Using Relative Rates to solve Rate Distance problems

[kapoor.sam]

Hi, I am trying to solve the problem below using the concept of relative rates and am unable to do so. The solution is quite straightforward using the standard Rate/Time/Distance chart but I would like to understand how I can use the concepts of relative rates to solve this. Any light on this would be greatly appreciated. Here is the problem from the Advanced Quant Strategy Supplement from MGMAT.

"Half an hour after Car A started traveling from Newtown to Oldtown, a distance of 62 miles, Car B started traveling along the same road from Oldtown to Newtown. The cars met each other on the road 15 minutes after Car B started it's trip. If Car A traveled at a constant rate that was 8 miles per hour greater than Car B's constant rate, how many miles had Car B driven when they met?"
A) 14 miles
B) 12 miles
C) 10 miles
D) 9 miles
E) 8 miles
[Page 232 Workout Sets Chapter 9 - Advanced Quant Strategy Supplement]

Thanks in advance.

-Saket

[RonPurewal]

though not impossible, it's extremely awkward to use relative rates here. basically, there is just too much stuff that depends on the non-relative rates in the problem.
see, the problem is that car a travels for half an hour by itself before the whole "relative" thing starts to happen.
if you actually knew the speed of car a, this wouldn't be a problem -- you could just find out how long car A travels in half an hour, subtract that distance out of the total, and then use a relative rate for the rest of the distance. but, unfortunately, you don't know the speed of car A, so, at this point, it's impossible to tell the distance that car A travels by itself.

it's still possible to do this problem with relative rates -- but only if basically everything in the problem is a variable.
e.g., let's say that car A travels at a speed of "x" miles per hour.
* for the first half-hour, car A travels alone at speed x, covering a distance of x/2 miles.
* for the rest of the time, car A travels at x miles per hour, and car b travels at (x - 8) miles per hour. so, that's a relative rate of (2x - 8) miles per hour for 1/4 hour, for a total distance of (2x - 8)/4 miles.
* since the total distance is 62 miles, you can write x/2 + (2x - 8)/4 = 62. you can solve that for x = 64 miles per hour, from which you can find out anything else you might want to know.

this is probably not any more convenient than just doing the longhand algebra solution -- but it's another way to solve the problem, so it's certainly worth learning/reviewing.

[kapoor.sam]

Thanks Ron! Your explanations are impeccable as always :-)

[tim]

indeed!

[ereyhan]

Would you mind showing the steps on how to solve this problem without relative rates? I'm finding these questions, actually word problems in general to be quite challenging for me. Any help would be greatly appreciated. Thanks!

[jnelson0612]

Would you mind showing the steps on how to solve this problem without relative rates? I'm finding these questions, actually word problems in general to be quite challenging for me. Any help would be greatly appreciated. Thanks!

Sure! Let's repost the question:
"Half an hour after Car A started traveling from Newtown to Oldtown, a distance of 62 miles, Car B started traveling along the same road from Oldtown to Newtown. The cars met each other on the road 15 minutes after Car B started it's trip. If Car A traveled at a constant rate that was 8 miles per hour greater than Car B's constant rate, how many miles had Car B driven when they met?"
A) 14 miles
B) 12 miles
C) 10 miles
D) 9 miles
E) 8 miles

Okay, so there are 62 miles between N and O. Car A started traveling from N to O, and then 30 minutes later Car B started traveling from O to N. The cars met 15 minutes after Car B started the trip. A goes 8 miles per hour faster than B.

Since they are working together to meet we will figure the distance each one traveled and add the two distances together to equal 62 miles. When the two cars meet they will have together covered 62 miles.

Remember, the important formula here is Rate * Time = Distance. Let's figure out the individual components:
Car B's rate will be B.
Car A's rate will be B + 8

Car B's time will be .25 hours.
Car A's time will be .75 hours (Car B's time plus the extra 30 minutes)

So A's distance is (B+8) * .75 or .75(B+8).
B's distance is B * .25 or .25B.

Together, .75(B+8) + .25B = 62 miles.

Solve:
.75B + 6 + .25B = 62
B = 56

Thus, B's speed is 56 miles per hour and A's speed is 64 mph.

So in .75 hours A would go 48 miles and in .25 hours B would go 14 miles. 48 + 14 = 62 miles. Looks good!

How far had B driven when they met? Answer A, 14 hours.

[Beulah]

Hi,
Sorry if this sounds silly but I'm slightly confused here. Wouldn't the cars have both traveled the same distance when they had met?
Thanks
Beulah

[RonPurewal]

Hi,
Sorry if this sounds silly but I'm slightly confused here. Wouldn't the cars have both traveled the same distance when they had met?
Thanks
Beulah

they start from opposite sides of the road. so, no, there would be no reason to expect this.

[RonPurewal]

if this is not obvious, think about this:
• you are 100 yards (or 100 metres, or whatever) away from a snail.
• you and the snail start moving toward each other—perhaps at the same time, perhaps not.
• you will most certainly NOT meet the snail halfway!

[Beulah]

Thanks, this makes sense. For some reason I read it the other way round (i.e from Newtown to Oldtown for town B as well).

Thank you
Beulah

Hi,
Sorry if this sounds silly but I'm slightly confused here. Wouldn't the cars have both traveled the same distance when they had met?
Thanks
Beulah

they start from opposite sides of the road. so, no, there would be no reason to expect this.

[tim]

[RonPurewal]

Thanks, this makes sense. For some reason I read it the other way round (i.e from Newtown to Oldtown for town B as well).

Thank you
Beulah

you're welcome.