Grade 10 Math Worksheets and Problems: Linear Equations in Two ... - Free Printable

Problem Set: Linear Equations in Two Variables

Below, I will solve each problem step by step.

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#### Problem 1:
Find the value of \( k \) for which the equations \( kx + 2y - 1 = 0 \) and \( -6x + 4y - 2 = 0 \) will have infinitely many solutions.

Solution:
For two linear equations to have infinitely many solutions, their coefficients must be proportional. The general form of the equations is:
\[ a_1x + b_1y + c_1 = 0 \]
\[ a_2x + b_2y + c_2 = 0 \]

The condition for infinitely many solutions is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

Given equations:
\[ kx + 2y - 1 = 0 \]
\[ -6x + 4y - 2 = 0 \]

Here:
\[ a_1 = k, \quad b_1 = 2, \quad c_1 = -1 \]
\[ a_2 = -6, \quad b_2 = 4, \quad c_2 = -2 \]

We need:
\[ \frac{k}{-6} = \frac{2}{4} = \frac{-1}{-2} \]

First, calculate \( \frac{2}{4} \):
\[ \frac{2}{4} = \frac{1}{2} \]

Next, calculate \( \frac{-1}{-2} \):
\[ \frac{-1}{-2} = \frac{1}{2} \]

So, we have:
\[ \frac{k}{-6} = \frac{1}{2} \]

Solve for \( k \):
\[ k = \frac{1}{2} \times (-6) \]
\[ k = -3 \]

Answer:
\[ \boxed{-3} \]

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#### Problem 2:
In a two-digit number, the sum of its digits is 11. If 45 is added to the number, the digits interchange their places. Find the number.

Solution:
Let the two-digit number be \( 10x + y \), where \( x \) is the tens digit and \( y \) is the units digit.

From the problem:
1. The sum of the digits is 11:
\[ x + y = 11 \quad \text{(Equation 1)} \]

2. If 45 is added to the number, the digits interchange:
\[ 10x + y + 45 = 10y + x \]

Rearrange the second equation:
\[ 10x + y + 45 = 10y + x \]
\[ 10x - x + y - 10y = -45 \]
\[ 9x - 9y = -45 \]
\[ x - y = -5 \quad \text{(Equation 2)} \]

Now solve the system of equations:
\[ x + y = 11 \]
\[ x - y = -5 \]

Add the two equations:
\[ (x + y) + (x - y) = 11 + (-5) \]
\[ 2x = 6 \]
\[ x = 3 \]

Substitute \( x = 3 \) into Equation 1:
\[ 3 + y = 11 \]
\[ y = 8 \]

Thus, the number is:
\[ 10x + y = 10(3) + 8 = 38 \]

Answer:
\[ \boxed{38} \]

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#### Problem 3:
If twice the daughter's age in years is added to the father's age, the sum is 41. If twice the father's age is added to the daughter's age, the sum is 61. Find the age of the father and daughter.

Solution:
Let the daughter's age be \( d \) and the father's age be \( f \).

From the problem:
1. Twice the daughter's age plus the father's age is 41:
\[ 2d + f = 41 \quad \text{(Equation 1)} \]

2. Twice the father's age plus the daughter's age is 61:
\[ d + 2f = 61 \quad \text{(Equation 2)} \]

Solve the system of equations:
\[ 2d + f = 41 \]
\[ d + 2f = 61 \]

Multiply Equation 1 by 2:
\[ 4d + 2f = 82 \quad \text{(Equation 3)} \]

Subtract Equation 2 from Equation 3:
\[ (4d + 2f) - (d + 2f) = 82 - 61 \]
\[ 4d + 2f - d - 2f = 21 \]
\[ 3d = 21 \]
\[ d = 7 \]

Substitute \( d = 7 \) into Equation 1:
\[ 2(7) + f = 41 \]
\[ 14 + f = 41 \]
\[ f = 27 \]

Thus, the daughter's age is 7 and the father's age is 27.

