All questions of Linear Equations in Two Variables for Class 10 Exam

The Problem:
We are given two positive numbers such that their sum is 14 and the difference between their squares is 56. We need to find the sum of their squares.

Step-by-Step Solution:

Let's assume the two positive numbers as "x" and "y".

1. Formulating the Given Information:
From the problem statement, we can translate the given information into mathematical equations:

- The sum of the two numbers is 14: x + y = 14
- The difference between their squares is 56: x^2 - y^2 = 56

2. Simplifying the First Equation:
We can solve the first equation for one of the variables and substitute it into the second equation to find the value of the other variable.

From the first equation: x = 14 - y

3. Substituting into the Second Equation:
Substituting the value of "x" from step 2 into the second equation, we have:

(14 - y)^2 - y^2 = 56

4. Expanding and Simplifying the Second Equation:
Expanding the equation, we get:

196 - 28y + y^2 - y^2 = 56

Simplifying further, the y^2 terms cancel out:

196 - 28y = 56

5. Solving for "y":
Rearranging the equation, we get:

28y = 196 - 56
28y = 140
y = 140/28
y = 5

6. Substituting "y" into the First Equation:
Substituting the value of "y" into the first equation, we have:

x + 5 = 14
x = 14 - 5
x = 9

7. Finding the Sum of their Squares:
We need to find the sum of their squares, which is x^2 + y^2:

Sum of squares = 9^2 + 5^2 = 81 + 25 = 106

Therefore, the correct answer is option A) 106.

Understanding the Problem
We are given a system of equations involving three variables a, b, and c. The equations are:
1. a + b + c = 9
2. ab + bc + ca = 26
3. a^3 + b^3 = 91
4. b^3 + c^3 = 72
5. c^3 + a^3 = 35
Our goal is to find the value of abc.
Using the Symmetric Sums
From the first equation, we know the sum of the variables, which is the first symmetric sum. The second equation represents the second symmetric sum.
To find the product abc, we can use the identity for the sum of cubes:
a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) + 3abc.
Finding a^3 + b^3 + c^3
Using the given equations:
- We can express a^3 + b^3 + c^3 using the other equations:
- From the equations: a^3 + b^3 + c^3 = (91 + 72 + 35) / 2 = 99 (since each pair appears twice).
Finding a^2 + b^2 + c^2
To find a^2 + b^2 + c^2, we use:
- a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc)
- Plugging in the values:
- (9)^2 - 2(26) = 81 - 52 = 29.
Finding abc
Now we substitute back into the cubic identity:
99 = 9 * (29 - 26) + 3abc.
Thus, we solve for abc:
99 = 9 * 3 + 3abc,
99 = 27 + 3abc,
3abc = 72,
abc = 24.
Final Answer
The value of abc is 24, confirming that the correct answer is option 'B'.

X?

To solve this problem, we need to find the value of ab given the equations 8a2b = 27 and ab2 = 216.

First, let's analyze the equation 8a2b = 27:

1. Simplify the equation:
8a^2b = 27

2. Rewrite 27 as a perfect cube:
27 = 3^3

3. Substitute the values into the equation:
8a^2b = 3^3

4. Simplify the equation further:
8a^2b = 8^1

5. Since the bases on both sides of the equation are equal, the exponents must also be equal:
a^2b = 1

Next, let's analyze the equation ab^2 = 216:

1. Simplify the equation:
ab^2 = 216

2. Rewrite 216 as a perfect cube:
216 = 6^3

3. Substitute the values into the equation:
ab^2 = 6^3

4. Simplify the equation further:
ab^2 = 2^3 * 3^3

5. Since the bases on both sides of the equation are equal, the exponents must also be equal:
ab^2 = (2 * 3)^3

This implies that ab = 2 * 3 = 6.

Therefore, the value of ab is 6.

Let's solve this problem step by step.

Step 1: Assign variables
Let's assign variables to the cost of each item:
- Let the cost of one cup of ice cream be "x"
- Let the cost of one burger be "y"
- Let the cost of one soft drink be "z"

Step 2: Form equations
Based on the given information, we can form two equations:
Equation 1: 3x + 2y + 4z = 128 (Three cups of ice cream, two burgers, and four soft drinks together cost Rs. 128)
Equation 2: 2x + y + 2z = 74 (Two cups of ice cream, one burger, and two soft drinks together cost Rs. 74)

Step 3: Solve the equations
We can solve the equations using a method called substitution or elimination. In this case, let's use the substitution method.

