MAHATRANSCO AE Electrical Mock Test - 5 - Electrical Engineering (EE) MCQ

By Leibnitz rule :

f'(x) = [e^x(x - 1)(x - 2)] - 0 < 0 

= (x - 1)(x - 2) < 0

= x > 1 and x < 2

= x ∈ (1, 2)

The above expression is for XOR gate 

IC₁(P₁) = 0.008P₁ + 6 
IC₂(P₂) = 0.006P₂ + a

I_C1 = I_C2 = λ = 8
⇒ I_C1 = 8

0.008P₁ + 6 = 8
P₁ = 250 MW , P₂ = 300 MW

L - 5 = G
M - 5 = H
N - 5 = I
O - 5 = J
LMNO → GHIJ

Similarly,
S - 5 = N
T - 5 = O
U - 5 = P
V - 5 = Q
STUV → NOPQ

Letters of the word PREMONITION are coded as follows:

Letters → Code
P → 6
R → 8
E → 5
M → 3
O → 0
N → 4
I → 9
T → 2
I → 9
O → 0
N → 4

Therefore, code for the word MONITOR is 3049208

x>2, the value of x is greater than 2. the value goes till infinity.
if X is equal or greater than 2 then we use solid dot, but it is only greater than 2
so the option c is correct.

Numbers obtained in three throws are 3, 5 and 6.
Let the number obtained in fourth throw be x.
Now, Sum > 20
=> 5 + 3 + 6 + x > 20
=> 14 + x > 20
=> x > 20 – 14
=> x > 6 (no one wins)

X is towards the east wrt to P

(3 x 4) + 3 = 15

and (7 x 5) + 3 = 38

Therefore (3 x 5) + 3 = 18.

As 2012 was a leap year, so the number of days in each month from 23rd January, 2011 to 31st July, 2013 is given as follows.
Since 2011 is not a leap year = 365 - 22 = 343 days
Number of days in 2012 = 366 days (leap year)
Number of days in 2013:
January = 31 days
February = 28 days
March = 31 days
April = 30 days
May = 31 days
June = 30 days
July = 31 days
Total number of days in 2013 = 212
Therefore, total number of days from 23rd January, 2011 to 31st July, 2013 = 343 + 366 + 212 = 921

He returned after 8 o'clock but before 9 o'clock, at such a time when both the minute hand and the hour hand were found to be coinciding.
In a simple way, we need to calculate the time when the angle between the minute hand and the hour hand is zero.
Angle between the minute hand and the hour hand at 8 o'clock = 8 × 30 = 240°
We can reduce the difference of 5.5 degree in 1 min,
Required time = 8 hrs + 240 × 2 ÷ 11

To be together between 9 and 10 o'clock, the minute hand has to gain 45 minute spaces.
55 minute space is gained in 60 minutes.
45 minute space will be gained in 
At 49minutes past 9, the hands of the clock will be together.

According to given information:

Statements: M ≤ N = O ; M ≤ P ; O < Q

Combining the statement: P > M ≤ N = O < Q

Here, P is greater than or equal to M which is less than or equal to N equal to O which is less than Q.

Conclusions:

I.  Q > N → True (In the given statement, P > M ≤ N = O < Q ,It clearly mentioned Q is greater than O which is equal to N .So, Q is greater than N too. Thus, the conclusion I is true).

II. M ≥ O → False (In the given statement, M ≤ N = O, It is clearly mentioned that M is less than or equal to N which is equal to O. So, M can never be greater than O. Thus, conclusion II is a false).

Hence, "Option b" is the correct answer.

According to given information:

Statements: S = R ≥ Q ≤ P ; N ≤ P ; M < N

Combine the statement: S = R ≥ Q ≤ P  ≥ N > M

Here, S is equal to R which is greater than or equal to Q which is less than or equal to P which is greater than or equal to N which is greater than M.

Conclusions:

I. M < S → False (As S = R ≥ Q ≤ P ; N ≤ P ; M < N → there is no direct relation between M and S. So, conclusion I is false)

II. M ≥ S → False (As S = R ≥ Q ≤ P ; N ≤ P ; M < N → there is no direct relation between M and S. So, conclusion II is false)

Here, both conclusions I and II form a complementary pair as both have the same element i.e. M and S and all the three symbols (<, =, >) are present between M and S.

Hence, Either I or II follows.

Hence, "Option d" is the correct answer.

