### Discussion :: Time and Work - General Questions (Q.No.3)

Chirag Lakhani said: (Jun 14, 2010) | |

How step 3 is make 1/10+1/10? please explain it. |

Vinoth said: (Jun 17, 2010) | |

How do you know total working day is 3 (last step). |

Satyendra Sharma said: (Jun 29, 2010) | |

3 days work = 1/5 it means 1 day work =1/15 that is 1/15 work is done in 1 Day now 4/5 part of work that is (1-1/5) will be done in :15*4/5 =12 days.... Hence total work done will done in (12 + 3 ) days... |

Abhishek said: (Sep 3, 2010) | |

@chirag: Step 3 gives work done in first 3 day i.e. work done in 1st day + work done in 2nd day +work done in 3rd day = 1/20 (by A alone) + 1/20 (by B alone) + 1/10 (done by A, B and C together) Similarly they continue and as per the question on every 3rd day A is joined by B and C. |

Nitesh Nandwana said: (Oct 1, 2010) | |

Hi chirag don't go on step, here our aim is to compute total work done in 3 days once, first 2 days A is working alone and last 3rd day we are adding work's of B and C. |

Sushant said: (Nov 22, 2010) | |

(A + B + C)'s 1 day's work = 1 + 1 + 1 = 6 = 1 . 20 30 60 60 10 how did we get 60 |

Bala said: (Nov 24, 2010) | |

I can't get this. |

Vikas said: (Dec 9, 2010) | |

Answer Explained: A is working alone for two days, 3rd day he is assisted by B and C. A's 1 day work=1/20 A working on 2 days=2*1/20=1/10 A+B+C working on 3rd day, so 1 day of working together=1/20+1/30+1/60=6/60=1/10 So total work done till 3rd day=1/10+1/10=2/10=1/5 So if in 3 days = 1/5 of work is completed.... Than, 3*5 days = 1/5*5 work will be completed. =15days |

Dhanam said: (Dec 20, 2010) | |

Why are you taking 2 days for first step? |

Vishnu said: (Dec 25, 2010) | |

Why you have taken the a+b+c? A is assisted by only B, C know. |

Anbu said: (Dec 29, 2010) | |

How you got 6/60? I think the answer is 12. |

Beeribrahim said: (Dec 31, 2010) | |

How it is user Now, 1 work is done in 3 days. 5 Whole work will be done in (3 x 5) = 15 days. |

Shreekanth said: (Jan 3, 2011) | |

@Vishnu... As per the statement A will do the work three days continuosly but only on third day B & C will assist, so A's thrid day work must also include with both B & C that's why... (A + B + C)'s 1 day's work is taken |

Vijay said: (Jan 7, 2011) | |

1/20+1/(30X3)+1/(60X3) = 1/15 Is my way of solving correct? |

Nikita Prasad said: (Jan 7, 2011) | |

There's a simple logic friends, If a person wrks for n days, his 1 day's work is 1/n. or the other way round, if 1/n is the 1 day's work then he works for n days. & it can be very well known if 1 day's work is known it can be multiplied with any numbers of days you wish to. Similarely, Here last statement tell the same i.e. 3days work is 1/5, so to calculate 1 day's work we need to divide it by 3 wiz gives 1/15. so if 1/15 is 1 day's work then they together work for 15 days. & the logic behind 2 days & 3 dats work is : A's 2 days work is (1/20)*2, i.e 1 day's multiplied by nth day. but on the "3rd day" they all A,B&C work together wiz equals to 1/10. Therefore 2 days + 3rd day's wrk gives 3 day's work. |

Chitra.G said: (Jan 8, 2011) | |

Why we take 2 days? Please help. |

Roten said: (Jan 8, 2011) | |

@vikas. Superb explanation. :-). |

Baskar said: (Jan 21, 2011) | |

@chitra assisted by B and C on every third day it means A's 2 days + (B&C)'s 1 day from d 1st explanation u got Work done in 3 days=(1/10 + 1/10)=1/5 dat is in 3 days=1/5 works in ques dey ask "In how many days can A do the work" now cross multiply.. (5*3) days=1 work.. 1 work = 15 days... now u got it ah? |

Samathu said: (Jan 28, 2011) | |

Nice explanation baskar. |

Munish said: (Jan 30, 2011) | |

Thank you Vikas for your information, I have understood. |

Karan Singh said: (Feb 5, 2011) | |

Indeed, the explanation of baskar is correct. |

Karan Singh said: (Feb 5, 2011) | |

Is Vijay's formuala correct? In real aptitude test, the quick formula will be handy and faster to calculate. |

Manasa said: (Feb 6, 2011) | |

@nikita prasad. You just rock. Good explanation with logic. :-) ). |

Poonam said: (Feb 9, 2011) | |

Please give altarnate methad to solve this question, then we know wether the ans you got is right or wrong. |

Rajasekhar said: (Feb 11, 2011) | |

The question does'nt mean to find the 3 days work how we can solve this for 3 days work?. |

