Discussion
Car X traveled from city A to city B in 30 minutes. The first half of the distance was covered at 50 miles per hour, and the second half of the distance was covered at 60 miles per hour. What was the average speed of car X?
*This question is included in Nova Math - Problem Set X: Word Problems, question #6
| (A) | 200/11 |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 09/28/2012 13:22
How did you get 300/11 for the distance by solving the second equation?
Posted: 09/28/2012 14:49
Jennifer, here's how to set up this algebra problem:
d = distance = d1 + d2 = 50 mph * t1 + 60mph * t2;
We are also told that d1 = d2;
So: 50 t1 = 60t2;
t = 30 minutes = .5 hour;
t1 + t2 = .5 hour;
So: t1 = .5 -t2;
Solve 2 equations, 2 unknowns:
(1) 50t1 = 60t2 and (2) t1 = .5 - t2.
Multiply both sides of equation 2 by 50 to get: 50t1 = 25 - 50t2;
Subtract that from equation 1 >> 0 = 25 - 110 t2; t2 = 25/110
t1 = .5 - t2 = 55/110 - 25/110 = 30 /110 = 3/11;
Ave speed = d / t = (50t1 + 60t2) / .5
= (50 * 3/11 + 60*25/110 ) /.5
= (150/11 + 150/11) / .5
= (300/11)/ .5
= 600/11
d = distance = d1 + d2 = 50 mph * t1 + 60mph * t2;
We are also told that d1 = d2;
So: 50 t1 = 60t2;
t = 30 minutes = .5 hour;
t1 + t2 = .5 hour;
So: t1 = .5 -t2;
Solve 2 equations, 2 unknowns:
(1) 50t1 = 60t2 and (2) t1 = .5 - t2.
Multiply both sides of equation 2 by 50 to get: 50t1 = 25 - 50t2;
Subtract that from equation 1 >> 0 = 25 - 110 t2; t2 = 25/110
t1 = .5 - t2 = 55/110 - 25/110 = 30 /110 = 3/11;
Ave speed = d / t = (50t1 + 60t2) / .5
= (50 * 3/11 + 60*25/110 ) /.5
= (150/11 + 150/11) / .5
= (300/11)/ .5
= 600/11