A boat travels from point A to point B upstream and returns from point

GIVEN: Speed of stream = 10 kmh
Let b = the speed of the boat in STILL (current-less) water

So, speed while travelling UPSTREAM = b - 10
And speed while travelling DOWNSTREAM = b + 10

GIVEN: round trip takes the boat 5 hours and the distance between point A and point B is 120 kms
Let's start with a "word equation"

(travel time UPSTREAM) + (travel time DOWNSTREAM) = 5 hours
time = distance/ speed
So we can write: (120/(b - 10)) + (120/(b + 10)) = 5

Multiply both sides by (b - 10) to get: 120 + (120)(b - 10)/(b + 10) = 5(b - 10)
Multiply both sides by (b + 10) to get: 120(b + 10) + (120)(b - 10) = 5(b - 10)(b + 10)
Expands to get: 120b + 1200 + 120b - 1200 = 5b² - 500
Simplify to get: 240b = 5b² - 500
Rearrange to get: 5b² - 240b - 500 = 0
Divide both sides by 5 to get: b² - 48b - 100 = 0
Factor to get (b - 50)(b + 2) = 0

So, EITHER b = 50 OR b = -2
Since the speed cannot be negative, we know that b = 50 kmh

How long did the upstream journey take?
The boat's UPSTREAM speed = b - 10
= 50 - 10
= 40 kmh

time = distance/ speed
So, the boat's travel time upstream = 120/40 = 3 hours

Answer: B

Cheers,
Brent
_________________

Solution

• Let the time taken by boat to travel from point A to point B upstream be t hours and the speed of the boat in still water be s km/hour

o So, the time taken by boat to travel to point A from the point B, downstream \(= 5-t\) hours
• While traveling from point A to point B upstream, speed of the stream is opposing the speed of boat.

o Hence, the resultant speed of the boat = s -10 km/hour

 t = distance between A and B/ resultant speed of the boat \(= \frac{120}{s-10}\)
 Or, \(s -10 = \frac{120}{t }⟹ s = \frac{120}{t }+ 10 …………..Eq.(i)\)
• While traveling from point B to point A, downstream, speed of the stream is supporting the speed of boat.

o Hence, the resultant speed of the boat = s +10 km/hour

 5 -t = distance between B and A /resultant speed of the boat \(=\frac{120}{s+10}\)
 Now, substituting the value of s from Eq.(i) into the above equation, we get

• \(5 -t = \frac{120}{(120/t +10+10)}\)
• Or, \(5-t = \frac{6t }{(6+t)}\)
• Or, \((6+t) (5-t) = 6t\)
• Or, \(-t^2 -t +30 = 6t\)
• Or , \(t^2 +7t – 30 = 0\)
• Or, \((t+10)(t-3) = 0\)
• So, \(t = -10\) or \( 3\)

Since, time cannot be negative, hence t = 3 hours.
Thus, the correct answer is Option B.
_________________

Say speed of the boat is B.
Total time taken (Upstream Time + Downstream Time) = 5 hrs

\(\frac{120}{(B - 10)} + \frac{120}{(B + 10)} = 5\)

Upstream time will be more than downstream time. Try 3 + 2. If upstream time is 3, B = 50 which gives downstream time as 2. Works.
(In an equation like this, try to plug in.)

Answer (B)
_________________

Solution:

We let r = the speed of the boat in still water. Thus, the speed of the boat when it is going downstream is (r + 10), and its upstream speed is (r - 10). Since distance / rate = time, we can create the equation for the sum of the combined time of the downstream and upstream legs of the journey as:

120 / (r + 10) + 120 / (r - 10) = 5

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

120(r - 10) + 120(r + 10) = 5(r + 10)(r - 10)

120r - 120 + 120r + 120 = 5(r^2 - 100)

240r = 5r^2 - 500

r^2 - 48r - 100 = 0

(r - 50)(r + 2) = 0

r = 50 or r = -2

Since r can’t be negative, r = 50 km/hr. Since the speed of the boat going upstream is r - 10 = 50 - 10 = 40 km/hr, the time it took to go upstream is 120/40 = 3 hours.

Alternate Solution:

Let t be the time, in hours, the boat spends traveling upstream. Notice that if r is the speed of the boat in still water, the speed of the boat traveling upstream is r - 10 and the speed of the boat traveling downstream is r + 10; hence, the speed of the boat traveling downstream is r + 10 - (r - 10) = 20 km/h faster than the speed of the boat traveling upstream. In terms of t, the speed of the boat traveling downstream is 120/(5 - t), and the speed of the boat traveling upstream is 120/t; hence we can create the following equation:

120 / (5 - t) - 120 / t = 20

Multiplying the equation by t(5 - t), we have:

120t - 120(5 - t) = 20t(5 - t)

120t - 600 + 120t = 100t - 20t^2

20t^2 + 140t - 600 = 0

Dividing each side by 20, we obtain:

t^2 + 7t - 30 = 0

(t - 3)(t + 10) = 0

t = 3 or t = -10

Since t cannot be negative, the boat spends t = 3 hours traveling upstream.

Answer: B
_________________

When the boat travels downstream -- WITH the current -- the speed of the stream increases the travel rate by 10 kph.
When the boat travels upstream -- AGAINST the current -- the speed of the stream decreases the travel rate by 10 kph.
Implication:
The downstream rate is 20 kph greater than the upstream rate.

We can PLUG IN THE ANSWERS, which represent the time upstream.
When the correct answer is plugged in, the total travel time = 5 hours.

B: 3 hours
Since the time to travel 120 km upstream = 3 hours, the upstream rate \(= \frac{distance}{time} = \frac{120}{3} = 40\) kph.
Since the downstream rate is 20 kph greater than the upstream rate, the downstream rate \(= 40+20 = 60\) kph.
Time to travel 120 km downstream at a rate of 60 kph \(= \frac{distance}{rate} = \frac{120}{60} = 2\) hours.
Total travel time = (3 hours upstream) + (2 hours downstream) = 5 hours.
Success!


_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________