Hamsa said: (Feb 21, 2011) | |

How did you get 6/60 could you please clarify? |

Divya said: (Mar 8, 2011) | |

Chitra its very clear. |

Suneel Kumar said: (Mar 28, 2011) | |

Why you multiply 3*5? |

Manoj Shukla said: (Apr 6, 2011) | |

Why 3*5? |

Niket said: (Apr 14, 2011) | |

Actually in 3rd step work done by A on 3rd day was not added that's why the answer became 15 days. |

Sindu said: (Apr 26, 2011) | |

Why in the last step 3 is multiplied with 5? |

Nitinkumar said: (May 4, 2011) | |

A+B+C=(1/20)+(1/30)+(1/60)=1/10 ON 3 RD DAY HE IS TAKING WORK FROM BOTH B&C HIS 2DAYS WORK +A+B+C WORK= 1/10+1/10 = 2/10 =1/5 MULTIPLY WIT 1/3 U LL GET 1/15 15 DAYS |

Ramesh said: (May 9, 2011) | |

Nice explanation nikitha:-) |

Shreya said: (May 23, 2011) | |

I dint get how did you multiply 5*3. It was 1/5 83? |

Rose said: (May 26, 2011) | |

Work done in 3days=1/10(A'S 2DAYS WORK)+1/10(1DAY WORK FROM ABC) =2/10=1/5 NOW CROSS MULTIPLY FOR 1/5 3DAYS ARE REQUIRED FOR 5/5 DAYS HOW MANY DAYS ARE REQUIRED NOW 5/5=1 1*3/1/5= 3*5/1=15 |

Fathi said: (Jun 6, 2011) | |

Awesome explanation nikitha! this is Wat I was searching for:-). |

Balaram said: (Jun 27, 2011) | |

Thank you everybody for participating in the discussion and making clarification for all my unanticipated doubts.... Nikita Prasad gave good explanation. |

Arpit Mathur said: (Jul 2, 2011) | |

Hey please explain why A's 2 day work is taken? |

Kumar said: (Jul 4, 2011) | |

Can you please tell me how you get 6/60? |

Sindhu said: (Jul 5, 2011) | |

@Baskar Can you please explain me why do you cross multiply at the end for the answer? |

Benny said: (Jul 6, 2011) | |

Here my simple ans: A is working continue 2 day's.So he's 2day's work is:1/20*2=1/10; in 3rd day he join other two person B and C.So here 3 member work: 1/20+1/30+1/60=1/10; 3 day's work is :1/0+1/10=1/5 So total work done by them is 10+5=15; |

Sathish said: (Jul 9, 2011) | |

Hi friends, we know that whole work completed means we will denote it as 1(full).... 2=1(in 2days full work get completed) 5=1(in 5days full work get completed) just simple 5=1/2 =>5*2=1(just cross multiple the 2) |

Preethi said: (Jul 24, 2011) | |

1/5 of work being done in 3 days. But how do we arrive at the total no. Of days worked by A? what is the logic behind? |

Neetu Singh said: (Aug 5, 2011) | |

@benny-reli nice explanation. |

Harry said: (Aug 9, 2011) | |

1.)days worked by A is 2; 1 day helped by B and C and not A, 2.)given is A's 3 day work 3.)total 1 day work of A,B,C is find 4.)A's 2 day work is found Result calculation: A's 3day work=>(1/10)+(1/10)=>A 2day work+A,B,C's 1days) =>1/5(of whole work) [5 is n] A's whole work=3*5=15days |

Manoj said: (Aug 24, 2011) | |

i.e; 3 days =1/5 6 days=2/5 9 days=3/5 12 days=4/5 15 days=5/5(1)==work completed |

Anil said: (Sep 3, 2011) | |

Thanks benny. |

Poornima said: (Sep 18, 2011) | |

How can you taking that 3 days? and 1/10+1/10 what is it ? |

Ganbrave said: (Sep 23, 2011) | |

@Poornima A works for 2 days and B,C joins him on 3rd day. so total combination of them is now 3days Based on these 3days we can furthur calculate..ok?? A's 2 days work=1/10 and on 3rd day A,B,C together work=1/10 So sum of them gives u total 3days of work done by all of them. |

Imraz said: (Sep 24, 2011) | |

Here 15 days is required for the completion of the work by A, B, C. But in question it is asked 'in how many days can A do the work' - which means the number of days A worked. So to get this we need to divide the 15 by 3 and get '5' as answer. Can anyone please make me clear if my interpretation is incorrect ? |

Rohit said: (Sep 24, 2011) | |

1st day A's + 2nd day A's work = (1/20 + 1/20) = 2/20 = 1/10--(1) 3rd day work is added by B's and C's work so, work done by A+B+C= (1/20+1/30+1/60) = 1/10--(2) from (1) and (2)= 1/10+1/10 =1/10 |

Balalakshmi said: (Sep 26, 2011) | |

Answer explained: A is working for 2 days alone so A's two day work is first calculated as A's 2 day work is =1/20*2 = 1/10 Now B and C work with A for only one day(ie 3rd day) and A also working along with B and C in the same 3rd day so A,B and C one day work(ie 3rd day work) is calculated as=(1/20+1/30+1/60)=1/5 Now A's 2 day work+ ABC one day work(ie 3rd day work is calculated)= (1/10+1/10)=1/5 Therefore in 3 days only 1/5 of work is completed Thus to complete full work 3*5=15 days (3days=1/5 of full wok completed 3*5 days=full work completed 15 days = full work completed) |

