CBSE Class 10 Mathematics Quadratic Equation Worksheet Set S

Question. In a class test, the sum of Arun's marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects.
Answer: Let Arun marks in hindi be \( x \) and marks in english be \( y \).
Then, according to question, we have
\( x + y = 30 \) ...(i)
and \( (x + 2) (y - 3) = 210 \) ...(ii)
from equation (i) put \( x = 30 - y \) in equation (ii)
\( (30 - y + 2) (y - 3) = 210 \)
\( (32 - y) (y - 3) = 210 \)
\( 32y - 96 - y^2 + 3y = 210 \)
\( y^2 - 35y + 306 = 0 \)
\( y^2 - 18y - 17y + 306 = 0 \)
\( y (y - 18) - 17 (y - 18) = 0 \)
\( (y - 18) (y - 17) = 0 \)
\( y = 18, 17 \)
Put \( y = 18 \) and 17 in equation (i), we get
\( x = 12, 13 \)
Hence his marks in hindi can be 12 or 13 and in english his marks can be 18 or 17.

Question. In a flight of 600 km, an aircraft was slowed down due to bad weather. The average speed of the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the duration of flight.
Answer: Let the original speed of the aircraft be \( x \) km/hr.
So, time taken to cover 600 km = \( \frac{600}{x} \) hours
Now, Reduced speed = \( (x - 200) \) km/hr.
So, time taken to cover 600 km at reduced speed = \( \frac{600}{(x - 200)} \) hours
According to the question,
\( \Rightarrow \frac{600}{x - 200} - \frac{600}{x} = \frac{1}{2} \)
\( \Rightarrow \frac{1}{x - 200} - \frac{1}{x} = \frac{1}{1200} \)
\( \Rightarrow \frac{x - (x - 200)}{x(x - 200)} = \frac{1}{1200} \)
\( \Rightarrow \frac{200}{x^2 - 200x} = \frac{1}{1200} \)
\( \Rightarrow x^2 - 200x = 240000 \)
\( \Rightarrow x^2 - 200x - 240000 = 0 \)
\( \Rightarrow x^2 - 600x + 400x - 240000 = 0 \)
\( \Rightarrow x(x - 600) + 400 (x - 600) = 0 \)
\( \Rightarrow (x - 600) (x + 400) = 0 \)
\( \Rightarrow x - 600 = 0 \) or \( x + 400 = 0 \)
\( \Rightarrow x = 600 \) or \( x = -400 \)
\( \Rightarrow x = 600 \) [\( \because \) speed cannot be negative]
\( \therefore \) Original speed of aircraft = 600 km/hr
\( \therefore \) Original duration of flight = \( \frac{600}{600} = 1 \) hour
And, Increased duration of flight = \( \left(1 + \frac{1}{2}\right) \) hours = 1.5 hours.

Question. Two taps running together can fill a cistern in \( 2\frac{8}{11} \) minutes. If one tap takes 1 minute more than the other to fill the cistern, find the time in which each tap separately can fill the cistern.
Answer: Let the time taken by one tap to fill the tank be \( x \) minutes.
Then, other tap will fill the tank in \( (x + 1) \) minutes
Now, according to the question,
\( \frac{1}{x} + \frac{1}{x + 1} = \frac{1}{2\frac{8}{11}} \)
\( \Rightarrow \frac{x + 1 + x}{x (x + 1)} = \frac{1}{30/11} \)
\( \Rightarrow \frac{2x + 1}{x (x + 1)} = \frac{11}{30} \)
\( \Rightarrow (2x + 1)30 = 11x(x + 1) \)
\( \Rightarrow 60x + 30 = 11x^2 + 11x \)
\( \Rightarrow 11x^2 - 49x - 30 = 0 \)
\( \therefore x = \frac{-(-49) \pm \sqrt{(-49)^2 - 4 \times 11 \times (-30)}}{2 \times 11} \)
\( = \frac{49 \pm \sqrt{2401 + 1320}}{22} \)
\( = \frac{49 \pm \sqrt{3721}}{22} = \frac{49 \pm 61}{22} \)
\( \Rightarrow x = \frac{49 + 61}{22} \) or \( x = \frac{49 - 61}{22} \)
\( \Rightarrow x = \frac{110}{22} \) or \( x = -\frac{12}{22} \)
\( \Rightarrow x = 5 \) or \( x = -\frac{12}{22} \)
\( \Rightarrow x = 5 \) [\( \because \) Time cannot be negative]
Hence, the two taps will take 5 minutes and 6 minutes respectively.

