Attached is a question from GMAT Prep Test #2.
What is the most efficient way of achieving the result?
Answer: B
Thanks,
K
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GMAT Prep #2_DS Rates, Distance #33
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Anurag@Gurome
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Let us assume that the average speed of the car = s km/hr, andkwah wrote:Attached is a question from GMAT Prep Test #2.
What is the most efficient way of achieving the result?
Answer: B
Thanks,
K
time taken to cover 400 km = t hours
Then distance = speed * time implies 400 = st or s = 400/t.
(1) The car traveled the first 200 km in 2.5 hours but we don't know the time taken by the car to travel the second 200 km, NOT sufficient.
(2) If the car's average speed had been 20km per hour greater than it was, it would have traveled the 400km in 1 hour less than it did.
(s + 20)(t - 1) = 400
st + 20t - s - 20 = 400, and st = 400
400 + 20t - s = 420
20t - s = 20
20t - 400/t = 20
20t² - 400 = 20t
20t² - 20t - 400 = 0
t² - t - 20 = 0
(t - 5)(t + 4) = 0
t = 5; SUFFICIENT.
The correct answer is B.
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We know that D=RT. From the stem, we know D=400, so our equation is 400=RT. To solve, we'll need some info about the rate.
S1) We have a partial rate, but it's not enough. We could have traveled the remaining 200 km at the same rate, a faster rate, or a slower rate, which means we can come up with different times for the entire trip. Insufficient.
S2) We can set up a new equation. Distance is the same, the rate is 20 km/h faster, and time is 1 hour less:
400=(r+20(t-1)
We now have 2 equations and 2 variables, so we can solve. Sufficient.
S1) We have a partial rate, but it's not enough. We could have traveled the remaining 200 km at the same rate, a faster rate, or a slower rate, which means we can come up with different times for the entire trip. Insufficient.
S2) We can set up a new equation. Distance is the same, the rate is 20 km/h faster, and time is 1 hour less:
400=(r+20(t-1)
We now have 2 equations and 2 variables, so we can solve. Sufficient.
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