Carl drove from his home to the beach at an average speed of 80 kilometers per hour and returned home by the same route at an average speed of 70 kilometers per hour. If the trip home took hour longer than the trip to the beach, how many kilometers did Carl drive each way?
(A) 350
(B) 345
(C) 320
(D) 280
(E) 240
I assumed the distance to be constant as "d" The time for the first journey is "x" so the time for the second jounery is 3x/2. I got stuck here. Is there another method whereby both the average speeds are used to find the average speed for the journey as a whole?
Speed distance and time
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You're missing a key piece of information above. Is it supposed to be 1/2 hour?RiyaR wrote:Carl drove from his home to the beach at an average speed of 80 kilometers per hour and returned home by the same route at an average speed of 70 kilometers per hour. If the trip home took hour longer than the trip to the beach, how many kilometers did Carl drive each way?
(A) 350
(B) 345
(C) 320
(D) 280
(E) 240
I assumed the distance to be constant as "d" The time for the first journey is "x" so the time for the second jounery is 3x/2. I got stuck here. Is there another method whereby both the average speeds are used to find the average speed for the journey as a whole?
Cheers,
Brent
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Based on the answer choices, I'm assuming the question SHOULD be...
One option is to TEST the answer choices.
We want time to travel at 70 kmh to be 1/2 hour MORE than time to travel at 80 kmh
time = distance/speed
(A) 350
time traveling at 70kmh = 350/70 = 5 hours
time traveling at 80kmh = 350/80 = 35/8 = 4 1/8 hours
Here, the times do NOT differ by 1/2 hour
ELIMINATE A
(B) 345
time traveling at 70kmh = 345/70 STOP. This will be an UGLY fraction, as will the one below.
time traveling at 80kmh = 345/80
Since both fractions are so ugly, it's unlikely that the times will differ by something as "pretty" as 1/2 hour
SKIP B for now.
NOTE: At this point, I might scan the answer choices to see if any of them will work NICELY with 70kmh and 80kmh. I spot 280, which seems to work well with both speeds. So, let's check that next..
(D) 280
time traveling at 70kmh = 280/70 = 4 hours
time traveling at 80kmh = 280/80 = 28/8 = 7/2 = 3 1/2 hours
Here, the times DO differ by 1/2 hour
Answer: C
Cheers,
Brent
RiyaR wrote:Carl drove from his home to the beach at an average speed of 80 kilometers per hour and returned home by the same route at an average speed of 70 kilometers per hour. If the trip home took 1/2 hour longer than the trip to the beach, how many kilometers did Carl drive each way?
(A) 350
(B) 345
(C) 320
(D) 280
(E) 240
One option is to TEST the answer choices.
We want time to travel at 70 kmh to be 1/2 hour MORE than time to travel at 80 kmh
time = distance/speed
(A) 350
time traveling at 70kmh = 350/70 = 5 hours
time traveling at 80kmh = 350/80 = 35/8 = 4 1/8 hours
Here, the times do NOT differ by 1/2 hour
ELIMINATE A
(B) 345
time traveling at 70kmh = 345/70 STOP. This will be an UGLY fraction, as will the one below.
time traveling at 80kmh = 345/80
Since both fractions are so ugly, it's unlikely that the times will differ by something as "pretty" as 1/2 hour
SKIP B for now.
NOTE: At this point, I might scan the answer choices to see if any of them will work NICELY with 70kmh and 80kmh. I spot 280, which seems to work well with both speeds. So, let's check that next..
(D) 280
time traveling at 70kmh = 280/70 = 4 hours
time traveling at 80kmh = 280/80 = 28/8 = 7/2 = 3 1/2 hours
Here, the times DO differ by 1/2 hour
Answer: C
Cheers,
Brent
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We can also start with a "word equation"Carl drove from his home to the beach at an average speed of 80 kilometers per hour and returned home by the same route at an average speed of 70 kilometers per hour. If the trip home took 1/2 hour longer than the trip to the beach, how many kilometers did Carl drive each way?
(A) 350
(B) 345
(C) 320
(D) 280
(E) 240
We have (trip time at 70 kmh) = (trip time at 80 kmh) + 1/2
time = distance/speed, so if we let d = the distance each way, we get:
d/70 = d/80 + 1/2
Since the LCM of 70, 80 and 2 is 560, we'll multiply both sides by 560 to get..