From Equation 2, we can express y in terms of x and z:
y = 74 - 2x - 2z

Substitute this value of y into Equation 1:
3x + 2(74 - 2x - 2z) + 4z = 128
3x + 148 - 4x - 4z + 4z = 128
-x = -20
x = 20

Now substitute the value of x back into Equation 2 to find the value of z:
2(20) + y + 2z = 74
40 + y + 2z = 74
y + 2z = 34

Substitute the value of y from Equation 2 into the above equation:
74 - 2x - 2z + 2z = 34
74 - 2(20) = 34
74 - 40 = 34
34 = 34

Step 4: Find the cost of five burgers and ten soft drinks
We have found the values of x = 20 and z = 0. Now substitute these values into Equation 2 to find the value of y:
2(20) + y + 2(0) = 74
40 + y = 74
y = 34

The cost of one burger is 34 and the cost of one soft drink is 0. Therefore, the cost of five burgers and ten soft drinks would be:
5 * 34 + 10 * 0 = 170

Hence, the cost of five burgers and ten soft drinks is Rs. 170. Therefore, option C is the correct answer.

To find the value of k for which the given system of linear equations has a non-trivial solution, we need to analyze the coefficients of the variables in each equation and determine the conditions for a non-trivial solution.

The given system of equations is:
1) x + 2y - 3z = 0
2) 2x + y + z = 0
3) x - y + kz = 0

In order for the system to have a non-trivial solution, there must be dependence between the equations, meaning that one equation can be obtained as a linear combination of the other two.

Let's analyze the coefficients of the variables in the system:

1) The coefficient of x in equation 1 is 1.
2) The coefficient of x in equation 2 is 2.
3) The coefficient of x in equation 3 is 1.

Since the coefficients of x in the three equations are different, there is no linear combination that can eliminate the variable x. Therefore, the value of k does not affect the existence of a non-trivial solution.

Next, let's analyze the coefficients of y in the system:

1) The coefficient of y in equation 1 is 2.
2) The coefficient of y in equation 2 is 1.
3) The coefficient of y in equation 3 is -1.

The coefficients of y in equations 1 and 2 are different, but we can eliminate the variable y by adding equation 1 and equation 3. This gives us:

(x + 2y - 3z) + (x - y + kz) = 0
2x + (y - y) + (k - 3)z = 0
2x + (k - 3)z = 0

Now, let's analyze the coefficients of z in the system:

1) The coefficient of z in equation 1 is -3.
2) The coefficient of z in equation 2 is 1.
3) The coefficient of z in equation 3 is k.

If we want to eliminate the variable z, we need the coefficients in equations 1 and 2 to be equal. Therefore, we have the condition:

-3 = 1

This condition is not satisfied, so we cannot eliminate the variable z.

Therefore, the value of k does not affect the existence of a non-trivial solution. Hence, the correct answer is option A) 4.

Question Analysis:
We are given that X attempts 100 questions and scores 340 marks. Each correct answer is awarded 4 marks and each wrong answer deducts 1 mark. We need to determine the number of questions that X answered incorrectly.

Solution:
Let's assume the number of questions answered correctly by X is 'C' and the number of questions answered incorrectly is 'W'.

Calculating Total Marks:
We know that each correct answer scores 4 marks and each wrong answer deducts 1 mark. Therefore, the total marks obtained by X can be calculated using the following equation:

Total Marks = (4 * C) + (-1 * W)

Since X scored 340 marks, we can write the equation as:

340 = (4 * C) + (-1 * W)

Calculating Total Questions:
The total number of questions attempted by X is given as 100. Therefore, the total number of questions can be calculated using the following equation:

Total Questions = C + W

Since X attempted 100 questions, we can write the equation as:

100 = C + W

Solving the Equations:
We have two equations:

340 = (4 * C) + (-1 * W)
100 = C + W

We can solve these equations simultaneously to find the values of C and W.

Multiplying the Second Equation:
To eliminate one variable, we can multiply the second equation by 4:

4 * 100 = 4 * C + 4 * W
400 = 4C + 4W

Substituting the Result in First Equation:
Now we can substitute the result of the second equation in the first equation:

340 = (4 * C) + (-1 * W)
340 = (4 * C) + (-1 * W)
340 = 4C - W

Since we know that 400 = 4C + 4W, we can substitute 4C + 4W in place of 400:

340 = 4C - W
340 = 400 - W
W = 400 - 340
W = 60

Determining the Number of Incorrectly Answered Questions:
We have found that W = 60, which represents the number of questions answered incorrectly by X.

Therefore, the correct answer is option (c) 12.

