratic equations with rational coefficients, irrational roots occur in conjugate pairs. If one root $= 5+\sqrt{7}$, then the second root $= 5-\sqrt{7}$. Hence, both Assertion and Reason are true, and the Reason is the correct explanation for the Assertion.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): One solution of the quadratic equation $(x-1)^{2}-5(x-1)-6=0$ is 0.
Reason (R): Other solution of the equation $(x-1)^{2}-5(x-1)-6=0$ is 7.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): One solution of the quadratic equation $(x-1)^{2}-5(x-1)-6=0$ is 0.
Reason (R): Other solution of the equation $(x-1)^{2}-5(x-1)-6=0$ is 7.
Answer
Option (A) is correct.
Expanding: $(x-1)^{2}-5(x-1)-6=0$ $$x^{2}-7x+6-6=0 \implies x^{2}-7x=0 \implies x(x-7)=0 \implies x=0 \text{ or } 7$$ Assertion is true (one root is 0). Reason is also true (the other root is 7). Reason is the correct explanation of Assertion.
Expanding: $(x-1)^{2}-5(x-1)-6=0$ $$x^{2}-7x+6-6=0 \implies x^{2}-7x=0 \implies x(x-7)=0 \implies x=0 \text{ or } 7$$ Assertion is true (one root is 0). Reason is also true (the other root is 7). Reason is the correct explanation of Assertion.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The positive root of $\sqrt{3x^{2}+6}=9$ is 5.
Reason (R): The negative root of $\sqrt{3x^{2}+6}=9$ is 4.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The positive root of $\sqrt{3x^{2}+6}=9$ is 5.
Reason (R): The negative root of $\sqrt{3x^{2}+6}=9$ is 4.
Answer
Option (C) is correct.
Squaring both sides: $3x^{2}+6=81 \implies 3x^{2}=75 \implies x^{2}=25 \implies x= \pm 5$.
So the positive root is 5 — Assertion is true.
However, the negative root is $-5$, not 4 — Reason is false.
Squaring both sides: $3x^{2}+6=81 \implies 3x^{2}=75 \implies x^{2}=25 \implies x= \pm 5$.
So the positive root is 5 — Assertion is true.
However, the negative root is $-5$, not 4 — Reason is false.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): A two-digit number is four times the sum of the digits. If it is also equal to 3 times the product of the digits, then the number is 25.
Reason (R): The denominator of a fraction is one more than twice its numerator. If the sum of the fraction and its reciprocal is $2\frac{16}{21}$, then the fraction is $\frac{3}{7}$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): A two-digit number is four times the sum of the digits. If it is also equal to 3 times the product of the digits, then the number is 25.
Reason (R): The denominator of a fraction is one more than twice its numerator. If the sum of the fraction and its reciprocal is $2\frac{16}{21}$, then the fraction is $\frac{3}{7}$.
Answer
Option (D) is correct.
Assertion: Let the unit’s digit be $x$ and ten’s digit be $y$. Then $10y+x=4(y+x) \implies 2y=x$. Also $10y+x=3xy$. Substituting $x=2y$: $12y=6y^{2} \implies y=2,\ x=4$. Number $= 24$. So Assertion is false (number is 24, not 25).
Reason: Let numerator $= x$, so fraction $= \dfrac{x}{2x+1}$.
$$\frac{x}{2x+1}+\frac{2x+1}{x}=\frac{58}{21} \implies 11x^{2}-26x-21=0 \implies (x-3)(11x+7)=0 \implies x=3$$ Fraction $= \dfrac{3}{7}$. Reason is true.
Assertion: Let the unit’s digit be $x$ and ten’s digit be $y$. Then $10y+x=4(y+x) \implies 2y=x$. Also $10y+x=3xy$. Substituting $x=2y$: $12y=6y^{2} \implies y=2,\ x=4$. Number $= 24$. So Assertion is false (number is 24, not 25).
Reason: Let numerator $= x$, so fraction $= \dfrac{x}{2x+1}$.
$$\frac{x}{2x+1}+\frac{2x+1}{x}=\frac{58}{21} \implies 11x^{2}-26x-21=0 \implies (x-3)(11x+7)=0 \implies x=3$$ Fraction $= \dfrac{3}{7}$. Reason is true.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If the coefficient of $x^{2}$ and the constant term have the same sign and if the coefficient of the $x$ term is zero, then the quadratic equation has no real roots.