Pavethra said: (Sep 29, 2011) | |

I tell that third day alone A join them so on third day it is 1/20 +1/30+1/60=1/10(here a,b,c all work together) then 2 days A never join them then B and C work together then SO 1/20 +1/30=1/5 then make it ulta 10+5=15days easy na. |

Yashraj said: (Oct 3, 2011) | |

(1/20) + (1/30) + (1/60) = 6/60. Please explain me this step. |

Xyz said: (Oct 4, 2011) | |

@Yashraj When we want to do (1/20)+(1/30)+(1/60) generally, we take the help of L.C.M, But in above case Highest number is 60 which is Divisible by all the remaining number i.e. 20 and 30. Hence we take L.C.M. as 60 i.e. (3+2+1)/60 becomes 6/60. Build your Basic maths Strong My friend. |

Rehan said: (Oct 14, 2011) | |

We can consider the avg of the 3 days and calculate for total work only if we know the number of days of their work is a multiple of 3. I mean if they had to work for 14 days, b and c would have assisted them only till day 12. On day 13 and 14 again a would have to work alone. So how can we take 1/5th of work is done in 3 days? |

Mohosin said: (Oct 31, 2011) | |

We can solve it in this way also: In 1 day, A alone can do 1/20th f the work, So, in 2 days A can do (1/20 * 2)= 1/10th f the work On 3rd day A,B and C together do= (1/20 + 1/30 + 1/60)= 6/60= 1/10 ___________________________________________________________ Now, after 3 days, total work done= (1/10 + 1/10)=2/10=1/5 so, in 3 days they can complete 1/5th of the work. so, for completing the whole work, they need (5*3) days= 15days |

Kv Manikanta said: (Nov 2, 2011) | |

How do you say that A's work is 2 days it means how it came 2 (1/20) = 1/10. |

Hitesh said: (Nov 8, 2011) | |

work done by A per day=1/20; work done by B per day=1/30; work done by C per day=1/60; Let the work is completed in 3x days So, 1/20*3x+1/30*x+1/60*x=1 => x=5 => No. of days required to complete the work=3*5=15days |

Anitha said: (Nov 12, 2011) | |

Why we have to calculate A's days work? Explain any one please. |

Siddu said: (Nov 21, 2011) | |

Please explain how we will get 6/60 please? |

Kamalakannan said: (Dec 5, 2011) | |

Could you tell me? why you take A has done by 2 days work? |

Eswarijana said: (Dec 22, 2011) | |

I can't understand the last step. |

Renjini said: (Jan 4, 2012) | |

How could get 1/5? I can't understand. |

Tania said: (Jan 7, 2012) | |

@kamalakannan Because A alone does the work in FIRST day and in 3rd day he takes the help of B, C. So 1st A's 2daz work is calculated. |

Naari said: (Jan 12, 2012) | |

@vikas. Good work dude. |

Sindhuja said: (Feb 5, 2012) | |

@monoj. Superb explanation. |

Emmanuel said: (Feb 10, 2012) | |

Can someone help me out? why are they multiplying 3 and 5 but not any number for example 3 and 6 making 18 ? |

Raman(Uiet,Kuk) said: (Feb 17, 2012) | |

Guys check dis out........ as we know A will be assist by B & C every third day, so first two day A will work again A's 1st day of work = 1/20 A's 2nd day of work = 1/20 (A+B+C)'s 3rd day of work = 1/20+1/30+1/60=1/10 total work of 3 days = 1/20+1/20+1/10=1/5 hence, 3 days of work = 1/5 1 day of work will be=1/5*3 =1/15 so work will complete in 15 days |

Sai said: (Feb 19, 2012) | |

I still can't get clarification. In the question they asked that in every 3rd day of A's work. But in explanation how can you say for only 3 days. |

Subasish said: (Feb 19, 2012) | |

A's 1 day work = 1/20 B's 1 day work = 1/30 C's 1 day work = 1/60 since on 3rd day A get assistance of B & C so 3 days work = A's 3 days work + ( B's + C's )1 day work = 2*(1/20)+ 1/30 + 1/60 = 12/60 = 1/5 1/5 parts is done by 3 days 1 or whole part is done by (5*3)= 15 days |

Shro said: (Feb 20, 2012) | |

Sorry Guys. Can somebody please explain me more clearly. My doubt is " when we say A+B+C =1/10. it means we have considered 1 day's work of A,B,C and the answer 1/10 is total of all 3 together working days" " so when we say A's 2 day's work = 1/10 it means we are considering only A's 2day's work " So,finally when we add for 3 days= (1/10) + (1/10) here 1/10 is A's 2day's work, but when we take 2nd time 1/10 then does'n it mean that we are considering all 3 working together's total 1/10 ???? As the given question is In how many days can A do the work if he is assisted by B and C on every third day? Hope u can udnerstand what I wrote above ... |

