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Word Problems

by akash singhal » Wed Oct 14, 2015 9:05 pm
A boat traveled upstream a distance of 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 half an hour longer than the trip downstream, 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

OE A
Well my doubt is when I tried to solve both equation making time constraint equal it was not successful but when i made distance constraint equal I got the answer.Why it didnt work out with time?

ALso my approach with distance constraint was very long(usual formula based).Can any1 advice some smaller process??
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by Brent@GMATPrepNow » Wed Oct 14, 2015 9:45 pm
A boat travelled upstream a distance of 90 miles at an average speed of (v-3) miles per hour and then travelled downstream at an average speed of (V+3) miles per hour. If the trip upstream took half an hour longer than the trip downstream, 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
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

Cheers,
Brent
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by Matt@VeritasPrep » Wed Oct 14, 2015 10:38 pm
The easiest way is probably just trying the answers.

Since 2.5 is the friendliest answer, let's start there. (If 2.0 was included, that would've been even friendlier, but we got lucky here.)

if t = 2.5, then we have

Distance downstream = Rate downstream * 2.5
90 = 2.5r
36 = r

Now let's check this to see if it works going upstream. We know that we're going 6 mph SLOWER upstream, so that r = 30.

Distance upstream = 30 * Time upstream
90 = 30*(Time up)
3 = Time up

Success! Time upstream needed to be .5 greater than 2.5, and this answer worked, so we must pick A.
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by Matt@VeritasPrep » Wed Oct 14, 2015 10:41 pm
A more algebraic approach would be to use ratios.

We know that Distance Down = Distance Up, so Rate Down * Time Down = Rate Up * Time Up.

This gives us (v + 3) * t = (v - 3) * (t + .5), or

(v + 3)/(v - 3) = (t + .5)/t, or

1 + 6/(v - 3) = 1 + .5/t, or

6/(v - 3) = .5/t, or

6t = .5(v - 3), or

12t = (v - 3)

Now we know that the Rate Upstream (v - 3) is equal to 12t, so ...

Distance Up = Rate Up * Time Up
90 = 12t * (t + .5)

and t = 2.5.
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by [email protected] » Thu Oct 15, 2015 9:04 am
Hi akash singhal,

This is a layered story-problem and takes a lot of effort to solve using a traditional "math approach" (Brent notes it in his explanation: "lots of work here"). Here's how you can solve it with a bit of logic and TESTing THE ANSWERS:

From the prompt, we can create 2 equations:

D = R x T

90 = (V-3)(T + 1/2)
90 = (V+3)(T)

We're asked for the value of T.

From the prompt, I find it interesting that the distance is a nice, round number (90).... because when looking at the answer choices, most of them are NOT nice decimals. When multiplying two values together (as we do in BOTH equations), if you end up with a round number, chances are that either....

1) both numbers are round numbers
2) one of the numbers includess a nice fraction (e.g. 1/2) which can be multiplied and the end result will be a round number.

This gets me thinking that 2.5 is probably the answer, but I still have to prove it....I'm going to plug in THAT value for T and see what happens to the 2 equations....

90 = (V-3)(3)
90 = (V+3)(2.5)

30 = (V-3)
36 = (V+3)

33 = V
33 = V

Notice how both values of V are THE SAME? That means that we have the solution. V=33 and T=2.5

Final Answer: A

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