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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.
name
R
T,P,R,Q,S
Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.
125, 123, 120, 115, 108, 100, 84
b
A’s speed=6 mph ,B’s speed=8 mph
Let, after x hrs, they will meet.
so, the distance traveled by A in x hrs should be the same as the distance traveled by B in (x-1/2)hrs [as B started the journey after 30 min of A]
Thus, 6x=8(x-1/2)[as distance=speed*time]
=>8x-6x=4
=>2x=4
=>x=2
after 2 hrs they will meet so time=(9+2)=11.00 a.m
237.6 Kmh
7500
30
0.000256
Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders)
= (208) – (11) = 197.
the answer is (a)40