A train traveled the first d miles of its journey it an average speed of 60 miles per hour, the next d miles of its journey at an average speed of y miles per hour, and the final d miles of its journey at an average speed of 160 miles per hour. If the train's average speed over the total distance was 96 miles per hour, what is the value of y?
A. 68
B. 84
C. 90
D. 120
E. 135
The OA is D.
I think that I can solve this PS question of the following way,
I know the total average speed over the total distance, 96 miles per hour.
Also I know that the total distance is, 3d.
The total time will be,
$$Total\ Time=d\left(\frac{1}{60}+\frac{1}{y}+\frac{1}{160}\right)$$
The total average speed can be determine by,
$$Total\ Speed=\frac{Total\ Dist}{Total\ Time}=\frac{3d}{d\left(\frac{1}{60}+\frac{1}{y}+\frac{1}{160}\right)}=96$$
$$\frac{3}{96}=\frac{1}{60}+\frac{1}{y}+\frac{1}{160}==>\frac{1}{y}=\frac{1}{32}-\frac{1}{60}-\frac{1}{160}\ Finally,\ \ y=120\ miles\ per\ hour$$
Is there a strategic approach to this PS question? Can any experts help, please? Thanks!
A. 68
B. 84
C. 90
D. 120
E. 135
The OA is D.
I think that I can solve this PS question of the following way,
I know the total average speed over the total distance, 96 miles per hour.
Also I know that the total distance is, 3d.
The total time will be,
$$Total\ Time=d\left(\frac{1}{60}+\frac{1}{y}+\frac{1}{160}\right)$$
The total average speed can be determine by,
$$Total\ Speed=\frac{Total\ Dist}{Total\ Time}=\frac{3d}{d\left(\frac{1}{60}+\frac{1}{y}+\frac{1}{160}\right)}=96$$
$$\frac{3}{96}=\frac{1}{60}+\frac{1}{y}+\frac{1}{160}==>\frac{1}{y}=\frac{1}{32}-\frac{1}{60}-\frac{1}{160}\ Finally,\ \ y=120\ miles\ per\ hour$$
Is there a strategic approach to this PS question? Can any experts help, please? Thanks!
