WBBSE Solutions For Class 10 Maths Chapter 1 Quadratic Equations In One Variable Exercise 1.3

WBBSE Solutions For Class 10 Maths Chapter 1 Quadratic Equations In One Variable Exercise 1.3

Question 1: The difference between two positive whole numbers is 3 and the sum of their squares is 117 by calculating, let us write the two numbers.

Solution: Let one positive number is x & the other is x + 3.

According to the problem,

\(x^2+(x+3)^2=117\)

⇒ \(x^2+x^2+6 x+9-117=0\)

⇒ \(2 x^2+6 x-108=0\)

⇒ \(2 x\left(x^2+3 x-54\right)=0\)

⇒ \(x^2+3 x-54=0\)

⇒ \(x^2+9 x-6 x-54=0\)

⇒ x(x + 9) – 6(x + 9) = 0

(x – 6)(x + 9) = 0

when x – 6 = 0

∴ x = 6

when x + 9 = 0

∴ x = -9 (not possible)

 

Chapter Name Solutions
Chapter 1 Quadratic Equations in One Variable Exercise 1 Exercise 2 Exercise 3 Exercise 4 Exercise 5
Chapter 2 Simple Interest Exercise 1 Exercise 2
Chapter 3 Theorems related to Circle Exercise 1 Exercise 2
Chapter 4 Rectangular Parallelopiped or Cuboid Exercise 1 Exercise 2
Chapter 5 Ratio and Proportion Exercise 1 Exercise 2 Exercise 3
Chapter 6 Compound Interest and Uniform Rate of Increase or Decrease Exercise 1 Exercise 2
Chapter 7 Theorems Related To Angles In A Circle Exercise 1 Exercise 2 Exercise 3 Exercise 4 Exercise 5
Chapter 8 Right Circular Cylinder Exercise 1 Exercise 2
Chapter 9 Quadratic Surd Exercise 1 Exercise 2 Exercise 3
Chapter 10 Theorems Related To Cyclic Quadrilateral Exercise 1
Chapter 11 Construction Of Circumcircle and Incircle Of A Traingle Exercise 1 Exercise 2
Chapter 12 Sphere Exercise 1 Exercise 2
Chapter 13 Variation Exercise 1 Exercise 2
Chapter 14 Partnership Business Exercise 1
Chapter 15 Theorems Related To Tangent To A Circle Exercise 1 Exercise 2
Chapter 16 Right Circular Cone Exercise 1
Chapter 17 Construction Of Tangent To A Circle Exercise 1
Chapter 18 Similarity Exercise 1 Exercise 2 Exercise 3 Exercise 4
Chapter 19 Problems Related To Different Solid Objects Exercise 1
Chapter 20 Trigonometry: Concept of Measurement Of Angle Exercise 1
Chapter 21 Construction: Determination Of Mean Proportional Exercise 1
Chapter 22 Pythagoras Theorem Exercise 1
Chapter 23 Trigonometric Ratios And Trigonometric Identities Exercise 1 Exercise 2 Exercise 3
Chapter 24 Trigonometric Ratios Of Complementary Angle Exercise 1
Chapter 25 Application Of Trigonometric Rations: Heights & Distances Exercise 1
Chapter 26 Statistics: Mean, Median, Ogive, Mode Exercise 1 Exercise 2 Exercise 3 Exercise 4

∴ One number is 6 & the other number is 6 + 3 = 9.

Read and Learn More WBBSE Solutions For Class 10 Maths

Question 2. The base of a triangle is 18m. more than two times its height, if the area of the triangle is 360 sq.m, then let us determine its height of it.

Solution: Let the height of the triangle is xm.

∴ Base of the triangle = (2x + 18)m.

