Engg math's practice question by satya prakash 1

Practice Question Given By Satya Prakash Yadav (Engg Math’s and Aptitude Faculty) truthlight1989@gmail.com
[1]
I humbly request you to not distribute this sheet
among non-registered students they are your
competitors so don’t give chance them to compete you
Topic: Probability
1. Satya Prakash and Minakshi, husband and wife
both appear for a interview. The probability that
Satya Prakash get selected is and of minakshi is
What is probability that
(A) Only one of them selected
(B) Both of them selected
(C) None of them is selected
1. A = , B = , C =
2. A = B = , C =
3. A = , B = , C =
4. A = , B = , C =
Solution :-
Option (1) is correct.
A) P(A) = P(H) . P( ̅ ) P(W) . P(̅)
= =
B) P(H W) = P(H) . P(W)
=
C) P(̅ ̅ ) = P(̅ . P( ̅̅̅̅̅
= ( )
=
2. One of ten keys opens the door. If we try the keys
one after another. What is the probabilities that
the door is opened on the second attempt.
(a) (b)
(c) (d)
Solution :-
Option (d) is correct.
P (Second attempt) = P(̅̅̅̅
= (1 ( )
=
3. When four dice are thrown what is the probability
that the same number appears on each of them?
(a) (b)
(c) (d)
Solution :-
Option (c) is correct.
n(s) = 6
= 64
The chances that all the dice show same number
(1, 1, 1, 1), (2, 2, 2, 2), (3, 3, 3, 3), (4, 4, 4, 4),
(5, 5, 5, 5), (6, 6, 6, 6)
n(E) = 6
P(E) = = =
4. Two squares are chosen at random on a
chessboard what is the probability that they have
a side in common?
(a) (b)
(c) (d)
Solution :-
Option (a) is correct.
Number of ways of selection of first
square = 64
Number of ways of selection of second
square = 63
n(s) = 63 64
= 4032
If the first square happen to be any of the four
corner ones then second can be chosen in two
ways.
So n(E1) = 4
= 8
If the first square happens to be any of the 24
squares on the side of the chess board then
second square can be chosen in 3 ways
P(E) =

[2]
n(E2) = 24 3
= 72
If the first square happen to be any of the 36
remaining squares then second square can be
chosen in 4 ways
If the first square happens to be any of the 36
remaining squares, the second square can be
chosen in 4 ways.
n(E3) = 36 4
= 144
n(E) = n (E1) n (E2) n(E3)
= 8 72 144
n(E) = 224.
P(E) = =
5. Seven car accidents occur in a week what is
probability that they all occurred on Monday
(a) (b)
(c) (d) None
6. E1 and E2 are events in a probability space
satisfying the following constraints P(E1) = P(E2)
P (E1 E2) = 1
E1 and E2 are independent the value of P (E1) is
(a) 0 (b)
(c) (d) 1
Solution :-
P(E1) = P (E2) = x
P(E1) P(E2) P (E1 E2) = P (E1 E2)
x2
(x = 0
(P(E1 E2) = P(E1).P(E2)
because
E1 and E2 are independent
7. A biased die is given in which 6 appears three
times as often as such other number but that of
other five outcomes are equally likely.
What is probability that an even number appears
When the die is rolled
(a) (b)
(c) (d)
Solution:-
Option (b) is correct.
P(1) P(2) P(3) ........... P(6) = 1
8x = 1
x =
P(E) =
= = ( )
8. Consider two events A and B such that
P(A B) = 2/3
P(A| = 1/2
P(A B) = 1/6 determine P(A) ?
(a) (b)
(c) (d)
Solution :-
P(P| =
P(A B) = P(A) P(B) P(A B)
9. In a Poisson distribution, the first two
frequencies are 150 and 90, then the third
frequency is____
(a) 27 (b) 29
(c) 28 (d) 30
Solution :-
Option (a) is correct
T0 = = = 150 ...(i)
T1 = = = 90 ...(ii)
Solving (i) and (ii)
λ = 0.6
x = 1
P(A) =

