Answer: Option C. -> 2: 3
:
C
Number of people , who likes at least one movie = 2000 - 120 = 1880

x + 40 + x - 120 + x - 200 + 640 - x + 560 - x + 600 - x + x = 1880
x = 360
160: 240 = 2: 3. Hence option (c)

Answer: Option A. -> 100
:
A
As, we know {A $\cup$ B} = A + B - (A $\cap$ B)
A $\cup$ B = 500 , A = 250 and B = 350
500 = 250 + 350- (A $\cap$ B)
(A $\cap$ B) = 100
100 are both Engineers and MBAs.
Hence option (a)

Answer: Option D. -> 400
:
D
People who are only Engineer = 250
People who are only MBA = 150
No. of people who are either only MBA or only Engineer = 250 + 150 = 400
Hence option (d)

Answer: Option E. -> 167
:
E
As per the informationgivenin the question, we get the venn diagram as:

We get the equations as:
T + 32 = V + 34 = G + 35
and V + T + G + 46 = 278 => V + T + G = 232
So, by solving the equations we get, no. of students who have occupied seat in Violin Class or Guitar Class, but not inTabla Class = 77 + 14 + 76 = 167. Hence option (e)

Answer: Option D. -> 20   miles
:
D
The ship is 18 miles from the shore, as the speed is in the ratio of 1:10 the distance will also be in the ratio of 1:10.
When the ship travels 1 mile the plane will travel 10 miles
At this point the ship is 19 miles from the shore and the plane is 10 miles away from the shore.
Next when the ship moves by 1 mile covering 20 miles totally the plane will also cover the same distance of 20 miles hence will meet the ship

Answer: Option B. -> Decrease the speed
:
B
(b) At a speed of 40 km/hr,60 km/hr, and 80 km/hr distance travelled in 1 litre of petrol.
$=\frac{40}{2.5}=16km,\frac{60}{4}=15km,\frac{80}{7.9}=10.1km.$ respectively.
Hence at lower speed fuel consumption is less.Hence in order to minimize the fuel consumption, the speed should be decreased.

Answer: Option B. -> 22.5km
:
B
Approach 1: Convetional :
(b) Time taken by A in covering (27-x) km is same as time taken by Bin covering (27+x) km.
$\frac{27-x}{5}=\frac{27+x}{7}$
$x=\frac{27}{6}$
$\therefore XZ=27-\frac{27}{6}=27\times \frac{5}{6}=22.5km.$
Approach 2 : Alternate Approach (Using Ratios):
The total distance they will have to travel will be 2d or 54 kms. As their speeds are in the ratio of 5:7 the distances will also be in the same ratio. Hence,54 kms will have to be split in the ratio of 5:7. (i.e.,)22.5 and 41.5,hence they will meet at 22.5km.
Approach 3 : Going from answer options :
The total distance they will have to travel will be 2d or 54kms.
Now,going by the answer options :
Let us assume 25 to be the right option. Hence,A will take 5 hrs in the same time B would have covered a distance of 35 kms.hence the total distance covered will be 60,which is not what is given in the question (we need to get 54.)
Now assume 22.5 to be the right answer,hence A will take 4.5 hrs in the same time B would have covered a distance of 31.5kms,the tally of which is 54 hence that is the right option.

Answer: Option B. -> 12 km/hr
:
B
Using $\frac{1}{x}\&\frac{1}{x+1}$
We know that the speed is increasing by $\frac{1}{2}$ hence the time has to decrease by $\frac{1}{3}$ (using the $\frac{1}{x}\&\frac{1}{x+1})$
shortcut (refer demo tutorial) but the time is actually decreasing by 2 hrs.
Hence the actual time of travel is 6 hrs hence he leaves at 7:00 AM.
The total distance travelled is $10\times 6=60kms$
To reach at noon he will have to cover 60 km in 5 hours. i.e., he has to travel at 12 km/hr.
Approach 2: Conventional Approach:
Let the person take t hrs to cover the distance at 10 km/hr.
If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m, which means he takes (t-2)hr.
Let the distance covered be D.
So we have $15\times (t-2)=10\times \left(t\right)$
T=6 , or he started 6 hours before 1, i.e., at 7.
Hence the total distance now is $10\times 6=60$ km.
To reach at noon he will have to cover 60 km in 5 hours. He was to travel at 12 km/hr.

Answer: Option A. -> 112
:
A
Approach 1:
Option (a) Time taken by B to cover 60 km $=\frac{60}{50}=\frac{6}{5}$ hours.
Time taken by Y at station C $=\frac{15}{60}=\frac{1}{4}$ hours.
Now distance travelled bt train X in $(\frac{6}{5}+\frac{1}{4})=\frac{29}{20}$ hours $=70\times \frac{29}{20}=101.5$ km.
Distance between X and Y when Y starts from station C =180 -(101.5 +60)=18.5 km.
Relative speed =(70+50)=120 km/hr.
Hence, time taken by them in crossing one another $=\frac{18.5}{120}=0.15$ hours. Now, distance travelled by X in 0.15 hour $=70\times 0.15=10.5$ km.
Therefore, distance of X from station A, when they meet =(101.5+10.5)=112 km.
Approach 2:
By using a table to check the distance travelled by each of X and Y:
$\begin{array}{cccc}Text& Xdistance& Ydistance& Total\\ 12:00& 70& 50& 120\\ 12:12& 14& 10& 144\\ 12:27& 17.5& 0& 161.5\\ Tally& 101.5& & \end{array}$
The last 18.5 kms will have to be divided in the ratio of 70:50. Hence we see that the distance travelled by X is around 112 kms.

Answer: Option B. -> 45s
:
B
Let the speed of escalator be y ft per second speed of man's be x ft per second .Let us assume that length of the escalator be 90 ft.
Then, $x+y=\frac{90}{30}\to x+y=3............\left(i\right)$
And $x-y=\frac{90}{90}\to x-y=1..................\left(ii\right)$
$\to$ x=25 ft/sec.And time taken by the man to walk up bt the escalator when it is not moving $=\frac{90}{2}=45$ second.
Also note : Options a and d can directly be eliminated.