Bralow said: (Feb 29, 2012) | |

Why we multiply 3*5 in last step? |

Sandya said: (Mar 8, 2012) | |

@ nikita prasad Its really good explanation. I had lot more confusion at this step. Now I'm clear. ThanQ. |

Yogendra said: (Mar 19, 2012) | |

Thanks nikithas & bhaskar for your nice explanation. |

Preethi said: (Apr 5, 2012) | |

Why we are takng 3*5 in last step please expain. |

Raju said: (Apr 5, 2012) | |

Why we are takng 3*5 in last step please expain. |

Karthik said: (Apr 6, 2012) | |

@Preethi. 3*5 because 1/5*5 gives you 1 which means the work has been completed. So 1/5 is derived for work done in 3 days. Understood ah. |

Kranthi said: (Apr 26, 2012) | |

A's 1 day work = 1/20 B every 3 days = 3/30 C also every 3 days = 3/60 Total work = 1/20+3/30+3/60 = 3+6+3/60 = 12/60 =1/5 3*5 days=full work completed 15 Is it correct way |

Anonymous said: (Apr 27, 2012) | |

Why do you need to find A's work in 2 days? |

Kumar said: (Jun 5, 2012) | |

@Preethi Finally we got the answer 1/5->work completed in 3 days clear upto here.Coming to your doubt we have to know the full work completed in no of days 1/5th work == 3days=1/5th work Just make above Rhs as 1 Full work == 3*? =1/5*5 == 3*5 =1/5*5 1 Full work in 15 days == 15 =1 Hope your doubt is cleared. |

Tuhin said: (Jun 6, 2012) | |

EASY WAY FOR THIS QUESTION : Let total no. of days be= N Work done by A,B,C respectively per day are :1/20,1/30,1/60 As B and C join every third day there work done by them :(N/3)/30 and (N/3)/60 respectively. And for A : N/20 Total work= N/20+(N/3)/30+(N/3)/60=1 We get N as 15 |

Shakti said: (Jun 28, 2012) | |

Last step. 1/5th of work completed in 3 days. We have to find 1 or whole work can completed in how many days. 1/5th work= 3 days. 1 work= 3/ (1/5) = 3*5= 15 days. |

Vinay Vidhani said: (Jul 4, 2012) | |

Please tell me some one where I'm doing mistake. Hear A can do a work in 20 days B-30 days and C doing in 60 days. A's 3 day work is 1/20*3. B's work will be on 3rd day is 1/90 and C's - 1/180 they work together then answer is coming 6 days please tell about that's solution soon please. |

Narayana said: (Jul 5, 2012) | |

A work in 2 Days = 1/20*2 = 1/10. After 3rd day add in B, C. So A, B, C work in 1 day. = 1/20+1/30+1/60 = 1/10. 2days + 1 day = 1/10+1/10 => 1/5. Work completed = 1/5*3 = 15 days. |

Karthikeyan said: (Aug 19, 2012) | |

Please explain me the lcm method for this problem. |

Arun said: (Aug 21, 2012) | |

Can you explain the 3rd step please? |

Chaaya said: (Aug 22, 2012) | |

Can anybody explain me why 5 is multiplied by 3 at the end?? |

Nirav said: (Aug 23, 2012) | |

@chaaya: 1/5 work done in no. of days= 3 1 work done in no. of days= ? so it is 3/(1/5)= 3*5=15 |

Netra said: (Aug 24, 2012) | |

Please tell me why they have multiplied 3 day's work with 5. |

Trisha said: (Aug 30, 2012) | |

Whole work will be done in (3 x 5) = 15 days. I m unable to understand a last step. Explain me? How they multiply by 5? |

Durga said: (Sep 16, 2012) | |

LCM of 20,30,60 is 600 A's work is 600/20 if b and c on third day then 2 days(first and second) for A and 1 day for B and c(third day) so A's work is 600/(20*2)=600/40=15 |

Priti said: (Sep 24, 2012) | |

We have already calculate 1/5 for 3 days. Then why again multiplying by 3? please help. |

Naresh said: (Sep 28, 2012) | |

If we add the work of b and c for 1 day we get 1/30 + 1/60=1/20. First day = 1/20. Second day =2/20. Third day =4/20 // work of b & c is added 3/20 + 1/20. Every 3 days 4/20. For 6 days we get 8/20. For 9 days we get 12/20. For 12 days we get 16/20. For 15 days we get 20/20 = 1. Job over. |

Nitin.Killekar@Gmail.Com said: (Oct 11, 2012) | |

In the last step the multiplication of 3 and 5 is for the no.of persons (3) and amount of work done (5). |

Madhuri Choudhary said: (Dec 13, 2012) | |

I didn't understood how did they calculated the work done in 3 days. Can somebody explain me clearly? |

Asanka said: (Dec 13, 2012) | |

If x = A worked day {x*1/20 + x/3*(1/30+1/20)} = 1 x/20+x/60 = 1 x = 15. |

Saf said: (Mar 6, 2013) | |

There is confusion because the number is same for A working for 2 days (i.e.1/10 Days) and all three, A+B+C=1/10 days. Hence its. =2 working days of A 1/10+ All 3 together 1/10 days. i.e. (1/10+1/10). |

