1000 Numerical Methods MCQ (Multiple Choice Questions) - Sanfoundry (2022)

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

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

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=92x-3y+4z=133x+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

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

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

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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

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

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

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

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 = -5x+y-6z = -123x-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

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?
b) No labour of back substitution
c) More operations involved
d) Elimination is easier

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 = 9x + y + z = 6x – 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

Explanation: 2x – y + 3z = 9 ……….(i)
x + y + z = 6 ……………………(ii)
x – y + z = 2 ……………………(iii)

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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

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 $$\begin{bmatrix}1&-1&1\\1&1&1\\1&2&4\end{bmatrix}$$

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

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

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=96x-9y+12z=139x+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

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

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 = 6.6x – y + z = 2.2x + 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

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

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

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 = 856x + 15y + 2z = 72x + y + 54z = 110

a) 0
b) 1.92
c) 1.88
d) 1.22

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

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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

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

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

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

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

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

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

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

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

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

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

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

Our 1000+ MCQs focus on all topics of the Numerical Methods subject, covering 100+ topics. This will help you to prepare for exams, contests, online tests, quizzes, viva-voce, interviews, and certifications. You can practice these MCQs chapter by chapter starting from the 1st chapter or you can jump to any chapter of your choice. You can also download the PDF of Numerical Methods MCQs by applying below.

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1. Numerical Solution of Algebraic and Transcendental Equations
2. Numerical Solution of Simultaneous Algebraic Equations
3. Matrix Inversion and Eigen Value Problems
4. Empirical Laws and Curve Fitting
5. Finite Differences
6. Interpolation
7. Numerical Differentiation and Integration
8. Difference Equations
9. Numerical Solution of Ordinary Differential Equations
10. Numerical Solution of Partial Differential Equations
11. 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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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.

 Introduction to DeterminantsIntroduction to MatricesSolution of Linear Simultaneous Equation using Direct Methods Solution of Linear Simultaneous Equation using Iterative MethodsILL Conditioned EquationsSolution of Non-Linear Simultaneous Equations

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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.

 Finite DifferencesDifferences of a PolynomialFactorial Notation and Reciprocal Factorial FunctionInverse Operator of DeltaError in Difference Table Difference OperatorsRelation between the OperatorsFinding Missing Term using DifferencesApplication to Summation Series

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.

 Newton-Gregory Forward Interpolation FormulaApproximation of Functions using Least Square MethodNewton’s Backward Interpolation FormulaCentral Difference Interpolation FormulaGauss Forward Interpolation FormulaGauss Backward Interpolation FormulaStirling’s FormulaBessel’s FormulaLaplace-Everett’s FormulaChoice of an Interpolation FormulaInterpolation with Unequal Intervals Lagrange’s Interpolation FormulaInverse Interpolation Using Lagrange’s Interpolation FormulaDivided DifferencesNewton’s Divided Differences FormulaRelation between Divided and Forward DifferencesHermite’s Interpolation FormulaSpline InterpolationDouble InterpolationInverse InterpolationLagrange’s Method and Iterative Method

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.

 Numerical DifferentiationFormulas for DerivativesMaxima and Minima of a Tabulated FunctionNumerical IntegrationNewton-Cotes Quadrature FormulasErrors in Quadrature Formulas Romberg’s MethodEuler-Maclaurin FormulaMethod of Undetermined CoefficientsGaussian IntegrationNumerical Double Integration

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.

 Formation of Difference EquationsRules for Finding the Complementary FunctionRules for Finding the Particular Integral Difference Equations Reducible to Linear FormSimultaneous Difference Equations with Constant CoefficientsApplication to Deflection of a Loaded String

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.

 Picard’s Method of Successive ApproximationTaylor’s MethodEuler’s MethodModified Euler’s MethodRunge’s MethodRunge-Kutta MethodPredictor-Corrector MethodMilne’s MethodAdams-Bashforth-Moulton Method Simultaneous First Order Differential EquationSecond Order Differential EquationError AnalysisConvergence of a MethodStability AnalysisBoundary Value ProblemsFinite Difference MethodShooting Method

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.

 Classification of Second Order EquationFinite Difference Approximations to Partial DerivativesElliptic EquationsSolution of Laplace’s EquationSolution of Poisson’s EquationSolution of Elliptic Equations by Relaxation Method Parabolic EquationSolution of One Dimensional Heat EquationSolution of Two Dimensional Heat EquationHyperbolic EquationsNumerical Solutions to 1D Wave EquationNumerical Solution to 2D Wave Equation

11. MCQ on Engineering Mathematics I & II

The section contains questions and answers on numerical analysis and methods.

• Engineering Mathematics Questions and Answers
• 12. Probability and Statistics MCQ (Mathematics III / M3)

The section contains multiple choice questions and answers on probability and statistics.

• Probability and Statistics Questions and Answers
• If you would like to learn "Numerical Methods" thoroughly, you should attempt to work on the complete set of 1000+ MCQs - multiple choice questions and answers mentioned above. It will immensely help anyone trying to crack an exam or an interview.

Wish you the best in your endeavor to learn and master Numerical Methods!

• Numerical Methods Books
• Numerical Analysis Books
• Statistics and Numerical Methods Books
• Numerical Methods for Partial Differential Equations Books
• Engineering Mathematics Books
• Class 11 Maths MCQ
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