[Aptitude] Time n Distance: Early and late to office (shortcut using product consistency method) for SSC, IBPS, CSAT, CAT, CMAT

Aptitude47 Comments

  1. Case : Early Late
  2. Case #2: Tappu’s school
  3. Case: Pinku’s college (total time given)
  4. Mock Questions
  5. Answer and explanation

Before proceeding further, make sure your concept regarding “product-consistency method” Is clear. If not, then go through my previous article click me

Case : Early Late

Jethalal goes to shop at the speed 30 km/h, and he reaches six minutes early. Next day he goes at the speed of 24 km/h, and he reaches five minutes late. Find the distance between his home and shop.
This can be solved with any of the two approaches

  1. Approach #1: Product consistency
  2. Approach #2: STD table.

Approach #1: Product consistency

Let me rephrase the question:
Price of sugar is increased from 24 per kg to 30 per kg and now Jethalal is buying 11/60 kilograms sugar less (in the same budget). What was his original consumption?

Does it ring any bell with previous sums of product Consistency? Yep, that’s our approach.
Prepare this table, plug in the “speed” values in ascending order

Slow speedFast speed
Speed km/h2430
Ratio-reversed (Time)

What is the time difference between these two cases?
suppose on regular speed, Jethalal used to reach office @10 AM
on slow speed, he is 5 minutes late=10.05AM
on fast speed he is 6 minutes early=09.54 minutes
so the time difference between slow speed and fast speed = 11 minutes.
in the exam, just add the two minutes given to you (6+5)=11 and since speed is given in km/h, we’ve to convert 11 minutes into hours =11/60 hours.

Slow speedFast speed
Speed km/h2430
Ratio-reversed (Time)

Now apply the product consistency method:
Take ratio of 24/30
=(6 x 4)/ (6 x 5)
=4/5
Reverse it.=5/4. Update the table

Slow speedFast speed
Speed km/h2430
Ratio-reversed (Time)54

So, when speed is increased, what is the decrease in time?
5 to 4
=(5-4)/5 x 100
=20% (or just keep it in fraction form of 1/5)
Meaning new time is 20% less than time.
suppose during slow speed, he took “M” time.
Then in fast speed he’ll take M minus 20% of M time.
That means difference between two situations is 20% of m
but we’ve already inferred that time difference between two situations is 11/60 hours
therefore 20% of m=11/60
or in other words
1/5 x m=11/60
M=11×5/60
M=11/12 hours.
This is the time he takes during slow speed, to reach his destination
Now just apply STD formula
(slow) Speed x time = distance
24 x 11/12 = distance
Hence distance = 22 kms.
This technique looks “odd” but It is very fast once you practice.

Thought process in the exam

You don’t even need to draw table. Just think in your head, speed is decreased from 24 to 30 so reverse ratio is 30/24=5/4
And hence decrease from 5 to 4 is (5-4)/5=1/5.
It means 1/5th of (slow) time =(6+5)/60
Hence time = 11x 5/60
Hence distance = just multiple time with slow speed
=11 x 5 x (24)/60
=22 km.
Now let’s try solving It, using the

Approach #2 (STD Table)

Case 1Case 2
Speed2430
Time??
DistanceDD

We’ve ssumed that in both cases, he has to cover same distance “D” kms.
Apply STD formula in column 1 (case 1)
Speed x time = distance
Therefore time = distance / speed = D/24
Similarly for case2, we get time=D/30
Update table

Case 1Case 2
Speed2430
TimeD/24D/30
DistanceDD

From the question, we can infer that time difference between two cases is (6+5=11 minutes =11/60 hours)
Therefore
D/24-D/30=11/60
Simplify this equation and you get D=22 kms.
Please note: in the fractions, D/24 is >greater than> D/30
That’s why I did D/24-D/30=11/60
Let’s try second question with both methods

Case #2: Tappu’s school

Tappu walks from home to school @5kmph and reaches 15 minutes early. After the school is over, he walks back from school to home @3kmph and reaches 9 minutes late. Find distance between his home and school.

Approach #1: Product consistency

The question is talking about two times: 15 minutes early and 9 minutes late.
Therefore total time difference between two situation =15+9=24 minutes=24/60 hrs.

