Here are 1000 MCQs on Numerical Methods (Chapterwise).
1. Numerical techniques more commonly involve _______
a) Elimination method
b) Reduction method
c) Iterative method
d) Direct method
View Answer
Explanation: Numerical techniques more commonly involve an iteration method due to the degree of accuracy involved. This is because iterations reduce the approximation errors which may occur in numerical problems. They perform sequential operations which in turn increases the accuracy.
2. Which of the following is an iterative method?
a) Gauss Seidel
b) Gauss Jordan
c) Factorization
d) Gauss Elimination
View Answer
Explanation: Gauss seidal method is an iterative method. Gauss elimination is based upon the elimination of unknowns. Gauss Jordan is based on back substitution as well as elimination. Factorization is based upon formation of two triangular matrices with a matrix.
3. Solve the equations using Gauss Jordan method.
x+y+z=9 2x-3y+4z=13 3x+4y+5z=40
a) x=1, y=3, z=7
b) x=1, y=3, z=2
c) x=1, y=3, z=4
d) x=1, y=3, z=5
View Answer
Explanation: By Gauss Jordan method we get
\(\begin{bmatrix}
1 & 1 & 1\\
2 & -3 & 4\\
3 & 4 & 5\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
13\\
40\\
\end{bmatrix} \)
By R2-2R1 and R3-3R1
\(\begin{bmatrix}1 & 2 & 6\\
0 & -5 & 2\\
0 & 1 & 2\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
-5\\
13\\
\end{bmatrix} \)
5R3 and -R2
\(\begin{bmatrix}1 & 1 & 1\\
0 & 5 & -2\\
0 & 5 & 10\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
5\\
65\\
\end{bmatrix} \)
R3-R2 and R2+(1/6) R3, (1/12)R3
\(\begin{bmatrix}1 & 1 & 1\\
0 & 5 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
15\\
5\\
\end{bmatrix} \)
(1/2)R2
\(\begin{bmatrix}1 & 1 & 1\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
3\\
5\\
\end{bmatrix} \)
R1-R2-R3
\(\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
1\\
3\\
5\\
\end{bmatrix} \)
Hence x =1, y=3, z=5.
4. Errors may occur in performing numerical computation on the computer due to which of the following reasons?
a) Operator fatigue
b) Back substitution
c) Rounding errors
d) Power fluctuation
View Answer
Explanation: Rounding errors generally produces errors while performing numerical computation on the computer. Power fluctuation and operator fatigue are not the general problems which occur in performing numerical computations.
5. Which of the following is also known as the Newton Raphson method?
a) Chord method
b) Tangent method
c) Diameter method
d) Secant method
View Answer
Explanation: Newton Raphson’s method is also known as Tangent Method. It is carried out by drawing a tangent to the curve at the point of initial guess.
6. If approximate solution of the set of equations, 2x+2y-z = 6, x+y+2z = 8 and -x+3y+2z = 4, is given by x = 2.8 y = 1 and z = 1.8. Then, what is the exact solution?
a) x = 1, y = 3, z = 2
b) x = 2, y = 3, z = 1
c) x = 3, y = 1, z = 2
d) x = 1, y = 2, z = 2
View Answer
Explanation: Substituting the approximate values x’ = 2.8, y’ = 1 z’ = 1.8 in the given equations, we get
2(2.8) + 2(1) – 1.8 = 5.8 …………..(i)
2.8 + 1 + 2(1.8) = 7.4 ……………..(ii)
-2.8 +3(1) + 2(1.8) = 3.8 …………..(iii)
Subtracting each equation (i), (ii), (iii) from the corresponding given equations we obtain
2xe + 2ye – ze = 0.2 …………….(iv)
Xe + ye +2ze = 0.6 ………………(v)
-xe + 3ye + 2ze = 0.2 ……………(vi)
Where xe = x – 2.8, ye = y – 1, ze = z = 1.8.
Solving the equations (iv), (v), (vi), we get xe = 0.2, ye = 0, ze = 0.2.
This gives the better solution x = 3, y = 1, z = 2, which incidentally is the exact solution.
7. Which of the methods is a direct method for solving simultaneous algebraic equations?
a) Relaxation method
b) Gauss seidel method
c) Jacobi’s method
d) Cramer’s rule
View Answer
Explanation: Cramer’s rule is the direct method for solving simultaneous algebraic equations. In other methods, certain iterations are involved and that’s why the process becomes tedious and consequently indirect.
