# Help with GMAT Math Question

July 23, 2006 1:38 PM Subscribe

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. It the trip upstream took half an hour longer than the trip downstream, how many hours did it take the boat to travel downstream?

This is a practice GMAT problem. I am kicking myself for not being able to solve it, but I must be overlooking something obvious.

This is a practice GMAT problem. I am kicking myself for not being able to solve it, but I must be overlooking something obvious.

Best answer: Distance = Rate * Time, so...

90 = (v-3) * (t+1/2) - since it took half an hour longer to go upstream, and 90 = (v+3) * t.

Expand these out:

90 = vt - 3t + v/2 - 3/2

90 = vt + 3t

Subtract the first from the second:

0 = 6t - v/2 + 3/2

Solve for t in terms of v:

6t = v/2 - 3/2

t = v/12 - 1/4

Substitute into the second equation so we can solve for t:

90 = vt+3t = v(v/12-1/4)+3(v/12-1/4)

= v^2/12-v/4+v/4-3/4

= v^2/12 - 3/4

v = sqrt((90+3/4)*12)

= sqrt(363/4*12)

= sqrt(1089) = 33

And since t = v/12-1/4:

t = 33/12-1/4 = 33/12-3/12 = 30/12 = 5/2. It took 2 1/2 hours downstream and 3 hours upstream.

posted by wanderingmind at 1:58 PM on July 23, 2006

90 = (v-3) * (t+1/2) - since it took half an hour longer to go upstream, and 90 = (v+3) * t.

Expand these out:

90 = vt - 3t + v/2 - 3/2

90 = vt + 3t

Subtract the first from the second:

0 = 6t - v/2 + 3/2

Solve for t in terms of v:

6t = v/2 - 3/2

t = v/12 - 1/4

Substitute into the second equation so we can solve for t:

90 = vt+3t = v(v/12-1/4)+3(v/12-1/4)

= v^2/12-v/4+v/4-3/4

= v^2/12 - 3/4

v = sqrt((90+3/4)*12)

= sqrt(363/4*12)

= sqrt(1089) = 33

And since t = v/12-1/4:

t = 33/12-1/4 = 33/12-3/12 = 30/12 = 5/2. It took 2 1/2 hours downstream and 3 hours upstream.

posted by wanderingmind at 1:58 PM on July 23, 2006

Response by poster: All right, I'm almost there. I was setting the two equations equal to each other instead of getting rid of on of the variables by substitution.

After I've substituted one into the other, is there a way around having to solve a nasty quadratic?

posted by VillageLion at 2:00 PM on July 23, 2006

After I've substituted one into the other, is there a way around having to solve a nasty quadratic?

posted by VillageLion at 2:00 PM on July 23, 2006

Response by poster: Nevermind, wanderingmind did it for me, thanks!

posted by VillageLion at 2:01 PM on July 23, 2006

posted by VillageLion at 2:01 PM on July 23, 2006

This is a slightly different substitution with a little help on the factoring to find the solutions to the quadratic (I don't know if you will be able to take the square root of 1098 on the exam).

Starting with the two equations from aeighty where t is the time to go up the river, and v is the velocity of the boat:

90 = t * (v - 3)

90 = (t - 0.5) * (v + 3)

Solve for v with the first equation:

v = (90/t) + 3

Plug this into the second equation:

90 = (t - 0.5) * [(90/t) +3 +3]

Simplify the second expression and multiply the binomials:

90 = 90 + 6t - (45/t) -3

Subtract 90 from each side:

0 = 6t - 3 - (45/t)

Multiply each term by t to get rid of that pesky t in the denominator in the third term:

0 = 6t^2 -3t - 45

Divide all terms by three to make your life easier:

0 = 2t^2 - t - 15

To factor this quadratic, multiply the first and third term:

2t^2 * (- 15) = -30t^2

Examine factors of -30t^2 to find pair of factors that add up to the middle term, -t:

factors.... sum

1t - 30t = -29t [not it]

2t - 15t = -13t [not it]

3t - 10t = -7t [not it]

5t - 6t = -t [yes!]

5t and -6t are the pair that allow us to easily factor the quadratic. Re-write the quadratic but replace -t with -6t and 5t (this is okay because -6t + 5t = -t). This allows us to factor using grouping:

0 = 2t^2 - 6t + 5t - 15

Treat this as two group and factor separately:

0 = 2t (t - 3) + 5 (t - 3)

There is now a common factor (t - 3) that can be factored out, leaving:

0 = (2t + 5) (t - 3)

Solve for t; t = 3 and - 5/2. Disregard the negative solution. As everyone else has already said, the time required for the upstream trip is 3 hours, thus the downstream trip is 2.5 hours.

posted by peeedro at 5:01 PM on July 23, 2006

Starting with the two equations from aeighty where t is the time to go up the river, and v is the velocity of the boat:

90 = t * (v - 3)

90 = (t - 0.5) * (v + 3)

Solve for v with the first equation:

v = (90/t) + 3

Plug this into the second equation:

90 = (t - 0.5) * [(90/t) +3 +3]

Simplify the second expression and multiply the binomials:

90 = 90 + 6t - (45/t) -3

Subtract 90 from each side:

0 = 6t - 3 - (45/t)

Multiply each term by t to get rid of that pesky t in the denominator in the third term:

0 = 6t^2 -3t - 45

Divide all terms by three to make your life easier:

0 = 2t^2 - t - 15

To factor this quadratic, multiply the first and third term:

2t^2 * (- 15) = -30t^2

Examine factors of -30t^2 to find pair of factors that add up to the middle term, -t:

factors.... sum

1t - 30t = -29t [not it]

2t - 15t = -13t [not it]

3t - 10t = -7t [not it]

5t - 6t = -t [yes!]

5t and -6t are the pair that allow us to easily factor the quadratic. Re-write the quadratic but replace -t with -6t and 5t (this is okay because -6t + 5t = -t). This allows us to factor using grouping:

0 = 2t^2 - 6t + 5t - 15

Treat this as two group and factor separately:

0 = 2t (t - 3) + 5 (t - 3)

There is now a common factor (t - 3) that can be factored out, leaving:

0 = (2t + 5) (t - 3)

Solve for t; t = 3 and - 5/2. Disregard the negative solution. As everyone else has already said, the time required for the upstream trip is 3 hours, thus the downstream trip is 2.5 hours.

posted by peeedro at 5:01 PM on July 23, 2006

*Disregard the negative solution.*

Or don't. It's the other half of the solution ... if time had run backwards!

posted by gleuschk at 6:48 PM on July 23, 2006

This thread is closed to new comments.

Start with the basics, you know that distance (d) is time of travel (t) multiplied by velocity (v): d = tv

For going up the river then we have the following:

90 = t * (v - 3)

For going down the river then we have the following:

90 = (t - 0.5) * (v + 3)

We don't have to introduce a third unknown for the time taken to travel down the river because we know that "the trip upstream took half an hour longer than the trip downstream". So if upstream time was "t", than downstream time is "t - 0.5".

Now you have two linear equations, with two unknowns. Solve for one unknown in one equation, substitute it into the other one, and Bob's your uncle.

posted by aeighty at 1:49 PM on July 23, 2006