While on a straight road, car X and car Y

[avinash.sunnasy]

(DS Problem)

While on a straight road, car X and car Y are travelling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of car Y ?

(1) Car X is travelling at 50 miles per hour and car Y is travelling at 40 miles per hour.

(2) 3 minutes ago car X was 0.5 mile ahead of car Y.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH Statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

:------------------------------------------------------:

Hello all,

I've actually solved this problem correctly, but I believe it was a fluke, lol!

STEP 1:-

Simplify the question.

Rate Time Distance
X R1 T D+1
Y R2 T D

The question becomes what is R1(rate of car x)?

STEP 2:-

Analyse statement (1)

With the information provided the above chart becomes...

Rate Time Distance
X 50 T D+1
Y 40 T D

with an additional mile...

Rate Time Distance
X 50 T+X D+1
Y 40 T D

Therefore...

(EQ1) 50T = D+1

(EQ2) 50T+50X = D+2

(EQ2)-(EQ1) -> 50X=1 X= 1/50 1.2 minutes

STEP 3:-

Analyse statement (2)

In 3 mins car X travelled 0.5miles relative to car Y

3*20 = 60

0.5*20 = 10

Therefore Car X travels 10 mph faster than Car Y

So at 10mph how long will it take to travel a mile?

6 minutes

Since the answer are ultimately different I question my approach.

Some help/advice would be appreciated.

Regards,

A.

(PS Is there a way to change your username for this forum?)

[gokul_nair1984]

A much simpler way for Rate Time and Distance problems is to always consider relative speed ...In this case since the cars are travelling in the same direction the relative speed is 50-40= 10miles/hr(with respect to eachother)

Case 1:
S=10miles/hr(relative)
D(req)=2 miles but they are already 1 mile apart , thus resultant D=1 mile
Therefore, T=D/S=1/10hours or 6minutes

Case 2:
3min --> 0.5mile.
So 1 more mile in 6 minutes.


Thus Answer has to be D

[RonPurewal]

A much simpler way for Rate Time and Distance problems is to always consider relative speed ...

your solution above is correct, but i hope you didn't intend the word "always" literally. there will be all sorts of rate/time/distance problems that cannot be solved with the relative-rate approach -- in fact, the vast majority of them.

in other words, if you think that you can always use relative rates, you're incorrect.

if, on the other hand, you don't literally mean "always" -- i.e., you're just trying to say that relative rates are one of many useful techniques for solving rate problems -- then, yes, that's excellent advice.

[RonPurewal]

here's a non-relative-rate approach. remember, your goal is never to use more than 1 variable, unless absolutely necessary... and i've never seen a rate/time/distance problem that absolutely required the use of more than 1 variable.

While on a straight road, car X and car Y are travelling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of car Y ?

simplified:
after how long will car X have traveled exactly 1 mile more than car Y?

(1) Car X is travelling at 50 miles per hour and car Y is travelling at 40 miles per hour.

since both cars travel for exactly the same amount of time, we only need one variable.
time traveled by each car = t
--> distance traveled by car X = 50t
--> distance traveled by car Y = 40t

we want car X to travel exactly one more mile than car Y, so
50t = 40t + 1

SUFFICIENT (no reason to bother actually solving)

(2) 3 minutes ago car X was 0.5 mile ahead of car Y.

this tells us that car X gains half a mile on car Y every 3 minutes.
therefore, we will need six minutes.
SUFFICIENT

[gokul_nair1984]

Ron,

Ya... I did not mean the "Always" literally but I just suggested that on such questions using the relative speed approach is a better alternative.(Only on such questions)

[nehajadoo]

thanks for the clear and quick tip on this one. was flustered in my prac test and got it wrong

[RonPurewal]

glad it helped.

[rkafc81]

Hi

I'm trying to work out as many ways to do this one as possible, so whilst I understand the conceptual way of doing it, for the RTD method, is it ok to do it like this for statement (1) ? Also does it matter either way to assign a variable to T and derive the value of D vs. assign a variable for D and derive the value of T ?

SITUATION NOW
CAR --- R --- T -------- D
X --- 50 --- (d + 1)/50 -------- d + 1
Y --- 40 --- d/40 ------------ d


SITUATION FUTURE
CAR --- R --- T -------- D
X --- 50 --- (d + 2)/50 -------- d + 2
Y --- 40 --- d/40 ---- -------- d

... and then just do SITUATION_FUTURE.TIME - SITUATION_NOW.TIME = the answer ?

thanks!

[RonPurewal]

n2739178, can you explain your approach a little more? i'm a little bit confused by it, especially since your "present" and "future" cases seem to place the second car in exactly the same location.

also, i don't understand exactly what "SITUATION_FUTURE.TIME", etc. mean. if this is programming code (which it seems to resemble), then ... well, let's just say that computers and i are not on the best terms, lol.

[rkafc81]

hi Ron - I'm ok with this one now - thanks though!


yeah it was programming-type pseudo-code... lol shouldn't have done that, didn't even realise i did it!

[jlucero]

Glad you figured it out

[BernardK777]

Can you please explain how you would solve this problem using the RTD chart. When I see rates problem, I try to use the same approach (RTD chart) and couldn't figure out how to apply it to this problem. Thanks.

[tim]

Several posters have described an approach that uses an RTD chart; can you let us know what didn't work for you when you tried that approach? Perhaps it will help if you work through what you can and let us help you once you get stuck.