Given that the Average age of all the residents in the complex = 38 years
Number of people having ages 51 and above = 30
Number of people having ages below 51 ≤ 39
Since we need to maximize the average age of people below 51 and to maintain the combined average as 38 years, we need to consider the ages of people above 51 as 51 (Any more than that would lower the other value)
Also, We need to take the Number of people having ages below as 39 (Again, for maximizing the value)

So, 13 : 38-x = 39 : 30 = 13 : 10
38 - x = 10
x = 28
Average age of people whose ages are below 51 = 28 years

Given that out of 200 students, 105 like Pizza and 134 like Burger
In order to find the possible number of students who like only Burger, we need to find the range between the minimum and maximum possible number of students

Minimum possible number of Students who like only Burger:

For this, we must try to optimize the Venn diagram in such a way that it has maximum intersection
No of Students who like only Burger = 134 – 105= 29

Maximum possible number of Students who like only Burger:

For this, we must optimize the Venn diagram in such a way that it has minimum intersection
We know that total number of Students = 200
So, Minimum number of students who like both = 105 + 134 – 200 = 39 students
Students who like Burger = 134
Max. no of Students who like only Burger = 134 – 39 = 95
So range => 29 ≤ n ≤ 95
From the options, only (D) 93 lies within range

Given, Tank gets filled in 6 hours when 6 filling and 5 draining pipes are on 
Let, F be the rate at which a single filling pipe fills the tanks and D be the rate at which a single
draining pipe drains the pipe 
6F – 5D = 1616 th of the tank ---(1) 
Also, Tank gets filled in 60 hours when 5 filling and 6 draining pipes are on 
5F-6D = 160160 th of the tank ---(2) 
Solving both (1) and (2) we get, 
6F - 5D = 1/6 
(-) 50 F - 60 D = 1/6 
--------------------------- 
44F = 55D 
F:D = 5:4 
Replacing values in (1), 6F – 5D = 1/6 
15D – 10D = 1/3 
D = 1/15 and F = 1/12 
When two filling pipes and one draining pipes are on, 
2(1/6) – 1(/15) = (3/30) = 1/10 th of the tank 
Therefore, they can fill the tank in 10 hours 

Given that Chords lie on the same side of diameter with lengths 4 cms and 6 cms 
Draw a perpendicular from the origin to both the chords and mark the points of intersection as P
and Q respectively 
Consider radius of circle as ‘r’ and distance OP as ‘x’ 
Draw lines from origin to the end of the chord and mark the points as D and B respectively 
Thus, △OQB and △OPD form a right Triangle 
Applying Pythagoras Theorem on both triangles, 
(x+1) 2 +2 2 = r 2 ---(1) 
x 2 +3 2 = r 2 ---(2) 
We find that there is an increase and decrease by 1 in both equations 
So, x 2 = 2 2 , x=2 
r 2 = 2 2 +3 2 = 4+9 = 13 
r = √13 cms

Given Area (ABCD) = 72 sq cm, CD = 9cms, AD = 16cms
Area of Parallelogram = Base × Height
CD × 16 = 72
CD = 72167216 cms
We extend CD till P such that ∠APD = 90°
So, again by applying the area formula considering CD as base
CD × AP = Area of Parallelogram ABCD
9 × AP = 72
AP = 8 cms

We have AP = 8 cms, AD = 16 cms We observe that the ratio of the sides form a 1 : √3 : 2 triangle Therefore DP = 8√3 cms Area (△APD) = 1/2 × AD × DP = 1/2 × 8 × 8√3 sq cms Area (△APD) = 32 √3 sq cms

Let a Human complete ‘H’ units of work per day and a Robot complete ‘R’ units of work per day 
Given, 
15H + 5R = (1/30) th work per day --- (1) 
5H + 15R = (1/60) th work per day --- (2) 
Dividing (1) and (2) by 5 and solving, 
3H + R = (1/150) and H + 3R = (1/300) 
9H + 3R = 1/50 
(-) H + 3R = 1/300 
--------------------------- 
8H = 1/50 - 1/300 => 8H = 1/60 
H = 1/480 
Work done by 15 Humans in a day = 15/480 = 1/32 
No of days required for 15 humans to complete the work = 32 days