Slow speedFast speed
Speed km/h35
Ratio-reversed (Time)53

What is the percentage decrease in time?
(5-3)/5
=2/5 (=40% decrease)
That’s it. If time taken during slow speed =”m”
Then 2/5th of m=24/60 hours (the time difference between two cases)
Hence M=1 hour (=time taken during slow speed)
Now speed x time = distance
3 (slow speed) x1= distance
Therefore distance between Tappu’s school and home is 3 kms.

Approach #2 (STD Table)

Slow speedFast speed
Speed km/h35
Time????
DistanceDD

Apply STD formula in each column you get
Speed x time =distance
Time = distance / speed
Time = D/3 in first case and D/5 in second case update table

Slow speedFast speed
Speed km/h35
TimeD/3D/5
DistanceDD

The time difference between two situations is (15+9)=24 minutes=24/60 hours
Therefore
D/3 – D/5=24/60
Solve this equation and you get D=3 kms
Meaning distance between Tappu’s school and home is 3 kms
Now let’s try a bit complicated case

Case: Pinku’s college (total time given)

Pinku goes to college @ speed of 3 kmph and returns back @2kmph. He spends total 5 hours in walking. What is the distance between his home and college?
Slow speed , fast speed = 2 and 3 km respectively.

Slow speedFast speed
Speed km/h23
Ratio-reversed (Time)32

What is the decrease % in time? (3-2)/3= 1/3 (=33.33%)
It means if Pinku take “M” hours during slow speed.
He’d take M minus 33.33% of M hours during fast speed.
Therefore, total time (taken to goto college and come back)
=m + m -33.33% of m
=2m-m/3 (because 33.33%=1/3)
=(6m-m)/3
=5m/3
And we know that total time is 5 hours
therefore 5m/3=5 hours
hence m=3 hours. (time taken during slow speed)
Apply STD
Speed x time = distance
2 (slow speed) x 3 (time)=distance
Hence distance=6 km

Thought process in the exam

Speed increased from 2 to 3, therefore reverse ratio is 3/2 and %decrease in time is 1/3.
Pinku’s “Total” time is given 5 hours, therefore
M + m -(1/5)m=5 hours. Solve it and multiple with slow speed, you’ll get the distance.

Mock Questions

  1. Gogi walks from home to school @2.5kmph and he is 6 minutes late. Next day he increases speed by 1 kmph and reaches 6 minutes early. Find distance between home and school?
  2. Sonu walks @6kmph and late to college by 5 minutes. If she walks @5kmph, she is late by 30 minutes. Find total distance. (please note: since she’s late in both cases the time difference is 30-5=25 minutes. rest approach is same)

Answer and explanation

1. Gogi school

Question is talking about two speeds : 2.5 and (2.5+1.0)=3.5 kmphs

Slow speedFast speed
Speed km/h2.53.5
Ratio-reversed (Time)75

What’s the decrease in time %
From 7 to 5,
=(7-5)/7
=2/7
Suppose during slow speed case, Gogi takes “m” hours to reach school.
In fast case, he’ll do it in less time =m – 2/7 of m.
but from question, we already know that time difference between two cases =6+6=12 minutes=12/60 hrs
it means 2/7 of m=12/60 hours
therefore m=7/10 hours :This is the time taken during slow speed.
Multiply it with slow speed and you’ll get the distance.
Distance
= 7/10 x 2.5
= 7/4 kms.

2. Sonu college

Slow speedFast speed
Speed km/h56
Ratio-reversed (Time)65

So % decrease in time
=(6-5)/6
=1/6
Therefore 1/6 of slow time (m)= 25/60 hrs.
M=25 x 6/60 hrs
Multiply it with slow speed (5) and you get distance
Distance
=speed x time
=5 x 25 x 6/60
=25/2
=12.5 km distance between home and college.

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So far 47 Comments posted

  1. sanjay

    mrunal please suggest strategy for IFS(Forest) exam General studies paper.please help me

  2. manish

    murnal sir you have been doing an edifying service to the pupils and good thereof.moreover, i want you to tell me the name of the book covering the portion of number system ,taking in view the SSC exam.

    regards
    manish

    1. Mrunal

      In SSC exams, the questions on number system / remainder problem, are asked from very limited concepts and I aim to cover those concepts in upcoming articles. so no need to get any special book.
      There are books like Nishth Sinha’s “Demystifying number system” but it goes way too deep from CAT point of view.