Krats said: (Mar 28, 2013) | |

1/5 works is being done in 3 days. i.e 1 of the 5 parts of work is done in 3 days. So total there are 5 parts of work which will take 5*3=15 days. |

No One said: (Mar 31, 2013) | |

A's one day work is 1/20 , Therefore A's 2 day work will be (1/20)*2=(1/10), Now (A+B+C)'s one day work will be = (1/20)+(1/30)+(1/60)=(1/10). Since A is assisted on every third day of his work then total work done in three days will include A's 2 day's of work plus the work done on third day, that is (A+B+C)'s one day work, Therefore work done in 3 days will be,A's two days work (1/10 )+ A,B,C's one day work (1/10)= (1/5). Now in three days 1 out of 5 parts of work is done , and yet 4 parts are to be done there fore 4*3= 12 days and since we have calculated the earlier 1 parts requirement of days i,e 3 days total days will to finish 5 parts will be (12+3)= 15 days. |

Shazam said: (Jun 13, 2013) | |

Here's a simple explanation : WORK = (Rate)*(Time) ***time will cancel out*** First Day: Person A does all the work. His working rate is 1/20 so (1/20)*1day = 1/20 work. Second Day: Person A does all the work. His working rate is 1/20 so (1/20)*1day = 1/20 work. All in all in 2 days is 2/20 or 1/10. Third Day: Person A is assisted by B and C. (1/20)+(1/30)+(1/60) times 1 day = 1/10. All in all in 3 days is (1/10)+(1/10)= 1/5 work is done. So 4/5 remaining.. Question Says B and C assisted A EVERY third day. Repeat the process until you get 5/5 or 1 full work that would be on the 15th day, A will complete 5/5 or 1 full work. |

Divya said: (Jun 19, 2013) | |

Why we taken 1/10+1/10? |

Suresh Kumar said: (Jun 29, 2013) | |

@Divya. First Two days work done by A = (1/20)+(1/20)=> 2/20=> 1/10. Third day work done by A,B,C = (1/20)+(1/30)+(1/60)=> 6/60 => 1/10. Total work Done in 3 days= (1/10)+(1/10)=2/10 =>1/5. |

Anurag said: (Jul 7, 2013) | |

Let N be the total no.of days A has to work. So, a/q B and C works only for N/3 days. Total work = 1. Therefore, N*(1/20)+ (N/3)*((1/30)+(1/60)) = 1. Solving this , you will get N = 15. |

Seema Duhan said: (Jul 13, 2013) | |

First Two days work done by A = (1/20)+(1/20)=> 2/20=> 1/10. Third day work done by A, B, C = (1/20)+(1/30)+(1/60)=> 6/60 => 1/10. 3 days work done: 1/5. Then 1 day work done: (1/5)*(1/3) = 1/15. So the answer is 115 days. |

V.Madankumar said: (Aug 8, 2013) | |

A's 2 day's work = (1/20x2) = 1/10. (A + B + C)'s 1 day's work = [(1/20)+(1/30)+(1/60)] = 6/60 = 1/10. Work done in 3 days = [(1/10)+(1/10)] = 1/5. Now what they ask "How many days can "A" do the work?" Here revise the FORMULA NO:2. That is: Days from Work: If A's 1 day's work =1/n,then A can finish the work in n days. Here ===> [1 DAY's work = NO.of WORK / required "n" Day's do the work ]. We want no.of DAY's; So rearrange the formula; we get, No.of WORK = (1DAY's work * "n" DAY's) ------> take eqn no.1. Now, (1/5) work is done in 3 days. 3 = 1/5. So apply above the data's in eqn no.1. 1(No.of Work) = (Day's of Work )3 * ("n" DAY's)5. 1(No.of Work) = 15 Day's So "A" is require 15 Day's to do a 1 work. "I HOPE U WILL B UNDERSTAND" |

Sundas said: (Sep 4, 2013) | |

I didn't get step 1 how a 2 days work? I mean why we multiply with 2? |

Ganesh said: (Sep 5, 2013) | |

Guys, as per my understanding, 1st day A complete 1/20th of work. 2nd day A completes 2/20th of work. 3rd day A along with B and C completes (3/20+1/30+1/60) i.e., 1/5th of the work. Could some one explain how to calculate? 4th day of A's work ? 5th day of A's work? 6th day of A along with B and C ? |

Arijit Roy said: (Sep 29, 2013) | |

It would be better if it is done in this way. B, C assist A on the 3rd day so directly we get, 3/20(3 day's of A) + 1/30(assistance of B) + 1/60(assistance of C) = 1/5. So 1/5 done in 3 days whole is done in 15 days. |

Bond said: (Oct 8, 2013) | |

1/5 of work --->3 days. 1 complete work ---> ? = (1*3)/(1/5). = 15 days. |

Siddalingaswamy C said: (Oct 26, 2013) | |

1/20+1/30+1/60=6/60. Can any one say the 6 came in 6/60? |

Liza said: (Nov 16, 2013) | |

In 1/20+1/30+1/60 we take LCM of 20, 30, 60 i.e. 60 then, 60/20 = 3. 60/30 = 2. 60/60 = 1 then, 3+2+1 = 6. So 6/60. Got it. |

