PS: Boat upstream/downstream

Question

See attachment.. Can't seem to simplify this cleanly.

ALI1 · Answer

Let the time in hours be x

Downstream:

Distance = Speed*time
90=(v+3)*x

Upstream 

Time is 1/2 hours additional and speed is v-3

so 

90= (v-3)(x+1/2)

Comparing the equations as upstream and downstream distance is same
9v+3)*x= (v-3)(x+1/2)

Simplifying it will give x=5/2 or 2.5 hours

eastcoaster9 · Answer

Yes, got to this point but couldn't simplify down to 5/2. Will reexamine in the morning.

eastcoaster9 · Answer

Ok, after multiplying the terms on each side I get,

vx+3x=vx-3x+1/2v-3/2

The vx terms drop out giving you

6x=1/2v-3/2 then...?

richardj · Answer

It is easier with a change of variables.
Let u = v+3
So v-3 = u-6
Let t = downstream time.

So we have 90 = ut, t = 90/u, u = 90/t

Also from upstream information

(t + 1/2) = 90 / (u-6)
(t + 1/2)(u-6) = 90
ut + (1/2)u - 6t - 3 = 90

ut=90
so u/2 - 6t - 3 = 0
u - 12t - 6 = 0

u= 90/t
so 90/t - 12t - 6 = 0
90 - 12t^2 - 6t = 0
2t^2 - t + 15 = 0
(2t + 5)(t - 3) = 0
t = -2.5, t=3

t>0, so t = 3 hours

eastcoaster9 · Answer

You missed a sign somewhere.. I actually had done the v, v-6 as well. thanks, have 2.5 now.

richardj · Answer

You're quite right.

90 - 12t^2 - 6t = 0
2t^2 + t - 15 = 0
(2t - 5)(t + 3) = 0
t = +2.5, t= -3

t>0, so t = 2.5 hours

ranga41 · Answer

I get the same as most others 

2t^2+t-15 = 0

t = 2.5

richardj · Answer

I hate to recommend a non-algebraic approach but ...

This is actually a good time to use some common sense practical examples.

It is quite likely in a question like this that there will be some round numbers in the time, or numbers with lots of factors in the speed.

You might guess it would be say 20%-25% quicker to go downstream as upstream, so a first bash at 2 or 3 hours is not unreasonable.

Given also you have v+3, v-3 speeds (or u, u+6) it would not be a surprise if the speed was a multiple of 6mph like 24,30 or 36 - (reasonably consistent with the times above.)

The actual times and speeds of 2.5, 3 hours and 30, 24 mph are guessable reasonably quickly by trial and error. Probably quicker and more accurate than solving a quadratic equation.

FN · Answer

bumping this thread...is there a neater way of doing this...

I got this on my GMATPREP CAT and couldnt do this in 2-3 minutes...is there a short cut for such problems?

HongHu · Answer

Ok this may not be the easist but is straight forward in logic:

90/(v-3)-90/(v+3)=0.5
90(v+3-(v-3))=0.5(v^2-9)
12*90=v^2-9
121*9=v^2
v=11*3

Now we need to know 90/(v+3) = 90/(12*3)=10/4=2.5

Notice that I haven't multiplied some of the numbers out because it would be easier if you could cancel some terms out.

PS: Boat upstream/downstream

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