Exercise 2.8 Class 10 Complete Solutions
In this post, we present detailed solutions for Exercise 2.8 Class 10, the final exercise of Unit 2, which focuses on real-life word problems solved using quadratic equations. This exercise is crucial for helping students understand how to apply mathematical concepts to everyday situations, enhancing their problem-solving abilities. Each solution is carefully explained, providing clear steps to guide students through the process of formulating and solving quadratic equations in various contexts. Whether you’re preparing for exams or seeking to strengthen your grasp of these concepts, this post will support you in mastering Exercise 2.8 Class 10.
Question 1 of Exercise 2.8 Class 10.
\(\textbf{The product of two positive consecutive numbers is 182. Find the numbers.}\)
\(\textbf{Solution:} \)
Suppose the first positive number is \( x \). The very next positive number is \( x + 1 \). By the given condition:
\( x(x + 1) = 182 \)
\( x^2 + x – 182 = 0 \)
\( x^2 + 14x – 13x – 182 = 0 \)
\( (x + 14)(x – 13) = 0 \)
Either \( x + 14 = 0 \) or \( x – 13 = 0 \), so \( x = -14 \) or \( x = 13 \).
As \( x \) is a positive number, we take \( x = 13 \).
Thus, the two consecutive numbers are \( 13 \) and \( 14 \).
Question 2 of Exercise 2.8 Class 10. \(\textbf{The sum of the squares of three positive consecutive numbers is 77. Find them.}\)
\(\textbf{Solution:} \)
Let \( x \), \( (x + 1) \), and \( (x + 2) \) be the three consecutive positive numbers. By the given condition:
\( x^2 + (x + 1)^2 + (x + 2)^2 = 77 \)
Expanding the squares:
\( x^2 + [x^2 + 2x + 1] + [x^2 + 4x + 4] = 77 \)
\( 3x^2 + 6x + 5 = 77 \)
\( 3x^2 + 6x – 72 = 0 \)
\( x^2 + 2x – 24 = 0 \)
\( (x + 6)(x – 4) = 0 \)
Either \( x + 6 = 0 \) or \( x – 4 = 0 \), so \( x = -6 \) or \( x = 4 \). Since \( x \) is positive, we take \( x = 4 \).
Thus, the three consecutive numbers are \( 4 \), \( 5 \), and \( 6 \).
Question 3 of Exercise 2.8 Class 10. \(\textbf{The sum of five times a number and the square of the number is 204. Find the number.}\)
\(\textbf{Solution:} \)
Let the required number be \( x \). The given condition is:
\( x^2 + 5x = 204 \)
\( x^2 + 5x – 204 = 0 \)
\( x^2 + 17x – 12x – 204 = 0 \)
\( (x + 17)(x – 12) = 0 \)
Either \( x + 17 = 0 \) or \( x – 12 = 0 \), so \( x = -17 \) or \( x = 12 \).
Thus, the required number is \( 12 \) (as we take the positive value).
Question 4 of Exercise 2.8 Class 10. \(\textbf{The product of five less than three times a certain number}\)
\(\textbf{ and one less than four times the number is 7. Find the number.}\)
\(\textbf{Solution:} \)
Let the required number be \( x \). The conditions are:
Five less than three times the number is \( 3x – 5 \)
One less than four times the number is \( 4x – 1 \)
By the given condition:
\( (3x – 5)(4x – 1) = 7 \)
Expanding:
\( 12x^2 – 3x – 20x + 5 = 7 \)
\( 12x^2 – 23x + 5 = 7 \)
\( 12x^2 – 23x – 2 = 0 \)
\( (x – 2)(12x + 1) = 0 \)
Either \( x – 2 = 0 \) or \( 12x + 1 = 0 \), so \( x = 2 \) or \( x = \frac{-1}{12} \).
Thus, the required number is \( 2 \) or \( \frac{-1}{12} \).