According to the problem,

Area of the triangle = \(\frac{1}{2} x base x height\)

⇒ \(\frac{1}{2} \text { base } \times \text { height }=360\)

⇒ \(\frac{1}{2} \times x(2 x+18)=360\)

⇒ \(\frac{1}{2} \times 2 x(2 x+18)=360\)

⇒ \(x^2+9 x-360=0\)

⇒ \(x^2+24 x-15 x-360=0\)

⇒ x(x + 24) – 15(x + 24) = 0

∴ (x + 24)(x – 15) = 0

When x + 24 = 0

∴ x = -24 [it is not possible]

& when x – 15 = 0

∴ x = 15

∴ Height of the triangle is 15 m.

WBBSE Solutions For Class 10 Maths Chapter 1 Quadratic Equations In One Variable Exercise 1.3

Question 3. If 5 times of a positive whole number is less by 3 than twice of its square, then let us determine the number.

Solution: Let the required number is x.

According to the problem,

\(2 x^2-3=5 x\)

⇒ \(2 x^2-5 x-3=0\)

⇒ \(2 x^2-6 x+x-3=0\)

⇒ 2x(x – 3) + 1(x – 3) = 0

⇒ (x – 3)(2x + 1) = 0

When x – 3 = 0

∴ x = 3

& when 2x + 1 = 0

∴ x = -1/2 (Not possible)

∴ The required number = 3.

Question 4. The distance between two places is 200 km; the time taken by a motor car from one place to another is less by 2 hrs. than the time taken by a Jeep car. If the speed of the motor car is 5 km/hr. more than the speed of the Jeep car, then by calculating let us write the speed of the motor car.

Solution: Let the speed of the car = \(\frac{\mathrm{x} \mathrm{km}}{\mathrm{hr}}\)

& the speed of the Jeep = \(\frac{(x+5) \mathrm{km}}{\mathrm{hr}}\)

To go 200 km the car takes = \(\frac{200}{x}\) hr and the Jeep takes = \(\frac{200}{x+5} h r\)

∴ \(\frac{200}{x}-\frac{200}{x+5}=2\)

⇒ \(200\left(\frac{1}{x}-\frac{1}{x+5}\right)=2\)

⇒ \(\frac{x+5-x}{x(x+5)}=\frac{2}{200}\)

⇒ \(x^2+5 x=500\)

⇒ \(x^2+5 x-500=0\)

⇒ \(x^2+25 x-20 x-500=0\)

⇒ x(x + 25) – 20(x + 25) = 0

∴ (x + 25)(x – 20) = 0

Either x + 25 = 0

∴ x = -25 (Not possible)

or, x – 20 = 0

∴ x = 20

∴ Speed of the car = 20 km/hr.

Question 5. The area of Amita’s rectangular land is 2000 sq.m and the perimeter of it is 180 m. By calculating, let us write the length and breadth of the Amita’s land.

Solution: Let the length of the land xm.

∴ Breadth of the land = \(\frac{2000}{x} \mathrm{~m}\)

Perimeter = 180 m.

∴ \(2\left(x+\frac{2000}{x}\right)=180\)

⇒ \(x+\frac{2000}{x}=90\)

⇒ \(x^2+2000=90 x\)

⇒ \(x^2-90 x+2000=0\)

⇒ \(x^2-(50+40) x+2000=0\)

⇒ \(x^2-50 x-40 x+2000=0\)

⇒ x(x – 50) – 40(x – 50) = 0

∴ (x – 50)(x – 40) = 0

∴ Either x – 50 = 0

∴ x = 50

or, x – 40 = 0

∴ x = 40

∴ Length of the land = 50 m.

& Breadth of the land = \(\frac{2000}{50}=40 \mathrm{~m} \text {. }\)

Question 6. The tens digit of a two-digit number is less by 3 than the unit digit. If the product of the two digits is subtracted from the number, the result is 15, let us write the unit digit of the number by calculation.

Solution: In a two digit number the digit in the unit place is (x + 3).

∴ The digit in the tenth place is x & the number is 10x + (x + 3).