Nitish Mehta said: (Dec 3, 2013) | |

@Liza. What will you do after calculate 6/60? Complete your answer. |

Shakti said: (Dec 7, 2013) | |

Let total work = 60. A's effi.=3, same as B and C have 2 and 1. Then work 1st 2nd 3rd 3 3 6 Total work in 3 days 12 and hence they take 15 days to complete the work(60). |

Praveen.M said: (Dec 17, 2013) | |

Hey guys I have done this way, correct me if I'm wrong. Let total work be 100%. A does 5%work in a day (1/20*100). B does 1/30*100 i.e. 3.3%work. And like wise C does 1.6% of work. A completes 10% work in 2 days and on 3rd day he is assisted by B and C hence on 3rd day A does 15%of work and total work on 3rd day is 15%+3.3%+1.6% total is 19.9% so total time is 100/19.9*3 = 15. |

Sami said: (Dec 22, 2013) | |

Hi. Can you explain me that how do you get A's 2 days work? |

Jansi said: (Jan 2, 2014) | |

@Sami. Dear friend A is assisted with B & C every 3rd day, Means first 2 days A is doing work lonely. |

Amey said: (Jan 21, 2014) | |

Why we are multiplying 3 by 5 ? |

Randheer said: (Jan 30, 2014) | |

Please anyone can explain what formula we have to use for this problems? |

Lalas Hasan said: (Feb 11, 2014) | |

A Little Addition to @Vikas's explanation A is working alone for two days, 3rd day he is assisted by B and C. A's 1 day work=1/20. A working on 2 days=2*1/20=1/10. A+B+C working on 3rd day, so 1 day of working together = 1/20+1/30+1/60 = 6/60 = 1/10. So total work done till 3rd day=1/10+1/10=2/10=1/5. So if in 3 days = 1/5 of work is completed.... Total work is always 1; So in order to make the value of RHS 1 multiply both side by 5; Then, 3*5 days = 1/5*5 work will be completed. = 15 days. |

Ggg said: (Feb 26, 2014) | |

A can do 1/20 of the work per day. B can do 1/30 of the work per day. C can do 1/60 of the work per day. Together, they can do 1/20 + 1/30 + 1/60 of the work per day. But since B and C only help every third day, they can do, on average, 1/20 + 1/3 (1/30 + 1/60) of the work per day. = 1/20 + 1/3 (1/30 + 1/60). = 1/20 + 1/3 (3/60). = 1/20 + 1/3 (1/20). = 1/20 + 1/60. = 4/60. = 1/15. So if they can do 1/15 of the work per day, they can finish the job in 15 days. Answer: 15 days. |

Haphyz said: (Feb 26, 2014) | |

I think @Manoj made it simpler and direct. Get the portion of the job that would be completed in 3 days and then keep adding that portion in 3 days interval until you get 1. |

Tejal said: (Mar 4, 2014) | |

I can't understand that how the total work is calculated which is done by A. |

Hrmridha said: (Mar 27, 2014) | |

A work in 3 days 3/20 part of the work. Third day B and C work (1/30 + 1/60) = 1/20 part. So, after 3 days work done by A, B and C = (3/20 + 1/20) = 1/5 part. Now, 1/5 part done in 3 days. 1 part done in 5/1*3= 15 days (Ans). |

Neha Sharma said: (Apr 16, 2014) | |

I am Neha sharma I want to know, How comes 6/60 = 1/10. And 1/10 + 1/10 = 1/5. If in 3 days work = 1/5 then how is that in 1 day =1/15. Is it should be in 1 day 1/5 and by adding in 3 days =1/15. Please tell me I gonne be confused. |

Mahesh Dubey said: (Apr 30, 2014) | |

Its very easy. If we cancel 6/60 = 1/10. And 1/10 + 1/10 = 2/10 (because upper term is add and lower term is common, so when we cancel 2/10 then you got = 1/5. |

Varinder Singh said: (May 16, 2014) | |

Guys I want know that in last step why did the multiplied 5 and three. Please explain. |

Azam Rizwan said: (May 16, 2014) | |

Why are we multiplying it by 5 in the end? |

Zyr said: (May 27, 2014) | |

1work/5days = 3 days or 1/5=3 then cross multiply to get 15. |

Alan said: (Jun 6, 2014) | |

Work done by A+B+C for 1 day is 1/10. So work done by them for 3 days should 1/10 + 1/10 + 1/10 = 3/10. How come its just for 2 days giving 1/5. |

Eliz said: (Jun 13, 2014) | |

Day1: A works alone = 1/20. Day2: Again, A works alone = 1/20. Day3: Includes work of A, B and C = 1/20+1/30+1/60. =>6/60. =>1/10. NOW, work in all 3 days = 1/20 + 1/20+ 1/10. =>1/5. Work in 1 day =1/(3*5). =>1/15. Taking reciprocal now will give us the no: of days. i.e., 15. |

