Exercise 3.1 Solutions
Question 1
Simplify the following expression:
\(\frac{x^2 - 5x}{x^2 - 25}\)
Solution:
\(\frac{x(x - 5)}{(x - 5)(x + 5)} = \frac{x}{x + 5}\)
Question 2
Find the value of the following expression:
\(\frac{1}{4(1 - \sqrt{x})} - \frac{1}{4(1 + \sqrt{x})} + \frac{2\sqrt{x}}{4(1 - x)}\)
Solution:
Simplifying each term:
\(\frac{1}{4(1 - \sqrt{x})} = \frac{1}{4 - 4\sqrt{x}}\)
\(\frac{1}{4(1 + \sqrt{x})} = \frac{1}{4 + 4\sqrt{x}}\)
\(\frac{2\sqrt{x}}{4(1 - x)} = \frac{\sqrt{x}}{2(1 - x)}\)
Combining the terms:
\(\frac{1}{4 - 4\sqrt{x}} - \frac{1}{4 + 4\sqrt{x}} + \frac{\sqrt{x}}{2(1 - x)}\)
Question 3
Find the missing frequency in the given data set:
Given: Median = 35
Marks obtained: 20-25, 25-30, 30-35, 35-40, 40-45, 45-50
No. of students: 2, 5, 8, k, 4, 5
Solution:
Total number of students (N) = 24
Median class interval: 35-40
Lower limit (L) = 35
Cumulative frequency before median class (cf) = 15
Frequency of median class (f) = k
Length of median class interval (h) = 5
Median formula:
\(Md = L + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h\)
Substitute the values:
\(35 = 35 + \left(\frac{\frac{24}{2} - 15}{k}\right) \times 5\)
\(35 = 35 + \left(\frac{12 - 15}{k}\right) \times 5\)
\(0 = \left(\frac{-3}{k}\right) \times 5\)
\(k = 3\)
Therefore, the missing frequency (k) = 3
Question 4
Find the value of Q1 and Q3 from the data given below:
Height (cm): <125, <130, <135, <140, <145, <150, <155
No. of students: 0, 5, 11, 24, 45, 60, 72
Solution:
Q1 is the value at the 25th percentile:
Q1 = 130 cm (as the 25th percentile lies between 5 and 11 students)
Q3 is the value at the 75th percentile:
Q3 = 145 cm (as the 75th percentile lies between 45 and 60 students)
Exercise 3.1 Solutions
Question 5
Solve the equation: \(x + 2 = 7\)
Solution:
\(x + 2 = 7\)
Subtract 2 from both sides:
\(x = 7 - 2\)
\(x = 5\)
Question 6
Solve the equation: \(3x - 4 = 5\)
Solution:
\(3x - 4 = 5\)
Add 4 to both sides:
\(3x = 5 + 4\)
\(3x = 9\)
Divide both sides by 3:
\(x = \frac{9}{3}\)
\(x = 3\)
Question 7
Solve the equation: \(2x + 3 = 11\)
Solution:
\(2x + 3 = 11\)
Subtract 3 from both sides:
\(2x = 11 - 3\)
\(2x = 8\)
Divide both sides by 2:
\(x = \frac{8}{2}\)
\(x = 4\)
Question 8
Solve the equation: \(4x - 5 = 15\)
Solution:
\(4x - 5 = 15\)
Add 5 to both sides:
\(4x = 15 + 5\)
\(4x = 20\)
Divide both sides by 4:
\(x = \frac{20}{4}\)
\(x = 5\)
Here are the solutions for questions 9, 10, 11, and 12 from Exercise 3.1:
**Question 9**:
Find the median for the following data:
- Class intervals: 0-10, 10-20, 20-30, 30-40, 40-50
- Frequencies: 5, 10, 10, 20, 5
To find the median, follow these steps:
