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The speed of a Motor - Boat is that of the current of water as 36:5.The boat goes along with the current in 5 hours 10 Minutes,It will Come Back in which time?

I have Tried: Speed of Motor : Speed of Water = 36:5

Relative downstream time along with the water is 5 hours 10 Minutes

speed of Downstream along with the water ,let the distance be x, let y be any parts in the Ratio

x/36y-5y = 5 1/6 hr x/31y = 5 1/6 hr

speed of upstream along with the water

x/41y = t

divide two equations we get:

41y/31y = 31/6t

then we get t,

by adding t+5 hours 10 Minutes

we will get answer?

Can anyone Please Explain how to solve the above sum?I ave little bit confusions in solving above guide Me for the answer,if there is any Shortcut please explain the Logic

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  • $\begingroup$ I sent you the mental answer by telepathy $\endgroup$ – user354674 Sep 10 '16 at 14:08
  • $\begingroup$ @igael i didnot get your answer $\endgroup$ – cloud computing in salesforce Sep 10 '16 at 14:09
  • $\begingroup$ Another Question also i am facing some problem,let us ask here i will post in another Question ah? $\endgroup$ – cloud computing in salesforce Sep 10 '16 at 14:13
  • $\begingroup$ after getting a first answer, a new one is always better $\endgroup$ – user354674 Sep 10 '16 at 14:22
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Let the speed of motor boat and water be 36x km/minute and 5x km/minute respectively and the distance travelled by boat is d km.

Now along with the current:

d=(36x+5x)*310 (applying formula DISTANCE = SPEED * TIME)

= 41x*310 km

and against the current:

41x*310=(36x-5x) * t

solving for t , we get t=410 minute i.e. 6 hours and 50 minutes.

if you have any doubt,feel free to ask in comments.

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If the speed of the motor is $36x$ unit, the speed of the current of water will be $5x$ unit.

So, while going along with the current, effective speed $=(36x+5x)$ and while going against effective speed $=(36x-5x)$

If the required time is $t$ minutes $$\dfrac{36x+5x}{5\text{ hout }10\text{ min}}=\dfrac{36x-5x}t$$

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  • $\begingroup$ what will be the final answer $\endgroup$ – cloud computing in salesforce Sep 10 '16 at 14:10
  • $\begingroup$ @user36188, Have you understood the method? $\endgroup$ – lab bhattacharjee Sep 10 '16 at 14:13
  • $\begingroup$ @lal bhatttacharjee i have understand your method,i think you are equatting the Distance on both sides it is Correct,route to the final step, what will be the answer $\endgroup$ – cloud computing in salesforce Sep 10 '16 at 14:15
  • $\begingroup$ @user36188, Nice to hear that.But I believe that this is not the best forum for calculation. $\endgroup$ – lab bhattacharjee Sep 10 '16 at 14:16
  • $\begingroup$ I have a another doubt shall you please provide your chat link $\endgroup$ – cloud computing in salesforce Sep 10 '16 at 14:17

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