Solve the math questions in the image.

Steps to Solve

Here's a breakdown of how to solve each question from the image:

Question 3. Plotting lines and finding intersection

1. Create a table of values for $y = 3x + 1$ Choose a few values of $x$ and calculate corresponding $y$ values. For example: If $x = 0$, $y = 3(0) + 1 = 1$. If $x = 1$, $y = 3(1) + 1 = 4$. If $x = -1$, $y = 3(-1) + 1 = -2$.

2. Create a table of values for $x = 3y + 5$ Choose a few values of $y$ and calculate corresponding $x$ values. For example: If $y = 0$, $x = 3(0) + 5 = 5$. If $y = 1$, $x = 3(1) + 5 = 8$. If $y = -1$, $x = 3(-1) + 5 = 2$.

3. Plot the lines on a graph Plot the points from the tables on a graph and draw the lines.

4. Find the point of intersection Identify the coordinates $(x, y)$ where the two lines intersect on the graph or solve the equations simultaneously.

To solve simultaneously: $y = 3x + 1$ $x = 3y + 5$

Substitute the first equation into the second: $x = 3(3x + 1) + 5$ $x = 9x + 3 + 5$ $x = 9x + 8$ $-8x = 8$ $x = -1$

Now substitute $x = -1$ into $y = 3x + 1$: $y = 3(-1) + 1$ $y = -3 + 1$ $y = -2$

So, the point of intersection is $(-1, -2)$.

Question 4.1. Area of the Octagon

1. Divide the octagon The octagon can be divided into a rectangle and two trapeziums.

2. Calculate the area of the rectangle The rectangle has length 15cm and width 6cm. Area = length $\times$ width = $15 \times 6 = 90 \text{ cm}^2$.

3. Calculate the area of one trapezium The parallel sides of trapezium are 6cm and 8cm, and the height is 8cm. Area = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times (6 + 8) \times 8 = \frac{1}{2} \times 14 \times 8 = 7 \times 8 = 56 \text{ cm}^2$.

4. Calculate the total area of octagon Add the area of the rectangle and two trapeziums. Total Area $= 90 + 56 + 56 = 202 \text{ cm}^2$.

Question 4.2. Finding $x^2 + \frac{1}{x^2}$ and $x - \frac{1}{x}$

1. Find $x^2 + \frac{1}{x^2}$ We have $x^4 + \frac{1}{x^4} = 119$. Add 2 to both sides: $x^4 + 2 + \frac{1}{x^4} = 121$ $(x^2 + \frac{1}{x^2})^2 = 121$ $x^2 + \frac{1}{x^2} = \sqrt{121} = 11$

2. Find $x - \frac{1}{x}$ We know $x^2 + \frac{1}{x^2} = 11$. Subtract 2 from both sides: $x^2 - 2 + \frac{1}{x^2} = 9$ $(x - \frac{1}{x})^2 = 9$ $x - \frac{1}{x} = \sqrt{9} = 3$

Question 4.3. Plotting points and verifying if they lie on a line

1. Plot the points for (a) The points are (3, 0), (5, 0), (-4, 0). These points lie on the x-axis (y = 0). So, the equation of the line is $y = 0$.

2. Plot the points for (b) The points are (-3, 3), (0, 3), (4, 3). These points lie on a horizontal line where y is always 3. So, the equation of the line is $y = 3$.

Question 5.1. Solving linear equations by substitution

1. Given Equations $5x + 4y = 14$ and $2x - 3 = y$

2. Substitute y in the first equation $5x + 4(2x - 3) = 14$ $5x + 8x - 12 = 14$ $13x = 26$ $x = 2$

3. Substitute x to find y $y = 2(2) - 3$ $y = 4 - 3$ $y = 1$

The solution is $x = 2$, $y = 1$.

Question 5.2. Number of cubical blocks

1. Convert all measurements to cm Dimensions of rectangular piece of metal: 3m = 300cm, 75cm, 60cm. Side of cubical block = 30cm.