1. Compute the cumulative frequencies.
2. Find the median class where N/2 falls, where N is the total number of frequencies.
3. Use the median formula:
\[
\text{Median} = L + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h
\]
where:
- \( L \) = lower boundary of the median class
- \( N \) = total number of frequencies
- \( cf \) = cumulative frequency of the class before the median class
- \( f \) = frequency of the median class
- \( h \) = class width
**Solution**:
1. Calculate cumulative frequencies:
- 0-10: 5
- 10-20: 5+10 = 15
- 20-30: 15+10 = 25
- 30-40: 25+20 = 45
- 40-50: 45+5 = 50
2. Total \( N \) = 50. \( \frac{N}{2} \) = 25, which falls in the class 20-30.
3. \( L = 20 \), \( cf = 15 \), \( f = 10 \), \( h = 10 \)
\[
\text{Median} = 20 + \left( \frac{25 - 15}{10} \right) \times 10 = 20 + 10 = 30
\]
Median = 30
**Question 10**:
Given the following data, find the median:
- Class intervals: 0-50, 50-100, 100-150, 150-200
- Frequencies: 15, 25, 35, 25
**Solution**:
1. Calculate cumulative frequencies:
- 0-50: 15
- 50-100: 15+25 = 40
- 100-150: 40+35 = 75
- 150-200: 75+25 = 100
2. Total \( N \) = 100. \( \frac{N}{2} \) = 50, which falls in the class 100-150.
3. \( L = 100 \), \( cf = 40 \), \( f = 35 \), \( h = 50 \)
\[
\text{Median} = 100 + \left( \frac{50 - 40}{35} \right) \times 50 = 100 + 14.29 = 114.29
\]
Median = 114.29
**Question 11**:
Find the median for the following data:
- Class intervals: 10-20, 20-30, 30-40, 40-50, 50-60
- Frequencies: 6, 11, 15, 9, 5
**Solution**:
1. Calculate cumulative frequencies:
- 10-20: 6
- 20-30: 6+11 = 17
- 30-40: 17+15 = 32
- 40-50: 32+9 = 41
- 50-60: 41+5 = 46
2. Total \( N \) = 46. \( \frac{N}{2} \) = 23, which falls in the class 30-40.
3. \( L = 30 \), \( cf = 17 \), \( f = 15 \), \( h = 10 \)
\[
\text{Median} = 30 + \left( \frac{23 - 17}{15} \right) \times 10 = 30 + 4 = 34
\]
Median = 34
**Question 12**:
Find the median for the following data:
- Class intervals: 0-20, 20-40, 40-60, 60-80, 80-100
- Frequencies: 7, 10, 15, 8, 10
**Solution**:
1. Calculate cumulative frequencies:
- 0-20: 7
- 20-40: 7+10 = 17
- 40-60: 17+15 = 32
- 60-80: 32+8 = 40
- 80-100: 40+10 = 50
2. Total \( N \) = 50. \( \frac{N}{2} \) = 25, which falls in the class 40-60.
3. \( L = 40 \), \( cf = 17 \), \( f = 15 \), \( h = 20 \)
\[
\text{Median} = 40 + \left( \frac{25 - 17}{15} \right) \times 20 = 40 + 10.67 = 50.67
\]
Median = 50.67
### Question 13
**Given:**
- A parallelogram \(ABCD\)
- Diagonal \(AC\) intersects diagonal \(BD\) at point \(O\)
- \(AB = 10 \, \text{cm}\), \(AD = 8 \, \text{cm}\), \(AC = 12 \, \text{cm}\)
**To Find:**
- Area of \(\Delta AOD\)
- Area of parallelogram \(ABCD\)
**Solution:**
In a parallelogram, diagonals bisect each other. Therefore, \(O\) is the midpoint of \(AC\) and \(BD\).
1. **Finding Area of \(\Delta AOD\)**:
Using the properties of a parallelogram, the area of \(\Delta AOD\) can be found using the determinant method.
