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General Aptitude - 4 - GATE Chemistry MCQ


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20 Questions MCQ Test GATE Chemistry Mock Test Series - General Aptitude - 4

General Aptitude - 4 for GATE Chemistry 2024 is part of GATE Chemistry Mock Test Series preparation. The General Aptitude - 4 questions and answers have been prepared according to the GATE Chemistry exam syllabus.The General Aptitude - 4 MCQs are made for GATE Chemistry 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for General Aptitude - 4 below.
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General Aptitude - 4 - Question 1

Operators □ , ◊ and → are defined by:  Find the value (66 □ 6) → (66◊6).

Detailed Solution for General Aptitude - 4 - Question 1

Correct answer is option C) 1

General Aptitude - 4 - Question 2

The velocity V of a vehicle along a straight line is measured in m/s and plotted as shown with respect to time in seconds. At the end of the 7 seconds, how much will the odometer reading increase by (in m)?

Detailed Solution for General Aptitude - 4 - Question 2

The odometer reading increases from starting point to end because odometer measures distance not displacement.
So, all the areas should be counted postive only.
Area of the given diagram = Odometer reading
Area of the velocity and time graph per second
1st sec triangle = 1/2 * 1 * 1= 1/2
2nd sec square = 1 × 1 = 1
3rd sec square + triangle = 1 × 1 + 1/2 × 1 × 1 = 3/2
4th sec triangle =1/2 * 1 * 2 = 1
5th sec straight line = 0
6th sec triangle = 1/2 * 1 * 1 = 1/2
7th sec triangle = 1/2 * 1 * 1 = 1/2
Total Odometer reading at 7 seconds = 1/2 + 1 + 3/2 + 1 + 0 + 1/2 + 1/2 = 5m

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General Aptitude - 4 - Question 3

log tan 1o + log tan 2o +.....+ log tan 89o is _____.

Detailed Solution for General Aptitude - 4 - Question 3

As per trigonometric properties

tan θ = cot (90 - θ) and tan θ . cot θ = 1

Thus, tan 89° = cot (90 -89° ) =  cot 1°

Similarly, tan 88° = cot 2°, tan 87° = cot 3° ans so on...

Also from Logarith properties

log m +log n = log (m × n)

Thus, log tan 1° + log tan 89° = log (tan1° × tan89°)

= log (tan1° × cot1°)

= log 1 = 0

similarly, log tan 2° + log tan 88° = log (tan1° × cot1°) = log 1 = 0

Thus, log tan1° +log tan2° +……..+log tan 89° = 0

General Aptitude - 4 - Question 4

An electric bus has onboard instruments that report the total electricity consumed since the start of the trip as well as the total distance covered. During a single day of operation, the bust travels on stretches M, N, O and P, in that order. The cumulative distances traveled and the corresponding electricity consumption are shown in the Table below:
gate ece 2015 question paper with solutions
Q. The stretch where the electricity consumption per km is minimum is

Detailed Solution for General Aptitude - 4 - Question 4

General Aptitude - 4 - Question 5

If a2 + b2 + c2 = 1, then ab + ac + bc lies in the interval

Detailed Solution for General Aptitude - 4 - Question 5

Concept:

1. (a - b)2 = a2 - 2ab + b2

2. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

Where, (a + b + c)2 > 0

If a, b & c ∀ 0, then

(a + b + c)2  = 0

Calculation:

Given that,

a2 + b2 + c2 = 1   ----(1)

We know that, (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

⇒ (ab + bc + ca) ≥ -1/2

Again  ( b − c )2 + ( c − a )2 + ( a − b )2 ≥ 0

⇒ bc + ca + ab ≤ a2 + b2 + c2 = 1

Hence  − 1/2 ≤ bc + ca + ab ≤ 1

General Aptitude - 4 - Question 6

The Venn diagram shows the preference of the student population for leisure activities.

Q. From the data given, the number of students who like to read books or play sports is_____

Detailed Solution for General Aptitude - 4 - Question 6

According to the given Venn diagram number of students who like to read books are.

13 + 12 + 44 + 7 = 76

Number of students who like to play sports are,

44 + 7 + 15 + 17 = 83

But the number of students who like both to read books and play sports are,

44 + 7 = 51

Hence the number of students who like to read books or play sports are,

76 + 83 – 51 = 108.

Here formula used is:

(No. of students Read books ⋃ no. of students play sports) ⋂ No of students read books 2 play sports.

General Aptitude - 4 - Question 7

From a circular sheet of paper of radius 30cm, a sector of 10% area is removed. If the remaining part is used to make a conical surface, then the ratio of the radius and height of the cone is ________.