Reason (R): The equation $13\sqrt{3}x^{2}+10x+\sqrt{3}=0$ has no real roots.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If the coefficient of $x^{2}$ and the constant term have the same sign and if the coefficient of the $x$ term is zero, then the quadratic equation has no real roots.
Reason (R): The equation $13\sqrt{3}x^{2}+10x+\sqrt{3}=0$ has no real roots.
Answer
Option (A) is correct.
Assertion: If $b=0$, discriminant $D = -4ac < 0$ when $ac > 0$ (same sign). So no real roots — Assertion is true.
Reason: $a=13\sqrt{3},\ b=10,\ c=\sqrt{3}$.
$D = 100 – 4(13\sqrt{3})(\sqrt{3}) = 100-156 = -56 < 0$. So no real roots — Reason is true.
Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
Assertion: If $b=0$, discriminant $D = -4ac < 0$ when $ac > 0$ (same sign). So no real roots — Assertion is true.
Reason: $a=13\sqrt{3},\ b=10,\ c=\sqrt{3}$.
$D = 100 – 4(13\sqrt{3})(\sqrt{3}) = 100-156 = -56 < 0$. So no real roots — Reason is true.
Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
The value(s) of $k$ for which the quadratic equation $2x^{2}+kx+2=0$ has equal roots is:
Solution
Option (B) is correct.
Comparing $2x^{2}+kx+2=0$ with $ax^{2}+bx+c=0$: $a=2,\ b=k,\ c=2$.
For equal roots, $D=0$: $$k^{2}-4 \times 2 \times 2 = 0 \implies k^{2}-16=0 \implies (k+4)(k-4)=0 \implies k=\pm 4$$
Comparing $2x^{2}+kx+2=0$ with $ax^{2}+bx+c=0$: $a=2,\ b=k,\ c=2$.
For equal roots, $D=0$: $$k^{2}-4 \times 2 \times 2 = 0 \implies k^{2}-16=0 \implies (k+4)(k-4)=0 \implies k=\pm 4$$
Three students were asked how they would verify their solution of a quadratic equation, $(x-2)(x-5)=0$. Shown below are their responses.
Student 1 said, “In the first bracket, $x$ must equal 2, and in the second bracket, $x$ must equal 5. So, $(2-2)(5-5)=0$.”
Student 2 said, “In the first bracket, $x$ must equal 2, but in the second bracket, $x$ can have any real number value. For example, $(2-2)(3-5)=0$ or $(2-2)(10-5)=0$.”
Student 3 said, “Both brackets should always have the same $x$ value. So, $x$ is either 2 or 5 in both brackets. For example, $(2-2)(2-5)=0$ and $(5-2)(5-5)=0$.”
Whose response is correct?
(A) Only Student 1
(B) Only Student 3
(C) Only Students 1 and 2
(D) All students 1, 2 and 3
Student 1 said, “In the first bracket, $x$ must equal 2, and in the second bracket, $x$ must equal 5. So, $(2-2)(5-5)=0$.”
Student 2 said, “In the first bracket, $x$ must equal 2, but in the second bracket, $x$ can have any real number value. For example, $(2-2)(3-5)=0$ or $(2-2)(10-5)=0$.”
Student 3 said, “Both brackets should always have the same $x$ value. So, $x$ is either 2 or 5 in both brackets. For example, $(2-2)(2-5)=0$ and $(5-2)(5-5)=0$.”
Whose response is correct?
(A) Only Student 1
(B) Only Student 3
(C) Only Students 1 and 2
(D) All students 1, 2 and 3
Solution
Option (B) is correct.
$(x-2)(x-5)=0 \implies x-2=0$ or $x-5=0 \implies x=2$ or $x=5$.
The same $x$ value must be used in both brackets. Student 3’s reasoning is correct: both brackets take the same value of $x$, so $x$ is either 2 or 5 in both brackets simultaneously.
$(x-2)(x-5)=0 \implies x-2=0$ or $x-5=0 \implies x=2$ or $x=5$.
The same $x$ value must be used in both brackets. Student 3’s reasoning is correct: both brackets take the same value of $x$, so $x$ is either 2 or 5 in both brackets simultaneously.
Which of the following is not a quadratic equation?
(A) $x^2 + 3x – 5 = 0$
(B) $x^2 + x^3 + 2 = 0$
(C) $3 + x + x^2 = 0$
(D) $x^2 – 9 = 0$
(A) $x^2 + 3x – 5 = 0$
(B) $x^2 + x^3 + 2 = 0$
(C) $3 + x + x^2 = 0$
(D) $x^2 – 9 = 0$
Solution
Option (B) is correct.