Let's consider the coordinates of \(A\), \(D\), and \(O\):
- \(A = (0, 0)\)
- \(D = (8, 0)\)
- \(O = (6, 4)\) (Using midpoint formula since \(AC\) is 12 cm and \(O\) is the midpoint, we have \(AO = 6 \, \text{cm}\) and using Pythagoras theorem for triangle \(AOD\))
Area of \(\Delta AOD = \frac{1}{2} \times | x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) |\)
Area of \(\Delta AOD = \frac{1}{2} \times | 0(0-4) + 8(4-0) + 6(0-0) |\)
Area of \(\Delta AOD = \frac{1}{2} \times | 0 + 32 + 0 | = 16 \, \text{sq cm}\)
2. **Finding Area of parallelogram \(ABCD\)**:
The area of parallelogram \(ABCD\) is twice the area of \(\Delta AOD\):
Area of \(ABCD = 2 \times 16 = 32 \, \text{sq cm}\)
### Question 14
**Given:**
- A right-angled triangle \(ABC\) with \( \angle ABC = 90^\circ\)
- \(AB = 6 \, \text{cm}\), \(BC = 8 \, \text{cm}\)
**To Find:**
- Radius of the circle that circumscribes the triangle \(ABC\)
**Solution:**
In a right-angled triangle, the circumcircle's diameter is the hypotenuse of the triangle.
1. **Find the hypotenuse \(AC\)**:
Using the Pythagorean theorem:
\[
AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{cm}
\]
2. **Radius of the circumcircle**:
The radius \(R\) of the circumcircle is half the hypotenuse.
\[
R = \frac{AC}{2} = \frac{10}{2} = 5 \, \text{cm}
\]
So, the radius of the circle that circumscribes the triangle \(ABC\) is \(5 \, \text{cm}\).
Exercise 3.2
### Exercise 3.2 - Questions 1, 2, 3, 4
#### Question 1
Find the median from the given data:
(a) 2.5, 4.5, 3.6, 4.9, 5.4, 2.9, 3.1, 4.2, 4.6, 2.2, 1.5
**Solution:**
First, arrange the data in ascending order: 1.5, 2.2, 2.5, 2.9, 3.1, 3.6, 4.2, 4.5, 4.6, 4.9, 5.4.
The median is the middle value in an ordered list of numbers.
- Since there are 11 numbers, the median is the 6th number: **3.6**.
#### Question 2
Find the median from the given data:
(b) 100, 105, 104, 197, 97, 108, 120, 148, 144, 190, 148, 22, 169, 171, 92, 100
**Solution:**
First, arrange the data in ascending order: 22, 92, 97, 100, 100, 104, 105, 108, 120, 144, 148, 148, 169, 171, 190, 197.
The median is the middle value in an ordered list of numbers.
- Since there are 16 numbers, the median is the average of the 8th and 9th numbers: (108 + 120) / 2 = **114**.
#### Question 3
Find the median from the given data:
(c) Marks obtained: 18, 25, 28, 29, 34, 40, 44, 46
No. of students: 3, 6, 5, 7, 8, 12, 5, 4
**Solution:**
To find the median, calculate the cumulative frequency and determine the median class.
Cumulative frequency:
- 18: 3
- 25: 3 + 6 = 9
- 28: 9 + 5 = 14
- 29: 14 + 7 = 21
- 34: 21 + 8 = 29
- 40: 29 + 12 = 41
- 44: 41 + 5 = 46
- 46: 46 + 4 = 50
The total number of students is 50, so the median is at the 25.5th position.
The median class is the one containing the 25.5th student:
- Median class: 34 (since 21 < 25.5 < 29)
- Frequency (f): 8
- Cumulative frequency of preceding class (cf): 21
- Class interval size (h): Not needed since we are using a discrete set.
Median formula: \( Median = L + \left( \frac{N/2 - cf}{f} \right) \times h \)
For exact value:
\( Median = 34 \)
#### Question 4
Find the median from the given data:
(d) Class interval (x): 102, 105, 125, 140, 170, 190, 200
Frequency (f): 10, 18, 22, 25, 15, 12, 8
**Solution:**
To find the median, calculate the cumulative frequency and determine the median class.
Cumulative frequency:
- 102-105: 10
- 105-125: 10 + 18 = 28
- 125-140: 28 + 22 = 50
- 140-170: 50 + 25 = 75
- 170-190: 75 + 15 = 90
- 190-200: 90 + 12 = 102
The total number of frequencies is 102, so the median is at the 51st position.
The median class is the one containing the 51st student:
- Median class: 125-140 (since 50 < 51 < 75)
- Lower limit (L): 125
- Frequency (f): 22
- Cumulative frequency of preceding class (cf): 28
- Class interval size (h): 15 (assuming 140-125 = 15)
Median formula:
\( Median = L + \left( \frac{N/2 - cf}{f} \right) \times h \)
\( Median = 125 + \left( \frac{51 - 28}{22} \right) \times 15 \)
\( Median = 125 + \left( \frac{23}{22} \right) \times 15 \)
\( Median = 125 + 1.045 \times 15 \)
\( Median = 125 + 15.68 \)
\( Median = 140.68 \)
Thus, the median is approximately **140.68**.