Shruthi said: (Jul 23, 2014) | |

A can do 1/20 of the work per day. B can do 1/30 of the work per day. C can do 1/60 of the work per day. Together, they can do 1/20 + 1/30 + 1/60 of the work per day. But since B and C only help every third day, they can do, on average, 1/20 + 1/3 (1/30 + 1/60) of the work per day. 1/20 + 1/3 (1/30 + 1/60) = 1/20 + 1/3 (3/60) = 1/20 + 1/3 (1/20) = 1/20 + 1/60 = 4/60 = 1/15. So if they can do 1/15 of the work per day, they can finish the job in 15 days. |

Rajaguru said: (Jul 23, 2014) | |

@Manoj mass reply nice bro, 3 days = 1/5. 6 days = 2/5. 9 days = 3/5. 12 days = 4/5. 15 days = 5/5(1) == work completed. |

Arun said: (Jul 28, 2014) | |

How to make first step? |

Sudha said: (Aug 1, 2014) | |

Is there any other simple trick to solve this? |

Vishal Grade said: (Aug 25, 2014) | |

A's 1 day work = 5%. B's 1 day work = 3.33%. C's 1 day work = 1.66%. A, B and C' 1 day work = 5+3.3+1.6 = 10%. On 1st 2 days total work = 5+5 = 10%. On 3rd day total work(by all 3)=10%. In 3 day total work done=10+10=20%. 20% work done in ---- 3 days. 100% work done in --- 15 days. |

Divyanshu Dwivedi said: (Sep 2, 2014) | |

It is much easier by this method : 1/20 + 1/(30*3) + 1/(60*3) = 1/x. 1/20 + 1/90 + 1/180 = 1/x. 12/180 = 1/x. x = 15 Ans. |

Psyco said: (Sep 9, 2014) | |

1/5th work done in 3 days. 5/5th of work will be done in x days so, 1/5---3. 5/5---x. x=3*5*5/5. x=15. |

Golu said: (Oct 6, 2014) | |

Why we are calculating 2 days of work A? |

Chitra said: (Oct 14, 2014) | |

Work done in 3 days = 1/5. Work done in next 3 days = 1/5+1/5. Work done in 5 days = 1/5+1/5+1/5+1/5+1/5 = 5/5 = 1 (task completed). Hence 3*5 is used. |

Mir Sakib said: (Oct 14, 2014) | |

What is its mean every third day? |

Ajit said: (Oct 27, 2014) | |

Why multiply 1/20*2? |

Vijay said: (Oct 29, 2014) | |

B and C joined with A for every 3 days. So their (A,B,C) 3 days work is == (A's 2 days work i.e. [1/10]) + (A+B+C 's 1 day work i.e [1/10]). = 1/10+1/10 = 1/5. So for 3 days 1/5 work is done. So total work will complete in 15 days. |

Deepti said: (Dec 9, 2014) | |

Its written every 3rd day NOT First 3 days. Therefore the answer is as below: (A+B+C)'s 1 day work = (1/20 + 1/30 + 1/60) = 1/10. => They all take 10 days to complete their work together. Out of those 10 days, 3th, 6th and 9th are the 3rd days. In total B and C can help A for only 3 days. Therefore, (A's 1 days work) + (B+C) 1 day work*3 = 1/20 + (1/30+1/60) *3 = 4/20 = 1/5. If 1/5 is the work in 3 days, Then complete work done = 5*3 = 15. Ans: 15 days. |

Nandha said: (Dec 12, 2014) | |

@Ajit. A do a work two days independently, so (1/20+1/20) = 1/10 instead of they taken (1/20*2). |

Ranju said: (Dec 12, 2014) | |

(A+B+C)'s 1 day of work is (1/20+1/30+1/60) = 1/10. A's 2 days of work is (1/20 X 2) = 1/10. As on the third day A is assisted by B and C. => A's 2 days of work + On 3rd day A is assisted by B and C i.e. (A+B+C) 's 1 day of work. 1/10 + 1/10 = 2/10 = 1/5. As 1/5 part of work is done in 3 days. Therefore 1 part of work is done in 3 X 5 = 15 days. |

Garima Yadav said: (Dec 26, 2014) | |

I haven't got it. Can anyone please help me for this? |

Ashwini said: (Dec 28, 2014) | |

Can you explain how 6/60 came? |

Animesh said: (Jan 4, 2015) | |

A 20--------------------3. B 30 ---LCM 60 -------2. C 60--------------------1. Total = 3+2+1 = 6. So answer = 60/6 = 10. |

Sindhu said: (Jan 5, 2015) | |

@Ashwini. As they mentioned in the question A is assisted B and C on every third day means on every third day A B C do the work together. That means you have to add A B C 's one day work. i.e 1/20+1/30+1/60 = 6/60 = 1/10. |

Trinadh said: (Jan 6, 2015) | |

1/20, 1/30, 1/60. Add 1/3 to the ends of both B and C. = 1/20+1/30*3+*1/60*3. = 1/15. |