A quadratic equation is of the form $ax^{2}+bx+c=0\ (a \neq 0)$ where the highest power of $x$ must be 2.
(A) $x^{2}+3x-5=0$ — highest power 2 → quadratic.
(B) $x^{2}+x^{3}+2=0$ — highest power 3 → cubic, not quadratic.
(C) $3+x+x^{2}=0$ — can be rewritten as $x^{2}+x+3=0$ → quadratic.
(D) $x^{2}-9=0$ — highest power 2 → quadratic.
A quadratic equation is of the form $ax^{2}+bx+c=0\ (a \neq 0)$ where the highest power of $x$ must be 2.
(A) $x^{2}+3x-5=0$ — highest power 2 → quadratic.
(B) $x^{2}+x^{3}+2=0$ — highest power 3 → cubic, not quadratic.
(C) $3+x+x^{2}=0$ — can be rewritten as $x^{2}+x+3=0$ → quadratic.
(D) $x^{2}-9=0$ — highest power 2 → quadratic.
The quadratic equation has degree:
Solution
Option (C) is correct.
The degree of a polynomial equation is the highest power of the variable present. A quadratic equation is of the form $ax^{2}+bx+c=0\ (a \neq 0)$. The highest power of $x$ is 2.
The degree of a polynomial equation is the highest power of the variable present. A quadratic equation is of the form $ax^{2}+bx+c=0\ (a \neq 0)$. The highest power of $x$ is 2.
The polynomial equation $x(x+1)+8=(x+2)(x-2)$ is:
Solution
Option (A) is correct.
LHS: $x(x+1)+8 = x^{2}+x+8$
RHS: $(x+2)(x-2) = x^{2}-4$
$$x^{2}+x+8 = x^{2}-4 \implies x+12=0$$ The highest power of $x$ in the final equation is 1, so it is a linear equation.
LHS: $x(x+1)+8 = x^{2}+x+8$
RHS: $(x+2)(x-2) = x^{2}-4$
$$x^{2}+x+8 = x^{2}-4 \implies x+12=0$$ The highest power of $x$ in the final equation is 1, so it is a linear equation.
The equation $(x-2)^{2}+1=2x-3$ is a:
Solution
Option (B) is correct.
Expanding LHS: $(x-2)^{2}+1 = x^{2}-4x+4+1 = x^{2}-4x+5$
$$x^{2}-4x+5 = 2x-3 \implies x^{2}-6x+8=0$$ The highest power of $x$ is 2, so it is a quadratic equation.
Expanding LHS: $(x-2)^{2}+1 = x^{2}-4x+4+1 = x^{2}-4x+5$
$$x^{2}-4x+5 = 2x-3 \implies x^{2}-6x+8=0$$ The highest power of $x$ is 2, so it is a quadratic equation.
The roots of the quadratic equation $6x^{2}-x-2=0$ are:
Solution
Option (C) is correct.
Splitting the middle term: $$6x^{2}-4x+3x-2=0 \implies 2x(3x-2)+1(3x-2)=0 \implies (3x-2)(2x+1)=0$$ $$x=\frac{2}{3} \quad \text{or} \quad x=-\frac{1}{2}$$
Splitting the middle term: $$6x^{2}-4x+3x-2=0 \implies 2x(3x-2)+1(3x-2)=0 \implies (3x-2)(2x+1)=0$$ $$x=\frac{2}{3} \quad \text{or} \quad x=-\frac{1}{2}$$
The quadratic equation whose roots are $1$ and $-\dfrac{1}{2}$ is:
Solution
Option (B) is correct.
Let $\alpha=1,\ \beta=-\dfrac{1}{2}$.
Sum of roots $= 1-\dfrac{1}{2}=\dfrac{1}{2}$; Product of roots $= 1 \times \left(-\dfrac{1}{2}\right)=-\dfrac{1}{2}$.
Equation: $x^{2}-\dfrac{1}{2}x-\dfrac{1}{2}=0$. Multiplying by 2: $$2x^{2}-x-1=0$$
Let $\alpha=1,\ \beta=-\dfrac{1}{2}$.
Sum of roots $= 1-\dfrac{1}{2}=\dfrac{1}{2}$; Product of roots $= 1 \times \left(-\dfrac{1}{2}\right)=-\dfrac{1}{2}$.
Equation: $x^{2}-\dfrac{1}{2}x-\dfrac{1}{2}=0$. Multiplying by 2: $$2x^{2}-x-1=0$$
The roots of the quadratic equation $x+\dfrac{1}{x}=3,\ x \neq 0$ are:
Solution
Option (C) is correct.