[3]
T2 = T1
= = 27
10. A determinant is chosen at random from the
set of all determinants of order 2 with elements
0 and 1 only. The probability that the value of
determinant chosen is a positive number is
(a) (b)
(c) (d)
Solution :-
& The positive determinants are
| | | | | | that is 3
Total number of possible determinants are 16
The required probability =
Topic : Numerical Methods & Numerical Integration
1. The equation x3 3x 4=0 has only one real root.
What is the first approximation obtained by the
method of false position in ( )?
(a) (b) 2.125
(c) 2.812 (d) 2.812
Solution :-
xn+2 = xn
( ̅ )
g(xn)
x2 = xD (x0)
x0 = 3
x1 =
f(x) = x3 3x 4
f(x0) = f( 3) =
f(x1) = f ( 2) = 2
x2 = x0
2. Using the Newton-Rapt ion method, to find
√
......
(a) 0.2286 (b) 0.2280
(c) 0.2290 (d) 0.2270
Solution :-
x =
√
x2 =
f(x) = x2
f '(x) = 2x
The positive number nearest to 12 which is a
perfect square is 9 and
√
= ( ) x0 = ( )
x1 = x0 [
( )
( ) ( )
=
x2 = x1 [ ]
x2 = .29154
x3 = .28883
3. What is the minimum number of iterations of
the bisection method that will produce the root
f(x) = 0 in the internal (0, 1) with error guarteed
to be no more than 10-3
(a) 10 (b) 9
(c) 11 (d) 8
Solution :-
n log2 ( )
n log2
n log2 10-3
n 9.986
4. Match list - I with list -II and select the correct
answer using the codes given below
List- I List-II
A. Bisection 1. X1+1 = x1
B. Newton 2. Xi+1 =
x2 = 2.125
x1 = .29154
x2 = .28883
x3 = .2886
n 10

[4]
C. Secant 3. Xi+1 =
Codes :
A B C
(a) 3 1 2
(b) 2 3 1
(c) 1 2 3
(d) 3 2 1
5. Evaluate √ using Newton-Raphson method
starting with x0 = 3 After 3 iterations, the root is
(a) 3.002 (b) 0.0151
(c) 3.0122 (d) 3.4641
Solution :-
x = √
x2 = 12
f(x) = x2 12
f '(x) = 2x
f(3) =
f '(3) = 6
x1 = x0
= 3
= 3.5
Similarly
x2 = 3.4643
x3 = 3.4641
6. What is the maximum error bound when
integrating f(x) = from 0 to 1 using 3 points
Simpson rule
(a) 0.0012 (b) 0.0001
(c) 0.053 (d) 0.0083
Solution: -
Option (d) is correct
TE = Ni ( )
Here Ni = 2, h = 0.5
fiv(x) =
max TE(bound) = | | | |
Where 0
= | |
=
= 0.0083
7. Apply Jacobi's method to the following system
10x y 3= 18 -------------(i)
x 15y 3 = 12 ----------(ii)
x y 203 = 17-----------(iii)
Starting with (0, 0, 0) as an initial solution , use
two iterations to find the solution. The solution is
(a) 1.965, 0.9767, 0.98
(b) 1.8, 0.8, 085
(c) 1.99, 0.99, 0.99
(d) 2.0, 1.0, 1
8. Evaluate ∫ using four intervals
(a) 0.277 (using Simpson Rule)
(b) 0.477 (c) 0.677 (d) 0.437
9. If bisection method is used to find the root of
f(x) = 0 in [0, 1] after 5 iterations, what is the
maximum size of error?
(a) (b)
(c) (d)
Solution :-
For bisection method, maximum error after n
iterations is given by
ϵn =
| |
Here a = 0
b = 1
ϵ5 =
| |
=
10. If the step size is halved keeping the points of
integration same while evaluating integral using
composite trapezoidal and Simpson’s rule, the
errors are reduced respectively by
(a) and (b) and th
(c) and (d) and th
Solution: -
For composite trapezoidal rule truncation
error for Ni intervals is given by
TE = Ni f" ( )
Ni =

[5]
TE = ( ) f" ( )
i.e. TE ∝ h2
Similarly for Simpson’s rule
TE = Ni fiV ( )
= fiV ( )
TE ∝ h4
= =
( )
( )
So error will be reduce to and th
In trapezoidal and Simpson .

Engg math's practice question by satya prakash 1

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