# 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:

\[ x^2 + (21 - x)^2 = 261 \]

\[ x^2 + 441 - 42x + x^2 = 261 \]

\[ 2x^2 - 42x + 441 = 261 \]

\[ 2x^2 - 42x + 180 = 0 \]

\[ x^2 - 21x + 90 = 0 \]

\[ a = 1, b = -21, c = 90 \]

\[ x = \frac{21 \pm \sqrt{(-21)^2 - 4 \cdot 1 \cdot 90}}{2 \cdot 1} \]

\[ x = \frac{21 \pm \sqrt{441 - 360}}{2} \]

\[ x = \frac{21 \pm \sqrt{81}}{2} \]

\[ x = \frac{21 \pm 9}{2} \]

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 \( x + 4 \).

\[ x(x + 4) = 221 \]

\[ x^2 + 4x - 221 = 0 \]

\[ a = 1, b = 4, c = -221 \]

\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-221)}}{2 \cdot 1} \]

\[ x = \frac{-4 \pm \sqrt{16 + 884}}{2} \]

\[ x = \frac{-4 \pm \sqrt{900}}{2} \]

\[ x = \frac{-4 \pm 30}{2} \]

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:

\[ x(22 - x) = 120 \]

\[ 22x - x^2 = 120 \]

\[ x^2 - 22x + 120 = 0 \]

\[ a = 1, b = -22, c = 120 \]

\[ x = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 1 \cdot 120}}{2 \cdot 1} \]

\[ x = \frac{22 \pm \sqrt{484 - 480}}{2} \]

\[ x = \frac{22 \pm \sqrt{4}}{2} \]

\[ x = \frac{22 \pm 2}{2} \]

So,

\[ x = \frac{24}{2} = 12 \]

\[ x = \frac{20}{2} = 10 \]

Therefore, the ages are 12 and 10.