Assingnment class ix | PDF - Free Printable

Let's solve the problems step by step. I'll provide detailed explanations for each problem.

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Problem 1: Find the value of \( k \) in the equation \( 2x + y = k \) if \( x = 3 \) and \( y = 5 \).

Solution:
Substitute \( x = 3 \) and \( y = 5 \) into the equation \( 2x + y = k \):

\[
2(3) + 5 = k
\]

\[
6 + 5 = k
\]

\[
k = 11
\]

Answer:
\[
\boxed{11}
\]

---

Problem 2: Express \( x \) in terms of \( y \) given that \( \frac{x}{3} + 2y = 5 \). Check whether \( (3, 2) \) is a solution of the given equation.

Step 1: Express \( x \) in terms of \( y \):

Start with the equation:

\[
\frac{x}{3} + 2y = 5
\]

Subtract \( 2y \) from both sides:

\[
\frac{x}{3} = 5 - 2y
\]

Multiply both sides by 3:

\[
x = 3(5 - 2y)
\]

\[
x = 15 - 6y
\]

So, \( x \) in terms of \( y \) is:

\[
x = 15 - 6y
\]

Step 2: Check if \( (3, 2) \) is a solution:

Substitute \( x = 3 \) and \( y = 2 \) into the original equation \( \frac{x}{3} + 2y = 5 \):

\[
\frac{3}{3} + 2(2) = 5
\]

\[
1 + 4 = 5
\]

\[
5 = 5
\]

Since the equation holds true, \( (3, 2) \) is a solution.

Answer:
\[
x = 15 - 6y, \quad \text{(3, 2) is a solution.}
\]

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Problem 3: For what value of \( p \) does the point \( (p, 4) \) lie on the line \( 3x + y = 10 \)?

Solution:
Substitute \( x = p \) and \( y = 4 \) into the equation \( 3x + y = 10 \):

\[
3(p) + 4 = 10
\]

Subtract 4 from both sides:

\[
3p = 6
\]

Divide by 3:

\[
p = 2
\]

Answer:
\[
\boxed{2}
\]

---

Problem 4: Give the geometric representation of \( 2x + 7 = 0 \):

Step 1: Solve for \( x \):

\[
2x + 7 = 0
\]

Subtract 7 from both sides:

\[
2x = -7
\]

Divide by 2:

\[
x = -\frac{7}{2}
\]

This represents a vertical line where \( x = -\frac{7}{2} \).

Step 2: Geometric representation:

- In one variable: The equation \( x = -\frac{7}{2} \) represents a single point on the number line at \( x = -\frac{7}{2} \).
- In two variables: The equation \( x = -\frac{7}{2} \) represents a vertical line parallel to the \( y \)-axis passing through \( x = -\frac{7}{2} \).

Answer:
\[
\text{I. A point at } x = -\frac{7}{2} \text{ on the number line.}
\]
\[
\text{II. A vertical line } x = -\frac{7}{2} \text{ in the coordinate plane.}
\]

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Problem 5: Write two solutions of the equation \( 5x + 4y = 0 \).

Solution:
To find solutions, choose values for \( x \) and solve for \( y \), or vice versa.

1. Let \( x = 0 \):

\[
5(0) + 4y = 0
\]

\[
4y = 0
\]

\[
y = 0
\]

So, one solution is \( (0, 0) \).

2. Let \( y = 5 \):

\[
5x + 4(5) = 0
\]

\[
5x + 20 = 0
\]

\[
5x = -20
\]

\[
x = -4
\]

So, another solution is \( (-4, 5) \).

Answer:
\[
\boxed{(0, 0), (-4, 5)}
\]

---

Problem 6: Draw the graph of \( 2x + y = 8 \).

Solution:
To draw the graph, find two points that satisfy the equation and plot them.

