A boat traveled upstream 90 miles at an average speed of (v-3) miles

BTGModeratorVI wrote: ↑

Fri Jun 05, 2020 11:47 am

A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?

A. 2.5
B. 2.4
C. 2.3
D. 2.2
E. 2.1

Answer: A
Source: GMAT prep

I like to begin with a "word equation."
We can write:
travel time upstream = travel time downstream + 1/2

Time = distance/rate
So, we can replace elements in our word equation to get:
90/(v-3) = 90/(v+3) + 1/2

Now solve for v (lots of work here)
.
.
.
v = 33

So, travel time downstream = 90/(v+3)
= 90/(33+3)
= 90/36
= 5/2
= 2 1/2 hours
Answer: A

Cheers,
Brent

BTGModeratorVI wrote: ↑

Fri Jun 05, 2020 11:47 am

A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?

A. 2.5
B. 2.4
C. 2.3
D. 2.2
E. 2.1

Answer: A
Source: GMAT prep

Since time = distance/rate, the time going upstream = 90/(v – 3) and the time going downstream = 90/(v + 3). Since the time going upstream is ½ hour more than the time going downstream, we add ½ hour to the downstream trip to make the two times equal, and then we can set up an equation as follows:

90/(v - 3) = 90/(v + 3) + 1/2

Let’s multiply the equation by 2(v - 3)(v + 3) to eliminate the denominators:

2(90)(v + 3) = 2(90)(v - 3) + (v - 3)(v + 3)

180v + 540 = 180v - 540 + v^2 - 9

540 = v^2 - 549

v^2 = 1089

v = √1089

v = 33

Since the time going downstream = 90 / (v + 3), and v = 33, the time going downstream = 90 / (33 + 3) = 90/36 = 5/2 = 2.5 hours.

Answer: A