Answer: Option D. -> 240
:
D
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
The number of people who like RNBDJ and Singh is King but not Ghajni= (600 - 360) = 240. Hence option (d)

Answer: Option B. -> 54%
:
B
240 + 200 + 280 + 360 = 1080 = 54% Hence option (b)

Answer: Option A. -> 30
:
A
Option (a)

Answer: Option D. -> 70
:
D
No. of students who failed only in Maths= 20 (Given)
According to condition given,
Students who failed in Maths and English only + Students who failed in Science and Maths only + Students who failed in all three papers= 30-20= 10.
Therefore, 6+8-x=10 (x being the number of students who failed in all three subjects)
Solving the above equation gives x=4
Student who failed in:-
All three subjects=4
Maths and English only=2
Maths and Science only=4
Science and English only=6
English only=16
Science only=18
Therefore total no. of students= 16+18+20+2+4+6+4=70
Hence answer option d.

Answer: Option B. -> 4 : 1
:
B
Approach 1: Conventional Approach :
Let the speed of the train be T, and let it be at a distance of X from the tunnel AB.
Let the speed of the catbe C, and let the distance of the tunnel be Y, hence thecat is at a distance of $\frac{3}{8}Y$ from A.
Now 2 conditions are given:
When the train meets the cat at B it would have travelled a distance of X+Y while the cat will travel $\frac{5}{8}Y$ at its respective speeds,
$\frac{X+Y}{T}=\left(\frac{5}{8}\right)\frac{Y}{C}.......\left(1\right)$
When the train meets the cat at A it would have travelled a distance of X while the cat will travel $\frac{3}{8}Y$ at its respective speeds.
Also, $\frac{X}{Y}=\frac{\left(\frac{5}{8}\right)Y}{C}.....\left(2\right)$
From (1)
$\frac{X}{T}+\frac{Y}{T}=\frac{5Y}{5C}$
Sub form (2)
$\frac{\left(\frac{5}{8}\right)Y}{C}+\frac{Y}{T}=\frac{5Y}{8C}$
$=\frac{Y}{T}=\frac{2Y}{8C}$
or $T:C=4:1$
Approach 2: Shortcut- Assumption
Assume the lengths of the tunnel to be 8 kmand the speed of the cat be 8 km/hr
Time taken for the cat to reach A is 3 hr.
Time taken for the cat to reach B is 5 hr.
The difference in time of A and B gives the time taken by the train, which is 2 hours.
Hence the cat takes 8 hours while the train takes 2 hours,
Hence T:C = 4:1

Answer: Option B. -> 48 meters
:
B
Approach 1: Conventional Approach:
(b) Let the distance of race be x metres which is covered by A in t seconds. Then, in the same time B covers (x-12)metres and C covers (x-18)metres.
$\therefore SpeedofA=\frac{x}{t}m/sec,$
$SpeedofB=\frac{x-12}{t}m/sec,$
$SpeedofC=\frac{x-18}{t}m/sec.$
Time taken by B to finish the race $=\frac{x}{\frac{x-12}{t}}=\frac{xt}{x-12}$ sec.
Now distance travelled by C in this time,
$=x\times t\times \frac{x-18}{(x-12)t}=x-8$
$\to \frac{x(x-18)}{x-12}=x-8\to x=48m.$
Approach 2: Reverse Gear Approach
Going from answer options:
Note the ratio of distance covered should be the same:
Assume option (c) to be correct: then we get the distances travelled to be in the ratio of60:48:40 and for B and C we get the ratio to be 60:52, here we observe for B and C that the ratio is not the same.
For option B we get A:B:C = 48:36:30 or 8:6:5 for B and C we get 48:40 or 6:5 we see that for both the options we have the same ratio of distance travelled hence this the correct answer option!

Answer: Option C. -> 9 : 8
:
C
(c) Time is inversely proportional to the speed.
Hence ratio of time taken by B and D = 18 : 16 = 9 : 8

Answer: Option D. -> 900
:
D
Let workers on leave: present workers = x: y.
0.04x come to work and 0.06y take leave on the second day. As there will be no difference in value of x and y after this exchange: 0.04x = 0.06y
x: y = 6: 4 or 3: 2.
Number of workers on leave $=\left(\frac{3}{5}\right)\times 1500=900$
Hence option (d)
Alternatively, go from answer options
$2^{nd}$ day number joining and number on leave should be the same
This happens only when the number of leave = 900
$2^{nd}$ day = $4\%of900-6\%of600=0$ . Option (d)

Answer: Option C. -> 6
:
C
It is given that simple interest for 2 years = Rs. 720
So, for 1 year = Rs. $\frac{720}{2}$ = Rs. 360 (because it is SI)
Compound Interest for 2 year = Rs. 756
Difference in the interest = Rs. 36 (756-720)
This difference is the interest on the interest of the first year.
So rate of interest $=\left(\frac{36}{360}\right)\times 100=10\%$
Now the interest for $1^{st}$Year = Rs. 360 at $10\%$ SI
So, principal for $1^{st}$ year = Rs. 3600
We have, $SI=\frac{(P\times R\times T)}{100}$
$\frac{3600\times k^2}{100}=1296$
k = 6

Answer: Option C. -> 60000
:
C
Using alligation

He should invest in the ratio 2:3
He should invest 60000