- Case : Early Late
- Case #2: Tappu’s school
- Case: Pinku’s college (total time given)
- Mock Questions
- 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

- Approach #1: Product consistency
- 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 speed | Fast speed | |

Speed km/h | 24 | 30 |

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 speed | Fast speed | |

Speed km/h | 24 | 30 |

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 speed | Fast speed | |

Speed km/h | 24 | 30 |

Ratio-reversed (Time) | 5 | 4 |

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 1/5^{th} 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 1 | Case 2 | |

Speed | 24 | 30 |

Time | ? | ? |

Distance | D | D |

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 1 | Case 2 | |

Speed | 24 | 30 |

Time | D/24 | D/30 |

Distance | D | D |

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 speed | Fast speed | |

Speed km/h | 3 | 5 |

Ratio-reversed (Time) | 5 | 3 |

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 2/5^{th} 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 speed | Fast speed | |

Speed km/h | 3 | 5 |

Time | ?? | ?? |

Distance | D | D |

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 speed | Fast speed | |

Speed km/h | 3 | 5 |

Time | D/3 | D/5 |

Distance | D | D |

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 speed | Fast speed | |

Speed km/h | 2 | 3 |

Ratio-reversed (Time) | 3 | 2 |

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

- 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?
- 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 speed | Fast speed | |

Speed km/h | 2.5 | 3.5 |

Ratio-reversed (Time) | 7 | 5 |

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 speed | Fast speed | |

Speed km/h | 5 | 6 |

Ratio-reversed (Time) | 6 | 5 |

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.

## So far 47 Comments posted

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

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

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.

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

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.

Please notify for new post and comments.

just fillup this form and you’ll get free email notifications, whenever new article is posted

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Thank you..bt i thnk ssc cgl is going to be more tougher this tym

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!!!

yeah definitely bro

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.

Either book is fine. It depends on your career choice.

if it is SSC-CMAT-CAT type then go for Sarvesh Kumar

If it is SSC-IBPS-LIC type then go for Rajesh Verma.

Regarding maths, start with NCERT textbooks first and get concept clarity, then maths wouldn’t look that tough.

read this article for more:

http://mrunal.org/2013/01/studyplan-ssc-cgl-maths-quantitative-aptitude-algebra-trigonometry-approach-booklist-sources-free-studymaterial-combined-graduate-level-exam-tier-1-2.html#105

sir,

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

it is given in the http://www.Mrunal.org/aptitude

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.

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…

STD Approach is fantastic.

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).

give me a sample question.

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 :

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

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??

the best simple rule …

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

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…

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

sir plz upload some cubes problems also.

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

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.

sir,

plz help me with the information (necessary link) about cds ssb interview..

sir plz reply….

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.

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

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:

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

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.

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

D=6

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

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..

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

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

Hello Sir,

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

hi mrunal sir,

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

Thanks

thanku this is very helpful mrunal sir.

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

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 ?/…

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?

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

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