Question. A passenger while boarding a plane hurt herself and the captain called for immediate medical attention. Thus, the plane left 30 minutes behind schedule. In order to reach its destination 1500 km away on time, the speed was increased by 100 km/hr from its usual speed. Find the usual speed.
Answer: Let the usual speed be \( x \) km/hr.
We know, \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
So, Usual time taken = \( \frac{1500}{x} \) hr.
Increased speed = \( (x + 100) \) km/hr
Thus, new time taken = \( \frac{1500}{x + 100} \)
Now, according to the question,
\( \frac{1500}{x + 100} + \frac{1}{2} = \frac{1500}{x} \)
\( \Rightarrow \frac{1500}{x} - \frac{1500}{x + 100} = \frac{1}{2} \)
\( \Rightarrow 1500 \left( \frac{1}{x} - \frac{1}{x + 100} \right) = \frac{1}{2} \)
\( \Rightarrow 1500 \left( \frac{x + 100 - x}{x(x + 100)} \right) = \frac{1}{2} \)
\( \Rightarrow 1500 \left( \frac{100}{x(x + 100)} \right) = \frac{1}{2} \)
\( \Rightarrow x(x + 100) = 300000 \)
\( \Rightarrow x^2 + 100x - 300000 = 0 \)
\( \Rightarrow x^2 + 600x - 500x - 300000 = 0 \)
\( \Rightarrow x(x + 600) - 500(x + 600) = 0 \)
\( \Rightarrow (x + 600) (x - 500) = 0 \)
\( \Rightarrow x = 500 \) and \( -600 \)
As speed cannot be negative, so the original speed of the plane is 500 km/hr.

Question. The sum of the areas of 2 squares is 400 \( m^2 \). If the difference of their perimeters is 16 m, then find the sides of the two squares.
Answer: Let the sides of the two squares be \( x \) m and \( y \) m respectively.
Thus, Area of first square = \( x^2 \) \( m^2 \) And Perimeter of first square = \( 4x \) m
Also, Area of second square = \( y^2 \) \( m^2 \) and Perimeter of second square = \( 4y \) m
Now, \( x^2 + y^2 = 400 \) ...(i)
and \( 4(x - y) = 16 \)
\( \Rightarrow y = x - 4 \) ...(ii)
From equation (i),
\( x^2 + y^2 = 400 \)
\( \Rightarrow x^2 + (x - 4)^2 = 400 \) [From (ii)]
\( \Rightarrow x^2 + x^2 - 8x + 16 = 400 \)
\( \Rightarrow 2x^2 - 8x - 384 = 0 \)
\( \Rightarrow x^2 - 4x - 192 = 0 \)
\( \Rightarrow x^2 - 16x + 12x - 192 = 0 \)
\( \Rightarrow x(x - 16) + 12(x - 16) = 0 \)
\( \Rightarrow (x + 12)(x - 16) = 0 \)
\( \Rightarrow x = -12 \) and 16
As length cannot be negative so we take \( x = 16 \) m
and \( y = 16 - 4 = 12 \) m
Thus, the sides of the two squares are 16 m and 12 m long.

Question. A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of slower train is 10 km/hr less than that of faster train, find the speed of two trains.
Answer: Let the speed of slower train be \( x \) km/hr.
Then, speed of faster train = \( (x + 10) \)km/hr.
We know, \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
So, according to the question,
\( \frac{200}{x} - \frac{200}{x + 10} = 1 \)
\( \Rightarrow 200 \left[ \frac{1}{x} - \frac{1}{x + 10} \right] = 1 \)
\( \Rightarrow 200 \left[ \frac{x + 10 - x}{x(x + 10)} \right] = 1 \)
\( \Rightarrow \frac{2000}{x(x + 10)} = 1 \)
\( \Rightarrow x^2 + 10x = 2000 \)
\( \Rightarrow x^2 + 10x - 2000 = 0 \)
\( \Rightarrow x^2 + (50 - 40)x - 2000 = 0 \)
\( \Rightarrow x^2 + 50x - 40x - 2000 = 0 \)
\( \Rightarrow x(x + 50) - 40(x + 50) = 0 \)
\( \Rightarrow (x - 40) (x + 50) = 0 \)
\( \Rightarrow x - 40 = 0 \) or \( x + 50 = 0 \)
\( \Rightarrow x = 40 \) or \( x = -50 \)
Since, speed cannot be negative
\( \therefore x = 40 \)
Hence, the speed of two trains are 40 km/hr and 50 km/hr respectively.

