- 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.

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

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 ?/…

Also this is my question. Same problem sir.

D=(DELTA T*S1*S2)/S1 DIFFERENCE S2)

DELTA T=40MIN EARLT+40MIN LATE=80MIN

S1=3KMPH S2=2KMPH

D=(80MIN*3KMPH*2KMPH)/(1KMPH) 80MIN=80/60 HOURS

=(80*3*2)/(60*1)

=8KM

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?

10 km

D=(DELTA T*S1*S2)/S1 DIFFERENCE S2)

DELTA T=2MIN EARLY+1MIN LATE=3MIN

S1=30KMPH S2=40KMPH

D=(3MIN*30KMPH*40KMPH)/(10KMPH) 3MIN=3/60 HOURS

=(3*30*40)/(60*10)

=6KM

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

19 min