How much time did it take a certain car to travel 400

[GMAT 5/18]

Source: Gmat Prep, mba.com, Test II

Hi, I was hoping to get some feedback on the method I used to solve this problem. Please let me know if you think my method is good and/or if there is a better (faster) way to solve. Thanks!

How much time did it take a certain car to travel 400 kilometers?

(1) The car travelled the first 200 km in 2.5 hrs.
(2) If the car's average speed had been 20 km/h greater than it was, it would have travelled the 400 km in 1 hr less than it did.

My method:

Step 1: Noticed Statement I was clearly easier, I started there and deemed it insufficient, so I crossed off AD, leaving BCE.
Step 2: Used statement II to try to see if it was sufficient (this is where I would like feedback)

rt = 400 (from question)
r= 400/t

(r+20)(t-1) = 400 (from statement II)
rt - r +20t - 20 = 400
rt - r +20t = 420
r(t-1) + 20t = 420

Input r = 400/t (from question) for r above:

400/t (t-1) + 20t = 420

400t^2 - 400/t + 20t = 420

This is where I stopped and felt that as I could solve for t, Statement II was sufficient. Therefore, I chose B.

Please let me know if you see any flaws in my method or if you have a faster/better method. Thanks!

[StaceyKoprince]

Your reasoning is fine but there is a flaw in your final equation (maybe it's just a typo). You went from:
400/t (t-1) + 20t = 420

400t^2 - 400/t + 20t = 420

But that second eqn should be
400 - 400/t + 20t = 420
To solve the rest (in case anyone cares) then you'd move 420 over to get
-20 - 400/t + 20t = 0 then multiply each term by t
-20t - 400 + 20t^2 = 0 divide everything by 20
t^2 - t - 20 = 0
(t-5)(t+4) = 0
t = 5 or -4. Can't have negative time, so one solution.

You could also notice that the problem seems to be dealing with nice numbers so maybe you could just play with some numbers and see what you come up with - but only do that on this one if either (a) you're REALLY good at just "seeing" what numbers would make a problem work or (b) you can't figure out the algebra so you try some numbers as a fallback to see if you get lucky.

[victorgsiu]

Did this for to save a couple of steps:

(1) RT=400
(2) R=400/T

(R+20)(T-1) = 400
(R+20)(T-1) = RT - Plug in (1)

RT-R+20T-20 = RT
-R+20T-20 =0 - Plug in (2)

-400/T+20T-20=0

[RonPurewal]

the big waste of time here is the use of two variables.

i still haven't EVER seen a gmat rate problem that requires 2 variables; this one is no exception. you should use the relationships in the problem to write everything with one variable.

i'm not saying that two-variable solutions won't work - they will - but they are a profligate waste of time.

use "r" for the ACTUAL speed; then 400/r is the time taken to travel 400 km.
then "r + 20" is the HYPOTHETICAL speed, and so 400/(r + 20) is the time taken to travel 400 km at that speed.

the problem tells us that this hypothetical time is 1 hour less, so
400/(r + 20) = 400/r - 1

go from there. you'll still get a quadratic, but you don't have to waste your time dealing with a second variable.

[pritesh.suvarna]

Ron/Stacey - why shouldnt (1) work here?

D = 400

1st 200 miles took 2.5 hrs

D1 = R1 * T1
200/2.5 = 80 = R1

D2 = R2 * T2
200/R2 = T2 or 200/T2 = R2

T1+T2 = D1+D2/R1+R2

2.5 + T2 = 400/80+(200/T2).

We find T2, So total time take is T1+T2

[Ben Ku]

T1+T2 = D1+D2/R1+R2

2.5 + T2 = 400/80+(200/T2).

We find T2, So total time take is T1+T2

I'm unsure how you went from one step to the next above. Basically, in the first equation,
D1 = 200
D2 = 200
R1 = 80
T1 = 2.5

You can't find out T2 because you don't have R2.

The first statement is insufficient because we only know about the first half the trip; we have no information about the second half.