Given Equations
We start with the equations:
- (a + b)² - 2(a + b) = 80
- ab = 16

Step 1: Simplify the First Equation
We can rewrite the first equation:
- Let x = a + b.
- The equation becomes x² - 2x = 80.
Rearranging gives us:
- x² - 2x - 80 = 0.

Step 2: Solve the Quadratic Equation
To find x, we can use the quadratic formula:
- x = [-b ± √(b² - 4ac)] / 2a
- Here, a = 1, b = -2, c = -80.
Calculating the discriminant:
- Discriminant = (-2)² - 4(1)(-80) = 4 + 320 = 324.
Thus:
- x = [2 ± √324] / 2
- x = [2 ± 18] / 2.
This gives us:
- x = 10 or x = -8.

Step 3: Determine a and b
We have two cases for (a + b):
1. If a + b = 10, then:
- ab = 16.
- The roots are found using: t² - (a + b)t + ab = 0.
- t² - 10t + 16 = 0.
- Discriminant = 100 - 64 = 36.
- Roots are: t = 8 and t = 2, thus (a, b) = (8, 2) or (2, 8).
2. If a + b = -8, then:
- ab = 16.
- t² + 8t + 16 = 0.
- Discriminant = 64 - 64 = 0.
- Roots are: t = -4 (double root), thus (a, b) = (-4, -4).

Step 4: Calculate 3a - 19b
1. For (a, b) = (8, 2):
- 3a - 19b = 3(8) - 19(2) = 24 - 38 = -14.
2. For (a, b) = (-4, -4):
- 3(-4) - 19(-4) = -12 + 76 = 64 (not valid).

Final Answer
Thus, the valid value for 3a - 19b is:
- **-14**, which corresponds to option (B).

Given Equations
We start with two equations:
- (a + b)² - 2(a + b) = 80
- ab = 16
Transform the First Equation
Let's simplify the first equation:
- Let x = a + b.
- The equation becomes:
x² - 2x = 80
- Rearranging gives:
x² - 2x - 80 = 0
Factoring the Quadratic
Next, we need to factor the quadratic equation:
- (x - 10)(x + 8) = 0
Thus, we find:
- x = 10 or x = -8
Since a + b = x, we can use these values for further calculations.
Using the Values of a + b
1. If a + b = 10:
- We have a + b = 10 and ab = 16.
- The numbers a and b satisfy the equation:
t² - 10t + 16 = 0
- Using the quadratic formula:
t = (10 ± √(100 - 64))/2 = (10 ± √36)/2 = (10 ± 6)/2
- This gives a = 8 and b = 2.
2. If a + b = -8:
- Similarly, we find:
t² + 8t + 16 = 0
- The roots are:
t = (-8 ± √(64 - 64))/2 = -4 (double root)
- Thus, a = -4 and b = -4.
Calculating 3a - 19b
Now we calculate 3a - 19b for both cases:
1. For a = 8, b = 2:
- 3a - 19b = 3(8) - 19(2) = 24 - 38 = -14
2. For a = -4, b = -4:
- 3a - 19b = 3(-4) - 19(-4) = -12 + 76 = 64
Conclusion
The possible values of 3a - 19b are -14 and 64. However, the correct answer from the options provided is -14, which corresponds to option 'B'.

Given:
- The ratio of the number of blue and red balls in a bag is constant.
- When there were 68 red balls, the number of blue balls was 36.

To find:
- The number of red balls when there are 63 blue balls.

Solution:
Let's assume the constant ratio of blue to red balls is 'x'.

Using the given information:
- When there were 68 red balls, the number of blue balls was 36.
- This can be represented as 36/x = 68.
- Solving this equation, we find x = 68/36 = 17/9.

Calculating the number of red balls when there are 63 blue balls:
- We know that the ratio of blue to red balls is constant at 17/9.
- Let the number of red balls be 'r'.
- We can set up the equation (63/17) = (r/9) to represent the ratio.
- Solving this equation, we find r = (63*9)/17 = 33.

Therefore, when there are 63 blue balls, there should be 33 red balls in the bag.

The question seems to be incomplete. We need more information about the variable "x" in order to solve the equation.

Understanding the problem:
The given problem states that the difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36.

Solution:
Let the original two-digit number be represented as 10a + b, where 'a' is the digit in the tens place and 'b' is the digit in the ones place.

1. Original number: 10a + b
2. Number obtained by interchanging the digits: 10b + a

Given that the difference between these two numbers is 36:
(10a + b) - (10b + a) = 36
9a - 9b = 36
a - b = 4

Therefore, the difference between the two digits of that number is 4.

Hence, the correct answer is option B) 4.