Answer:
\[ \boxed{27, 7} \]

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#### Problem 4:
Which of the following equations has a unique solution?
1. \( 4x + 2y + 3 = 0 \)
2. \( 12x + 6y + 8 = 0 \)
3. \( 4x + 2y + 3 = 0 \)
4. \( 8x + 4y + 6 = 0 \)

Solution:
For a pair of linear equations to have a unique solution, the ratios of their coefficients must not be equal:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]

Check each pair:

1. \( 4x + 2y + 3 = 0 \)
2. \( 12x + 6y + 8 = 0 \)

Here:
\[ \frac{4}{12} = \frac{1}{3}, \quad \frac{2}{6} = \frac{1}{3} \]
Since \( \frac{4}{12} = \frac{2}{6} \), these equations are dependent and do not have a unique solution.

3. \( 4x + 2y + 3 = 0 \)
4. \( 8x + 4y + 6 = 0 \)

Here:
\[ \frac{4}{8} = \frac{1}{2}, \quad \frac{2}{4} = \frac{1}{2} \]
Since \( \frac{4}{8} = \frac{2}{4} \), these equations are dependent and do not have a unique solution.

None of the given pairs have a unique solution.

Answer:
\[ \boxed{\text{None}} \]

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#### Problem 5:
There are two numbers. If four times the larger of two numbers is divided by the smaller one, we get 7 as quotient and 0 as remainder. If six times the smaller of two numbers is divided by the larger one, we get 3 as quotient and 15 as remainder. Find the numbers.

Solution:
Let the larger number be \( L \) and the smaller number be \( S \).

From the problem:
1. Four times the larger number divided by the smaller number gives a quotient of 7 and a remainder of 0:
\[ 4L = 7S \quad \text{(Equation 1)} \]

2. Six times the smaller number divided by the larger number gives a quotient of 3 and a remainder of 15:
\[ 6S = 3L + 15 \quad \text{(Equation 2)} \]

Solve the system of equations:
From Equation 1:
\[ 4L = 7S \]
\[ L = \frac{7S}{4} \quad \text{(Equation 3)} \]

Substitute \( L = \frac{7S}{4} \) into Equation 2:
\[ 6S = 3\left(\frac{7S}{4}\right) + 15 \]
\[ 6S = \frac{21S}{4} + 15 \]

Multiply through by 4 to clear the fraction:
\[ 4(6S) = 4\left(\frac{21S}{4}\right) + 4(15) \]
\[ 24S = 21S + 60 \]
\[ 24S - 21S = 60 \]
\[ 3S = 60 \]
\[ S = 20 \]

Substitute \( S = 20 \) into Equation 3:
\[ L = \frac{7S}{4} \]
\[ L = \frac{7(20)}{4} \]
\[ L = \frac{140}{4} \]
\[ L = 35 \]

Thus, the numbers are 35 and 20.

Answer:
\[ \boxed{35, 20} \]

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#### Problem 6:
12 chairs and 9 tables cost Rs 4950, and 5 chairs and 6 tables cost Rs 2850. Find the cost of one chair and one table separately.

Solution:
Let the cost of one chair be \( C \) and the cost of one table be \( T \).

From the problem:
1. \( 12C + 9T = 4950 \quad \text{(Equation 1)} \)
2. \( 5C + 6T = 2850 \quad \text{(Equation 2)} \)

To eliminate one variable, we can use the method of elimination. First, make the coefficients of \( T \) the same in both equations. Multiply Equation 1 by 2 and Equation 2 by 3:
\[ 2(12C + 9T) = 2(4950) \]
\[ 24C + 18T = 9900 \quad \text{(Equation 3)} \]

\[ 3(5C + 6T) = 3(2850) \]
\[ 15C + 18T = 8550 \quad \text{(Equation 4)} \]

Subtract Equation 4 from Equation 3:
\[ (24C + 18T) - (15C + 18T) = 9900 - 8550 \]
\[ 24C - 15C + 18T - 18T = 1350 \]
\[ 9C = 1350 \]
\[ C = 150 \]

Substitute \( C = 150 \) into Equation 2:
\[ 5(150) + 6T = 2850 \]
\[ 750 + 6T = 2850 \]
\[ 6T = 2850 - 750 \]
\[ 6T = 2100 \]
\[ T = 350 \]

Thus, the cost of one chair is Rs 150 and the cost of one table is Rs 350.