8. What is the value of the determinant \(\begin{pmatrix}3&5&2\\7&4&5\\1&2&3\end{pmatrix}\)?
a) -56
b) -58
c) -54
d) -66
View Answer
Explanation: \(\begin{pmatrix}3&5&2\\7&4&5\\1&2&3\end{pmatrix}\) = 2 \(\begin{vmatrix}4&5\\2&3\end{vmatrix}\) -5 \(\begin{vmatrix}7&5\\1&3\end{vmatrix}\) + 1 \(\begin{vmatrix}7&4\\1&2\end{vmatrix}\)
= 3(12-10) – 5(21-5) + 2(14-4)
= 6 – 80 + 20
= -54.
9. If EF exists, then (EF)-1 will be equal to which of the following?
a) F-1 E-1
b) E-1 F-1
c) EF
d) FE
View Answer
Explanation: Inverse acts on product inversely. The inverse operation acting on the product first applies on the last matrix and then moves towards the first matrix if product is of more than two matrices.
10. Solve the following equations by Gauss Elimination Method.
x+4y-z = -5 x+y-6z = -12 3x-y-z = 4
a) x = 1.64491, y = 1.15085, z = 2.09451
b) x = 1.64691, y = 1.14095, z = 2.08461
c) x = 1.65791, y = 1.14185, z = 2.08441
d) x = 1.64791, y = 1.14085, z = 2.08451
View Answer
Explanation: By Gauss Elimination method we get
\(\begin{bmatrix}
1 & 4 & -1\\
1 & 1 & -6\\
3 & -1 & -1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
-5\\
-12\\
4\\
\end{bmatrix} \)
By R2-R1 and R3-3R1
\(\begin{bmatrix}
1 & 4 & -1\\
0 & -3 & -5\\
0 & -13 & 2\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
-5\\
-7\\
19\\
\end{bmatrix} \)
R3-(-13/-3)*R2
\(\begin{bmatrix}
1.0000 & 4.0000 & -1.0000\\
0 & -3.0000 & -5.0000\\
0 & 0 & 23.6667\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
-5\\
-7\\
49.33333\\
\end{bmatrix} \)
x+4y-z = -5
-3y-5z = -7
23.6667z = 49.3333
Hence, z = 2.08451
-3y = -7+5z
Hence, y = -1.14085
x = -4y+z-5
Hence, x = 1.64791.
11. Which of the following is the advantage of using the Gauss Jordan method?
a) Additional Calculations
b) No labour of back substitution
c) More operations involved
d) Elimination is easier
View Answer
Explanation: The advantage of using Gauss Jordan method is that it involves no labour of back substitution. Back substitution has to be done while solving linear equations formed during solving the problem.
12. Apply Gauss Elimination method to solve the following equations.
2x – y + 3z = 9 x + y + z = 6 x – y + z = 2
a) X = -13, y = 1, z = -8
b) X = 13, y = 1, z = -8
c) X = -13, y = 4, z = 15
d) X = 5, y = 14, z = 5
View Answer
Explanation: 2x – y + 3z = 9 ……….(i)
x + y + z = 6 ……………………(ii)
x – y + z = 2 ……………………(iii)
To eliminate x, operate (ii) – (iii)
y = 4
Now, operate (i) – 2(ii),
-3y + z = 3
Now by back substitution,
Z = 15
X + 4 +15 = 6
X = -13.
13. Matrix which does not have an inverse by solving it, is classified as which of the following?
a) singular matrix
b) non-singular matrix
c) linear matrix
d) unidentified matrix
View Answer
Explanation: This is because singular matrices do not have inverse, because the value of their determinant is zero.
14. Find the inverse of the following matrix?
\(\begin{bmatrix}1&-1&1\\1&1&1\\1&2&4\end{bmatrix}\)
a) 1/6 \(\begin{bmatrix}1&1&1\\0&-1&1\\1&3&4\end{bmatrix}\)
b) 1/6 \(\begin{bmatrix}1&-1&1\\8&1&6\\1&2&4\end{bmatrix}\)
c) 1/6 \(\begin{bmatrix}1&1&1\\5&-1&6\\1&6&4\end{bmatrix}\)
d) 1/6 \(\begin{bmatrix}1&-1&1\\1&1&1\\1&2&4\end{bmatrix}\)
View Answer
Explanation: We have, the minor of the (1, 1) entry is \(\begin{vmatrix}1&1\\2&4\end{vmatrix}\)
The minor of the entry (1, 2) is\(\begin{vmatrix}1&1\\1&4\end{vmatrix}\)
Repeating this process, the matrix of minors is
\(\begin{pmatrix}2&3&1\\-6&3&3\\-2&0&2\end{pmatrix}\).
