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.
X-340 = X x 4 x 8/ 100
100X – 34000 = 32X
68X = 34000
X = 34000/68
X= 500.
B.25
57 and 3
12 x 57 = 684
-8 x 3 = -24
684 – 24 = 660
660/ 60 = 11
45(deg) 31' 4.3"
THEN WHAT WILL BE THE DEGREE WHEN TIME IS 3 O CLOCK
90 degree
15/46
Let:
1. speed of boat in still water = x
2. speed of stream = y;
Then net speed of boat upstream = x – y = 16km / 4 h = 4kmph
Net speed of boat downstream = x + y = 16 / 2 = 8kmph
Now solve for x to get x = 6kmph in still water.
1/8
79.2
Indira Gandhi