Case Study

A ball is thrown vertically upwards, from a roof top, which is 70 m above the ground. The ball once thrown, reach a maximum vertical height and then fall back again to the ground. The height of the ball from the ground at the time ‘t’ is ‘h’, which is given in the form of a quadratic equation represents as : \( h = -16t^2 + 64t + 80 \)

Question. What is the height reached by the ball after 1 second?
(a) 128 m
(b) 130 m
(c) 125 m
(d) 160 m
Answer: (a) After 1 second, height of the ball is
\( P(t) = h = -16t^2 + 64t + 80 \)
\( P(1) = h = -16(1)^2 + 64(1) + 80 \)
\( = -16 + 144 = 128 \)

Question. Calculate the maximum height reached by the ball.
(a) 144 m
(b) 150 m
(c) 124 m
(d) 164 m
Answer: (a) Rearrange the given equation by completing the square method, we get
\( h = -16(t^2 - 4t - 5) \)
\( = -16[(t - 2)^2 - 9] \)
\( = -16(t - 2)^2 + 144 \)
Height is maximum, when \( t = 2 \) sec then, \( h = 144 \) m

Question. Find the roots of the given quadratic equation.
(a) 0, – 5
(b) 0, 5
(c) 5, 1
(d) 5, – 1
Answer: (d) Given, \( P(t) = -16t^2 + 84t + 80 = 0 \)
Then, \( t^2 - 4t - 5 = 0 \)
\( \Rightarrow t^2 - 5t + t - 5 = 0 \)
\( \Rightarrow t(t - 5) + 1(t - 5) = 0 \)
\( \Rightarrow t = 5, -1 \)

Question. What will be height of the ball at t = 4 sec?
(a) 30
(b) 50
(c) 80
(d) 120
Answer: (c) \( P(t) = -16t^2 + 64t + 80 \)
\( P(4) = -16(4)^2 + 64 \times 4 + 80 \)
\( = -256 + 256 + 80 = 80 \)

Question. The shape of the graph of quadratic polynomial is a :
(a) straight line
(b) circle
(c) parabola
(d) ellipse
Answer: (c) Parabola.

A company wants to make frames as part of a new product that they are launching. This frame is rectangular in shape, and cut out of a sheet of steel. In this frame a smaller rectangle will be cut out from the bigger sheet of steel, so as to lower down its weight. The inner dimensions of the steel frame are 6 cm × 11 cm and the width of the frame is ‘x’ cm. The final area of the frame is \( 28\text{ cm}^2 \).

Question. What is the dimensions of outer frame in terms of ‘x’?
(a) 6 – 2x, 11 – 2x
(b) 11 + 2x, 6 – x
(c) 11 + 2x, 6 + 2x
(d) 11 – x, 6 – x
Answer: (c) Length of outer frame = \( (11 + 2x) \) cm
Breadth of outer frame = \( (6 + 2x) \) cm

Question. Form a quadratic equation in terms of x, if the final area of the frame is \( 28\text{ cm}^2 \).
(a) \( 2x^2 + 17x - 14 = 0 \)
(b) \( x^2 + 17x - 14 = 0 \)
(c) \( x^2 + 15x - 15 = 0 \)
(d) \( 2x^2 + 15x - 14 = 0 \)
Answer: (a) Here, Area of frame = Area of outer rectangle – Area of Inner rectangle
\( = (6 + 2x)(11 + 2x) - 6 \times 11 \)
\( = 66 + 22x + 12x + 4x^2 - 66 \)
\( = 34x + 4x^2 \)
\( \Rightarrow 34x + 4x^2 = 28 \)
\( \Rightarrow 2x^2 + 17x - 14 = 0 \)

Question. What is the width of the frame?
(a) 2 cm
(b) 1 cm
(c) 0.5 cm
(d) 0.25 cm
Answer: (b) On the basis of equation obtained in part (ii), we get
\( 2x^2 + 17x - 14 = 0 \)
\( \therefore x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-17 \pm \sqrt{289 + 112}}{2 \times 2} \)
\( = \frac{-17 \pm \sqrt{401}}{4} \approx \frac{-17 \pm 20.02}{4} = 0.8 \) or \( -9.3 \)
\( \approx 1 \)

