The time it took car A to travel 400 miles was 2 hours less than the

[BTGModeratorVI]

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

Answer: C
Source: GMAT paper tests

[Brent@GMATPrepNow]

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

Answer: C
Source: GMAT paper tests

Car A's average speed was 10 miles per hour greater than that of car B
Let x = Car B's average speed
So, x + 10 = Car A's average speed

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
In other words: (car A's travel time) = (car B's travel time) - 2

time = distance/speed
So, we get: 400/(x + 10)) = 400/x - 2

NOTE: At this point, we can either test the five answer choices to see which one satisfies the above equation, or we can solve the equation algebraically.
Although testing the answer choices is probably faster, let's solve the equation algebraically
Multiply both sides of the equation by (x + 10) to get: 400 = 400(x + 10)/x - (x + 10)(2)
Multiply both sides of the equation by x to get: 400x = 400(x + 10) - (x)(x + 10)(2)
Expand: 400x = 400x + 4000 - 2x² - 20x
Subtract 400x from both sides to get: 0 = 4000 - 2x² - 20x
Multiply both sides by -1 to get: 0 = -4000 + 2x² + 20x
Rewrite as follows: 2x² + 20x - 4000 = 0
Divide both sides by 2 to get: x² + 10x - 2000 = 0
Factor: (x + 50)(x - 40) = 0
So, EITHER x = -50 OR x = 40
Since the speed can't be negative, it must be the case that x = 40

Answer: C

Cheers,
Brent

[Scott@TargetTestPrep]

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

Answer: C

Solution:

We are given that cars A and B both traveled 400 miles and that car A’s average speed was 10 mph greater than that of car B. We can let the rate of car B = r and the rate of car A = r + 10.

Since both cars traveled 400 miles and time = distance/rate, the time of car A is 400/(r+10).

We are also given that it took car A 2 hours less than it took car B to travel the 400 miles. We can set up the following equation:

400/(r+10) + 2 = 400/r

Multiplying the entire equation by r(r+10), we have:

400r + 2(r)(r+10) = 400(r+10)

400r + 2r^2 + 20r = 400r + 4000

2r^2 + 20r - 4000 = 0

r^2 + 10r - 2000 = 0

(r + 50)(r - 40) = 0

r = -50 or r = 40

Since r must be positive, then r = 40.

Answer: C