Multiplying both sides by $x$: $x^{2}+1=3x \implies x^{2}-3x+1=0$.
Using the quadratic formula: $$x = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$$
Multiplying both sides by $x$: $x^{2}+1=3x \implies x^{2}-3x+1=0$.
Using the quadratic formula: $$x = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$$
The roots of the quadratic equation $2x^{2}-2\sqrt{2}\,x+1=0$ are:
Solution
Option (A) is correct.
$a=2,\ b=-2\sqrt{2},\ c=1$.
$$x = \frac{2\sqrt{2} \pm \sqrt{(-2\sqrt{2})^{2}-4(2)(1)}}{4} = \frac{2\sqrt{2} \pm \sqrt{8-8}}{4} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$$ The discriminant is zero, so the equation has equal roots both equal to $\dfrac{1}{\sqrt{2}}$.
$a=2,\ b=-2\sqrt{2},\ c=1$.
$$x = \frac{2\sqrt{2} \pm \sqrt{(-2\sqrt{2})^{2}-4(2)(1)}}{4} = \frac{2\sqrt{2} \pm \sqrt{8-8}}{4} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$$ The discriminant is zero, so the equation has equal roots both equal to $\dfrac{1}{\sqrt{2}}$.
The speed of a motor boat in still water is 20 km/hr. To cover a distance of 15 km, the boat takes 1 hour more while going upstream than while going downstream.
(i) Let the speed of the stream be $x$ km/hr. The speed of the motorboat in upstream will be:
(A) 20 km/hr
(B) $(20+x)$ km/hr
(C) $(20-x)$ km/hr
(D) 2 km/hr
(ii) What is the relation between speed, distance, and time?
(A) $\text{speed} = \dfrac{\text{distance}}{\text{time}}$
(B) $\text{distance} = \dfrac{\text{speed}}{\text{time}}$
(C) $\text{time} = \text{speed} \times \text{distance}$
(D) $\text{speed} = \text{distance} \times \text{time}$
(iii) Which is the correct quadratic equation for the speed of the current?
(A) $x^{2}+30x-200=0$
(B) $x^{2}+20x-400=0$
(C) $x^{2}+30x-400=0$
(D) $x^{2}-20x-400=0$
(iv) What is the speed of the current?
(A) 20 km/hour
(B) 10 km/hour
(C) 15 km/hour
(D) 25 km/hour
(v) What is the speed of the current? (confirm)
(A) 20 km/hour
(B) 10 km/hour
(C) 15 km/hour
(D) 25 km/hour
(i) Let the speed of the stream be $x$ km/hr. The speed of the motorboat in upstream will be:
(A) 20 km/hr
(B) $(20+x)$ km/hr
(C) $(20-x)$ km/hr
(D) 2 km/hr
(ii) What is the relation between speed, distance, and time?
(A) $\text{speed} = \dfrac{\text{distance}}{\text{time}}$
(B) $\text{distance} = \dfrac{\text{speed}}{\text{time}}$
(C) $\text{time} = \text{speed} \times \text{distance}$
(D) $\text{speed} = \text{distance} \times \text{time}$
(iii) Which is the correct quadratic equation for the speed of the current?
(A) $x^{2}+30x-200=0$
(B) $x^{2}+20x-400=0$
(C) $x^{2}+30x-400=0$
(D) $x^{2}-20x-400=0$
(iv) What is the speed of the current?
(A) 20 km/hour
(B) 10 km/hour
(C) 15 km/hour
(D) 25 km/hour
(v) What is the speed of the current? (confirm)
(A) 20 km/hour
(B) 10 km/hour
(C) 15 km/hour
(D) 25 km/hour
Answer
(i) Option (C) is correct.
When moving upstream, the stream opposes motion. Speed upstream $= (20-x)$ km/hr.
(ii) Option (A) is correct.
$\text{speed} = \dfrac{\text{distance}}{\text{time}}$
(iii) Option (C) is correct.
Time upstream $= \dfrac{15}{20-x}$; Time downstream $= \dfrac{15}{20+x}$.
$$\frac{15}{20-x}-\frac{15}{20+x}=1 \implies \frac{30x}{400-x^{2}}=1 \implies 30x=400-x^{2}$$ $$x^{2}+30x-400=0$$
(iv) Option (B) is correct.
$(x+40)(x-10)=0 \implies x=10$ (speed cannot be negative).
Speed of current $=$ 10 km/hour.
