# MTH603 Grand Quiz Solution and Discussion

• Grand Quiz Total Questions : 30
Quiz will be based upon Multiple Choice Questions (MCQs).

You have to attempt the quiz online. You can start attempting the quiz any time within given date(s) of a particular subject by clicking the link for Quiz in VULMS.

Each question has a fixed time of 90 seconds. So you have to save your answer before 90 seconds. But due to unstable internet speeds, it is recommended that you should save your answer within 60 seconds. While attempting a question, keep an eye on the remaining time.

Attempting quiz is unidirectional. Once you move forward to the next question, you can not go back to the previous one. Therefore before moving to the next question, make sure that you have selected the best option.

DO NOT press Back Button / Backspace Button while attempting a question, otherwise you will lose that question.

DO NOT refresh the page unnecessarily, specially when following messages appear
Saving…

Javascript MUST be enabled in your browser; otherwise you will not be able to attempt the quiz.

If for any reason, you lose access to internet (like power failure or disconnection of internet), you will be able to attempt the quiz again from the question next to the last shown question. But remember that you have to complete the quiz before expiry of the deadline.

If any student failed to attempt the quiz in given time then no re-take or offline quiz will be held.

Start Quiz

• A series 16+8+4+2+1 is replaced by the series 16+8+4+2, then it is called

Each number in the sequence is half the value of the number receding it. So the common difference in the series is dividing by two.

16÷2=8

8÷2=4

4÷2=2

2÷2=1

1÷2=½

The answer is ½ or 0.5

When you keep dividing by two, you will notice an interesting pattern: the denominator continues to increase by two, while the numerator value remains the same. That’s fascinating because in natural, whole numbers the numbers in the series would decrease by two.

1/4 , 1/8 , 1/16 etc.

• A series 16+8+4+2+1 is replaced by the series 16+8+4+2, then it is called

• More MCQs for Numerical Analysis click here

1. Solve the system of equations by Jacobi’s iteration method.

10a - 2b - c - d = 3

• 2a + 10b - c - d = 15
• a - b + 10c - 2d = 27
• a - b - 2c + 10d = -9
a) a = 1, b = 2, c = 3, d = 0
b) a = 2, b = 1, c = 9, d = 5
c) a = 2, b = 2, c = 9, d = 0
d) a = 1, b = 1, c = 3, d = 5

Explanation: Rewriting the given equations as
a = 110(3 + 2b + c + d)
b = 110(15 + 2z + c + d)
c = 110(27 + a + b + 2d)
d = 110(-9 + a + b + 2d)
We start from an approximation a = b = c = d = 0.
First iteration: a = 0.3, b = 1.5, c = 2.7, d = -0.9
Second iteration:
a = 110[3 + 2(1.5) + 2.7 + (-0.9)] = 0.78
b = 110[15 + 2(0.3) + 2.7 + (-0.9)] = 1.74
c = 110[27 + 0.3 + 1.5 + 2(-0.9)] = 2.7
d = 110[-9 + 0.3 + 1.5 + 2(-0.9)] = -0.18
Proceeding in this way we get,
Third iteration, a = 0.9, b = 1.908, c = 2.916, d = -0.108
Fourth iteration, a = 0.9624, b = 1.9608, c = 2.9592, d = -0.036
Fifth iteration, a = 0.9845, b = 1.9848, c = 2.9851, d = -0.0158
Sixth iteration, a = 0.9939, b = 1.9938, c = 2.9938, d = -0.006
Seventh iteration, a = 0.9939, b = 1.9975, c = 2.9976, d = -0.0025
Eighth iteration, a = 0.999, b = 1.999, c = 2.999, d = -0.001
Ninth iteration, a = 0.9996, b = 1.9996, c = 2.9996, d = -0.004
Tenth iteration, a = 0.9998, b = 1.9998, c = 2.9998, d = -0.0001
Hence, a = 1, b = 2, c = 3, d = 0.

1. Solve the system of equations by Jacobi’s iteration method.

10a - 2b - c - d = 3

• 2a + 10b - c - d = 15
• a - b + 10c - 2d = 27
• a - b - 2c + 10d = -9
a) a = 1, b = 2, c = 3, d = 0
b) a = 2, b = 1, c = 9, d = 5
c) a = 2, b = 2, c = 9, d = 0
d) a = 1, b = 1, c = 3, d = 5

1. Solve the system of equations by Jacobi’s iteration method.

10x = y – x = 11.19
x + 10y + z = 28.08
-x + y + 10z = 35.61

correct to two decimal places.
a) x = 1.00, y = 2.95, z = 3.85
b) x = 1.96, y = 2.63, z = 3.99
c) x = 1.58, y = 2.70, z = 3.00
d) x = 1.23, y = 2.34, z = 3.45