2. Calculate the volume of the rectangular piece of metal Volume $= 300 \times 75 \times 60 = 1350000 \text{ cm}^3$.

3. Calculate the volume of one cubical block Volume $= 30 \times 30 \times 30 = 27000 \text{ cm}^3$.

4. Calculate the number of cubical blocks Number of blocks $= \frac{\text{Volume of rectangular metal}}{\text{Volume of one cubical block}} = \frac{1350000}{27000} = 50$.

Question 5.3. Divide by Factorization

1. Factorize the expression $33(x^3 - 2x^2 - 15x) = 33x(x^2 - 2x - 15)$ $33x(x^2 - 2x - 15) = 33x(x - 5)(x + 3)$

2. Divide $\frac{33x(x - 5)(x + 3)}{11x(x - 5)} = \frac{33}{11} \times \frac{x}{x} \times \frac{(x - 5)}{(x - 5)} \times (x + 3) = 3(x + 3) = 3x + 9$

Question 6.1. Simple Interest

1. Formula for Simple Interest $SI = \frac{P \times R \times T}{100}$, where P = Principal, R = Rate, T = Time

2. Calculate Simple Interest $SI = \frac{15000 \times 15 \times 3}{100} = 150 \times 15 \times 3 = 6750$

3. Calculate the Amount Amount = Principal + Simple Interest $= 15000 + 6750 = 21750$

Question 6.2. Ages of Astha and Shiv

1. Set up the equations Let Astha's age be 5x and Shiv's age be 6x. Five years from now, Astha's age will be $5x + 5$ and Shiv's age will be $6x + 5$. The ratio of their ages will be $\frac{5x + 5}{6x + 5} = \frac{6}{7}$

2. Solve for x $7(5x + 5) = 6(6x + 5)$ $35x + 35 = 36x + 30$ $x = 5$

3. Find their present ages Astha's age $= 5x = 5(5) = 25$. Shiv's age $= 6x = 6(5) = 30$.

Question 6.3. Construct a quadrilateral

1. Draw base PQ = 4.5cm
2. From P, draw an arc of radius 6.5cm (PR)
3. From Q, draw an arc of radius 4cm (QR) intersecting the previous arc at R
4. From P, draw an arc of radius 3cm (PS)
5. From R, draw an arc of radius 6.5cm (RS) intersecting the previous arc at S
6. Join PQ, QR, RS, and SP to complete the quadrilateral PQRS

Question 7.1. Solve the equation

1. Equation $\frac{x+2}{3} - \frac{x-2}{4} = \frac{x+1}{5} - 1$

2. Multiply by the LCM of 3, 4, and 5, which is 60 $60(\frac{x+2}{3}) - 60(\frac{x-2}{4}) = 60(\frac{x+1}{5}) - 60(1)$ $20(x+2) - 15(x-2) = 12(x+1) - 60$ $20x + 40 - 15x + 30 = 12x + 12 - 60$ $5x + 70 = 12x - 48$ $7x = 118$ $x = \frac{118}{7}$

Question 7.2. Area of the label

1. Height of the label The label is placed 2cm from the top and bottom, so the height $= 20 - 2 - 2 = 16$cm.

2. Radius of the container The radius is 7cm.

3. Area of the label The label is wrapped around the curved surface, so area = curved surface area of cylinder = $2\pi rh = 2 \times \frac{22}{7} \times 7 \times 16 = 2 \times 22 \times 16 = 704 \text{ cm}^2$.

Question 7.3. What sum will amount to ₹6060

1. Formula for Amount $A = P(1 + \frac{RT}{100})$

2. Values given $A = 6060$, $T = 6.5$ years, $R = 5\frac{1}{3} = \frac{16}{3}%$

3. Substitute and solve $6060 = P(1 + \frac{16}{3} \times \frac{6.5}{100})$ $6060 = P(1 + \frac{16 \times 6.5}{300})$ $6060 = P(1 + \frac{104}{300})$ $6060 = P(1 + \frac{26}{75})$ $6060 = P(\frac{75 + 26}{75})$ $6060 = P(\frac{101}{75})$ $P = \frac{6060 \times 75}{101} = 60 \times 75 = 4500$ So, $P = 4500$.