(v) Option (B) is correct.
Confirmed from part (iv): Speed of current $=$ 10 km/hour.
When moving upstream, the stream opposes motion. Speed upstream $= (20-x)$ km/hr.
(ii) Option (A) is correct.
$\text{speed} = \dfrac{\text{distance}}{\text{time}}$
(iii) Option (C) is correct.
Time upstream $= \dfrac{15}{20-x}$; Time downstream $= \dfrac{15}{20+x}$.
$$\frac{15}{20-x}-\frac{15}{20+x}=1 \implies \frac{30x}{400-x^{2}}=1 \implies 30x=400-x^{2}$$ $$x^{2}+30x-400=0$$
(iv) Option (B) is correct.
$(x+40)(x-10)=0 \implies x=10$ (speed cannot be negative).
Speed of current $=$ 10 km/hour.
(v) Option (B) is correct.
Confirmed from part (iv): Speed of current $=$ 10 km/hour.
Some students planned a picnic to Wayanad as part of their Scout and Guide activities. The total budget for the picnic was ₹2000 per student. However, 5 students failed to attend, and the contribution for each remaining student increased by ₹20.
(i) If $x$ is the number of students planned for the picnic, write the quadratic equation that describes the situation.
(ii) Find the number of students planned for the picnic.
(iii) Find the contribution by each student when the picnic was planned.
(i) If $x$ is the number of students planned for the picnic, write the quadratic equation that describes the situation.
(ii) Find the number of students planned for the picnic.
(iii) Find the contribution by each student when the picnic was planned.
Answer
(i) After 5 students dropped out, new contribution $= \dfrac{2000}{x-5}$.
Given the contribution increased by ₹20: $$\frac{2000}{x-5} = \frac{2000}{x}+20 \implies 20x^{2}-100x-10000=0 \implies \boxed{x^{2}-5x-500=0}$$
(ii) $(x-25)(x+20)=0 \implies x=25$ (reject negative value).
Number of students planned $=$ 25.
(iii) Initial contribution per student $= \dfrac{2000}{25} = $ ₹80.
Given the contribution increased by ₹20: $$\frac{2000}{x-5} = \frac{2000}{x}+20 \implies 20x^{2}-100x-10000=0 \implies \boxed{x^{2}-5x-500=0}$$
(ii) $(x-25)(x+20)=0 \implies x=25$ (reject negative value).
Number of students planned $=$ 25.
(iii) Initial contribution per student $= \dfrac{2000}{25} = $ ₹80.
An auditorium was booked for School Annual Day Celebrations, and the seats are arranged such that the number of rows equals the number of seats in each row. When the number of rows was doubled and the number of seats per row was reduced by 10, the total number of seats increased by 300.
(i) If $x$ is the number of rows in the original arrangement, which quadratic equation describes the situation?
(ii) How many rows are there in the original arrangement?
(iii) How many seats are there in the auditorium in the original arrangement?
(i) If $x$ is the number of rows in the original arrangement, which quadratic equation describes the situation?
(ii) How many rows are there in the original arrangement?
(iii) How many seats are there in the auditorium in the original arrangement?
Answer
(i) Original seats $= x^{2}$. New arrangement: $2x$ rows, $(x-10)$ seats per row.
$$2x(x-10) = x^{2}+300 \implies 2x^{2}-20x=x^{2}+300 \implies \boxed{x^{2}-20x-300=0}$$
(ii) $(x-30)(x+10)=0 \implies x=30$ (reject negative value).
Number of rows in original arrangement $=$ 30.
(iii) Total seats $= 30 \times 30 = $ 900.
$$2x(x-10) = x^{2}+300 \implies 2x^{2}-20x=x^{2}+300 \implies \boxed{x^{2}-20x-300=0}$$
(ii) $(x-30)(x+10)=0 \implies x=30$ (reject negative value).
Number of rows in original arrangement $=$ 30.
(iii) Total seats $= 30 \times 30 = $ 900.
Noida Authority decided to make a park for the people. A grassy park is in the form of a rectangle having length 20 m and breadth 14 m. At the centre of the park, there is a rectangular pool. Around the pool, there is a path of equal width, and the area of the path is 120 m².
(i) If the pool is at a distance of $x$ metres from all sides of the park, the length and breadth of the pool (in metres) will be:
(A) $(20-4x),\ (14-4x)$
(B) $(20-x),\ (14-x)$
(C) $(20-2x),\ (14-2x)$
(D) $(20+2x),\ (14+2x)$
(ii) If the area of the path is 120 m², the quadratic equation in $x$ is:
(A) $x^{2}-17x+48=0$
(B) $x^{2}-17x+30=0$
(C) $x^{2}-17x+36=0$
(D) $x^{2}-16x+138=0$
(iii) Find the nature of the roots of the equation from part (ii).