1. Let \( x = 0 \):

\[
2(0) + y = 8
\]

\[
y = 8
\]

Point: \( (0, 8) \)

2. Let \( y = 0 \):

\[
2x + 0 = 8
\]

\[
2x = 8
\]

\[
x = 4
\]

Point: \( (4, 0) \)

Plot the points \( (0, 8) \) and \( (4, 0) \) and draw a straight line through them.

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Problem 7: If the point \( (3, 4) \) lies on the graph of the equation \( 3y = ax + 7 \), find the value of \( a \).

Solution:
Substitute \( x = 3 \) and \( y = 4 \) into the equation \( 3y = ax + 7 \):

\[
3(4) = a(3) + 7
\]

\[
12 = 3a + 7
\]

Subtract 7 from both sides:

\[
5 = 3a
\]

Divide by 3:

\[
a = \frac{5}{3}
\]

Answer:
\[
\boxed{\frac{5}{3}}
\]

---

Problem 8: By drawing the graph of \( 3x - 2y = 4 \) and \( x + y - 3 = 0 \), find the value of \( x \) and \( y \).

Solution:
Solve the system of equations algebraically:

1. Equation 1: \( 3x - 2y = 4 \)
2. Equation 2: \( x + y - 3 = 0 \) or \( x + y = 3 \)

From Equation 2, solve for \( y \):

\[
y = 3 - x
\]

Substitute \( y = 3 - x \) into Equation 1:

\[
3x - 2(3 - x) = 4
\]

\[
3x - 6 + 2x = 4
\]

\[
5x - 6 = 4
\]

Add 6 to both sides:

\[
5x = 10
\]

Divide by 5:

\[
x = 2
\]

Substitute \( x = 2 \) back into \( y = 3 - x \):

\[
y = 3 - 2
\]

\[
y = 1
\]

Answer:
\[
\boxed{(2, 1)}
\]

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Problem 9: Give the equation of two lines passing through \( (3, 4) \). How many more such lines are there and why?

Solution:
1. Equation of a line passing through \( (3, 4) \):

The general form of a line is \( y = mx + c \). Since the line passes through \( (3, 4) \), substitute \( x = 3 \) and \( y = 4 \):

\[
4 = 3m + c
\]

Rearrange to express \( c \) in terms of \( m \):

\[
c = 4 - 3m
\]

So, the equation of the line is:

\[
y = mx + (4 - 3m)
\]

Choose two different values for \( m \):

- Let \( m = 1 \):

\[
y = 1x + (4 - 3 \cdot 1)
\]

\[
y = x + 1
\]

- Let \( m = -1 \):

\[
y = -1x + (4 - 3 \cdot -1)
\]

\[
y = -x + 7
\]

So, two lines are:

\[
y = x + 1 \quad \text{and} \quad y = -x + 7
\]

2. How many more such lines are there?

There are infinitely many lines passing through a single point because the slope \( m \) can take any real value. Each value of \( m \) gives a different line.

Answer:
\[
\boxed{y = x + 1, \, y = -x + 7, \, \text{infinitely many more lines}}
\]

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Problem 10: Find the cost of a car if the selling price of the car is Rs. 72000 and the profit is 20%.

Solution:
Let the cost price of the car be \( C \). The selling price is given as Rs. 72000, and the profit is 20%.

The relationship between cost price, selling price, and profit is:

\[
\text{Selling Price} = \text{Cost Price} + \text{Profit}
\]

\[
\text{Profit} = 20\% \text{ of Cost Price} = 0.2C
\]

So,

\[
72000 = C + 0.2C
\]

\[
72000 = 1.2C
\]

Divide by 1.2:

\[
C = \frac{72000}{1.2}
\]

\[
C = 60000
\]

Answer:
\[
\boxed{60000}
\]

---

Problem 11: The perimeter of a rectangular plot is 120 m. If the length of the plot is decreased by 5 m and the breadth is increased by 5 m, the area is increased by 75 sqm. Find the length and breadth of the rectangular plot.

Solution:
Let the length of the rectangle be \( l \) and the breadth be \( b \).