Next we negate every other entry, according to the pattern
\(\begin{pmatrix}+&-&+\\-&+&-\\+&-&+\end{pmatrix}\).
The matrix of minors becomes
\(\begin{pmatrix}2&-3&1\\6&3&-3\\-2&0&2\end{pmatrix}\).
Transposing, we get \(\begin{pmatrix}2&6&-2\\-3&3&0\\1&-3&2\end{pmatrix}\).
In order to divide by the determinant we must first compute it.
We can compute that det = 6.
Hence,
A-1 = \(\frac{1}{6}\begin{pmatrix}2&6&-2\\-3&3&0\\1&-3&2\end{pmatrix}\).
15. Which of the following transformations are allowed in the Gauss Jordan method?
a) Swapping a column
b) Swapping two rows
c) Swapping two columns
d) Swapping a row
View Answer
Explanation: Only row transformations are allowed in Gauss Jordan method. Swapping of two rows doesn’t affect the determinant value hence it is allowed here.
16. The Gauss Jordan method reduces a original matrix into a _____________
a) Skew Hermitian matrix
b) Non-symmetric matrix
c) Identity matrix
d) Null matrix
View Answer
Explanation: The Gauss Jordan method reduces an original matrix into a Row Echelon form. In Row Echelon form the matrix is identity.
17. Solve the given equations using Gauss Jordan method.
3x+3y+3z=9 6x-9y+12z=13 9x+12y+15z=40
a) x=1, y=0.33, z=1.67
b) x=0.33, y=1, z=1.67
c) x=0.33, y=1.67, z=1
d) x=1.67, y=1, z=0.33
View Answer
Explanation: By Gauss Jordan method we get
\(\begin{bmatrix}
3 & 3 & 3\\
6 & -9 & 12\\
9 & 12 & 15\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
13\\
40\\
\end{bmatrix} \)
By R2-2R1 and R3-3R1
5R3 and -R2
R3-R2 and R2+(1/6) R3, (1/12)R3
(1/2)R2
R1-R2-R3
\(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
0.33\\
1\\
1.67\\
\end{bmatrix} \)
Hence x=0.33, y=1, z=1.67.
18. What is the other name for factorization method?
a) Muller’s Method
b) Decomposition Method
c) Lin Bairstow Method
d) Doolittle’s Method
View Answer
Explanation: Another name for factorization method is Doolittle’s Method as Doolittle’s method is basically an algorithm of Factorization method.
19. Apply Cramer’s rule to solve the following equations.
x + y + z = 6.6 x – y + z = 2.2 x + 2y + 3z = 15.2
a) x = 1.2, y = 2.2, z = -3.2
b) x = 1.2, y = 2, z = 3.2
c) x = 1.5, y = 2.2, z = -0.5
d) x = 1.5, y = 2.2, z = -0.5
View Answer
Explanation:
∆ = \(\begin{pmatrix}1&1&1\\1&-1&1\\1&2&3\end{pmatrix}\) = -4
X = (1/∆) = \(\begin{pmatrix}6.6&1&1\\2.2&-1&1\\15.2&2&3\end{pmatrix}\) = -4.8/-4 = 1.2
Y = (1/∆) = \(\begin{pmatrix}1&6.6&1\\1&2.2&1\\1&15.2&3\end{pmatrix}\) = -8.8/-4 = 2.2
Z = (1/∆) = \(\begin{pmatrix}1&1&6.6\\1&-1&2.2\\1&2&15.2\end{pmatrix}\) = -12.8/-4 = 3.2
Hence, x = 1.2, y = 2.2, z = 3.2.
20. Cramer’s Rule fails for ___________
a) Determinant = 0
b) Determinant = non-real
c) Determinant < 0
d) Determinant > 0
View Answer
Explanation: This is because Cramer’s rule involves division by determinant which should never be equal to 0 leading to not defined numbers.
21. Gauss Seidal method is also termed as a method of _______
a) Iterations
b) False positions
c) Successive displacement
d) Eliminations
View Answer
Explanation: Gauss Seidal method is also termed as a method of successive displacement because with the iterations, we keep on displacing the values we select for the unknowns which are used in the subsequent steps.