Answer:
\[ \boxed{150, 350} \]

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#### Problem 7:
Three years ago, Shyam was six times older than his daughter. After three years, Shyam will be 3 years more than three times the age of his daughter. Find the present age of Shyam and his daughter.

Solution:
Let the present age of Shyam be \( S \) and the present age of his daughter be \( D \).

From the problem:
1. Three years ago, Shyam was six times older than his daughter:
\[ S - 3 = 6(D - 3) \quad \text{(Equation 1)} \]

2. After three years, Shyam will be 3 years more than three times the age of his daughter:
\[ S + 3 = 3(D + 3) + 3 \quad \text{(Equation 2)} \]

Simplify Equation 1:
\[ S - 3 = 6D - 18 \]
\[ S = 6D - 15 \quad \text{(Equation 3)} \]

Simplify Equation 2:
\[ S + 3 = 3D + 9 + 3 \]
\[ S + 3 = 3D + 12 \]
\[ S = 3D + 9 \quad \text{(Equation 4)} \]

Equate Equation 3 and Equation 4:
\[ 6D - 15 = 3D + 9 \]
\[ 6D - 3D = 9 + 15 \]
\[ 3D = 24 \]
\[ D = 8 \]

Substitute \( D = 8 \) into Equation 4:
\[ S = 3(8) + 9 \]
\[ S = 24 + 9 \]
\[ S = 33 \]

Thus, the present age of Shyam is 33 and the present age of his daughter is 8.

Answer:
\[ \boxed{33, 8} \]

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#### Problem 8:
The pair of equations \( 3p - q - 1 = 0 \) and \( 6p - 4q - 4 = 0 \) have:
a. infinitely many solutions
b. a unique solution
c. exactly two solutions
d. no solution

Solution:
Rewrite the equations in standard form:
1. \( 3p - q - 1 = 0 \)
2. \( 6p - 4q - 4 = 0 \)

Compare the coefficients:
\[ \frac{3}{6} = \frac{1}{2}, \quad \frac{-1}{-4} = \frac{1}{4} \]

Since \( \frac{3}{6} \neq \frac{-1}{-4} \), the lines are parallel and do not intersect. Therefore, there are no solutions.

Answer:
\[ \boxed{\text{d}} \]

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#### Problem 9:
Usha has only Rs.10 and Rs.1 notes with her. If the total number of notes that she has is 20 and the amount of money with her is Rs. 146, then the number of Rs.10 and Rs.1 notes are, respectively:
a. 14 and 6
b. 16 and 4
c. 13 and 7
d. 12 and 8

Solution:
Let the number of Rs.10 notes be \( x \) and the number of Rs.1 notes be \( y \).

From the problem:
1. The total number of notes is 20:
\[ x + y = 20 \quad \text{(Equation 1)} \]

2. The total amount of money is Rs. 146:
\[ 10x + y = 146 \quad \text{(Equation 2)} \]

Solve the system of equations:
From Equation 1:
\[ y = 20 - x \quad \text{(Equation 3)} \]

Substitute \( y = 20 - x \) into Equation 2:
\[ 10x + (20 - x) = 146 \]
\[ 10x + 20 - x = 146 \]
\[ 9x + 20 = 146 \]
\[ 9x = 126 \]
\[ x = 14 \]

Substitute \( x = 14 \) into Equation 3:
\[ y = 20 - 14 \]
\[ y = 6 \]

Thus, the number of Rs.10 notes is 14 and the number of Rs.1 notes is 6.

Answer:
\[ \boxed{\text{a}} \]

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Final Answers:

1. \(\boxed{-3}\)
2. \(\boxed{38}\)
3. \(\boxed{27, 7}\)
4. \(\boxed{\text{None}}\)
5. \(\boxed{35, 20}\)
6. \(\boxed{150, 350}\)
7. \(\boxed{33, 8}\)
8. \(\boxed{\text{d}}\)
9. \(\boxed{\text{a}}\)