Detailed Solution for General Aptitude - 4 - Question 7

Area of Circle= πr2 = π.(30)2

Remaining Area = 0.9π(30)2 = 810π

If cone is made of some radius 'r'

Then Lateral Surface Area = πrl

Now 810π = π.rl where l = 30cm

810= r x 30

r = 27

h = √171

so r/h = 2.064

General Aptitude - 4 - Question 8

A tiger is 50 leaps of its own behind a deer. The tiger takes 5 leaps per minute to the deer’s 4. If the tiger and the deer cover 8 meters and 5 meters per leap respectively. What distance in metres will be tiger have to run before it catches the deer?

Detailed Solution for General Aptitude - 4 - Question 8

From The Question

A tiger is 50 leaps of its own from a deer.

Tiger takes 5 leaps per minute.

Deer takes 4 leaps per minute.

Tiger covers 8 meters per leap.

Deer covers 5 meters per leap.

Therefore, the tiger covers 5 x 8 = 40 meters/minute.
The deer covers 4 x 5 = 20 meters/minute.
Hence, the relative speed is 20 meters/minute.
The tiger is behind the deer by 50 x 8 = 400 meters.
Therefore, the time taken = distance/speed = 400/20 = 20 minutes.
Therefore, in the 20 minutes, the tiger will cover 20 x 40 = 800 meters.

General Aptitude - 4 - Question 9

The statistics of runs scored in a series by four batsmen are provided in the following table:

Q. Who is the most consistent batsman of these four?

Detailed Solution for General Aptitude - 4 - Question 9

Average only gives the mean value, Standard Deviation gives how close to mean value (consistency) of a sample population distribution.
A standard deviation close to 0 means very close to mean value of a distribution.
Here K has the lowest SD (5.21)

General Aptitude - 4 - Question 10

A train that is 280 metres long, travelling at a uniform speed, crosses a platform in 60 seconds and passes a man standing on the platform in 20 seconds. What is the length of the platform in metres?

Detailed Solution for General Aptitude - 4 - Question 10

The train with length 280m passes by a man standing on platform in 20s.
The speed of train is V = 280/20 = 14m/s
Let the length of platform be L
(L + 280 )/60 = 14
L + 280 = 60×14
L = 840 - 280
L = 560 metres

General Aptitude - 4 - Question 11

The sum of eight consecutive odd numbers is 656. The average of four consecutive even numbers is 87. What is the sum of the smallest odd number and second largest even number?

Detailed Solution for General Aptitude - 4 - Question 11

The sum of 8 consecutive odd numbers is 656
⇒ Average of the 8 odd numbers = 656/8 = 82
the odd numbers immediately preceding and succeeding 82 are 81 are 83 respectively and these will be the middle numbers.
Therefore, we see that 4th and 5th odd numbers are 81 and 83 respectively.
Thus we can easily obtain the odd numbers as 75, 77, 79, 81, 83, 85, 87, 89 and smallest odd number as 75
the average of four consecutive even numbers is 87
As seen earlier, 86 and 88 will be 2nd and 3rd even numbers.
i.e., 84, 86, 88 and 90 will be the even numbers and the second largest even number is 88
Therefore required sum is 75 + 88 = 163

General Aptitude - 4 - Question 12

A man can row at 8 km per hour in still water. If it takes him thrice as long to row upstream, as to row downstream, then find the stream velocity in km per hour.

Detailed Solution for General Aptitude - 4 - Question 12

Speed of man (m) = 8km/h
Let the speed of stream be s
According to the question:
Speed of man upstream = S1= m − s
Speed of man downstream = S= m + s
Speed = Distance/Time
Here, since the distance D is same
D = S× T1
D = S× T2
S× T= S× T2
S1/S= T2/T= 3
m + s = 3(m − s)
or, 8 + s = 3(8 − s)
⇒ s = 4km/h

General Aptitude - 4 - Question 13

A car travels 8 km in the first quarter of an hour, 6 km in the second quarter and 16 km in the third quarter. The average speed of the car in km per hour over the entire journey is

Detailed Solution for General Aptitude - 4 - Question 13

This is a simple formula based one.
Average speed = total distance/ total time
Total distance = 8+6+16 = 30 km
Total time taken = 3 quarters of an hour = ¾ hr
Average speed = total distance/ total time = 30/(3/4) = 40 kmph

General Aptitude - 4 - Question 14

You are given three coins: one has heads on both faces, the second has tails on both faces, and the third has a head on one face and a tail on the other. You choose a coin at random and toss it, and it comes up heads. The probability that the other face is tails is