22. Solve the following equation by Gauss Seidal Method up to 2 iterations and find the value of z.
27x + 6y - z = 85 6x + 15y + 2z = 72 x + y + 54z = 110
a) 0
b) 1.92
c) 1.88
d) 1.22
View Answer
Explanation: From the given set of equations-
x=\(\frac{(85-6y+z)}{27}\)
y=\(\frac{(72-6x-2z)}{15}\)
y=\(\frac{(110-x-y)}{54}\)
1st iteration:
y=0, z=0
x=\(\frac{85}{27}\)=3.14
x=3.14, z=0
y=\(\frac{72-(6×3.14)-2×(0)}{15}\)=3.54
x=3.14, y=3.54
z=\(\frac{110-3.14-3.54}{54}\)=1.91
2nd iteration:
z=1.91, y=3.54
x=\(\frac{85-(6×3.54)+(1.91)}{27}\)=2.43
z=1.91, x=2.43
y=\(\frac{72-(6×2.43)-2×(1.91)}{15}\)=3.57
y=3.57, x=2.43
z=\(\frac{110-2.43-3.57}{54}\)=1.92
Thus, after the second iteration
x=2.43, y=3.57, z=1.92.
23. What is the condition applied in the factorization method?
a) There must exist a diagonal matrix form of the given matrix
b) Matrix should not be singular
c) All principal minors of the matrix should be non-singular
d) Back substitution should be done
View Answer
Explanation: The necessary condition for factorization method is that all principal minors of the matrix should be non-singular. Otherwise, there will be no formation of lower and upper triangular matrix.
24. A quadratic equation x4-x-8=0 is defined with an initial guess of 1 and 2. Find the approximated value of x2 using Secant Method.
a) 1.571
b) 7.358
c) 7.853
d) 7.538
View Answer
Explanation: By Secant Method the iterative formula is given as:
x(n+1)=\(\frac{x(n-1)f[x(n)]-x(n)f[x(n-1)]}{[f[x(n)]-f[x(n-1)]]} \)
For n=1
x(2)=\(\frac{x(0)f[x(1)]-x(1)f[x(0)]}{[f[x(1)]-f[x(0)]]} \)
x(0)=1 and x(1)=2, f(x0)=-8 and f(x1)=6
Hence x2= 1.571.
25. If the equation y = aebx can be written in linear form Y=A + BX, what are Y, X, A, B?
a) Y = logy, A = a, B=logb and X=x
b) Y = y, A = a, B=logb and X=logx
c) Y = y, A = a, B=b and X=x
d) Y = logy, A = loga, B=b and X=x
View Answer
Explanation: The equation is
y = aebx.
Taking log to the base e on both sides,
we get logy = loga + bx.
Which can be replaced as Y=A+BX,
where Y = logy, A = loga, B = b and X = x.
26. Secant Method is also called as?
a) 5-point method
b) 2-point method
c) 3-point method
d) 4-point method
View Answer
Explanation: Secant Method is also called as 2-point method. In Secant Method we require at least 2 values of approximation to obtain a more accurate value.
27. Find f(0.18) from the following table using Newton’s Forward interpolation formula.
| x | 0 | 0.1 | 0.2 | 0.3 | 0.4 |
| f(x) | 1 | 1.052 | 1.2214 | 1.3499 | 1.4918 |
a) 0.8878784
b) 1.9878785
c) 1.18878784
d) 1.8878784
View Answer
Explanation: Here,
x0 = 0
x = 0.18
h = 0.1
x = x0 + nh,
0.18 = 0 + n(0.1)
n = 1.8
| x | y | Δy | Δ2y | Δ3y | Δ4y |
| 0 | 1 | 0.052 | 0.1174 | -0.1583 | 0.2126 |
| 0.1 | 1.052 | 0.1694 | -0.0409 | 0.0543 | |
| 0.2 | 1.2214 | 0.1285 | 0.0134 | ||
| 0.3 | 1.3499 | 0.1419 | |||
| 0.4 | 1.4918 |
y0 is 1 since it is forward interpolation formula.
Δy0 = 0.052
Δ2y0 = 0.1174
Δ3y0 = -0.1583
Δ4y0 = 0.2126
Substituting in the formula,
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3! +….
