Two planes left simultaneously from the same airport and headed in the same direction towards another airport 1800 km away. The speed of one of the planes was 100 km/hour slower than the speed of the other plane, and so it arrived at its destination 36 minutes after the faster plane. Find the speeds of both planes.
speed of faster plane --- x km/h
speed of slower plane ---- x - 100 km/h
time for faster plane = 1800/x
time for slower plane = 180/(x-100)
1800/(x-100) - 1800/x = 36/60 = 3/5
multiply each term by 5x(x-100) , the LCD
9000x - 9000(x-100) = 3x(x-100)
9000x - 9000x + 900000 = 3x^2 - 300x
3x^2 - 300x - 900000 = 0
x^2 - 100x - 300000 = 0
(x - 1000)(x +
x^2 - 100x = 300000
x^2 - 100x + 2500 = 300000+2500
(I completed the square)
(x - 50)^2 = 302500
x - 50 = ±√302500 = ± 550
x = 600 or a negative
the faster plane went 600 km/h
the slower plane went 500 km/h
check:
time of faster = 1800/600 = 3 hrs = 180 minutes
time of slower plane = 1800/500 = 3.6 hrs = 216 minutes
difference = 216-180= 36 minutes
My answer is correct
your answer is not correct
Let's assume the speed of the faster plane is "x" km/hour.
According to the problem, the speed of the slower plane is "x - 100" km/hour.
We need to find the values of "x" and "x - 100".
We know that the distance covered by both planes is the same, which is 1800 km.
The time taken by the faster plane to cover this distance is "t" hours.
The time taken by the slower plane is "t + 36/60" hours.
Using the formula distance = speed * time, we can set up the following equations:
For the faster plane:
1800 = x * t ----(1)
For the slower plane:
1800 = (x - 100) * (t + 36/60) ----(2)
Let's solve these equations to find the values of "x" and "x - 100".
From equation (1), we have:
t = 1800 / x
Substituting this value of "t" in equation (2), we get:
1800 = (x - 100) * (1800 / x + 36/60)
Simplifying this equation, we get:
1800 = (x - 100) * (1800x + 36x) / 60
Multiplying both sides by 60 to get rid of the denominator, we have:
60 * 1800 = (x - 100) * (1800x + 36x)
108000 = (x - 100) * (1836x)
Dividing both sides by (x - 100), we get:
108000 / (x - 100) = 1836x
Simplifying further, we obtain:
180 / (x - 100) = 3x
Cross-multiplying, we get:
180 * x - 180 * 100 = 3x * (x - 100)
180x - 18000 = 3x^2 - 300x
Rearranging this equation, we have:
3x^2 - 480x + 18000 = 0
Dividing both sides by 3, we get:
x^2 - 160x + 6000 = 0
We can now solve this quadratic equation using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = -160, and c = 6000.
Substituting these values, we get:
x = (-(-160) ± sqrt((-160)^2 - 4(1)(6000))) / (2 * 1)
Simplifying further, we find:
x = (160 ± sqrt(25600 - 24000)) / 2
x = (160 ± sqrt(1600)) / 2
x = (160 ± 40) / 2
So, we have two possible values for "x":
1. x = (160 + 40) / 2 = 200 / 2 = 100 km/h
2. x = (160 - 40) / 2 = 120 / 2 = 60 km/h
Therefore, the speeds of the two planes are:
1. Faster plane: 100 km/h
2. Slower plane: 100 - 100 = 0 km/h
However, since the speed of the slower plane is negative (0 km/h), it is not physically possible. Therefore, the speed of the slower plane should be positive.
Hence, the speeds of the two planes are:
1. Faster plane: 100 km/h
2. Slower plane: 100 - 100 = 0 km/h (Not considered valid)
Let's assume that the speed of the faster plane is x km/hour.
Since the speed of the slower plane is 100 km/hour slower than the speed of the faster plane, the speed of the slower plane is (x - 100) km/hour.
We are given that the planes left simultaneously, and the slower plane arrived at its destination 36 minutes (which is 36/60 = 0.6 hours) after the faster plane.
We know that the distance covered by both planes is the same (1800 km).
We can use the formula distance = speed × time to set up two equations based on the given information:
1. For the faster plane: distance = speed × time
1800 = x × t1, where t1 is the time for the faster plane to complete the journey.
2. For the slower plane: distance = speed × time
1800 = (x - 100) × t2, where t2 is the time for the slower plane to complete the journey.
We also know that the slower plane arrived 36 minutes (or 0.6 hours) after the faster plane, so t1 = t2 + 0.6.
Now, let's rearrange the equations to solve for x, the speed of the faster plane.
From equation 1: t1 = 1800 / x
Substituting t2 + 0.6 for t1 in equation 1:
t2 + 0.6 = 1800 / x
From equation 2: t2 = 1800 / (x - 100)
Now we can substitute this value of t2 in the previous equation:
1800 / (x - 100) + 0.6 = 1800 / x
To solve this equation, we can cross-multiply and simplify:
1800x + 1800(0.6)(x - 100) = 1800(x - 100)
1800x + 1800(0.6)x - 1800(100) = 1800x - 180,000
1080x + 1080(0.6)x = 1800x - 180,000
1080x + 648x = 1800x - 180,000
1728x = 1800x - 180,000
1800x - 1728x = 180,000
72x = 180,000
x = 180,000 / 72
x ≈ 2500
So, the speed of the faster plane is approximately 2500 km/hour.
Now we can substitute this value of x into equation 2 to find the speed of the slower plane:
1800 = (2500 - 100) × t2
1800 = 2400 × t2
t2 = 1800 / 2400
t2 = 0.75
Therefore, the speed of the slower plane is 2400 km/hour.
To summarize, the speed of the faster plane is approximately 2500 km/hour, and the speed of the slower plane is 2400 km/hour.