      1. Avish

        Sir,how does the ratio of speed ..gives the increase and decrease in tym ??

  3. Anshuman Rai

    Thanks sir. I found the second method of solving it better. quickly D/Time and equating it to time gap. Thanks for both the methods.

  4. Pratik Kumbhani

    Please notify for new post and comments.

  5. Anuraag Singh

    Hi Mrunal Good Job Dear!!
    I have been following your blog since 2 years now. 2 months ago someone told me about the Featured Articles Section of the Press Information Bureau. I have started reading the articles on a daily basis, apart from your articles. can you tell me if this is going to do any good?? As the Articles are written by serving Bureaucrats and Freelance writers, they give you very important info-nuggets as well as the viewpoint of the Government on a variety of issues related to Governmental activities. Please Check them Out and let me know. This query is not only for Mrunal. any advice or suggestion is welcome.
    Waiting for your reply!!!

    1. Abhimanyu

      yeah definitely bro

  6. neha

    hi murnal sir
    plz help me. im confused as to which book should i choose, RAJESH VERMA or SARVESH KUMAR for quantitative aptitude of SSC CGL. im very poor in maths.

  7. izaz

    sir,
    please can you give me tricks to solve the profit and loss tuff problems and also percentages.

  8. Lady Nightingale

    other way to solve such question is to follow question language .like in topmost question regarding jetalal..
    let distance be x km and speed be u km/hr.

    from 1st part of question-
    s/u-s/30=6/60………….eqn-1

    from 2nd part of quesn
    s/24-s/u=5/60…..eqn-2

    now from eqn-1…put s/u value in eqn-2 and we get
    s/24- s/30- 6/60= 5/60
    s(5-4)/120= 11/60
    s=11*120/60= 22kms…. is the ans.

  9. anonymous kashmiri

    sir is pearson’s book fr csat better than tmh?
    Ther hv been sm negatv reviwz abou TMH,,,so i m n a fix wch book to buy?
    I hv ample tym left…so plz help me out
    if anybdy else wnts to giv hz/hr sugstn n ds matr i wl b thankful…

  10. yogesh

    STD Approach is fantastic.

  11. Ankit SIngh

    Sir,,,
    plzzz also explain the questions in which speed is not given and only told that speed is increased or decreased ,,,, also late or early time is given and we have to find the distance ….

    solution Like above methods(atleast by first method).

      1. ankit singh

        A man covered a certain distance at some speed.Had he moved 3 kmph faster,he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in Km) is :

        1. Sudeep

          Let distance = D, speed= S, time taken = T (formula, we know S*T=D)

          Lets assume the man covered the distance twice(2D), one D with S+3 kmph and another D with S-2 kmph

          So, D + D = (S+3) (T-40) + (S-2) (T+40)

          2D = ST + 3T – 40S – 120 + ST – 2T + 40S – 80
          ==> 2D = 2ST + T -200
          But ST = D,
          So T-200 = 2D-2D = 0
          ==> T = 200 min

          Now with S+3 speed, time taken = 200-40 = 160
          and with S-2 speed, time taken = 200+40 = 240… now we have problem in our earlier format.. solve it.. You will get D, distance = 40 km

          1. sankar

            A man covered a certain distance at some speed.Had he moved 3 kmph faster,he would have taken 30 minutes less. If he had moved 2 kmph slower, he would have taken 20 minutes more. The distance (in Km) is ///now would you solve in above method??

  12. amit

    the best simple rule …

    [(24*30)*(5+6)]/[(24~30)*60]=22

  13. lily dutta

    thanks 4 ur article sir,m little bit weak in English and arithmetic,plz suggest me which book i’ll usefull 4 me to improve in these area,as m preparing 4 competitive xam…

    1. G.A.suryaYour Good Name (आपका शुभ नाम?)

      11×11=right side fist number ×right firstnumber.this result left side 11 1=12 1 =121 . if 12×12 you can add the last number 2 to the left side of the number.this is surya-9705701141

  14. deepa

    sir plz upload some cubes problems also.

  15. Raghav

    Sir Pls guide me regarding SBI PO, here they are including marketing, computer, data interpretation etc beside other usual stuff

  16. Yagya

    Assume distance = x
    Time = y

    Follow the question, if ‘t’ min early then time = y-t
    If late then = y t
    Make 2 equations in form of ‘distance=speed*time’ then,

    We know distance will remain same,so equal both equations.
    You will get ans.

