# Engineering Mathematics (Test 5)

## Gate Exam : Ee Electrical Engineering

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Engineering Mathematics
| Engineering Mathematics |
Q.1
The current erection cost of a structure is Rs. 13,200. If the labour wages per day increase by 1/5 of the current wages and the working hours decrease by 1/24 of the current period, then the new cost of erection in Rs. is
A. 16,500
B. 15,180
C. 11,000
D. 10,120
Explaination / Solution:

Let ‘W’ be the labour wages, and ‘T’ be the working hours.
Now, total cost is a function of W × T
Increase in wages = 20%
∴ Revised wages = 1.2 W
Decrease in labour time = (100/24)%
∴ Re vised time = (1-(1/24))T = (23/24)T
Re vised Total cost = 1.2 × (23/24)WT = 1.15WT
= 1.15 × 13200 = 15180

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Q.2
Consider the binary relation R = {(x, y), (x, z), (z, x), (z, y)} on the set {x, y, z}. Which one the following is TRUE?
A. R is symmetric but NOT antisymmetric.
B. R is NOT symmetric but antisymmetric
C. R is both symmetric and antisymmetric
D. R is neither symmetric nor antisymmetric
Explaination / Solution:
No Explaination.

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Q.3
A square matrix B is skew-symmetric if
A. B= -B
B. B= B
C. B-T = B
D. B-1 BT
Explaination / Solution:
No Explaination.

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Q.4
The Laplace Transform of f(t) = e2t sin(5t) u(t) is
A.
B.
C.
D.
Explaination / Solution:

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Q.5
The standard normal probability function can be approximated as

where xN = standard normal deviate. If mean and standard deviation of annual precipitation are 102 cm and 27 cm respectively, the probability that the annual precipitation will be between 90 cm and 102 cm is
A. 66.7 %
B. 20.0 %
C. 33.3 %
D. 16.7 %
Explaination / Solution:
No Explaination.

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Q.6
If three coins are tossed simultaneously, the probability of getting at least one head is
A. 1/8
B. 3/8
C. 1/2
D. 7/8
Explaination / Solution:
No Explaination.

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Q.7
Probability density function of a random variable X is given below

A. 3/4
B. 1/2
C. 1/4
D. 1/8
Explaination / Solution:

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Q.8
An input to a 6-level quantizer has the probability density function f(x) as shown in the figure. Decision boundaries of the quantizer are chosen so as to maximize the entropy of the quantizer output. It is given that 3 consecutive decision boundaries are' −1'. '0 ' and '1'.

The values of a and b are
A.
B.
C.
D.
Explaination / Solution:

Area under the pdf curve must be unity 2a + 4a + 4b = 1 2a + 8b = 1.................................................(1) For maximum entropy three region must by equivaprobable thus 2a = 4b = 4b...............................................(2) From (1) and (2) we get b = 1/12 and a = 1/6

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Q.9
The area enclosed between the curves y2 =4x and x2 = 4y is
A. 16/3
B. 8
C. 32/3
D. 16
Explaination / Solution:
No Explaination.

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Q.10
What is the possible number of reflexive relations on a set of 5 elements?

A. 210
B. 215
C. 220
D. 225