According to the problem,

10x + (x + 3) – x(x + 3) = 15

⇒ \(11 x+3-x^2-3 x-15=0\)

⇒ \(-x^2+8 x-12=0\)

⇒ \(x^2-8 x+12=0\)

⇒ \(x^2-6 x-2 x+12=0\)

⇒ x(x – 6) – 2(x – 6) = 0

⇒ (x – 6)(x – 2) = 0

When x – 6 = 0

∴ x = 6

& when x – 2 = 0

∴ x = 2

∴ the digit in unit place 6 + 3 = 9 or 2 + 3 = 5.

Question 7. There are two pipes in the water reservoir of our school. Two pipes together take 11 1/9 minutes to fill the reservoir. If the two pipes are opened separately, then one pipe would take 5 minutes more time than the other pipe. Let us write by calculating the time taken to fill the reservoir separately by each of the pipes.

Solution: Let the two pipes take x min & (x + 5)min.

To fill the tank respectively.

∴ In 1 min the two pipes fill the portion of the tank \(\frac{1}{x}\) part & \(\frac{1}{x+5}\) part, respectively

∴ In 1 min they together fill \(\left(\frac{1}{x}+\frac{1}{x+5}\right)\) part of the tank.

∴ \(\left(\frac{1}{x}+\frac{1}{x+5}\right) \times 11 \frac{1}{9}=1\)

⇒ \(\frac{x+5+x}{x(x+5)} \times \frac{100}{9}=1\)

⇒ (2x + 5) x 100 = x(x + 5) x 9

⇒ \(9 x^2+45 x=200 x+500\)

⇒ \(9 x^2+45 x-200 x-500=0\)

⇒ \(9 x^2-155 x-500=0\)

⇒ \(9 x^2-180 x+25 x-500=0\)

⇒ 9x(x – 20) + 25(x – 20) = 0

∴ (x – 20)(9x + 25) = 0

Either x – 20 = 0

∴ x = 20

or, 9x + 25 = 0

∴ x = –\(\frac{25}{9}\) (Not possible)

∴ 1st pipe takes 20 min & the 2nd pipe take 25 min.

Question 8. Porna and Pijush together complete a work in 4 days. If they work separately, then the time taken by Porna would be 6 days more than the time taken by Pijush. Let us write by calculating the time taken by Porna alone to complete the work.

Solution: Let Porna alone can do a work in x days &

Pijush alone can do the work in (x + 6) days.

∴ In 1 day, Parna can do \(\frac{1}{x}\) if the work &

Piyush can do \(\frac{1}{x+6}\) par of the work

∴ They together can do in 1 day

\(\left(\frac{1}{x}+\frac{1}{x+6}\right) \text { part }=\frac{x+6+x}{x(x+6)}=\frac{2 x+6}{x(x+6)}\)

∴ They together can do in 4 days = \(\frac{2 x+6}{x(x+6)} \times 4\)

According to the problem,

\(\frac{4(2 x+6)}{x^2+6 x}=1\)

⇒ \(x^2+6 x-8 x-24=0\)

⇒ \(x^2+6 x=8 x+24\)

⇒ \(x^2-2 x-24=0\)

⇒ (x – 6)(x + 4) = 0

Either x – 6 = 0

∴ x = 6

Or, x + 4 = 0

∴ x = -4

∴ Porna alone can do the work in 6 days.

Question 9. If the price of 1 dozen pens is reduced by Rs. 6, then 3 more pens will be purchased in Rs. 30. Before the reduction of price, let us calculate the price of 1 dozen pen.

Solution: Let previous price of 1 dozen pen is Rs. x.

∴ Now price of 1 dozen pen is Rs. (x – 6)

Previously in Rs. 30 no. of pens available = \(\frac{1}{x} \times 30 \text { dozen }=\frac{30}{x} \text { dozen }\)

At present in Rs. 30, no. of pen available = \(\frac{1}{x-6} \times 30 \text { dozen }=\frac{30}{x-6} \text { dozen }\)

According to the problem,

\(\frac{30}{x-6}-\frac{30}{x}=\frac{3}{12} \quad\left[3=\frac{3}{12} \text { dozen }\right]\)

⇒ \(30\left(\frac{x-x+6}{x(x-6)}\right)=\frac{1}{4}\)

⇒ \(x^2-6 x=720\)

⇒ \(x^2-6 x-720=0\)

⇒ \(x^2-30 x+24 x-720=0\)

⇒ x(x – 30) + 24(x – 30) = 0

∴ (x – 30)(x + 24) = 0

Either x – 30 = 0

or, x + 24 = 0

∴ x = 30

Previous price of 1 dozen pens is Rs. 30.