  17. santu

    sir,
    plz help me with the information (necessary link) about cds ssb interview..
    sir plz reply….

  18. smit

    if we take the problem as..
    appu walks from home to school @5kmph and reaches 9 minutes late. After the school is over, he walks back from school to home @3kmph and reaches 15 minutes late. Find distance between his home and school.

  19. Abubakkar Sithik

    Hi Mrunal, Thank above has been given all information .This site is ever use in my life.this site not only improving exam but also increase our knowledge

  20. ISSHU

    A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:

    1. Mudasir

      Dear ISSUE,
      Let total distance covered be ‘D’,
      total time taken ‘T’ and average speed be ‘S’
      Now, S=D/T or
      D =S*T
      Ist case:
      ‘D’distance is same,speed increased from ‘S’ to (S+3) and time decreased from ‘T’to (T-40/60)=T-2/3=(3T-2)/3
      D=(S+3)(3T-2/3)
      3D=3ST-2S+9T-6
      3D-3ST=-2S+9T-6
      0+2S=9T-6
      Bcoz, 3 speed*time=3 distance
      2S=(9T-6) —-i
      2nd case
      distance is same ‘D’,speed reduced to (S-2), time increased to (3T+2)/3
      D=(S-2)(3T+2)/3
      3D=3ST+2S-6T-4
      3D-3D-2S=-6T-4
      (-2S=-6T-4)*-
      S=(6T+4)/2 —-ii
      put the value of S in =n i ,we get
      S=12 and T=10/3
      D=12*10/3
      D=40 ans

  21. john pk

    M + m -(1/5)m=5 hours. Solve it and multiple with slow speed, you’ll get the distance.

    sir i think there is a error it should be M + m -(1/3)m=5 hours. Solve it and multiple with slow speed, you’ll get the distance.

    1. Mudasir

      Another way to do it is D/2+D/3=5
      D=6

  22. Rahul

    guys y complicate things use this

    (x/5) + ( 15 / 60 ) = (x/3) – (9/ 60) . solve for x..thats your ans for tappu

    (x/30) + ( 6 / 60 ) = (x/24) – (5/60) . solve for x..thats your ans for Jethalal goes to shop

  23. ARJUN SINGH

    Sir it can be solved by different formula which goes to..S1.S2/S1-S2.T1+T2/60….where s1,s2 r speeds,n t1,t2 time for late and early..n it consume lots of timee during exam..

  24. Dharamvirsinh

    Sir, I tried a lot but couldn’t found std table and pipe cistern article….can u please proved me a link for that……?

  25. sib

    Mrunal Ji u saved a lot of bucks of the poor…u are doing a marvellous and kind job..If there are websites like yours then surely all coaching business will be at dawn…just take care of u as u might have made a lot of enemies in the country

  26. Akash

    Hello Sir,
    I applied STD table in Pinku’s college question but not working out to be a correct answer. Please help

  27. hafis

    hi mrunal sir,

    please direct me by giving the link for pipes and cistern article.
    Thanks

  28. amrit

    thanku this is very helpful mrunal sir.

  29. Illaventhan

    When Kiran drives a car at a s peed of 55 km/hr towards his office, he reaches l ate by 20 minutes ;
    but if he driv es at a s peed of 70 km/hr, he reaches earl y by 15 minutes . Find the usual time that
    he takes to reach his office exactly on time? please anyone slove this

  30. Ajay Kumar

    Sir help for this question

    a man covered a certain distance at some speed. had he moved 3kmph faster, he would have taken 40 minutes less. if he had moved 2 kmph slower, he would have taken 40 minutes more. the distance is ?/…

  31. asha

    a persong is going in a car at 30kmph reaches his ofdice t1 minutes late. if he goes at 40kmph reaches there t2 minutes earlier. how far is the office from his house?

  32. gaurav

    dear sir can you post ODIA literature optional subject note please .

  33. vimal

    If a train runs at 40 km/h, it reaches the destination late by 11 mins. But if it runs at 50 km/h, it is late by 5 mins only. The correct time for the train to complete its journey is ? use product constancy mehod

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