(A) Real and equal
(B) Real and distinct
(C) Imaginary
(D) None of these
(iv) Width of the path is:
(A) 6 m
(B) 2 m
(C) 12 m
(D) 7 m
(v) The area of the rectangular pool is:
(A) 30 m²
(B) 40 m²
(C) 46 m²
(D) 160 m²
(i) If the pool is at a distance of $x$ metres from all sides of the park, the length and breadth of the pool (in metres) will be:
(A) $(20-4x),\ (14-4x)$
(B) $(20-x),\ (14-x)$
(C) $(20-2x),\ (14-2x)$
(D) $(20+2x),\ (14+2x)$
(ii) If the area of the path is 120 m², the quadratic equation in $x$ is:
(A) $x^{2}-17x+48=0$
(B) $x^{2}-17x+30=0$
(C) $x^{2}-17x+36=0$
(D) $x^{2}-16x+138=0$
(iii) Find the nature of the roots of the equation from part (ii).
(A) Real and equal
(B) Real and distinct
(C) Imaginary
(D) None of these
(iv) Width of the path is:
(A) 6 m
(B) 2 m
(C) 12 m
(D) 7 m
(v) The area of the rectangular pool is:
(A) 30 m²
(B) 40 m²
(C) 46 m²
(D) 160 m²
Answer
(i) Option (C) is correct.
Since the pool is $x$ m from all sides, both length and breadth reduce by $2x$:
Length of pool $= (20-2x)$, Breadth $= (14-2x)$.
(ii) Option (B) is correct.
Area of park $= 280$ m². Area of pool $= (20-2x)(14-2x)$.
$$280-(20-2x)(14-2x)=120 \implies 68x-4x^{2}=120$$ $$\implies x^{2}-17x+30=0$$
(iii) Option (B) is correct.
$D = 17^{2}-4(1)(30) = 289-120 = 169 > 0$ and is a perfect square → roots are real and distinct.
(iv) Option (B) is correct.
$(x-15)(x-2)=0 \implies x=15$ or $x=2$. Only $x=2$ m is valid (must be less than half the breadth).
(v) Option (D) is correct.
Length of pool $= 20-4 = 16$ m; Breadth $= 14-4=10$ m.
Area $= 16 \times 10 = $ 160 m².
Since the pool is $x$ m from all sides, both length and breadth reduce by $2x$:
Length of pool $= (20-2x)$, Breadth $= (14-2x)$.
(ii) Option (B) is correct.
Area of park $= 280$ m². Area of pool $= (20-2x)(14-2x)$.
$$280-(20-2x)(14-2x)=120 \implies 68x-4x^{2}=120$$ $$\implies x^{2}-17x+30=0$$
(iii) Option (B) is correct.
$D = 17^{2}-4(1)(30) = 289-120 = 169 > 0$ and is a perfect square → roots are real and distinct.
(iv) Option (B) is correct.
$(x-15)(x-2)=0 \implies x=15$ or $x=2$. Only $x=2$ m is valid (must be less than half the breadth).
(v) Option (D) is correct.
Length of pool $= 20-4 = 16$ m; Breadth $= 14-4=10$ m.
Area $= 16 \times 10 = $ 160 m².
In a cricket match of World Cup 2016, Ashwin took 2 wickets less than twice the number of wickets taken by Ishant. The product of the wickets taken by Ashwin and Ishant is 24.
(i) If Ishant took $x$ wickets, then wickets taken by Ashwin are:
(A) $(2x+2)$
(B) $(2x-2)$
(C) $(2x-1)$
(D) $(2x-5)$
(ii) The given statement represents in equation form as:
(A) $x^{2}-3x+10=0$
(B) $x^{2}+x-12=0$
(C) $x^{2}-x-12=0$
(D) $x^{2}-2x+12=0$
(iii) If a quadratic equation has real and equal roots, then the condition of discriminant $D$ is:
(A) $D<0$
(B) $D=0$
(C) $D>0$
(D) $D \geq 0$
(iv) The nature of roots of the equation formed by the given statement is:
(A) real and equal
(B) real and distinct
(C) imaginary
(D) real and imaginary
(i) If Ishant took $x$ wickets, then wickets taken by Ashwin are:
(A) $(2x+2)$
(B) $(2x-2)$
(C) $(2x-1)$
(D) $(2x-5)$
(ii) The given statement represents in equation form as:
(A) $x^{2}-3x+10=0$
(B) $x^{2}+x-12=0$
(C) $x^{2}-x-12=0$
(D) $x^{2}-2x+12=0$
(iii) If a quadratic equation has real and equal roots, then the condition of discriminant $D$ is:
(A) $D<0$
(B) $D=0$
(C) $D>0$
(D) $D \geq 0$
(iv) The nature of roots of the equation formed by the given statement is:
(A) real and equal
(B) real and distinct
(C) imaginary
(D) real and imaginary
Answer
(i) Option (B) is correct.
