The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car A’s average speed was 10 miles per hour greater than that of car B, what was car B’s average speed, in miles per hour?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 80
GMAT Forum
4 postsPage 1 of 1
- Guest
The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car A’s average speed was 10 miles per hour greater than that of car B, what was car B’s average speed, in miles per hour?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 80
time taken by Car B = x hrs
time taken by car A = x-2 hrs
speed of car A = 400/(x-2)
speed of car B = 400/x
given speedA- speedB = 10mph
so 400/(x-2) - 400/x = 10
400x- 400x + 800 = 10x(x-2)
800 = 10x^2 - 20x
x^2 - 2x - 80 = 0
(x+8)(x-10) = 0
clearly x = 10
so the answer is 400/10 = 40 (C)
Is that the OA?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 80
time taken by Car B = x hrs
time taken by car A = x-2 hrs
speed of car A = 400/(x-2)
speed of car B = 400/x
given speedA- speedB = 10mph
so 400/(x-2) - 400/x = 10
400x- 400x + 800 = 10x(x-2)
800 = 10x^2 - 20x
x^2 - 2x - 80 = 0
(x+8)(x-10) = 0
clearly x = 10
so the answer is 400/10 = 40 (C)
Is that the OA?
- RonPurewal
- Students
- Posts: 19744
- Joined: Tue Aug 14, 2007 8:23 am
in this problem, you have a choice between designating a variable to represent time (as done by the poster above) and designating a variable to represent speed. either will work.
just for the sake of completeness, here's a solution in which the variable represents speed instead of distance:
car A speed = x + 10
car B speed = x
now, rt = d. therefore, t = d/r.
therefore,
car A time = 400 / (x + 10)
car B time = 400 / x
per the problem statement, car B took two hours longer to travel the distance than did car A.
therefore,
400/(x + 10) + 2 = 400 / x
at this point, you could solve this equation with quadratic techniques and factoring, as in the previous example.
but this approach is slightly superior to that one, in that you can solve at this point by just plugging the answer choices into the equation.
you can do so in this case because the answer choices represent the average speed of car B, the quantity for which variables were chosen in the first place.
this is faster - not to mention much easier - than quadratic techniques (unless you're lightning fast at quadratics).
plugging in choice (c), the correct answer, yields
400/50 + 2 = 400/40
8 + 2 = 10
since this is a true statement, (c) is the correct answer.
--
moral of the story:
on problems such as this one - on which there is absolutely no difference, algebra-wise, between the different options for a variable (here, you're going to wind up with fractions --> quadratic either way), you should try to pick a variable for which you can plug in the answer choices if you get stuck.
for this problem, the answer choices represent speeds. therefore, if you're algebraically indifferent between speeds and times, you should use your variable to represent speed, so that you can plug in if you come to an impasse and/or don't feel like solving a quadratic with fractions.
just for the sake of completeness, here's a solution in which the variable represents speed instead of distance:
car A speed = x + 10
car B speed = x
now, rt = d. therefore, t = d/r.
therefore,
car A time = 400 / (x + 10)
car B time = 400 / x
per the problem statement, car B took two hours longer to travel the distance than did car A.
therefore,
400/(x + 10) + 2 = 400 / x
at this point, you could solve this equation with quadratic techniques and factoring, as in the previous example.
but this approach is slightly superior to that one, in that you can solve at this point by just plugging the answer choices into the equation.
you can do so in this case because the answer choices represent the average speed of car B, the quantity for which variables were chosen in the first place.
this is faster - not to mention much easier - than quadratic techniques (unless you're lightning fast at quadratics).
plugging in choice (c), the correct answer, yields
400/50 + 2 = 400/40
8 + 2 = 10
since this is a true statement, (c) is the correct answer.
--
moral of the story:
on problems such as this one - on which there is absolutely no difference, algebra-wise, between the different options for a variable (here, you're going to wind up with fractions --> quadratic either way), you should try to pick a variable for which you can plug in the answer choices if you get stuck.
for this problem, the answer choices represent speeds. therefore, if you're algebraically indifferent between speeds and times, you should use your variable to represent speed, so that you can plug in if you come to an impasse and/or don't feel like solving a quadratic with fractions.
4 posts Page 1 of 1