I Need Help With An Algebra Word Problem
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I Need Help With An Algebra Word Problem

[From: ] [author: ] [Date: 11-09-11] [Hit: ]
Tips would be helpful, I might need to know how to do this on my test !The distance between the college and the home is 40 miles.Best tip is identify the givens and varibles and assign a varible to the unit u need to find which in above case is distance.Formulate an appropriate eqn read the question again to get a good idea of the given situation.HOPE IT HELPS!......
The word problem reads, "If a student is driving home from college at 40 mph then he is late. However, if he makes the same trip at 50 mph her is 12 minutes early. What is the distance between the college and his home?

I have to write an equation and then solve.
Tips would be helpful, I might need to know how to do this on my test !

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Let the distance be x

Time taken while traveling with a speed of 40mph
t1=x/40hrs

time taken while traveling with a speed of 50mph
t2=x/50hrs

Given that he is 12 min or 12/60 hrs early in second case and time taken in first case is more so

t1-t2=12/60
x/40-x/50=12/60
x=40 miles

The distance between the college and the home is 40 miles.

Best tip is identify the givens and varibles and assign a varible to the unit u need to find which in above case is distance.

Formulate an appropriate eqn read the question again to get a good idea of the given situation.

HOPE IT HELPS!!

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D/40-D/50 = 12/60
5D-4D=40
D = 40 miles
Hope it helps!

Report Abuse


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This is a poorly written problem. One cannot derive a single distance with the information given in the problem. For example, if the student is driving at 49 mph, will he be also late? The 12 minutes early is relative to being on time, or is it suppose to be earlier.

I'll assume 12 minutes earlier to make the problem simpler.

First, you need to determine the distance the student could have traveled in that extra 12 minutes. So 12 minutes is equal to 0.2 hours (12minutes / 60 minutes). So the extra distance is 0.2 hours * 50mph = 10 miles.

Second, let the distance be D. Then in the same amount of time it takes the student to D miles at 40 mph, he could have traveled (D+10) miles at 50 mph. So:
time = distance/speed = D/40 = (D+10)/50.

Solve the equation to find D.

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as given, the q can't be solved because it's not given how late he is @ 40 mph
let me solve it assuming 15 mins late @ 40 mph

d = rt, so t = d/r and

d/40 - 1/4 = d/50 + 1/5

multiply acrosss by 400
10d - 100 = 8d + 80
2d = 180
d = 90 mi
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D/40-D/50 = 12/60
5D-4D=40
D = 40 miles
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