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Two trains start from City A and travel towards City B, which is at a distance of 100 miles from City A, at the same time with an
average speeds of 60 miles per hour and 80 miles per hour respectively. A train starts from City B at the same time and travels towards City A at an average speed of 70 miles per hour. How far from city B is the slower train starting from City A when the faster train starting from City A meets the train starting from City B?
  • a)
    30 miles
  • b)
    40 miles
  • c)
    50 miles
  • d)
    60 miles
  • e)
    70 miles
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Two trains start from City A and travel towards City B, which is at a ...
Problem:

Two trains start from City A and travel towards City B, which is at a distance of 100 miles from City A, at the same time with an average speeds of 60 miles per hour and 80 miles per hour respectively. A train starts from City B at the same time and travels towards City A at an average speed of 70 miles per hour. How far from city B is the slower train starting from City A when the faster train starting from City A meets the train starting from City B?

Solution:

Let the distance of the slower train from City B be x miles.

The time taken by both the trains starting from City A to meet is given by:
t = d / (v1 + v2)
Where d is the distance between the cities, v1 is the speed of the slower train, and v2 is the speed of the faster train.

So, the time taken by both the trains starting from City A to meet is:
t = 100 / (60 + 80) = 5/6 hours

In this time, the faster train starting from City A covers a distance of:
d1 = v2 * t = 80 * 5/6 = 400/6 miles

The slower train starting from City A covers a distance of:
d2 = v1 * t = 60 * 5/6 = 50 miles

Now, the train starting from City B is also traveling towards the slower train starting from City A. So, we need to find the time taken by the train starting from City B to meet the slower train starting from City A.

The time taken by the train starting from City B to meet the slower train starting from City A is given by:
t' = x / (v1 + v3)
Where v3 is the speed of the train starting from City B.

In the same time t', the train starting from City B covers a distance of:
d3 = v3 * t' = 70 * t'

Now, the distance covered by the slower train starting from City A in the same time t' is:
d4 = v1 * t' = 60 * t'

So, when the faster train starting from City A meets the train starting from City B, the slower train starting from City A has covered a distance of:
x + d4 = x + 60 * t'

But we know that the time taken by both the trains starting from City A to meet is 5/6 hours, which is the same as the time taken by the train starting from City B to meet the slower train starting from City A. So, we can equate the distances covered by the train starting from City B and the slower train starting from City A:

d3 = x + 60 * t'

Substituting the value of t' from the above equation, we get:
70 * t' = x + 60 * t'
Or, x = 10 * t'

Substituting the value of t' from the above equation, we get:
x = 10 * (d3 / 70)

Substituting the value of d3, we get:
x = 10 * (400 / 7)

Simplifying, we get:
x = 57.14 miles

Therefore, the slower train starting from City A is 57.14 miles from City B when the faster train starting from City A
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Two trains start from City A and travel towards City B, which is at a distance of 100 miles from City A, at the same time with anaverage speeds of 60 miles per hour and 80 miles per hour respectively. A train starts from City B at the same time and travels towards City A at an average speed of 70 miles per hour. How far from city B is the slower train starting from City A when the faster train starting from City A meets the train starting from City B?a)30 milesb)40 milesc)50 milesd)60 milese)70 milesCorrect answer is option 'D'. Can you explain this answer?
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