WBBSE Class 10 Quadratic Equation In One Variable Exercise 1.3 Multiple Choice Questions

Question 1. The number of roots in a quadratic equation is

1. One

2. Two

3. Three

4. None of them

Answer: 2. Two

Question 2. If ax2 + bx + c is a quadratic equation then

1. b + 0

2. c + 0

3. a + 0

4. None of these

Answer: 3. a + 0

Question 3. The highest power of the variable of a quadratic equation is

1. 1

2. 2

3. 3

4. None of these

Answer: 2. 2

Question 4. The equation 4 (5x2-7x+2) = 5 (4x2-6x + 3) is

1. Linear

2. Quadratic

3. 3rd degree

4. None of these

Answer: 1. Linear

Question 5. The root/roots of equation, x2/x = 6 are:

1. 0

2. 6

3. 0 & 6

4. -6

Answer: 3. 0 & 6

Chapter 1 Quadratic Equations In One Variable Exercise 1.3 True or False

Question 1. (x-3)2= x2-6x + 9 it is a quadratic equation

Answer: False

Question 2. Equation x2 = 25 has only one root (s)

Answer: False

Chapter 1 Quadratic Equations In One Variable Exercise 1.3 Let Us Fill In The Blank :

1. If a = 0 & b + 0, in the equation ax2 + bx + c = 0, then the equation is a Linear equation.

2. If the two roots of a quadratic equation are equal and equal to 1, then the equation is    x2 – 2x + 1 = 0.

3. The two roots of the x2-6x are   0   &   6

Chapter 1 Quadratic Equations In One Variable Exercise 1.3 Short Answers

Question 1. Let us find the value of an if one root of equation x2 + ax + 3 = 0 is 1.

Solution:

If x2 + ax + 3 = 0 has one root = 1,

the other root is 12+ a.1 +3 = 0

⇒ a+ 4 = 0

⇒  a = -4

Question 2. Let us find the value of the other root if one root of the equation x2 – (2 + b) x + 6 = 0 is 2.

Solution: If one root of x2 – (2 + b) x + 6 = 0 is 2, the other root

⇒ (2)2 (2+b) 2+6= 0

⇒ 4-4-2b+6=0

⇒ 2b = 6

⇒ b = 3

The other root = 3

Question 3. Let us write the value of the other root if one root of equation 2x2 + kx + 4 = 0 is 2.

Solution: \(2 x^2+k x+4=0\)

As one root of this equation is 2,

then \(2(2)^2+k \cdot 2+4=0\)

or 8 + 2k + 4 = 0

∴ k = -6

∴ Equation is \(2 x^2-6 x+4=0\)

or, \(x^2-3 x+2=0\)

or, \(x^2-2 x-x+2=0\)

or, x(x – 2) – 1(x – 2) = 0

(x – 2)(x – 1) = 0

∴ One root is 2 and the other root = 1.

Question 4. Let us write the equation if the difference between a proper fraction and its reciprocal is 9/20.

Solution: Let the proper fraction is x & its reciprocal =  \(\frac{1}{x}\).

∴ \(\frac{1}{x}-x=\frac{9}{20}\)

Question 5. Let us write the values of a and b if the two roots of the equation ax2 + bx +35 = 0 are -5 and -7.

Solution: When one root of the equation ax2 + bx + 35 = 0, is (-5)a(-5)2 + (-5)a + 36 = 0

or, 25a – 5b + 35 = 0

or, 5a – b + 7 = 0     ……(1)

Again, when the other root is -7

∴ a(-7)2 + (-7)b + 35 = 0

or, 49a – 7b + 35 = 0

or, 7a – b + 5 = 0              …..(2)

by solving equations (1) & (2) we get, a = 1 & b = 12.