\(f(0.18)=1+(1.8)(0.052)+\frac{(1.8)(0.8)(0.1174)}{2}+\frac{(1.8)(0.8)(-0.2)(-0.1583)}{6}\)
\(+\frac{(1.8)(0.8)(-0.2)(-1.2)(0.2126)}{24} \)
f(0.18) = 1.18878784.
28. Which of the following is an assumption of Jacobi’s method?
a) The coefficient matrix has zeroes on its main diagonal
b) The coefficient matrix has no zeros on its main diagonal
c) The rate of convergence is quite slow compared with other methods
d) Iteration involved in Jacobi’s method converges
View Answer
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.
29. Find the approximated value of x till 6 iterations for x3-4x+9=0 using Bisection Method. Take a = -3 and b = -2.
a) -0.703125
b) -3.903125
c) -1.903125
d) -2.703125
View Answer
Explanation: Iteration table is given as follows:
| No. | a | b | c | f(a)*f(c) |
| 1 | -3.0 | -2.0 | -2.5 | (-) |
| 2 | -3.0 | -2.5 | -2.75 | (-) |
| 3 | -2.75 | -2.5 | -2.625 | (-) |
| 4 | -2.75 | -2.625 | -2.6875 | (-) |
| 5 | -2.75 | -2.6875 | -2.71875 | (+) |
| 6 | -2.71875 | -2.6875 | -2.703125 | (-) |
Hence we stop the iterations after 6. Therefore the approximated value of x is -2.703125.
30. Find the positive root of the equation x3 – 4x – 9 = 0 using Regula Falsi method and correct to 4 decimal places.
a) 2.7506
b) 2.6570
c) 2.7065
d) 2.7605
View Answer
Explanation: f(2) = -9
f(3) = 6
Therefore, root lies between 2 and 3
a = 2; f(a) = -9
b = 3; f(b) = 6
Substituting the values in the formula,
\(x = \frac{bf(a)-af(b)}{f(a)-f(b)}\),
we get \(x1 = \frac{3(-9)-2(6)}{-9-6}\)=2.6; f(x1) = -1.824
Therefore, x1 becomes a to find the next point.
\(X2 =\frac{3(-1.824)-2.6(6)}{-1.824-6}\)= 2.693251534; f(x2) = -0.23722651
Therefore, x2 becomes a to find the next point.
\(X3 = \frac{3(-0.23722651)-2.693251534(6)}{-0.23722651-6}\)=2.704918397; f(x3) = -0.028912179
Therefore, x3 becomes a to find the next point.
\(X4 = \frac{3(-0.028912179)-2.704918397(6)}{-0.028912179-6}\)=2.70633487; f(x4) = -3.495420729*10-3
Therefore, x4 becomes a to find the next point.
\(X5 = \frac{3(-3.495420729*10^{-3})-2.70633487(6)}{(-3.495420729*10^{-3})-6}\)=2.706505851; f(x5) = -3.973272762*10-4
Therefore x5 becomes a to find the next point.
\(X6 = \frac{3(-3.973272762*10^{-4})-2.706505851(6)}{(-3.973272762*20^{-4})-6}\)=2.706525285.
Therefore, the positive root corrected to 4 decimal places is 2.7065.
31. The equation f(x) is given as x3+4x+1=0. Considering the initial approximation at x=1 then the value of x1 is given as _______________
a) 1.85
b) 1.86
c) 1.87
d) 1.67
View Answer
Explanation: Iterative formula for Newton Raphson method is given by
x(1)=x(0)+\(\frac{f(x(0))}{f’x(x(0))}\)
Hence x0=1 (initial guess), f(x0)=6 and f’(x0)= 7.
Substituting the values in the equation we get x1=1.857 which is 1.86 rounded to 2 decimal places.
32. In Newton Raphson method if the curve f(x) is constant then __________
a) f(x)=0
b) f’(x)=c
c) f’’(x)=0
d) f’(x)=0
View Answer
Explanation: If the curve f(x) is constant then the slope of the tangent drawn to the curve at an initial point is zero. Hence the value of f’(x) is zero.
33. Which of these methods is named after the mathematician Carl Friedrich Gauss?
a) Gauss Jordan method
b) Runge Kutta method
c) Secant method
d) Newton Raphson method
View Answer
Explanation: Gauss Jordan method is named after the mathematician Carl Friedrich Gauss. Its algorithm is used to solve linear equations.
