neoreaves wrote:A crew can row a certain course up stream in 84 minutes; they can row the same course down stream in 9 minutes less than they can row in still water. How long would they take to row down with the stream.

a) 45 minutes
b) 63 minutes
c) 60 minutes
d) 19 minutes
e) 12 minutes


I know how to form the algebraic equations for this. My question is : Is there a faster way to solve this ?

This question is unfathomable algebraically. So there is no point in forming the algebraic equations for this. Only trick and that again in the presence of the flawlessly prejudiced answer choices, may probably work here.

If the crew can row the course up stream in 84 minutes, then the given distance is a certain multiple of 84. Hence, if the time taken to row down with the stream is an integer (as the choices suggest); then this time must also be a factor of 84. No choice other than [spoiler]E[/spoiler] is a factor of 84.

[spoiler]E[/spoiler]

based on this 75 can also be a multiple of d. But its still a multiple! doesnt give us anything ...Moreover, E is not the OA

neoreaves wrote:based on this 75 can also be a multiple of d. But its still a multiple! doesnt give us anything ...Moreover, E is not the OA

I didn't get you here. What is d, and how can 75 be a multiple of this? What's OA then, would you say it's [spoiler]B[/spoiler]? Do you believe that forming algebraic equations may solve this for you?

Let me try with equations Let the distance be D
stream speed - s and boat speed d

D/(d-s) =84
D/d- D/(d+s) = 9

We need to find out D/(d+s)

D= 84(d-s)
D = 9 (d+s)*d/s

9 (d+s)*d = 84(d-s)s
9(d^2+sd) = 84 (sd - s^2)

3d^2 - 25 sd + 28s^2 =0

(d-7s)(3d-4s) =0
Since d is greater than s
d=7s
d/s =7

D = 9 (d+s)*d/s
D= 9* (d+s)*7

[spoiler]D/(d+s) = 63, B ( I was not able to solve it non-algebraically, posting the solution for the benefit of others)[/spoiler]

ajith wrote:Let me try with equations Let the distance be D
stream speed - s and boat speed d

D/(d-s) =84
D/d- D/(d+s) = 9

We need to find out D/(d+s)

D= 84(d-s)
D = 9 (d+s)*d/s

9 (d+s)*d = 84(d-s)s
9(d^2+sd) = 84 (sd - s^2)

3d^2 - 25 sd + 28s^2 =0

(d-7s)(3d-4s) =0
Since d is greater than s
d=7s
d/s =7

D = 9 (d+s)*d/s
D= 9* (d+s)*7

[spoiler]D/(d+s) = 63, B ( I was not able to solve it non-algebraically, posting the solution for the benefit of others)[/spoiler]

If d > s, then 3 d = 4 s can also be taken into account, which proves E to be the answer then. Only one thing about the required time in the question can be said with certainty, and that is that it is less than 75 minutes. Algebra won't work here, I have told them.

Time Upstream = Tu = 84
Time Downstream = Td = ?

Effect of flow Upstream = Fu = ?
Effect of flow downstream = Fd = 9

Effect of flow is proportional to Time Taken, therefore:

(Tu/Fu) = (Td/Fd)

If no flow, it would take the same amount of time in both directions. Therefore:

Tu-Fu = Td+Fd

Substituting in all of the options gives b as the only possible answer.

Even this trial & error method took me a good few minutes - I'd hate to get this question!

yeahdisk wrote:Time Upstream = Tu = 84
Time Downstream = Td = ?

Effect of flow Upstream = Fu = ?
Effect of flow downstream = Fd = 9

Effect of flow is proportional to Time Taken, therefore:

(Tu/Fu) = (Td/Fd) __________(1)

If no flow, it would take the same amount of time in both directions. Therefore:

Tu-Fu = Td+Fu__________(2)

Substituting in all of the options gives b as the only possible answer.

Even this trial & error method took me a good few minutes - I'd hate to get this question!

On which hypothesis the bold (1) is based, I'm not sure. What pick-n-plug policy was used with the bold (2), only yeahdisk knows. I only wish yeahdisk to be sure there's no typo in his work.

1) Effect of flow must be proportional to the time taken. Pretty straight forward.

2) Straight forward conclusion.

...what pick & plug policy...

Take answer a) 45 as a possible for Td.

Plugging this into equation 1 (remember, we already know Tu and Fd) means Fu would have to equal 16.8. Decimals obviously won't work with equation 2.

So, try answer b) 63 as a possible for TD. Plugging into equation 1 gives us a Fu of 12.

Subbing into equation 2 gives 84 = 63 + 12 + 9. Which works.

The same method disproves c) d) and e).

If you have a quicker way please share

*edit* ok it doesn't actually disprove E, so using my method could only lead you to 2 possible answers. hmmm.

yeahdisk wrote:1) Effect of flow must be proportional to the time taken. Pretty straight forward.

2) Straight forward conclusion.

...what pick & plug policy...

Take answer a) 45 as a possible for Td.

Plugging this into equation 1 (remember, we already know Tu and Fd) means Fu would have to equal 16.8. Decimals obviously won't work with equation 2.

So, try answer b) 63 as a possible for TD. Plugging into equation 1 gives us a Fu of 12.

Subbing into equation 2 gives 84 = 63 + 12 + 9. Which works.

The same method disproves c) d) and e).

If you have a quicker way please share

I'm sorry for still not getting you so properly. If Fu = 12 and the bold (2) has no typo, then [spoiler]C[/spoiler] becomes the answer.

Let neoreaves share the OA first, then I would definitely hammer my mind out for you.

Plugging 60 into formula 1 gives us a Fu of 12.6

Plugging 60 and 12.6 into formula 2 doesn't compute.

I agree with you in seeing the OA though; I think my method would lead you to 2 answers:[spoiler] b) or e)[/spoiler]

yeahdisk wrote:Plugging 60 into formula 1 gives us a Fu of 12.6

Plugging 60 and 12.6 into formula 2 doesn't compute.

I agree with you in seeing the OA though; I think my method would lead you to 2 answers:[spoiler] b) or e)[/spoiler]

Please try to get what I'm insisting on

Tu-Fu = Td+Fu__________(2)

and that

Time Upstream = Tu = 84
Time Downstream = Td = ?

Effect of flow Upstream = Fu = 12 (if taken)
Effect of flow downstream = Fd = 9

now, the bold (2) becomes

84 - 12 = Td + 12

or Td = 60

You have one more chance to admit there's no typo in bold (2).

yes there's a typo.

good work.