Question 5 of Exercise 2.8 Class 10.\(\textbf{The difference of a number and its reciprocal is } \frac{15}{4}. \textbf{ Find the number.}\)
\(\textbf{Solution:} \)
Let the required number be \( x \).
The reciprocal of the number is \( \frac{1}{x} \).
By the given condition:
\( x – \frac{1}{x} = \frac{15}{4} \)
Multiplying both sides by \( x \), we get:
\( x^2 – 1 = \frac{15x}{4} \)
Multiplying both sides by 4:
\( 4(x^2 – 1) = 15x \)
\( 4x^2 – 4 = 15x \)
\( 4x^2 – 15x – 4 = 0 \)
Factoring the quadratic equation:
\( (4x + 1)(x – 4) = 0 \)
Either \( x – 4 = 0 \) or \( 4x + 1 = 0 \), so \( x = 4 \) or \( x = \frac{-1}{4} \).
Thus, the required numbers are \( 4 \) or \( \frac{-1}{4} \).
Question 6 of Exercise 2.8 Class 10. \(\textbf{The sum of the squares of two digits of a positive integral number}\)
\(\textbf{ is 65 and the number is 9 times the sum of its digits. Find the number.}\)
\(\textbf{Solution:} \)
Let the digits at the unit’s place of the number be \( x \), and the digit at the ten’s place be \( y \).
The required number is \( 10y + x \).
By the first condition:
\( x^2 + y^2 = 65 \) \(\text{(i)}\)
By the second condition:
\( 10y + x = 9(x + y) \)
Simplifying:
\( 10y + x = 9x + 9y \)
\( 10y + x – 9y = 9x \)
\( y = 8x \) \(\text{(ii)}\)
Substitute the value of \( y \) from equation (ii) into equation (i):
\( x^2 + (8x)^2 = 65 \)
\( x^2 + 64x^2 = 65 \)
\( 65x^2 = 65 \)
\( x^2 = 1 \)
\( x = \pm 1 \)
Since \( x \) represents a digit, which is always positive, we take \( x = 1 \).
Put \( x = 1 \) in equation (ii):
\( y = 8(1) \)
\( y = 8 \)
So, the required number is \( 10(8) + 1 = 81 \).
Question 7 of Exercise 2.8 Class 10. \(\textbf{The sum of the coordinates of a point is 9 and the sum}\)
\(\textbf{ of their squares is 45. Find the coordinates of the point.}\)
\(\textbf{Solution:} \)
Let \( (x, y) \) be the coordinates of the required point.
By the given conditions:
\( x + y = 9 \) \(\quad \text{(i)}\)
\( x^2 + y^2 = 45 \) \(\quad \text{(ii)}\)
From equation (i):
\( x = 9 – y \) \(\quad \text{(iii)}\)
Substituting \( x = 9 – y \) into equation (ii):
\( (9 – y)^2 + y^2 = 45 \)
Expanding:
\( 81 – 18y + y^2 + y^2 = 45 \)
Simplifying:
\( 2y^2 – 18y + 81 = 45 \)
\( 2y^2 – 18y + 36 = 0 \)
Dividing the whole equation by 2:
\( y^2 – 9y + 18 = 0 \)
Now, solving this quadratic equation using the quadratic formula:
\( y = \frac{-(-9) \pm \sqrt{(-9)^2 – 4(1)(18)}}{2(1)} \)
\( y = \frac{9 \pm \sqrt{81 – 72}}{2} \)
\( y = \frac{9 \pm \sqrt{9}}{2} \)
\( y = \frac{9 \pm 3}{2} \)
So, we get two values of \( y \):
\( y = \frac{9 + 3}{2} = 6 \quad \text{or} \quad y = \frac{9 – 3}{2} = 3 \)
Using equation (iii) \( x = 9 – y \):
When \( y = 6 \), \( x = 9 – 6 = 3 \)
When \( y = 3 \), \( x = 9 – 3 = 6 \)
Thus, the coordinates of the point are \( (3, 6) \) or \( (6, 3) \).