Explanation: Rewriting the equations as,
x = 110 (11.19 – y + z)
y = 110 (28.08 – x – z)
z = 110 (35.61 + x – y)
We start from an approximation, x = y = z = 0.
First iteration, x = 1.119, y = 2.808, z = 3.561
Second iteration,
x = 110 (11.19 – 2.808 + 3.651) = 1.19
y = 110 (28.08 – 1.119 – 3.561) = 2.34
z = 110 (35.61 + 1.119 – 2.808) = 3.39
Third Iteration:
x = 1.22, y = 2.35, z = 3.45
Fourth iteration:
x = 1.23, y = 2.34, z = 3.45
Fifth iteration:
x = 1.23, y = 2.34, z = 3.45
Hence, x = 1.23, y = 2.34, z = 3.45.

1. Solve the system of equations by Jacobi’s iteration method.

10x = y – x = 11.19
x + 10y + z = 28.08
-x + y + 10z = 35.61

correct to two decimal places.
a) x = 1.00, y = 2.95, z = 3.85
b) x = 1.96, y = 2.63, z = 3.99
c) x = 1.58, y = 2.70, z = 3.00
d) x = 1.23, y = 2.34, z = 3.45

1. Solve the system of equations by Jacobi’s iteration method.
20x + y – 2z = 17
3x + 20y – z = -18
2x – 3y + 20z = 25

a) x = 1, y = -1, z = 1
b) x = 2, y = 1, z = 0
c) x = 2, y = 1, z = 0
d) x = 1, y = 2, z = 1

Explanation: We write the equations in the form
x = 120 (17 – y +2z)
y = 120 (-18 -3x + z)
z = 120 (25 -2x +3y)
We start from an approximation x = y = z = 0.
Substituting these in the right sides of the equations (i), (ii), (iii), we get
First iteration:
x = 0.85, y = -0.9, z = 1.25
Putting these values again in equations (i), (ii), (iii), we obtain,
x = [17 – (-0.9) + 2(1.25)] = 1.02
y = [-18 -3(0.85) + 1.25] = -0.965
z = [25 – 2(0.85) + 3(-0.9)] = 1.03
Substituting these values again in equations (i), (ii), (iii), we obtain,
Second iteration:
x = 1.00125, y = -1.0015, z = 1.00325
Proceeding in this way, we get,
Third iteration:
x = 1.0004, y = -1.000025, z = 0.9965
Fourth iteration
x = 0.999966, y = -1.000078, z = 0.999956
Fifth iteration
x = 1.0000, y = -0.999997, z = 0.999992
The values in the last iterations being practically the same, we can stop.
Hence the solution is
x = 1, y = -1, z = 1.

1. Solve the system of equations by Jacobi’s iteration method.
20x + y – 2z = 17
3x + 20y – z = -18
2x – 3y + 20z = 25

a) x = 1, y = -1, z = 1
b) x = 2, y = 1, z = 0
c) x = 2, y = 1, z = 0
d) x = 1, y = 2, z = 1

1. Which of the following is another name for Jacobi’s method?
a) Displacement method
b) Simultaneous displacement method
c) Simultaneous method
d) Diagonal method

Explanation: Jacobi’s method is also called as simultaneous displacement method because for every iteration we perform, we use the results obtained in the subsequent steps and form new results.

1. Which of the following is another name for Jacobi’s method?
a) Displacement method
b) Simultaneous displacement method
c) Simultaneous method
d) Diagonal method

1. The Jacobi’s method is a method of solving a matrix equation on a matrix that has no zeros along its main diagonal.
a) True
b) False

Explanation: The Jacobi’s method is a method of solving a matrix equation on a matrix that has no zeros along its main diagonal because the desirable convergence of the answer can be achieved only for a matrix which is diagonally dominant and a matrix that has no zeros along its main diagonal can never be diagonally dominant.

1. How many assumptions are there in Jacobi’s method?
a) 2
b) 3
c) 4
d) 5

Explanation: There are two assumptions in Jacobi’s method.

1. How many assumptions are there in Jacobi’s method?
a) 2
b) 3
c) 4
d) 5

1. Which of the following is an assumption of Jacobi’s method?
a) The coefficient matrix has no zeros on its main diagonal
b) The rate of convergence is quite slow compared with other methods
c) Iteration involved in Jacobi’s method converges
d) The coefficient matrix has zeroes on its main diagonal

Explanation: This is because it is the method employed for solving a matrix such that for every row of the matrix, the magnitude of the diagonal entry in a row is larger than or equal to the sum of the magnitudes of all the other (non-diagonal) entries in that row. This helps in converging the result and hence it is an assumption.  1

2

1

3

3

5

65
• ## MTH102 Assignment 2 Solution and Discussion MTH102 - Basic Algebra and Trigonometry • mth102 assignment 2 solution discussion fall 2019 • • zareen

2  | |