Ashwin’s wickets $= 2x-2$.
(ii) Option (C) is correct.
Product of wickets $= 24$: $$x(2x-2)=24 \implies 2x^{2}-2x-24=0 \implies x^{2}-x-12=0$$
(iii) Option (B) is correct.
For real and equal roots: $D = b^{2}-4ac = 0$.
(iv) Option (B) is correct.
For $x^{2}-x-12=0$: $D = (-1)^{2}-4(1)(-12) = 1+48 = 49 > 0$.
So the roots are real and distinct.
Ashwin’s wickets $= 2x-2$.
(ii) Option (C) is correct.
Product of wickets $= 24$: $$x(2x-2)=24 \implies 2x^{2}-2x-24=0 \implies x^{2}-x-12=0$$
(iii) Option (B) is correct.
For real and equal roots: $D = b^{2}-4ac = 0$.
(iv) Option (B) is correct.
For $x^{2}-x-12=0$: $D = (-1)^{2}-4(1)(-12) = 1+48 = 49 > 0$.
So the roots are real and distinct.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The equation $x^{2}+3x+1=(x-2)^{2}$ is a quadratic equation.
Reason (R): Any equation of the form $ax^{2}+bx+c=0,\ a \neq 0$, is called a quadratic equation.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The equation $x^{2}+3x+1=(x-2)^{2}$ is a quadratic equation.
Reason (R): Any equation of the form $ax^{2}+bx+c=0,\ a \neq 0$, is called a quadratic equation.
Answer
Option (D) is correct.
Expanding: $x^{2}+3x+1 = x^{2}-4x+4 \implies 7x-3=0$.
This is a linear equation, so Assertion (A) is false.
The Reason (R) is true — it correctly defines a quadratic equation.
Expanding: $x^{2}+3x+1 = x^{2}-4x+4 \implies 7x-3=0$.
This is a linear equation, so Assertion (A) is false.
The Reason (R) is true — it correctly defines a quadratic equation.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The value of $k$ is 2, if one root of the quadratic equation $6x^{2}-x-k=0$ is $\dfrac{2}{3}$.
Reason (R): The quadratic equation $ax^{2}+bx+c=0,\ a \neq 0$ has two roots.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The value of $k$ is 2, if one root of the quadratic equation $6x^{2}-x-k=0$ is $\dfrac{2}{3}$.
Reason (R): The quadratic equation $ax^{2}+bx+c=0,\ a \neq 0$ has two roots.
Answer
Option (B) is correct.
Substituting $x=\dfrac{2}{3}$ in $6x^{2}-x-k=0$: $$6\left(\frac{2}{3}\right)^{2}-\frac{2}{3}-k=0 \implies \frac{24}{9}-\frac{6}{9}-k=0 \implies \frac{18}{9}=k \implies k=2$$ Assertion is true. Reason is also true (a quadratic has two roots). However, Reason does not explain why $k=2$, so it is NOT the correct explanation.
Substituting $x=\dfrac{2}{3}$ in $6x^{2}-x-k=0$: $$6\left(\frac{2}{3}\right)^{2}-\frac{2}{3}-k=0 \implies \frac{24}{9}-\frac{6}{9}-k=0 \implies \frac{18}{9}=k \implies k=2$$ Assertion is true. Reason is also true (a quadratic has two roots). However, Reason does not explain why $k=2$, so it is NOT the correct explanation.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If one root of the quadratic equation $6x^{2}-x-k=0$ is $\dfrac{2}{3}$, then the value of $k=2$.
Reason (R): The quadratic equation $ax^{2}+bx+c=0,\ a \neq 0$ has at most two roots.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If one root of the quadratic equation $6x^{2}-x-k=0$ is $\dfrac{2}{3}$, then the value of $k=2$.
