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

October 8, 2017 | Author: Khalid Khan | Category: Speed, Odds, Trigonometry, Mathematics, Physics & Mathematics

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[Aptitude] Time n Distance: Early and late to office (shortcut using produc...

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APTITUDE

JANUARY 24TH, 2013

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

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.

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

Ratio-reversed (Time) 5

30 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

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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.

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

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

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.

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 2/5th 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.

Slow speed Fast speed Speed km/h 3

5

Time

??

??

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Distance

D

http://mrunal.org/2013/01/aptitude-time-n-distance-early-and-late-to-offic...

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

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

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.

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)

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

Ratio-reversed (Time) 7

3.5 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.

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Distance = 7/10 x 2.5 = 7/4 kms.

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.

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

manish Reply to this comment 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

Mrunal Reply to this comment 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.

Anshuman Rai Reply to this comment 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.

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

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

Anuraag Singh Reply to this comment 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!!!

Abhimanyu Reply to this comment yeah definitely bro

neha Reply to this comment 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.

Mrunal Reply to this comment 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-sourcesfree-studymaterial-combined-graduate-level-exam-tier-1-2.html#105

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

Mrunal Reply to this comment it is given in the http://www.Mrunal.org/aptitude

Lady Nightingale Reply to this comment 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 questions/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.

anonymous kashmiri Reply to this comment 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…

yogesh Reply to this comment STD Approach is fantastic.

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Ankit SIngh Reply to this comment 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).

Mrunal Reply to this comment give me a sample question.

ankit singh Reply to this comment 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 :

Sudeep Reply to this comment 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

amit Reply to this comment the best simple rule … [(24*30)*(5+6)]/[(24~30)*60]=22

lily dutta Reply to this comment 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…

deepa Reply to this comment sir plz upload some cubes problems also.

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

Yagya Reply to this comment 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.

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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.

Abubakkar Sithik Reply to this comment 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

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