[Aptitude] Boats and Streams made-easy using our STD-Table Method

Aptitude31 Comments

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  1. Case#1: Basics
  2. Case 2: Time same but Different distance covered in each case
  3. Case #3: unknown variables
  4. Previous Articles

I hope you’ve mastered the STD table method from earlier articles on Time and Work / Pipes and Cistern.
We can use that STD method even in boats and streams question as well.

But first some terminologies

Downstream
  • Boat is moving along with flow of river (B), so water stream (S) helps the boat to move faster.
  • It is same like “A and B work together”. So their speed will increase and we can do addition.
  • Hence Downstream speed = speed of Boat PLUS speed of stream (B+S)
Upstream
  • Boat is moving against the direction of river. It is same like “Pipe A can fill the tank in 2 hours while Pipe B can empty the tank in 1 hours”
  • In short they work against each other, hence final speed is decreases so we’ve to subtract. (B-S)
  • Upstream speed = Speed of Boat MINUS speed of stream (B-S)

Case#1: Basics
Question: A man can row Upstream @15kmph and downstream @21 kmph, what is the speed of water in river?
Now construct the usual STD table from the given data

B aloneS aloneDownstream (B+S)Upstream (B-S)
Speed??2115
Time
Distance

As we’ve seen in earlier Time and work problems, we can do addition and subtraction in the Speed cells directly.
B+S=21…..eq(1)
B-S=15……eq(2)
Add both equations
(B+S) PLUS (B-S) = 21 PLUS 15
2B=36
B=36/2
B=18 kmph
We already know that
B+S=21
SO IF B is 18 kmph then S = 21 minus 18 = 3 kmph.
Answer: Speed of water is 3 kmph.
[Alternatively: You can directly calculate speed of water by subtracting eq(1) from eq(2)]

Shortcut Method

Above “equations” only for showing you the concept.
Otherwise in the exam hall you can directly subtract column 4 from column 3, divide it by “2” you get speed of column 2.
then subtract column 2 from column 3 you get speed of column 1:See this Image

shortcut method

DONE! No lengthy calculation required. But mind it: In the STD table, We can do direct addition, subtraction only for the speed “cells” and not for time cells.
You can also verify the answer from the table.

B aloneS aloneDownstream (B+S)Upstream (B-S)
Speed1832115
Time
Distance

From the first two columns B minus S= 18 minus 3 = 15.
In the last column, you can see that Upstream (B-S) = 15.
So yes answer is correct.
Anyways this was quite easy no brainer. Time to make things a bit complicated with next question
Case 2: Time same but Different distance covered in each case
Q. A Man rows the boat downstream for 60 km and Upstream 36 km, taking 4 hours each time. What is the speed of this boat?
Fill up the STD table

B aloneS aloneDownstream (B+S)Upstream (B-S)
Speed??
Time44
Distance6036

Apply the STD formula on third column
Speed x Time = Distance
Speed x 4= 60
Speed = 60/4
Downstream speed =15 kmph
Same way calculate for fourth column, you’ll get upstream speed = 36/4=9kmph
Update the table

B aloneS aloneDownstream (B+S)Upstream (B-S)
Speed??159
Time44
Distance6036

Shortcut method revisited

Compare this with first case. We know the speeds of upstream and downstream, we can use the shortcut method.
shortcut method for  que2

For those still uncomfortable with shortcut method, just do it manually
B+S=15 ; from column 4
B-S=9 from column 3
add both equations
(B+S) PLUS (B-S) = 15 PLUS 9
2B=24
B==24/2 =12 kmph
Answer. Speed of boat is 12 km per hour.

Verify the answer

So far we’ve calculated that
B+S=15 and
B=12.
Hence S=15 minus B=15 minus 12 =3kmph
Update the table

B aloneS aloneDownstream (B+S)Upstream (B-S)
Speed123159
Time44
Distance6036

For the speed cells, Column 1 minus column 2 equals column 4. Hence answer is correct.