Chapter 1 Quadratic Equations In One Variable Application

Question 1. Let us check whether the two obtained values of x, that is, x = 10 and x = -7 satisfy the quadratic equation (1) or not.

Solution: x2 – 3x – 70 = 0

when x = 10

(10)2 – 3(10) – 70 = 0

100 – 30 – 70 = 0

& when x = -7,

(-7)2 – 3(-7) -70 + 49 + 21 + 70 = 0

∴ x = 10 & x = -7

satisfy the equation x2 – 3x – 70 = 0.

Question 2. The product of two consecutive positive odd numbers is 143. Let me construct the equation and write the two odd numbers by applying Sreedhar Acharyya’s formula. 

Solution: Let two consecutive positive numbers are (2x – 1) & (2x + 1).

∴ (2x – 1)(2x + 1) = 143

⇒ 4x2 – 1 = 143

⇒ 4x2 – 144 = 0

⇒ 4(x2 – 36) = 0

⇒ (x – 6)(x + 6) = 0

∴ x = 6 & x = -6 (Not possible)

∴ The numbers are (2x – 1) = 12 – 1 = 11 & (2x + 1) = 12 + 1 = 13.

Question 3. If the following quadratic equations have real roots, then let us determine their roots by applying Sreedhar Acharyya’s formula.

1. 1- x = 2x2

Solution: 1 – x = 2x2

or 2x2 + x – 1 = 0

Comparing this equation with ax2 + bx + c = 0,

∴ a = 2; b = 1; c = -1

∴ \(b^2-4 a c=(1)^2-4.2(-1)=1+8=9>0\)

∴ The roots of the equation are real.

The roots are = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-1 \pm \sqrt{(1)^2-4.2(1)}}{2 \times 1}=\frac{-1 \pm 3}{2} x=\frac{-1+3}{2}=\frac{2}{2}=1\)

& \(x=\frac{-1-3}{2}=\frac{-4}{2}=-2\)

∴ the roots are (1, -2).

2. 2x2 – 9x+7=0

Solution: 2×22 – 9x + 7 = 0

Comparing this equation with ax2 + bx + c = 0, a = 2; b = -9; c = 7

\(\mathrm{b}^2-4 \mathrm{ac}=(-9)^2-4 \cdot 2 \cdot 7=81-56=25>0\)

∴ The roots of the equation are real.

The roots are = \(\frac{-(-9) \pm \sqrt{(-9)^2-4.2 .7}}{2 x^2}=\frac{9 \pm \sqrt{25}}{4}=\frac{9 \pm 5}{4}\)

∴ \(x=\frac{9+5}{4}=\frac{14}{4}=\frac{7}{4}\)

& \(x=\frac{9-5}{4}=\frac{4}{4}=1\)

3. x2 – (√2+1) x + √2 =0

Solution: \(\mathbf{x}^2-(\sqrt{2}+1) \mathbf{x}+\sqrt{2}=\mathbf{0}\)

Comparing this equation with ax2 + bx + c = 0

∴ a = 1; \(b=-(\sqrt{2}+1) ; c=\sqrt{2}\)

∴ \(b^2-4 a c=\{-(\sqrt{2}+1)\}^2-4 \cdot 1 \cdot \sqrt{2}=2+1+2 \sqrt{2}-4 \sqrt{2}=3-2 \sqrt{2}>0\)

∴ The roots of the equation are real.

The roots are = \(\frac{(\sqrt{2}+1) \pm \sqrt{3-2 \sqrt{2}}}{2.1}=\frac{(\sqrt{2}+1) \pm(\sqrt{2}-1)}{2}\)

\(x=\frac{\sqrt{2}+1+\sqrt{2}-1}{2}=\frac{2 \sqrt{2}}{2}=\sqrt{2}\)

& \(x=\frac{\sqrt{2}+1-\sqrt{2}+1}{2}=\frac{2}{2}=1\)

∴ The roots are √2 & 1.

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