Here are the solutions for Exercise 3.2, Questions 5, 6, 7, and 8:
### Question 5
**Given:**
- Data: 5, 7, 8, 9, 10, 12, 14, 16, 18, 20
- Find the median.
**Solution:**
1. Arrange the data in ascending order (already sorted): 5, 7, 8, 9, 10, 12, 14, 16, 18, 20.
2. Since there are 10 data points (an even number), the median is the average of the 5th and 6th numbers.
Median = \( \frac{10 + 12}{2} = \frac{22}{2} = 11 \)
Therefore, the median is **11**.
### Question 6
**Given:**
- Data: 3, 7, 4, 9, 6, 10, 8
- Find the median.
**Solution:**
1. Arrange the data in ascending order: 3, 4, 6, 7, 8, 9, 10.
2. Since there are 7 data points (an odd number), the median is the middle number.
Median = 7 (the 4th number)
Therefore, the median is **7**.
### Question 7
**Given:**
- Data: 32, 36, 34, 38, 42, 44, 48, 52, 56, 60
- Find the median.
**Solution:**
1. Arrange the data in ascending order (already sorted): 32, 34, 36, 38, 42, 44, 48, 52, 56, 60.
2. Since there are 10 data points (an even number), the median is the average of the 5th and 6th numbers.
Median = \( \frac{42 + 44}{2} = \frac{86}{2} = 43 \)
Therefore, the median is **43**.
### Question 8
**Given:**
- Class intervals and frequencies:
- 10-20: 5
- 20-30: 10
- 30-40: 15
- 40-50: 10
- 50-60: 5
**To Find:**
- Median
**Solution:**
1. Calculate cumulative frequencies:
- 10-20: 5
- 20-30: 5 + 10 = 15
- 30-40: 15 + 15 = 30
- 40-50: 30 + 10 = 40
- 50-60: 40 + 5 = 45
2. Total number of observations \( N \) = 45.
3. Median is at the \( \frac{N}{2} \) position, which is the 22.5th observation.
The median class is the class interval containing the 22.5th observation:
- Median class: 30-40 (since 15 < 22.5 ≤ 30)
4. Use the median formula:
\[
\text{Median} = L + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h
\]
where:
- \( L \) = lower limit of the median class = 30
- \( cf \) = cumulative frequency before the median class = 15
- \( f \) = frequency of the median class = 15
- \( h \) = class width = 10
\[
\text{Median} = 30 + \left( \frac{22.5 - 15}{15} \right) \times 10
\]
\[
\text{Median} = 30 + \left( \frac{7.5}{15} \right) \times 10
\]
\[
\text{Median} = 30 + 5
\]
\[
\text{Median} = 35
\]
Therefore, the median is **35**.
Here are the solutions to Exercise 3.2, questions 9-14:
**Question 9:**
If the product of two consecutive even numbers is 80, find the numbers.
Let the two consecutive even numbers be \( x \) and \( x + 2 \).
\[ x(x + 2) = 80 \]
\[ x^2 + 2x - 80 = 0 \]
Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 1, b = 2, c = -80 \]
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-80)}}{2 \cdot 1} \]
\[ x = \frac{-2 \pm \sqrt{4 + 320}}{2} \]
\[ x = \frac{-2 \pm \sqrt{324}}{2} \]
\[ x = \frac{-2 \pm 18}{2} \]
So,
\[ x = \frac{16}{2} = 8 \]
\[ x = \frac{-20}{2} = -10 \] (not possible since we need positive even numbers)
Therefore, the two consecutive even numbers are 8 and 10.
**Question 10:**
If the product of two consecutive odd numbers is 225, find the numbers.
