Enercon Engineering Interview Questions, Process, and Tips

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petticoat

A. Rs. 410

Hard work would mean spending long hours to complete my work without any shortcuts. It definitely ensures results but the process is long and stressful. Smart work would be aiming for the same results but with planning and prioritization of tasks.

The local value of 7 in the number is at 10000 so 7×10000 and face of 7 is 7
So 70000 -7 so the answer is C 69993

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.

4 mins

4.224 days

20000

8:20

let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240