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Let the age of the man be x
Then age of his son becomes (x−24)
2 years later from now,
Age of man will be = x+2
and age of his son will be =(x−24+2)=x−22
According to question,
2(x−22)=x+2
i.e., 2x−44=x+2
i.e., x=46
Therefore ,
Present age of man=46 years
And present age of his son =46−24=22 years
In plan A 7500/- and other amounts in plan B
10, 14, 28, 32, 64, 68, 132
132
Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125
Length of train = 125Mtrs
Speed of man = 5Km/Hr
=(5*1000)/(1*3600)
=1.4M/s
Time taken for train to pass = 10s
Spd = D/T
Total Distance (Man distance+ Train Length)
#Distance covered by Man in 10s
D=Spd*T
= (1.4M/s*10s)
= 14Mtrs
#Train length = 125Mtrs
Total D = 14Mtrs + 125Mtrs
= 139Mtrs
Time taken = 10s
Sp = D/t
=139M/10s
=13.9M/s
Total distance to cover = 132 +110 = 242m
Speed = 72kn/hr = 20m/s
time = 242/20 = 12.1 sec.
Average = (1+2+3+4+5+6+7+8+9)/3
= 45/3 = 15
Every side will contain sum of 15.
1st side contains 8 and 7.
2nd side contains 9 and 6.
3rd side contains 1,2,3,4 and 5.
5/9
let the for digit number be = pqrs
p=q/3 ==>q = 3p
r=p+q=p+3p =4p
s=3q = 3(3P)=9p
number:
p 3p 4p 9p
let p=1 answer is 1349
if it 2 answer 2 6 8 18
so it becomes five digit number so correct answer is 1349