Let the two consecutive odd numbers be \( x \) and \( x + 2 \).
\[ x(x + 2) = 225 \]
\[ x^2 + 2x - 225 = 0 \]
Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 1, b = 2, c = -225 \]
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-225)}}{2 \cdot 1} \]
\[ x = \frac{-2 \pm \sqrt{4 + 900}}{2} \]
\[ x = \frac{-2 \pm \sqrt{904}}{2} \]
\[ x = \frac{-2 \pm 30}{2} \]
So,
\[ x = \frac{28}{2} = 14 \]
\[ x = \frac{-32}{2} = -16 \] (not possible since we need positive odd numbers)
Therefore, the two consecutive odd numbers are 15 and 17.
**Question 11:**
If the sum of a number and its reciprocal is \( \frac{10}{3} \), find the number.
Let the number be \( x \).
\[ x + \frac{1}{x} = \frac{10}{3} \]
Multiplying through by \( x \) to clear the fraction:
\[ x^2 + 1 = \frac{10}{3} x \]
\[ 3x^2 + 3 = 10x \]
\[ 3x^2 - 10x + 3 = 0 \]
Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 3, b = -10, c = 3 \]
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \]
\[ x = \frac{10 \pm \sqrt{100 - 36}}{6} \]
\[ x = \frac{10 \pm \sqrt{64}}{6} \]
\[ x = \frac{10 \pm 8}{6} \]
So,
\[ x = \frac{18}{6} = 3 \]
\[ x = \frac{2}{6} = \frac{1}{3} \]
Therefore, the number could be 3 or \( \frac{1}{3} \).
**Question 12:**
If the sum of two natural numbers is 21 and the sum of their squares is 261, find the numbers.
Let the numbers be \( x \) and \( y \).
\[ x + y = 21 \]
\[ x^2 + y^2 = 261 \]
From the first equation, \( y = 21 - x \).
Substituting into the second equation:
\[ x^2 + (21 - x)^2 = 261 \]
\[ x^2 + 441 - 42x + x^2 = 261 \]
\[ 2x^2 - 42x + 441 = 261 \]
\[ 2x^2 - 42x + 180 = 0 \]
\[ x^2 - 21x + 90 = 0 \]
Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 1, b = -21, c = 90 \]
\[ x = \frac{21 \pm \sqrt{(-21)^2 - 4 \cdot 1 \cdot 90}}{2 \cdot 1} \]
\[ x = \frac{21 \pm \sqrt{441 - 360}}{2} \]
\[ x = \frac{21 \pm \sqrt{81}}{2} \]
\[ x = \frac{21 \pm 9}{2} \]
So,
\[ x = \frac{30}{2} = 15 \]
\[ x = \frac{12}{2} = 6 \]
Therefore, the numbers are 15 and 6.
**Question 13:**
If the age difference between two brothers is 4 years and the product of their ages is 221, find their ages.
Let the ages be \( x \) and \( x + 4 \).
\[ x(x + 4) = 221 \]
\[ x^2 + 4x - 221 = 0 \]
Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 1, b = 4, c = -221 \]
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-221)}}{2 \cdot 1} \]
\[ x = \frac{-4 \pm \sqrt{16 + 884}}{2} \]
\[ x = \frac{-4 \pm \sqrt{900}}{2} \]
\[ x = \frac{-4 \pm 30}{2} \]
So,
\[ x = \frac{26}{2} = 13 \]
\[ x = \frac{-34}{2} = -17 \] (not possible since we need positive ages)
Therefore, the ages are 13 and 17.
**Question 14:**
The sum of the present age of two brothers is 22 and the product of their ages is 120. Find their present ages.
Let the ages be \( x \) and \( y \).
\[ x + y = 22 \]
\[ xy = 120 \]
From the first equation, \( y = 22 - x \).
Substituting into the second equation:
\[ x(22 - x) = 120 \]
\[ 22x - x^2 = 120 \]
\[ x^2 - 22x + 120 = 0 \]
Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 1, b = -22, c = 120 \]
\[ x = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 1 \cdot 120}}{2 \cdot 1} \]
\[ x = \frac{22 \pm \sqrt{484 - 480}}{2} \]
\[ x = \frac{22 \pm \sqrt{4}}{2} \]
\[ x = \frac{22 \pm 2}{2} \]
So,
\[ x = \frac{24}{2} = 12 \]
\[ x = \frac{20}{2} = 10 \]
Therefore, the ages are 12 and 10.