Question 8 of Exercise 2.8 Class 10. \(\textbf{Find two integers whose sum is 9 and the difference}\)
\(\textbf{ of their squares is also 9.}\)
\(\textbf{Solution:} \)
Suppose \( x \) and \( y \) are two integers.
By the given conditions:
\( x + y = 9 \) \(\quad \text{(i)}\)
\( x^2 – y^2 = 9 \) \(\quad \text{(ii)}\)
From equation (i):
\( x = 9 – y \) \(\quad \text{(iii)}\)
Substituting the value of \( x \) in equation (ii), we get:
\( (9 – y)^2 – y^2 = 9 \)
Expanding:
\( 81 – 18y + y^2 – y^2 = 9 \)
Simplifying:
\( 81 – 18y = 9 \)
\( -18y = 9 – 81 \)
\( -18y = -72 \)
\( y = \frac{-72}{-18} = 4 \)
Putting the value of \( y = 4 \) in equation (iii), we get:
\( x = 9 – 4 = 5 \)
So, the required integers are \( 4 \) and \( 5 \).
Question 9 of Exercise 2.8 Class 10. \(\textbf{ Find two integers whose difference is 4 and whose squares differ by 72.}\)
\(\textbf{Solution:} \)
Let \( x \) and \( y \) be two integers.
By the given conditions:
\( x – y = 4 \) \(\quad \text{(i)}\)
\( x^2 – y^2 = 72 \) \(\quad \text{(ii)}\)
From equation (i):
\( x = 4 + y \) \(\quad \text{(iii)}\)
Substituting the value of \( x \) in equation (ii), we get:
\( (4 + y)^2 – y^2 = 72 \)
Expanding:
\( 16 + 8y + y^2 – y^2 = 72 \)
Simplifying:
\( 16 + 8y = 72 \)
\( 8y = 72 – 16 \)
\( 8y = 56 \)
\( y = \frac{56}{8} = 7 \)
Putting the value of \( y = 7 \) in equation (iii), we get:
\( x = 4 + 7 = 11 \)
So, the required integers are \( 7 \) and \( 11 \).
Question 10 of Exercise 2.8 Class 10. \(\textbf{Find the dimensions of a rectangle, whose perimeter}\)
\(\textbf{ is 80 cm and its area is 375 cm(^2).}\)
\(\textbf{Solution:} \)
Let the width of the rectangle be \( x \) cm and the length be \( y \) cm.
By the given condition,
\( 2(x + y) = 80 \)
\( x + y = 40 \) \(\quad \text{(i)}\)
The area of the rectangle is given as:
\( xy = 375 \) \(\quad \text{(ii)}\)
From equation (i):
\( y = 40 – x \) \(\quad \text{(iii)}\)
Substituting \( y = 40 – x \) into equation (ii):
\( x(40 – x) = 375 \)
\( 40x – x^2 = 375 \)
\( x^2 – 40x + 375 = 0 \)
Now solving this quadratic equation:
\( x = \frac{-(-40) \pm \sqrt{(-40)^2 – 4(1)(375)}}{2(1)} \)
\( x = \frac{40 \pm \sqrt{1600 – 1500}}{2} \)
\( x = \frac{40 \pm \sqrt{100}}{2} \)
\( x = \frac{40 \pm 10}{2} \)
So, \( x = \frac{40 + 10}{2} = 25 \) or \( x = \frac{40 – 10}{2} = 15 \).
Thus, the width is 15 cm and the length is 25 cm (or vice versa).
In conclusion, these solutions to Exercise 2.8 class 10 provide a step-by-step guide to help you fully understand the methods and concepts involved. By working through each problem, you’ll gain a stronger grasp of solving quadratic equations and deepen your understanding of the techniques needed in similar questions. Keep practicing to build your confidence and master these essential skills. If you have any questions or need further clarification, feel free to explore other resources such as formulas of unit# 2 or more on to the solutions of next unit#3 or reach out for assistance. Here is a video lecture of this exercise. Further more here is a complete Playlist of whole unit 2 lectures.
