If a man buys 1 lt of milk for Rs.12 and mixes it with 20% water and sells it for Rs.15, then what is the percentage of gain?

Given, M₁ = 120, M₂ = 120 - n, D₁ = 64, D₂ = 60
W₁= 2/3 and W₂ = 1/3
According to the formula,
(M₁D₁) / W₁ = (M₂D₂) / W₂
? [120 x (64/2)] / 3 = [(120 - n) x 60] / (1/3)
? (120 x 64) / (2 x 60) = (120 - n)
? (120 - n) = 64
? n = 120 - 64 = 56
Now, 56 men can be discharged to finish the work in time.

One person can select one house out of 3= 3 C 1 ways =3.

Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.

Therefore, probability that all thre apply for the same house is 1/9

Total number of elementary events =  10 C 5

Number of ways of selecting exactly one defective bulb out of 3 and 4 non-defective out of 7 is  3 C 1 * 7 C 4

So,required probability = 3 C 1 * 7 C 4/ 10 C 5 = 5/12.

n(S) = 36

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

n A = 6 , n B = 5 , n A ? B = 1

?Required probability =  P A ? B

 =  P A + P B - P A ? B

=   6 36 + 5 36 - 1 36 =  5 18

Total number of balls = (8 + 7 + 6) = 21.

Let E = event that the ball drawn is neither red nor green 

            = event that the ball drawn is blue.

n(E) = 7.

P(E) = n(E)/n(S) = 7/21 = 1/3.

Let the work be finished in N days.
Then, A's N day's work + B's (N - 1) days + work + C's (N - 2) day's work = 1
? N/16 + (N - 1/32) + (N - 2/48) = 1
? N/1 + N - 1/2 + N - 2/3 = 16
? (6N + 3N - 3 + 2N - 4)/6 = 16
? 11N - 7 = 96
? 11N = 103
? N = 103/11 = 9⁴/₁₁ days

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:

=  6 * 6 * 6 * 6 = 6 4

n(S) =  6 4

Let X be the event that all dice show the same face. 

X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)}

n(X) = 6

Hence required probability =  n ( X ) n ( S ) = 6 6 4 = 1 216

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7 = 7 C 2 = 21

Let E = Event of drawing 2 balls, none of which is blue.

n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls = 5 C 2 = 10

Therefore, P(E) = n(E)/n(S) = 10/ 21.

Let the leak empties the full tank in x h, then
Part emptied in 1 by leak = 1/x
Part filled by inlet in 1 h = 1/20
According to the question,
1/20 + 1/x = 1/40
? 1/x = 1/40 - 1/20 = 1- 2/40 = -1/40 [-ve sign Indicates emptying.]
Clearly, leak will empty the full tank in 40 h.

P ( P ) = 1 4 , P ( Q ) = 1 5 , P ( R ) = 1 6 , P ( S ) = 1 7
All the events are mutually exclusive hence,

Required probability = P(P)+P(Q)+P(R)+P(S)

1 4 + 1 5 + 1 6 + 1 7= 319 420