Question. What is the value of ‘K’, if the given equation \( (9x^2 + 6kx + 4 = 0) \) has equal roots?
(a) + 2, – 2
(b) + 3, – 3
(c) + 4, – 4
(d) + 1, – 1
Answer: (a) Here, \( 9x^2 + 6kx + 4 = 0 \)
For equal roots, \( D = b^2 - 4ac = 0 \)
\( = (6k)^2 - 4 \times 9 \times 4 = 0 \)
\( \Rightarrow 36k^2 - 144 = 0 \)
\( \Rightarrow k^2 = 4 \)
\( \Rightarrow k = \pm 2 \)

Question. What is the highest degree of variable in the quadratic polynomial?
(a) 1
(b) 2
(c) 3
(d) 6
Answer: (b) Quadratic polynomial is of the form \( ax^2 + bx + c = 0 \).
So, the highest degree of quadratic polynomial is 2.

During digging by archaeologist they find some important things regarding the human activity and culture. During their digging they find an ancient coin. The coin is in the shape of a circle with a square inside it from the centre. This coin was handed over to a mathematician. So he formulated a problem regarding this coin, and asked to his students. By measuring he find out that diameter of a circle is 6 cm and he said that side of a square inside it is of length ‘x’ cm.

Question. If area of each face of coin \( 6\pi \) sq. cm. Formulate an equation in ‘x’.
(a) \( x^2 - 3\pi = 0 \)
(b) \( x^2 + y = 0 \)
(c) \( x^2 + \pi = 3 \)
(d) \( x^2 - \pi = 0 \)
Answer: (a) Area of each face of coin = Area of circle – Area of square
\( 6\pi = \pi(3)^2 - (x)^2 \)
\( \Rightarrow x^2 = 9\pi - 6\pi = 3\pi \)
\( \Rightarrow x^2 - 3\pi = 0 \)

Question. Solve the equation obtained in Q (i), to find the side of the square.
(a) 2 cm
(b) 4 cm
(c) 3 cm
(d) 1 cm
Answer: (c) \( x^2 = 3\pi \)
\( x = \sqrt{3\pi} \)
\( = \sqrt{3 \times 3.14} = \sqrt{9.42} \)
\( = 3.06 \approx 3 \) cm

Question. Calculate the ratio of perimeter of square and circle.
(a) 1 : 2
(b) \( \pi : 7 \)
(c) 5 : 2
(d) 2 : \( \pi \)
Answer: (d) \( \frac{\text{Perimeter of square}}{\text{Perimeter of circle}} = \frac{4 \times \text{Side}}{2\pi r} \)
\( = \frac{4 \times 3}{\pi \times 6} = \frac{2 \times 3}{\pi \times 3} = \frac{2}{\pi} \)

Question. A polynomial of ............... degree is called a quadratic polynomial.
(a) first
(b) second
(c) third
(d) fourth
Answer: (b) Second.

Question. Find the roots of the quadratic equation \( x(2x - 5) = 12 \).
(a) \( 2, -\frac{1}{2} \)
(b) 1, 1
(c) \( 4, -\frac{3}{2} \)
(d) \( 12, \frac{5}{2} \)
Answer: (c) \( 2x^2 - 5x - 12 = 0 \)
\( \Rightarrow 2x^2 - 8x + 3x - 12 = 0 \)
\( \Rightarrow 2x(x - 4) + 3(x - 4) = 0 \)
\( \Rightarrow (x - 4) (2x + 3) = 0 \)
\( \Rightarrow x = 4, -3/2 \)

Assertion and Reasoning Based Questions

DIRECTIONS : In the following questions, a statement 1 is followed by statement 2. Mark the correct choice as :
(A) If both statement 1 and statement 2 are true and statement 2 is the correct explanation of statement 1.
(B) If both statement 1 and statement 2 are true, but statement 2 is not the correct explanation of statement 1.
(C) If statement 1 is true, but statement 2 is false.
(D) If statement 1 is false, but statement 2 is true.

Question. Statement 1 : x = – 1 is the solution of the given quadratic equation \( x^2 - 3x + 2 = 0 \). Statement 2 : A real number \( \alpha \) is said to be zero of the quadratic polynomial \( p(x) = ax^2 + bx + c \), if \( p(\alpha) = 0 \).
Answer: (A) Both the statements 1 and 2 are correct and statement 2 is correct reason for statement 1.