Reason (R): The quadratic equation $ax^{2}+bx+c=0,\ a \neq 0$ has at most two roots.
Answer
Option (B) is correct.
Substituting $x=\dfrac{2}{3}$: $$6 \times \frac{4}{9} – \frac{2}{3} – k = 0 \implies \frac{8}{3}-\frac{2}{3}=k \implies k=\frac{6}{3}=2$$ Assertion is true. Reason is also true (a quadratic has at most two roots). However, Reason does not directly explain why $k=2$, so it is NOT the correct explanation of Assertion.
Substituting $x=\dfrac{2}{3}$: $$6 \times \frac{4}{9} – \frac{2}{3} – k = 0 \implies \frac{8}{3}-\frac{2}{3}=k \implies k=\frac{6}{3}=2$$ Assertion is true. Reason is also true (a quadratic has at most two roots). However, Reason does not directly explain why $k=2$, so it is NOT the correct explanation of Assertion.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): $(2x-1)^{2}-4x^{2}+5=0$ is not a quadratic equation.
Reason (R): An equation of the form $ax^{2}+bx+c=0,\ a \neq 0$, where $a,b,c \in \mathbb{R}$ is called a quadratic equation.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): $(2x-1)^{2}-4x^{2}+5=0$ is not a quadratic equation.
Reason (R): An equation of the form $ax^{2}+bx+c=0,\ a \neq 0$, where $a,b,c \in \mathbb{R}$ is called a quadratic equation.
Answer
Option (B) is correct.
Expanding: $(2x-1)^{2}-4x^{2}+5 = 4x^{2}-4x+1-4x^{2}+5 = -4x+6=0$.
This simplifies to a linear equation (degree 1), not quadratic.
So Assertion (A) is true. Reason (R) is also true (correct definition of quadratic equation). However, Reason does not directly explain why the given expression becomes linear — so it is NOT the correct explanation.
Expanding: $(2x-1)^{2}-4x^{2}+5 = 4x^{2}-4x+1-4x^{2}+5 = -4x+6=0$.
This simplifies to a linear equation (degree 1), not quadratic.
So Assertion (A) is true. Reason (R) is also true (correct definition of quadratic equation). However, Reason does not directly explain why the given expression becomes linear — so it is NOT the correct explanation.
Frequently Asked Questions
What is the chapter Quadratic Equations about in CBSE Class 10 Maths? ⌄
Quadratic Equations is Chapter 4 of CBSE Class 10 Maths. It introduces your child to polynomial equations of degree 2 of the form $ax^{2}+bx+c=0$ where $a \neq 0$. The chapter covers finding roots by factorisation, completing the square, and the quadratic formula, as well as determining the nature of roots using the discriminant. Students also explore real-life word problems involving areas, speeds, and time that lead to quadratic equations.
How many marks does Quadratic Equations carry in the Class 10 board exam? ⌄
Quadratic Equations typically carries around 10–12 marks in the CBSE Class 10 Maths board examination. Questions appear across multiple formats — MCQs (1 mark), Short Answer questions (2 marks), and Case Study Based Questions (4–5 marks). Mastering this chapter gives your child a strong advantage in securing marks from algebra-related sections of the paper.
What are the most important topics in Quadratic Equations for Class 10? ⌄
The key topics your child should focus on include: (1) Identifying whether an equation is quadratic, (2) Solving by factorisation method, (3) Completing the square, (4) Using the quadratic formula, (5) The discriminant ($D = b^2 – 4ac$) and the nature of roots — equal, distinct, or no real roots, and (6) Word problems based on speed, time, distance, areas, and numbers. Case study and assertion-reason questions are also increasingly important.
What are common mistakes students make in Quadratic Equations? ⌄
Students commonly make the following mistakes: forgetting to check whether an equation is actually quadratic before solving it (some equations simplify to linear equations after expansion), accepting negative values of physical quantities like speed, length, or time as valid roots, errors in splitting the middle term during factorisation, and sign mistakes when calculating the discriminant. Practising a wide variety of questions — including the assertion-reason type — helps your child avoid these pitfalls in the exam.
How does Angle Belearn help students master Quadratic Equations? ⌄
Angle Belearn’s CBSE specialists curate competency based questions that mirror the latest CBSE exam pattern — including MCQs, case studies, and assertion-reason questions — all with detailed step-by-step solutions. Our expert tutors help students understand concepts deeply rather than just memorising formulas, ensuring your child can tackle any variation of a Quadratic Equations problem in the board exam. One-on-one sessions and regular practice tests build lasting confidence.