Case #3: unknown variables
A boat sails downstream from point A to B, which is 10 km away from A, and then returns to A. If actual speed of the boat in still water is 3kmph, and the total upstream and downstream journey takes 12 hours. What must be the actual speed of boat for the trip from A to B to take exactly 100 minutes.
Difficulty of a question doesn’t depend on the length of question paragraph. Above sum has no ‘dum’ in it, just like our PM.  This can be solved using the universal “STD” method.
Given in the problem:
Speed of boat in still water (B alone) =3kmph
Length of river =10km
We know that total time  taken for upstream+ downstream=12 hours.
Suppose upstream takes journey takes “t” hours, then downstream journey takes= (12-t) hours. Fill up the table

From A to BFrom B to A
B aloneS aloneDownstream (B+S)Upstream (B-S)
Speed3??3+s3-s
Time12-tt
Distance1010

Apply the STD formula for both upstream and downstream columns
Speed x time = distance
(3+s)*(12-t)=10 →(t-12)=10/(3+s)…eq(1)
(3-s)*(t)=10→t=10/(3-s)…eq(2)
The total time taken in upstream + downstream journey
(12-t)+ t=12
Substitute the values of (12-t) and (t) with the things from eq 1 and 2

See this image for calculation
calculation of boat equations
Therefore
S2=9-5=4

MIND IT: Square Roots

S2=4 that doesn’t mean s=2 only.
Because square of (-2) =(-2)*(-2)=(+4)
When you take square root of 4, it can be (+2) or (-2)
But Since speed of water is a positive value, we’ll use  s=(+2). But keep this thing in mind especially for ‘data-sufficiency’ problems.
Update the table

From A to BFrom B to A
B aloneS aloneDownstream (B+S)Upstream (B-S)
Speed323+2=53-2=1
Time12-tt
Distance1010

We are not concerned with finding time in this question but still for practice :find Upstream time
Speed xtime = distance
1 x t =10
t = 10/1=10 hours.
Upstream time is 10 hours.
Similarly downstream time is 2 hours.  (apply STD or use 12-t, answer is 2 hours)
Coming to the ultimate question
What must be the actual speed of boat for the trip from A to B to take exactly 100 minutes.
Rephrase:
We want to go downstream for 10 kilometers. Speed of river is 2 kmph. We want to cover this distance in exactly 100 minutes, how fast should we run this boat?
Make a new column
Assume that new speed of boat should be (Bn)

From A to BFrom B to ASpecial case: A to B
B aloneS aloneDownstream (B+S)Upstream (B-S)Downstream (Bn+S)
Speed3251(Bn+2)
Time2 hrs10 hrs100 minutes
Distance101010

MIND IT: all units must be in same format

To get correct answers in STD formula, everything must be in same format.
Either “kilometer-hour format OR metre-second format”
Let’s stick to hours in this case.
Convert 100 minutes into hours
60 minutes =1 hour
1 minute =1/60 hour
Multiply both sides with 100
100 minutes = (100/60) hours.
(In short: When you want to convert minutes into hours, just divide minutes by 60)
Apply STD formula on last column

Speed  x time = distance
(Bn+2) x 100/60=10
(Bn+2) =60×10/100
(Bn+2) =6
Bn=6-2
Bn=4 kmph
Final answer: if we wish to cover 10 km downstream in 100 minutes, we must run the boat at the speed of 4kmph.
For previous Articles on Aptitude, visit Mrunal.org/aptitude

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So far 31 Comments posted

  1. vijayManikanta

    Hi
    In this article in case#3 I think question should be like following.
    A boat sails downstream from point A to B, which is 10 km away from A, and then returns to A. If actual speed of the boat in still water is 3kmph, and total time taken for the trip from A to B and B to A is 12 hours . What must be the actual speed of boat for the trip from A to B to take exactly 100 minutes.

    B'coz for the question you have given the times will be 't' from A to B and 't+12" from B to A.
    If I was wrong clarify

  2. Mrunal ♣ मृणाल ♣ મૃણાલ

    yes question itself had mistake in wording.
    but it is rectified now. Thanks for commenting.

  3. amanz

    really , well defined………..

  4. Rajani

    Hi
    Can anybody help me out by solving this problem in the method which Mrunal has said?

    A man can row 40km upstream and 55km downstream in 13 hours also, he can row 30km upstream and 44km down stream in 10 hours. Find the speed of the man in still water and the speed of the current.

    plzzzzzzz help me out ASAP as i am preparing for NICL AO exam.