Question. Statement 1 : The given equation is a quadratic equation \( 4x^2 + 8x + 2 = 0 \). Statement 2 : A polynomial of degree 2 is a quadratic equation.
Answer: (A) Both statement 1 and 2 are correct. Statement 2 is a correct reason for statement 1 as for an equation to be quadratic its degree should be 2.

Question. Statement 1 : The discriminant of the given quadratic equation \( 2x^2 + 5x + 5 = 0 \) is – 15. Statement 2 : If D < 0, then the quadratic equaton has no real roots.
Answer: (B) Both statements are correct. But statement 2 is not a correct explanation for statement 1. As in statement 1 we need to calculate discriminant, but in statement 2, reason for no real roots is given.

Question. Statement 1 : The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm then the quadratic equation to find the base is \( x^2 - 7x - 60 = 0 \). Statement 2 : In a triangle, the sum of all the three angles is 180°.
Answer: (B) Both the statements are correct. According to statement 1, if we form the equation by using Pythagoras theorem, it is correct. Statement 2 is also correct but is not the correct reason for statement 1.

Archives

Question. For what values of k, the roots of the equation \( x^2 + 4x + k = 0 \) are real?
Answer: \( k \leq 4 \)

Question. If \( x = -\frac{1}{2} \), is a solution of the quadratic equation \( 3x^2 + 2kx - 3 = 0 \), find the value of k.
Answer: \( -\frac{9}{4} \)

Question. If \( x = \frac{2}{3} \) and \( x = -3 \) are roots of the quadratic equation \( ax^2 + 7x + b = 0 \), find the values of \( a \) and \( b \).
Answer: \( a = 3, b = 6 \)

Question. Solve the following quadratic equation for x : \( 9x^2 - 6b^2x - (a^4 - b^4) = 0 \)
Answer: \( \frac{b^2 \pm a^2}{3} \)

Question. Solve the following quadratic equation for x : \( x^2 - 2ax - (4b^2 - a^2) = 0 \)
Answer: \( a \pm 2b \)

Question. If the quadratic equation \( px^2 - 2\sqrt{5}px + 15 = 0 \), has two equal roots then find the value of p.
Answer: 3

Question. Solve for x : \( \sqrt{6x + 7} - (2x - 7) = 0 \)
Answer: \( \frac{3}{2}, 7 \)

Question. Solve for x : \( \frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} = \frac{2}{3}, x \neq 1, 2, 3 \)
Answer: 0, 4

Question. Find x in terms of a, b and c : \( \frac{a}{x - a} + \frac{b}{x - b} = \frac{2c}{x - c}, x \neq a, b, c \)
Answer: \( \frac{ac + bc - 2ab}{a + b - 2c} \)

Question. Solve for x : \( x^2 + 5x - (a^2 + a - 6) = 0 \)
Answer: \( (a - 2), -(a + 3) \)

Question. If the roots of the equation \( (a^2 + b^2) x^2 - 2(ac + bd)x + (c^2 + d^2) = 0 \) are equal, prove that \( \frac{a}{b} = \frac{c}{d} \).
Answer: [Full solution text for proof omitted in original OCR result snippet]

Question. Solve the following quadratic equation for x : \( x^2 + \left( \frac{a}{a + b} + \frac{a + b}{a} \right)x + 1 = 0 \)
Answer: \( -\frac{a}{a + b}, -\frac{a + b}{a} \)

Question. Find that non-zero value of k, for which the quadratic equation \( kx^2 + 1 - 2(k - 1)x + x^2 = 0 \) has equal roots. Hence find the roots of the equation.
Answer: \( \frac{1}{2}, \frac{1}{2} \)

Question. Two water taps together can fill a tank in \( 1\frac{7}{8} \) 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.
Answer: 3 hr., 5 hr.

Question. A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it?
Answer: 12 days

Question. The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is \( \frac{29}{20} \). Find the original fraction.
Answer: \( \frac{7}{10} \)

Question. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metre more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find the length and breadth of the rectangular park.
Answer: 7 m, 4 m

Question. A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck.
Answer: 60 km/h

Question. A motor boat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return donwstream to the same spot. Find the speed of the stream.
Answer: 6 km/h