Detailed Solution for General Aptitude - 4 - Question 14

This problem is an application of bayes' theorem.
This theorem shold be applied when the outcome is already determined .
so in the problem we need to calculate all possible ways a head could show up.
We have HH TH TT
probability of head showing up on the first coin is chance of first coin getting picked x chance of head showing up = 1/3 x 1 = 1/3
Simlarly for the 2nd coin the probablility is 1/3  x 1/2
For the 3rd since it doesnt have a head we dont really care
Now to calculate the prob that the head came from the 2 nd coin applying bayes theorem
p(H) = 1/3(1/2) / 1/3 + 1/3(1/2) = 1/3

General Aptitude - 4 - Question 15

It takes 30 minutes to empty a half-full tank by draining it at a constant rate. It is decided to simultaneously pump water into the half-full tank while draining it. What is the rate at which water has to be pumped in so that it gets fully filled in 10 minutes?

Detailed Solution for General Aptitude - 4 - Question 15

Lets say capacity of tank is 1 litre
draining rate = 0.5litre/30minutes = 1/60 litre/min
let filling rate = x litre/min
in 1 min tank gets x - (1/60) litre filled.
to fill the remaining half part we need 10mins
x - 1/60 litre → 1min
0.5 litre → 10 mins
0.5/(x - 1/60) = 10
solving, we get x = 4/60
which is 4 times more than draining rate.

General Aptitude - 4 - Question 16

A firm producing air purifiers sold 200 units in 2012. The following pie chart presents the share of raw material, labour, energy, plant & machinery, and transportation costs in the total manufacturing cost of the firm in 2012. The expenditure on labour in 2012 is Rs. 4,50,000. In 2013, the raw material expenses increased by 30% and all other expenses increased by 20%. If the company registered a profit of Rs. 10 lakhs in 2012, at what price (in Rs.) was each air purifier sold?

Detailed Solution for General Aptitude - 4 - Question 16

Given that expenditure on labour in 2012 is Rs, 4,50,000 which is 15% of cost of total manufacturing cost of firm

Total Manufacturing Cost x 0.15 = 4,50,000

Total Mfg. Cost = 4,50,000 / 0.15

= 30,00, 000

Total Profit is = 10,00,000

Total Sale obtained from 200 purifier  = manufaturing cost + profit = 40,00,000

So cost of one purifier = 40,00,000 / 200

= Rs. 20,000

General Aptitude - 4 - Question 17

In the summer of 2012, in New Delhi, the mean temperature of Monday to Wednesday was 41 °C and of Tuesday to Thursday was 43 °C. If the temperature on Thursday was 15% higher than that of Monday, then the temperature in °C on Thursday was

Detailed Solution for General Aptitude - 4 - Question 17

Let the temperature on Monday, Tuesday, Wednesday and Thursday be respectively as TM, TTU, TW, TTH,
So, from the given data we have
The mean temperature of Monday to Wednesday was 41oC => (M+T+W)/3 = 41
 ...(1)
The mean temperature of Tuesday to Thursday was 43oC=> (T+W+Th)/3 = 43
and  ...(2)
Also, as the temperature on Thursday was 15% higher than that of Monday
i.e.,  …(3)
After Solving The above equations (1), (2) and (3) we obtain

Hence correct option is C.

General Aptitude - 4 - Question 18

Find the sum to n terms of the series 10 + 84 + 734 + _____.

Detailed Solution for General Aptitude - 4 - Question 18

Check from the options by substituting n = 1, 2…
For n = 1, all options are right
For n = 2, sum must be 10 + 84 = 94.
Only option D is correct for this

Given Series in AGP (Arithematic Geomatric Progression)

10 + 84 + 734 + ... = (91 +1) + (92 + 3) + (93 + 5)+...

For solving AGP progression we need to find

General Aptitude - 4 - Question 19

If x > y > 1, which of the following must be true?
(i) ln x > ln y
(ii) ex > ey
(iii) yx > xy
(iv) cos x > cos y

Detailed Solution for General Aptitude - 4 - Question 19

From the logarithmic property, 

when x > y 

ln x > ln y if x and y both greater than 1.

and x > y 

ln x < ln y, if both x and y are less than 1.

but when x > y 

ex > ey  for any value of x and y.

but when x > y, x2 > y2

 also cos x < cos y.

General Aptitude - 4 - Question 20

A batch of one hundred bulbs is inspected by testing four randomly chosen bulbs. The batch is rejected if even one of the bulbs is defective. A batch typically has five defective bulbs. The probability that the current batch is accepted is _________

Detailed Solution for General Aptitude - 4 - Question 20

Probability for one bulb to be non defective is 
∴ Probabilities that none of the bulbs is defectives 
Binomial Problem with n = 4 and p(not defective) = 0.95
P(x = 4 non defective) = 0.95 x 4 = 0.8145

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