1. Perimeter equation:

\[
2(l + b) = 120
\]

\[
l + b = 60 \quad \text{(Equation 1)}
\]

2. Area condition:

Original area: \( A = l \times b \)

New dimensions: Length \( l - 5 \), Breadth \( b + 5 \)

New area: \( (l - 5)(b + 5) \)

The new area is 75 sqm more than the original area:

\[
(l - 5)(b + 5) = lb + 75
\]

Expand the left side:

\[
lb + 5l - 5b - 25 = lb + 75
\]

Subtract \( lb \) from both sides:

\[
5l - 5b - 25 = 75
\]

Add 25 to both sides:

\[
5l - 5b = 100
\]

Divide by 5:

\[
l - b = 20 \quad \text{(Equation 2)}
\]

3. Solve the system of equations:

From Equation 1: \( l + b = 60 \)

From Equation 2: \( l - b = 20 \)

Add the two equations:

\[
(l + b) + (l - b) = 60 + 20
\]

\[
2l = 80
\]

\[
l = 40
\]

Substitute \( l = 40 \) into \( l + b = 60 \):

\[
40 + b = 60
\]

\[
b = 20
\]

Answer:
\[
\boxed{40, 20}
\]

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Problem 12: If a driver drives at a speed of 24 km/hr, he reaches his destination 5 minutes too late. But if he drives at the speed of 30 km/hr, he reaches his destination 4 minutes soon. How far is his distance?

Solution:
Let the distance to the destination be \( D \) km, and let the correct time to reach the destination be \( T \) hours.

1. Time taken at 24 km/hr:

\[
\text{Time} = \frac{D}{24}
\]

Since he is 5 minutes (or \( \frac{5}{60} = \frac{1}{12} \) hours) late:

\[
\frac{D}{24} = T + \frac{1}{12}
\]

2. Time taken at 30 km/hr:

\[
\text{Time} = \frac{D}{30}
\]

Since he is 4 minutes (or \( \frac{4}{60} = \frac{1}{15} \) hours) early:

\[
\frac{D}{30} = T - \frac{1}{15}
\]

3. Solve the system of equations:

From the first equation:

\[
\frac{D}{24} = T + \frac{1}{12}
\]

\[
T = \frac{D}{24} - \frac{1}{12} \quad \text{(Equation 1)}
\]

From the second equation:

\[
\frac{D}{30} = T - \frac{1}{15}
\]

\[
T = \frac{D}{30} + \frac{1}{15} \quad \text{(Equation 2)}
\]

Equate the two expressions for \( T \):

\[
\frac{D}{24} - \frac{1}{12} = \frac{D}{30} + \frac{1}{15}
\]

Find a common denominator (120):

\[
\frac{5D}{120} - \frac{10}{120} = \frac{4D}{120} + \frac{8}{120}
\]

Combine like terms:

\[
\frac{5D - 10}{120} = \frac{4D + 8}{120}
\]

Multiply through by 120:

\[
5D - 10 = 4D + 8
\]

Subtract \( 4D \) from both sides:

\[
D - 10 = 8
\]

Add 10 to both sides:

\[
D = 18
\]

Answer:
\[
\boxed{18}
\]

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Final Answers:

1. \(\boxed{11}\)
2. \(x = 15 - 6y\), \((3, 2)\) is a solution.
3. \(\boxed{2}\)
4. I. A point at \( x = -\frac{7}{2} \). II. A vertical line \( x = -\frac{7}{2} \).
5. \(\boxed{(0, 0), (-4, 5)}\)
6. Graph of \( 2x + y = 8 \).
7. \(\boxed{\frac{5}{3}}\)
8. \(\boxed{(2, 1)}\)
9. \(y = x + 1\), \(y = -x + 7\), infinitely many more lines.
10. \(\boxed{60000}\)
11. \(\boxed{40, 20}\)
12. \(\boxed{18}\)