8d = 7d + 280
Solve to get: [spoiler]d = 280[/spoiler]
Answer: D
Cheers,
Brent
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Alternate approach:RiyaR wrote:Carl drove from his home to the beach at an average speed of 80 kilometers per hour and returned home by the same route at an average speed of 70 kilometers per hour. If the trip home took 1/2 hour longer than the trip to the beach, how many kilometers did Carl drive each way?
(A) 350
(B) 345
(C) 320
(D) 280
(E) 240
RATE RATIO:
(rate to the beach) : (rate back home) = 80:70 = 8:7.
Since time and rate are RECIPROCALS, the time ratio is the reciprocal of the rate ratio.
TIME RATIO:
(time to the beach) : (time back home) = 7:8.
Implication:
If the trip to the beach takes 7 hours, then the trip back home takes 8 hours, for a difference of 1 hour.
Since the actual time difference is only 1/2 hour, the actual times must be 1/2 of 7 and 8:
Time to the beach = (1/2)(7) = 3.5 hours.
Time back home = (1/2)(8) = 4 hours.
Time difference = 4  3.5 = 0.5 hours.
Thus:
Distance between the beach and home = (rate back home)(time back home) = (70)(4) = 280 miles.
The correct answer is D.
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Hi RiyaR,
In your original post, you mentioned referring to the two "times" as X and 3X/2 but that is NOT correct.
The prompt states that the second time was 1/2 hour LONGER, which means that the two times should be:
X and (X + .5)
3X/2 would mean that the second time was 50% longer.
If you were looking for a "pure algebra" approach, you COULD set up a 3variable "system" of equations. It would take more work and time than TESTing THE ANSWERS, but here's what it would look like:
D = 80(T)
D = 70(V)
V = T + .5
From here, you can do a mix of substitution and combination to solve for D.
GMAT assassins aren't born, they're made,
Rich
In your original post, you mentioned referring to the two "times" as X and 3X/2 but that is NOT correct.
The prompt states that the second time was 1/2 hour LONGER, which means that the two times should be:
X and (X + .5)
3X/2 would mean that the second time was 50% longer.
If you were looking for a "pure algebra" approach, you COULD set up a 3variable "system" of equations. It would take more work and time than TESTing THE ANSWERS, but here's what it would look like:
D = 80(T)
D = 70(V)
V = T + .5
From here, you can do a mix of substitution and combination to solve for D.
GMAT assassins aren't born, they're made,
Rich
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Here's another approach that begins with a different word equation.Carl drove from his home to the beach at an average speed of 80 kilometers per hour and returned home by the same route at an average speed of 70 kilometers per hour. If the trip home took 1/2 hour longer than the trip to the beach, how many kilometers did Carl drive each way?
(A) 350
(B) 345
(C) 320
(D) 280
(E) 240
Distance traveled at 80 kmh = Distance traveled at 70 kmh
Distance = (speed)(time)
We know the speeds, but not the times.
So, let t = time spent driving at 80 kmh
This means that (t + 1/2) = time spent driving at 70 kmh [since that trip takes 1/2 hour longer to complete]
So, we get:
(80)(t) = (70)(t + 1/2)
Expand to get: 80t = 70t + 35
Subtract 70t from both sides to get: 10t = 35
Solve to get: t = 3.5
So, Carl spent 3.5 hours traveling at 80 kmh
So, the distance traveled = (speed)(time) = (80)(3.5) = 280
Answer: D
Cheers,
Brent

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If you want to use d = rt, here's how I'd do it.
Trip TO the beach:
Distance = d
Rate = 80
Time = t
Trip FROM the beach:
Distance = d
Rate = 70
Time = t + 1/2
Since d = 80t and d = 70(t + 1/2), we know that 80t = 70(t + 1/2), or t = 3.5
Hence Carl's trip to the beach had d = 80t = 80(3.5) = 280.
Trip TO the beach:
Distance = d
Rate = 80
Time = t
Trip FROM the beach:
Distance = d
Rate = 70
Time = t + 1/2
Since d = 80t and d = 70(t + 1/2), we know that 80t = 70(t + 1/2), or t = 3.5
Hence Carl's trip to the beach had d = 80t = 80(3.5) = 280.