General Aptitude - 4 - GATE Chemistry MCQ

Correct answer is option C) 1

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
1^st sec triangle = 1/2 * 1 * 1= 1/2
2^nd sec square = 1 × 1 = 1
3^rd sec square + triangle = 1 × 1 + 1/2 × 1 × 1 = 3/2
4^th sec triangle =1/2 * 1 * 2 = 1
5^th sec straight line = 0
6^th sec triangle = 1/2 * 1 * 1 = 1/2
7^th 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

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

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

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.

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

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.

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)

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

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 3^rd 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

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 = S₂ = m + s
Speed = Distance/Time
Here, since the distance D is same
D = S₁ × T₁
D = S₂ × T₂
S₁ × T₁ = S₂ × T₂
S₁/S₂ = T₂/T₁ = 3
m + s = 3(m − s)
or, 8 + s = 3(8 − s)
⇒ s = 4km/h

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

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

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.

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

Let the temperature on Monday, Tuesday, Wednesday and Thursday be respectively as T_M, T_TU, T_W, T_TH,
So, from the given data we have
The mean temperature of Monday to Wednesday was 41^oC => (M+T+W)/3 = 41
 ...(1)
The mean temperature of Tuesday to Thursday was 43^oC=> (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.

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

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.

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