the ratio of present age of x:y is 5:6
And,we have to find the present age of x.
so,the present age of X is definitely will be multiple of 5.
X=35
so,35 is the answer.
sister
1261
Let’s assume the length of each train is ‘L’ and the speeds of the two trains are ‘V₁’ and ‘V₂’ respectively.
When the trains are moving in the opposite direction, their relative speed is the sum of their individual speeds. The total distance they need to cover is the sum of their lengths. Since they cross each other completely in 5 seconds, we can set up the following equation:
(V₁ + V₂) × 5 = 2L
When the trains are moving in the same direction, their relative speed is the difference between their individual speeds. The total distance they need to cover is the difference between their lengths. Since they cross each other completely in 15 seconds, we can set up the following equation:
(V₁ – V₂) × 15 = 2L
Now, let’s solve these equations to find the ratio of their speeds.
From the first equation, we have:
(V₁ + V₂) × 5 = 2L
V₁ + V₂ = (2L) / 5
From the second equation, we have:
(V₁ – V₂) × 15 = 2L
V₁ – V₂ = (2L) / 15
Let’s add these two equations together:
V₁ + V₂ + V₁ – V₂ = (2L) / 5 + (2L) / 15
2V₁ = (6L + 2L) / 15
2V₁ = (8L) / 15
V₁ = (4L) / 15
So, the speed of the first train is (4L) / 15.
Now, let’s substitute this value back into the first equation to find V₂:
(4L) / 15 + V₂ = (2L) / 5
V₂ = (2L) / 5 – (4L) / 15
V₂ = (6L – 4L) / 15
V₂ = (2L) / 15
Therefore, the speed of the second train is (2L) / 15.
The ratio of their speeds is given by:
(V₁ / V₂) = ((4L) / 15) / ((2L) / 15)
(V₁ / V₂) = 4L / 2L
(V₁ / V₂) = 2
So, the ratio of their speeds is 2:1.
$150000
d
Let the journey distance be : x miles
Now x/40+x/30=8 (time=distance/speed)
Solving this for x we get. x=960/7
therefore, x=137(approx)
(137.14 exact)
8 games They played
A own 3 games = 18 Rs
B loss 3 Rs and own 1 games i,e 9(A own 3 Games)-6(B own
1 game)=3
c own 12 Rs and loss 4 games(A own 3+b own 1) and own 4
games i.e 24(own)-12(loss)
so totally A own 3 games
B own 1 game
c own 4 games
=8
Out of 10 persons, 4 are graduates; so, (10 – 4) = 6 are under-graduates.
If there is no restriction, any three can be chosen from the ten in (10C3) = 120 ways.
Now, if all three chosen are under-graduates; it can take place in (6C3) = 20 ways.
Therefore, the probability that there will be no graduate among the three chosen = (20 / 120) = (1 / 6).
Therefore, the probability that there will be at least one graduate among the three chosen = {1 – (1 / 6)} = (5 / 6) = 0.8333.