Chapterwise Multiple Choice Questions on Numerical Methods
- Numerical Solution of Algebraic and Transcendental Equations
- Numerical Solution of Simultaneous Algebraic Equations
- Matrix Inversion and Eigen Value Problems
- Empirical Laws and Curve Fitting
- Finite Differences
- Interpolation
- Numerical Differentiation and Integration
- Difference Equations
- Numerical Solution of Ordinary Differential Equations
- Numerical Solution of Partial Differential Equations
- Engineering Mathematics I & II
- Probability and Statistics (Mathematics III / M3)
1. Numerical Analysis MCQ on Algebraic and Transcendental Equations
The section contains Numerical Analysis multiple choice questions and answers on bisection method, regula falsi method, secant method, newton raphson method, transformation, polynomial synthetic division, iterative and iteration method, graphic solutions, convergence rate, mullers method, polynomial equations roots, multiple and complex roots using newton’s method, lin bairstow method and graeffe’s root squaring method.
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2. Numerical Methods MCQ on Simultaneous Algebraic Equations
The section contains Numerical Methods questions and answers on determinants and matrices basics, ill conditioned equations, linear Simultaneous Equation solution using direct and iterative methods.
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3. Numerical Analysis MCQ on Matrix Inversion and Eigen Value Problems
The section contains Numerical Analysis MCQs on matrix inversion, gauss elimination and jordan method, factorization, crout’s and gauss seidel methods, cramers rule, jacobi’s iteration method, partition and iterative method, eigen values and vectors, properties and bounds of eigen values, power method, given’s and house holder’s method.
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4. Numerical Methods MCQ on Empirical Laws and Curve Fitting
The section contains Numerical Methods multiple choice questions and answers on graphical methods basics, linear laws, least squares principle and methods, finding curve types, unknowns plausible values, group averages and moments method.
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5. Numerical Methods MCQ on Finite Differences
The section contains Numerical Methods questions and answers on finite differences, polynomial differences, factorial notation, reciprocal factorial function, delta inverse operator, difference table errors, difference operators, operators relation, finding the missing tem using differences and summation series application.
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6. Numerical Analysis MCQ on Interpolation
The section contains Numerical Analysis MCQs on newton-gregory forward and backward interpolation formula, functions approximation using least square method, central difference interpolation formula, gauss forward and backward interpolation formula, stirling’s, laplace-everett’s and bessel’s formula, lagrange’s interpolation formula, divided and forward differences relation, hermite’s interpolation formula, spline, double and inverse interpolation, lagrange’s and iterative method.
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7. MCQ on Numerical Differentiation and Integration
The section contains Numerical Methods multiple choice questions and answers on numerical differentiation, derivatives formulas, tabulated function maxima and minima, numerical integration, newton-cotes quadrature formulas, quadrature formulas errors, romberg’s method, euler-maclaurin formula, undetermined coefficients method, gaussian and numerical double integration.
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8. Numerical Analysis MCQ on Difference Equations
The section contains Numerical Analysis questions and answers on difference equations formation, rules for finding the complementary function and particular integral, simultaneous difference equations with constant coefficients and loaded string deflection application.
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9. Numerical Methods MCQ on Ordinary Differential Equations
The section contains Numerical Methods MCQs on picard’s method, taylor’s and euler’s method, runge’s and runge-kutta method, predictor-corrector, milne’s and adams-bashforth-moulton method, second order differential equation, error analysis, method convergence, stability analysis, boundary value problems, finite difference and shooting method.
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10. Numerical Methods MCQ on Partial Differential Equations
The section contains Numerical Methods multiple choice questions and answers on second order equation classification, partial derivatives approximations, elliptic equations, laplace’s and poisson’s equation solution, parabolic and hyperbolic equations, one and two dimensional heat equation solution, 1d and 2d wave equation numerical solutions.
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11. MCQ on Engineering Mathematics I & II
The section contains questions and answers on numerical analysis and methods.
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12. Probability and Statistics MCQ (Mathematics III / M3)
The section contains multiple choice questions and answers on probability and statistics.
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Wish you the best in your endeavor to learn and master Numerical Methods!
Important Links:
- Numerical Methods Books
- Numerical Analysis Books
- Advance Numerical Analysis Books
- Statistics and Numerical Methods Books
- Numerical Methods for Partial Differential Equations Books
- Engineering Mathematics Books
- Class 11 Maths MCQ
- Class 12 Maths MCQ
