Car A is 20 miles behind car B, which is traveling in the same directi

Relative speed of car A is 58-50=8 miles per hour, to catch up 20 miles and drive 8 miles ahead so to drive 28 miles it'll need 28/8=3.5 hours.

Answer: E.

P.S. The way you are doing it should be 20/8+8/8=2.5+1=3.5 hours.
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Using relative speed concept-
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Relative speed = the distance will shrink between the two cars effectively @ 8 miles per hour

Distance to be covered by A = Difference + Lead = 20 + 8 = 28 miles

Time = Distance / Relative Speed = 28 / 8 = 7/2 = 3.5 hrs.


using time distance concept -
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Let t hrs be the time to overtake

Distance traveled by A = 58t miles = 20 + Distance traveled by B = (20 + 50t)

=> 58t = 50t + 20

t = 2.5 hrs

Let k hrs be the time to cover extra 8 miles by A

Distance traveled by A = 58k miles = 8 + Distance traveled by B in k hrs = (8 + 50k) miles

=>8 + 50k = 58k
k = 1.0 hrs

Total time = t + k = 2.5 + 1.0 = 3.5 hrs

Question says that B gets a head start of 20 miles from A.

Say, B traveled x miles when A was already 8 miles ahead of B.

Thus,
Total distance traveled by B = x miles @ speed 50m/h
Total distance traveled by A = 20+x+8 miles @ speed 58m/h ("A" covered the 20m lag; covered the distance x that B covered and got a lead of 8 miles)

It is given that they both traveled these distances in the same time/duration;
Time spent by B = Time spent by A

Time = Distance/Speed

x/50 = (20+x+8)/58
58x = 1000+50x+400
8x = 1400
x = 175 miles

We know the value of x.

We need to find out the time taken by A to travel the distance = 20+x+8 = 20+175+8 = 203miles

A traveled the distance of 203 @ 58m/h in 203/58 h = 3.5h.

Ans: "E"

Just making some formulas :

1. If Objects move in samedirection : The relative speed can be Added / Substracted .
ie. If Car A is moving at X km/hr and Car B is moving at Y km/hr . X > Y
imagine car B to be stopped at car A moving at X-Y km/hr

2. If Objects moving in oppdirection (towards OR away from each other)
if If Car A is moving at X km/hr from Left to right And Car B @ Y km/hr from R2L
imagine car B stationary and A moving toward it at X+Y kms/hr


-Eski

car a 58mph...car b 50mph..so in 1hour..car goes 8mile ahead..to cover 20mile car a will take 2.5hr..hence an additional 1hour to go 8mile ahead..hence 3.5 in total

Car A need to drive 28 mile, so time for A= 28/58
But as car B is moving too 28/58-8/50=3
Can you help me, what i am missing?

Hi Marchikn, you are missing the fact that while a "catches up" B is still driving forward.

Here is my solution: \(speedA = 58, speedB=50\) this means that every hour A gains \(58-50=8\) miles on B.
In how many hours will A reach B?
space=time*speed, space = distance between cars, speed is the rate at which A is catching up (difference of the speeds A-B)
\(t=\frac{20}{8}=2.5h\) A will take 2.5 h to reach B.
But the question asks "when A will be 8 miles ahead", so we need to add to this time the time it will take A to do so.
\(8=t*8\), \(t=1\), 1h to get 8 miles ahead + 2.5 h to reach B = \(3.5h\)

Let me know if it's clear.

The concept being used here is relative speed.

Time taken by A = 28/(58 - 50) = 3.5 hrs
We subtract 50 because of what you said - B is moving too and B's speed is 50. The concept will be clear once you through the post.
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Let t be time car A needs to overtake car B and drive 8 miles ahead. Then:

58t - 20 = 50t + 8

t=28/8=3.5 hours

Hope this helps

Posted from my mobile device

Hi Karishma,
I read your document at below:

veritasprep/blog/2012/07/quarter-wit-quarter-wisdom-speeding-relatively/

Could you please share the next week doc which is mentioned there for some tougher relative speed questions?

"Next week, we will look at some tougher relative speed questions"

Check out the blog posts on the link given in my signature below.
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Hi Bunuel,

Can you please guide me to the location of additional relative rate problems?

Thanks!

I am sure Bunuel will give you a dozen links for practice questions but meanwhile, here are some of my posts with relative speed practice questions:

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2012/07 ... elatively/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2012/08 ... -speeding/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2012/08 ... -concepts/
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We can classify this problem as a “catch up and pass” rate question. This means that one car is catching up to the other car and passing it by some distance.

Since there is a change in rate as well as a change in distance between the two cars, we can use the formula:

time = (change in distance)/(change in rate)

We are given that car A is 20 miles behind car B and we need to determine the time when car A is 8 miles ahead of car B. Thus, we can say that the change in distance is 20 + 8 = 28 miles.

We are also given that car A travels at a constant speed of 58 mph and car B travels a constant speed of 50 miles per hour. Thus, we can say that the change in rate is 58 – 50 = 8 mph.

We can plug this information into our equation:

time = (change in distance)/(change in rate)

time = 28/8 = 7/2 = 3.5 hours

The answer is E.
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Hope this illustration helps

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If there is an initial difference between the 2 people, driving in the same direction, the time take for them to meet = initial difference in distance \ (speed of a-speed of b). But if want to know when A will overtake B by x miles, we can add that difference in the distance to the initial difference in the formula.

=> (initial difference in distance + miles to overtake)/(speed of a - speed of b)
=> (20+8)/(58-50)
=> 28/8
=> 3.5hr.

Need to know if the approach is right. got 20/8= 2.5 hrs (time taken to overtake B) and then for to A proceed for another 8 miles ahead of B I used: 2.5/20= x/8 which results in 1 hr. so, 2.5 +1 = 3.5 hr. does this make sense?

Dear ueh55406
your approach is interesting but correct
Just my five cents, if you already took the relative speed and distance\( \frac{20}{8}\) you can mere add additional 8 miles to get the total time= \( \frac{20+8}{8}\) = 3.5h. Because the time taken by cars A and B will be equal; the difference is only in miles.

Here is an illustration of what's happening:

As you can see, car B drives from line X to line Y, whereas car A drives from line W to line Z, which means car A drives the same distance that car B drives PLUS an additional 28 miles.

APPROACH #1: Start with a word equation
The question tells us that Car A's total travel distance is 28 miles greater than Car B's total travel distance.
So we can write: (Car A's travel distance) - (Car B's travel distance) = 28 miles

Let t = Car A's travel time, which means t = Car B's travel also.
Distance = (rate)(time)

Plug our values into the word equation to get: 58t - 50t = 28
Subtract 50t from both sides: 8t = 28
Solve: t = 28/8 = 7/2 = 3.5
Answer: E

APPROACH #2: Answer the question in two parts
58 mph - 50 mph = 8 mph.
So, the original 20 mile gap between the two cars shrinks at a rate of 8 mph.
Time to reduce the gap to zero = distance/rate = 20/8 = 2.5 hours.
Time to increase the gap from 0 to 8 miles = distance/rate = 8/8 = 1 hour.
So, the total time for car A go from being 20 miles behind to being 8 miles ahead = 2.5 hours + 1 hour = 3.5 hours
Answer: E
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