Let B's present age = x years. Then, A's present age = (x + 9) years.
(x + 9) + 10 = 2(x - 10)
=> x + 19 = 2x - 20
=> x =39.
Let the son's present age be x years .Then, (38 - x) = x => x= 19.
Son's age 5 years back = (19 - 5) = 14 years
In a leap year,there are 366 days=52 weeks and 2 days
Remaining favourable 2 days can be sunday and monday or saturday and sunday
Exhaustive number of cases =7
Favourable number of cases =2
So,required probability=2/7
P(black ball)=3/12
P(red ball)=5/12
P(black or red)=3/12+5/12=2/3
P(getting prize) = 10/ (10 + 25) =2/7
Let the present ages of son and father be x and (60 -x) years respectively.
Then, (60 - x) - 5= 4(x - 5)
55 - x = 4x - 20
5x = 75 => x = 15
Let the present age of the man be 'P' and son be 'Q',
Given, P + Q = 64 or Q = (64 - P)
Now the man says "I am five times as old as you were, when I was as old as you are",
So, P = 5[B - (P - Q)]
We get 6P = 10Q,
Substitute value for Q,
6P = 10(64 - P),
Therefore P = 40, Q = 24.
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