A cyclist and a jogger start from a town at the same time and head for a destination 11 mi away. The rate of the cyclist is twice the rate of the jogger. The cyclist arrives 1.1 h before the jogger. Find the rate of the cyclist.
Jogger's Rate = r mi/h.
Cyclist Rate = 2r mi/h.
Cyclist time = t hrs.
Jogger's time = (t + 1.1) hrs.
Eq1: 2r*t = 11 mi.
Eq2: r(t + 1.1) = 11 mi.
Solve Eq1 for r:
2r*t = 11,
r = 11 / 2t.
In Eq2, substitute 11/2t for r:
11/2t(t + 1.1) = 11,
(5.5/t)(t + 1.1) = 11,
5.5 + 6.05 / t = 11.
Common denominator = t.
(5.5t + 6.05) / t = 11,
Cross multiply:
5.5t + 6.05 = 11t,
11t - 5.5t = 6.05,
5.5t = 6.05,
t = 6.05 / 5.5 = 1.1 h.
Solve Eq1 for 2r;
2r * 1.1 = 11,
2r = 11 / 1.1 = 10 mi / h = cyclist rate.
To find the rate of the cyclist, we need to set up a system of equations based on the given information.
Let's assume the rate of the jogger is represented by r mph. Since the rate of the cyclist is twice the rate of the jogger, the rate of the cyclist would be 2r mph.
Let's also assume that the time taken by the jogger to reach the destination is t hours. This means that the cyclist would take t - 1.1 hours to reach the destination, as the cyclist arrives 1.1 hours before the jogger.
We know that distance = rate × time, so we can set up the following equations:
For the jogger:
Distance = r × t (Equation 1)
For the cyclist:
Distance = 2r × (t - 1.1) (Equation 2)
Since the distance is the same for both the cyclist and the jogger (11 miles), we can set Equation 1 equal to Equation 2:
r × t = 2r × (t - 1.1)
Now, let's solve for r:
rt = 2rt - 2.2r
rt - 2rt = -2.2r
-rt = -2.2r
t = 2.2
Now that we have the value of t, we can substitute it back into Equation 1 to find the rate of the jogger:
r × 2.2 = 11
2.2r = 11
r = 5 mph
Since the rate of the cyclist is twice the rate of the jogger, we can calculate it:
Rate of cyclist = 2 × 5 mph = 10 mph
Therefore, the rate of the cyclist is 10 mph.