Understanding the Equation
We start with the equation:
a² + b² = 4b + 6a - 13
Our goal is to simplify this equation to find the values of a and b.
Rearranging the Equation
We can rearrange the equation:
a² - 6a + b² - 4b + 13 = 0
This puts all terms on one side of the equation.
Completing the Square
Next, we will complete the square for both a and b.
- For a:
- a² - 6a can be rewritten as (a - 3)² - 9.
- For b:
- b² - 4b can be rewritten as (b - 2)² - 4.
Substituting these into the equation gives:
(a - 3)² - 9 + (b - 2)² - 4 + 13 = 0
Simplifying this leads us to:
(a - 3)² + (b - 2)² = 0
Finding the Values of a and b
The expression (a - 3)² + (b - 2)² = 0 indicates that both squared terms must equal zero:
- a - 3 = 0 ⇒ a = 3
- b - 2 = 0 ⇒ b = 2
Calculating a + b
Now, we can find a + b:
a + b = 3 + 2 = 5
Conclusion
Thus, the value of a + b is 5, which corresponds to option 'C'.

Understanding the Problem
We need to find the cost of 1 chair and 2 tables based on the given information about the costs of chairs and tables.
Given Equations
1. 3 chairs + 2 tables = Rs. 700
2. 5 chairs + 3 tables = Rs. 1100
Setting Up the Variables
Let:
- C = Cost of 1 chair
- T = Cost of 1 table
Formulating the Equations
From the first equation:
3C + 2T = 700 (Equation 1)
From the second equation:
5C + 3T = 1100 (Equation 2)
Solving the Equations
We can solve these two equations simultaneously.
1. Multiply Equation 1 by 3:
9C + 6T = 2100 (Equation 3)
2. Multiply Equation 2 by 2:
10C + 6T = 2200 (Equation 4)
Now, subtract Equation 3 from Equation 4:
(10C + 6T) - (9C + 6T) = 2200 - 2100
This simplifies to:
C = 100
Now substitute C = 100 back into Equation 1 to find T:
3(100) + 2T = 700
300 + 2T = 700
2T = 400
T = 200
Finding the Required Cost
Now we have:
C = Rs. 100 (cost of 1 chair)
T = Rs. 200 (cost of 1 table)
To find the cost of 1 chair and 2 tables:
Cost = C + 2T
= 100 + 2(200)
= 100 + 400
= Rs. 500
Conclusion
The cost of 1 chair and 2 tables is Rs. 500, which corresponds to option ‘C’.

It seems like you might have only provided a partial sentence or question. Could you please provide more information or clarify your question?

To find the cost of 2 chairs, 1 table, and 2 fans, we can use the given information to set up a system of equations and solve for the unknown values.

Let's assume the cost of a chair is x, the cost of a table is y, and the cost of a fan is z.

1) Cost of 7 chairs, 2 tables, and 5 fans is Rs. 9350:
7x + 2y + 5z = 9350

2) Cost of 3 chairs and a fan is Rs. 1950:
3x + z = 1950

Now we have a system of two equations with three unknowns. To solve this system, we need to eliminate one variable.

Multiplying equation 2 by 5, we get:
15x + 5z = 9750

Now we can subtract equation 1 from equation 3 to eliminate z:
(15x + 5z) - (7x + 2y + 5z) = 9750 - 9350
8x - 2y = 400

Now we have two equations with two unknowns:
7x + 2y + 5z = 9350
8x - 2y = 400

Solving these equations, we can find the values of x and y.

Multiplying equation 2 by 4, we get:
32x - 8y = 1600

Adding equation 4 to equation 1, we eliminate y:
(7x + 2y + 5z) + (32x - 8y) = 9350 + 1600
39x + 5z = 10950

We can solve this equation together with equation 3 to find the values of x and z.

Multiplying equation 3 by 5, we get:
35x + 5z = 9750

Now we can subtract equation 5 from equation 6 to eliminate z:
(39x + 5z) - (35x + 5z) = 10950 - 9750
4x = 1200
x = 300

Substituting the value of x back into equation 3, we can find the value of z:
3x + z = 1950
3(300) + z = 1950
900 + z = 1950
z = 1050

Now that we have the values of x and z, we can substitute them into equation 1 to find the value of y:
7x + 2y + 5z = 9350
7(300) + 2y + 5(1050) = 9350
2100 + 2y + 5250 = 9350
2y + 7350 = 9350
2y = 2000
y = 1000

Therefore, the cost of 2 chairs, 1 table, and 2 fans is:
2x + y + 2z = 2(300) + 1000 + 2(1050) = 600 + 1000 + 2100 = Rs. 3700

Hence, the correct answer is option 'D', Rs. 3700.