Given: A > B, B > C and C > D

After combining the operators,

we get: A > B > C > D

Now, checking the options:

Option A: A > D → True (As, A > B > C > D gives A > D)

Option B: A > C → True (As, A > B > C > D gives A > C)

Option C: D > A → False (As, A > B > C > D gives A > D)

Option D: B > D → True (As, A > B > C > D gives B > D)

So, Option C, D > A is definitely wrong.

Hence, the correct answer is "Option c".

As the result of Race is Fatigue similarly the result of Fast is Hunger.

As Tiger is found in Forest similarly Otter is found in the water.

As magnet has poles similarly battery has terminals.

A Priest wears cassock (a full-length garment worn by certain Christian clergy, members of church choirs, and others having an office or role in a church.) while Graduate wears gown.

The man in the photograph is the daughter of the only son of Lata's grandfather i.e., the girl is the daughter of Lata's father. Hence, the girl is the sister of Lata.

M x N → M is the mother of N

N % S → N is the wife of S

and S + T → is the father of T.

Hence, M is the maternal grandmother of T.

Rate downstream = 

Rate upstream = 2 km/hr.

Speed in still water = 

 Required time = 

Solve this question through options.
⇒  For instance, if he travelled at 25 km/h, his original speed would have been 24 km/h.
⇒ The time difference can be seen to be 6 minutes in this case = 60 / 24 – 60 / 25 = 0.1 hrs = 6 mins

Thus, 25 km/h is the correct answer. 

So Option A is correct

Mother's age when Ayesha's brother was born = 36 years.

Father's age when Ayesha's brother was born = (38 + 4) years = 42 years.

 Required difference = (42 - 36) years = 6 years.

D. 78%
Explanation: Consider Total no of students = 100 Ratio is 3:2 i.e Boys=60 and Girls=40 20% of boys who get scholarship = 60*20/100=12% 25% of girls who get scholarship = 40*25/100 =10% Therefore % of students who do not get scholarship =100-(12+10) = 78%

मन्वंतर या शब्दाची संधी मनु + अंतर ही होईल.

मन्वंतर या शब्दाची संधी उ किंवा ऊ पुढे विजातीय स्वर आल्यास उ किंवा ऊ बद्दल व याप्रमाणे होईल.

मन्वंतर हे यणादेश स्वरसंधीचे उदाहरण आहे.

मन्वंतर = मनु + अंतर = उ + अ = व् + अ = व

यणादेश स्वरसंधी - इ, उ, ऋ (ह्रस्व किंवा दीर्घ) यांच्या पुढे विजातीय स्वर आल्यास इ/ई बद्दल य्, उ/ऊ बद्दल व् तर ऋ बद्दल र् वर्ण येतात आणि त्यात पुढील स्वर मिसळून संधी तयार होते. अशाप्रकारे य्, व्, र् असे बदल होतात त्याला यणादेश असे म्हणतात.

Important Points

संधी - जोडशब्द तयार करताना पहिल्या शब्दातील शेवटचा वर्ण व दुसऱ्या शब्दातील पहिला वर्ण हे एकमेकांमध्ये मिसळतात व त्या दोन्हीबद्दल एक वर्ण तयार होतो. वर्णाच्या अशा एकत्र होण्याच्या प्रकाराला संधी असे म्हणतात.

संधीचे तीन प्रकार आहेत.

1. स्वरसंधी
2. व्यंजनसंधी
3. विसर्गसंधी

स्वरसंधी - एकमेकांच्या शेजारी येणारे वर्ण हे जर स्वरांनी जोडले असतील तर त्यांना स्वरसंधी असे म्हणतात.

स्वर + स्वर असे याचे स्वरूप असते.​ उदा. विद्या + अर्थी = विद्यार्थी  (आ + अ = आ)

व्यंजनसंधी - जवळ जवळ येणाऱ्या या दोन वर्णांपैकी दोन्ही वर्ण व्यंजने असतील किंवा पहिला वर्ण व्यंजन व दुसरा वर्ण स्वर असेल तर त्याला व्यंजनसंधी असे म्हणतात.

व्यंजन + व्यंजन किंवा स्वर असे त्याचे स्वरूप असते. उदा. सत् + आचार = सदाचार (त् + आ)

विसर्गसंधी - एकत्र येणाऱ्या वर्णातील पहिला विसर्ग व दुसरा वर्ण व्यंजन किंवा स्वर असेल तर तेव्हा त्यास विसर्ग संधी असे म्हणतात. 

विसर्ग + व्यंजन किंवा विसर्ग + स्वर असे त्याचे स्वरूप असते. उदा. नि: + अंतर = निरंतर