❨	√3 -	1	❩	2	simplifies to:
		√3

Work done by both the taps in 5 min
= 5 (1/20 +1/25) = (5 x 9)/100 = 9/20
Remaining part = (1 - 9/20) = 11/20
Now, 1/20 part is filled in 1 min.
So, 11/20 part will be filled in 11 min.
Hence, the tank will be full in 11 more min.

Let the principle and rate of interest be P and R%
Then, (P x R x 10)/100 = 600
? PR = 6000
Total interest at the end of year = 300 + (3P x R x 5)/100 = 300 + (15 x 6000)/100 = ? 1200

Given in the question,
n(M ? B) = 50, n(M) = 35, n(B) = 37, n(M ? B) =?
use the below formula
n(M ? B) = n(M) + n(B) ? n(M ? B)
We get 50 = 35 + 37 ? n(M ? B)
? n(M ? B) = 35 + 37 ? 50 = 72 ? 50 = 22
? 22 students have opted for both Mathematics and Biology.
Again number of students who have opted for only Mathematics = n(M) ? n(M ? B) = 35 ? 22 = 13.

Required probability = P(A) x P(B)
= (1 - 1/4) x (1 - 1/3) = 3/4 x 2/3 = 1/2

Relative speed, when the trains are running in the opposite direction = Sum of the speed

Therefore, Relative speed = 36 km/hr +72 km/hr = 90 km/hr
Relative speed in m/sec = 90 km/hr x 5/18= 25 m/sec

Distance covered = sum of the lengths of the train = 200 m+ 300 m = 500m
Therefore, time taken = 500m / 25 m/sec = 20 sec

Let sum = p
Then, after 15 yr
Sum = 8p

? SI = 8p - P = 7P
Now, 7P = (P x R x 15)/100
? 7 = 15R/100 = 3R/20
? R = (20 x 7)/3 = 140/3
= 46²/₃%

Total number of possible arrangement for 4 boy and 3 girls in a queue = 7!
when they occupy alternate position then the arrangement would be like
BGBGBGB. Thus the total number of possible arrangement = 4! x 3!
? Required probability = 4! x 3! / 7!
= (4 x 3 x 2 x 3 x 2) / (7 x 6 x 5 x 4 x 3 x 2) = 1/35

Work done by (A + B) in 1 day = 1/24
Work done by B alone in 1 day = 1/(3 x 12) = 1/36
? Work of A for 1 day = 1/24 - 1/36 = 1/72
After 12 days the remaining work = 1 - 1/3 = 2/3
? 1/72 work is done by A in 1 day
? 2/3 work is done by A in (1 x 72)/(1 x 2/3) = 48 days