The distance between 2 towns is 330km. A train travels between these two towns and returns on the same day. On the return journey, the train left late and so the driver decided to drive the train 5km/h fasterthan its inward journey, and it took 30 minutes less time to return. What is the speed on the return journey?
I used the folowing equation:
Let inward speed be x,
330/x - 330/x+5 = 1/2
However, I keep getting a squared x?
330/x - 330/(x+5) = 1/2
That's ok, we end up with a quadratic.
Multiply each term by 2x(x+5)
660(x+5) - 660x = x(x+5)
660x + 3300 - 660x = x^2 + 5x
x^2 + 5x - 3300 = 0
(x-55)(x+60) = 0
x = 55 or x = -60, we would reject the -60
So your inward speed is 55 km/h, the return is 60 km/h
check:
330/55 = 6 hrs
330/60 = 5.5 hrs, which is 1/2 hour less.
Thanks Reiny, appreciate it
To solve this problem, let's break it down step by step.
Let's start by defining the variables:
- Let's call the inward speed of the train "x" km/h.
- The return speed of the train would then be "x + 5" km/h, as it is 5 km/h faster.
The distance between the two towns is given as 330 km.
From the given information, we are told that on the return journey, the train took 30 minutes less time.
Now, to create an equation, we need to use the formula d = rt, where d is the distance, r is the rate (speed), and t is the time.
On the inward journey, the time would be: t1 = 330 / x hours.
On the return journey, the time would be: t2 = 330 / (x + 5) hours.
We are informed that the return journey took 30 minutes less than the inward journey. Since there are 60 minutes in an hour, we can convert the time difference of 30 minutes to hours by dividing it by 60: 30 / 60 = 0.5 hours.
So, the equation becomes:
t1 - t2 = 0.5
330 / x - 330 / (x + 5) = 0.5
Now, let's solve this equation without any squared terms for "x".
To remove the denominators, we can multiply the entire equation by x(x + 5):
[(330)(x + 5)] - [(330)(x)] = 0.5(x)(x + 5)
Now, let's simplify the equation:
1650 + 330x - 330x = 0.5x^2 + 2.5x
The 330x terms cancel out:
1650 = 0.5x^2 + 2.5x
Multiply the entire equation by 2 to eliminate the decimal:
3300 = x^2 + 5x
Rearrange the equation:
x^2 + 5x - 3300 = 0
Now we have a quadratic equation. We can solve this equation either by factoring, completing the square, or using the quadratic formula. To factor this quadratic equation, we need to find two numbers whose product is -3300 and whose sum is 5.
After examining the options, we can rewrite the equation as:
(x + 60)(x - 55) = 0
Setting each factor equal to zero:
x + 60 = 0 or x - 55 = 0
Solving for x:
x = -60 or x = 55
Since we are dealing with speed (which cannot be negative), we can discard the negative value of -60. Therefore, the inward speed of the train (x) is 55 km/h.
The speed on the return journey would then be x + 5:
55 + 5 = 60 km/h.
So, the speed on the return journey is 60 km/h.