  5. Rajani

    Can anybody plzzzzzzzzzzzz help me out with this question ASAP………

  6. shine

    make the distance covered in upstream same . that is multiply the second case by 4/3 then the situation changes to 40 km in upstream and 44* 4/3 km in downstream in 10* 4/3 hours. now compare with the first .take the differences. we get 176/3-55 km in downstream in 40/3 – 13 hours . That is 11/3 km in 1/3 hours. in 1 hour 11 kms downstream. So time taken to cover 55 km is 5 hours . The remaining 8 hours 40 km in upstream. So the speed in upstream is 5 km/hr. So the speed of the current is (11-5)/2 =3 km/hr
    speed of the man is 8 km /hr

    1. Rajani

      Better be late than never…….. any wayz if not helpful for the exam, atleast I got this problem solved and i got the method of solving it. Thank u so much……

  7. shine

    Upstream downstream time
    40 55 13
    30 *4/3 44* 4/3 10*4/3 subtract first from the second

    0 11/3 1/3
    therefore speed in downstream is 11 km/hr. 55kms in 5hr and remaining 8 hour 40 km in upstream. Speed in up stream is 5km/hr

  8. raj chauhan

    graet job man !!

    1. Shine Raj

      Thank You

  9. Mayank

    when i clicked Case#1: Basics it says you have not permission even though i have made new a/c.

  10. Chanveer

    Sir , U could also use the following to make it further simpler. (The last part 10 Km in 100 minutes part)
    Since by the time we reach to this part ,we know the speed of stream is 2km/hr.

    Part II – Actual Speed of boad for the trip from A to B to take exactly 100 minutes.

    i)100 minutes (speed of water ) effect = (2/60) * 100 = 3.33 Km
    ii)Boat to cover in 100 minutes = 10 – 3.33 = 6.67 Km
    iii) Therefore , Speed of the boat should be = (6.67/100) * 60 ={ 4 Km/hr } Thats your FINAL Answer.

  11. Anshu Sharma

    Thanks sir

  12. Dev

    this question is indeed complicated but it has been made more complicated by solving it in a lengthy way which makes it look very difficult to grasp. This type of question takes lot of time .

  13. ANANT RANA

    plz help me to solve the question -At noon ship A starts from a point P towards a point Q and at 1.OOPM ship B starts from Q towards P.If the ship A is expected to complete the voyage in 6Hrs and ship B is moving at a speed 2/3rd of that of ship ,at what time are the two ships expected to meet one another.

  14. rahul raj

    Awesome …

  15. san

    is the answer 4 p.m ?

  16. ABHISHEK SINGH

    A man can row 6 km/hr in still water.When the river is running at 1.2 km/hr, it takes him 1 hour to row to a place and back.How far is the place?

    By STD formula, B=6
    S=1.2
    What to do next? Please someone explain.

  17. prasanta

    good technique. i like it

  18. ria

    a man can row three- quarters of a km against the upstream in 45/4 min.the speed in( km/ hr) of the man in still water is?

  19. ranjitha

    even I have not understand big questions y

  20. Bluff master

    A swimmer swims from a point A against a current for 5 minutes and then swims backwards in favour of the current for next 5 minutes and comes to the point B. If AB = 100 metres, the speed of the current (in km per hour) is ? How to solve this ques using STD table…

  21. Aspirant

    The images embedded with this page or any page in aptitude section isn’t visible. Even when we hit the link of image, its not there. Without the image, very difficult to follow up. Kindly address this problem

  22. doit

    image loading problem

  23. yudi

    photos are not accessible

  24. varsha

    Can somebody please explain me the difference between speed of stream and the rate of stream..i m really confused in these terms.

  25. varsha

    Can somebody plz explain me the difference between rate of stream n speed of stream?

    1. ASHOK

      I guess rate of stream means vol/time ,speed of the stream means with what velocity the stream is flowing i.e, dist/time .,.,.,

  26. Sh. Sanathoi

    Sir, the image is not showing.

  27. supriya

    how to solve it by using STD table . A boatman goes 2 km against the current of the stream in 1 hr and goes 1 km along the cirrent in 10 min . How long will it take to go 5 km in stationary water ? plzzz ans it soon

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