Veda said: (Jan 20, 2015) | |

"A" can complete work in 20 days. A's 1 day work is 1/20. A's 2 days work is 2/20. According to the question A work for 2 day's and on the third day. B & C assists (**helps, not replaced). So it goes like this. A's first day + A's second day + (A + B + C) on third day, B & C helped. i.e 2 1/20+1/20/+(1/20+1/30+1/60). =>2/20+(1/20+1/30+1/60). =>2/20+1/10. = 1/5. So for first 2 days and combining B and C on third day they will work 1/5th of a work. All together (A+A+ (A+B+C) ) works for 3 days to get 1/5th of a work so. If 1/n work is done in 1 days total work is done in 1*n = n days. The same here, if 1/5 work is done in 3 days total work is done in 5*3 = 15 days. |

Ram said: (Feb 5, 2015) | |

Last step: 1/5 work = 3 days. W/5 = 3d. W = 3*5d. W = 15d. Full work = 15 days. |

A Govardhan Reddy said: (Apr 9, 2015) | |

A can do 20 days so = 1/20. A can do with B and C every third day = 20/3 = 18+2 (alone). A+B+C = (3/18+1/30+1/60) = 1/5. So the A work done in three days = 5*3 = 15 days. |

Farha said: (Apr 12, 2015) | |

The question is not about to ask the work of 3 days. Then why we are finding 3 days's work. Can anybody explain, I am so confused. |

Sunil said: (Apr 15, 2015) | |

@Nikita answer is fantastic. |

Raman said: (Apr 17, 2015) | |

"A" can complete work in 20 days. A's 1 day work is 1/20. A's 2 days work is 2/20. A's first day + A's second day + (A + B + C). => 1/20+1/20/+ (1/20+1/30+1/60). =>2/20+ (1/20+1/30+1/60). =>2/20+1/10 = 1/5. So 3 days work = 1/5 convert it to one day work. 3/1/5 = 5/1*3 = 5*3 = 15 days. Other way: = 1/5 in three days we want to finish work. = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 5/5 = work finished. |

Raj said: (Apr 21, 2015) | |

Question is on every 3rd day, it means from 20 days B and C will assist him on every 3rd day. So count every 3rd day till 20 days. So we get 6 days. It means B and C will assist him 6 days from 20 days i.e. on 3rd, 6th, 9th, 12th, 15th and 18th day. I think the answer is 16. |

Suribabu said: (Jun 7, 2015) | |

In this problem first we understand the work done till 3rd days. A can do his work for 2 days (2*(1/20)) and the 3rd day. A can do work with B & C i.e. 1/20+1/30+1/60 = 1/10 so up-to 3 days. A can do 1/10+1/10 = 1/5 work (3 days work) then total work done in 3*5 = 15 days. Hope understand. |

Daniyar said: (Jun 9, 2015) | |

For 1st 2 days A will do work himself. A's work in 1st 2 days: (1/20)*2 = 1/10 or 0.1. 0.1 is 10% of total work done which is 1 or 100%. On 3rd day B and C will assist A to do work. A's, B's, and C's work in 3rd day: (1/20+1/30+1/60)*1 = 1/10 or 0.1. Again in this day three of them could do only another 10% of work out of 100% or 0.1 out of 1. For 1st three days A, B, and C accomplished: 1/10+1/10 = 2/10, 0.2 or 20 % of total work out of 100% or 1. To finish they need to accomplish 100-20 = 80 % or 0.8 of work remaining. Having 20% or 3 days we need 15 (in which every 3rd day is assisted by B and C) more days (100%/20% = 5 => 5*3 = 15) to achieve total work accomplishment. |

Pawan Choudhary said: (Jun 13, 2015) | |

Why the first step I don't understand please explain me? |

Aby said: (Jun 13, 2015) | |

Please explain the 3rd step. |

Susi said: (Jun 20, 2015) | |

A's 2 days work: (1/20*2). Answer: 1/10. A, B, C add 1 days work. (1/20 add 1/30 add 1/60). Answer: (110/36000).(110/2/36000/600). Answer: (5 add 1/60). Answer: 1(1/10). Three days work (1/10 Add 1/10). Answer (1/5)*3. Answer: 16. |

Neha Sharma said: (Jul 2, 2015) | |

Why you are multiply 3*5? And in third step why 1/10 is taken in two times? Please exlain it? |

Saraswati said: (Jul 8, 2015) | |

Another method to solve this problem: A can do the work in 1 day = 1/20. A can do the work in 2 days = 1/20*2 = 1/10. B+C can do the work in 1 day = 1/30+1/60 = 1/20. B+C can do the work in 3 days = 1/20*3 = 3/20. A is assisted by B and C on every third day. So that 1/20+3/20 = 1/5 (1 days work of A+3 days work of B and C). Now, 1/5 work is done in 3 days, Therefore whole work will be done in (3*5) = 15 days. |

Sandhu said: (Jul 13, 2015) | |

Sir I can't understand the calculation of 1\20+1\30+1\60 please tell me. |

Bindu said: (Jul 15, 2015) | |

Here 1/5 is calculated considering ABC together but here we have to fine work done by A? |

Lallitha said: (Jul 15, 2015) | |

In given question, A is assisted on every 3rd days but you all doing for one 3rd day? Can anyone explain? |

Vipul said: (Jul 22, 2015) | |

Work of 1 day = 1/20+1/30+1/60= 6/60= 1/10. It means that 1/10th part of work is done in 1 day. (It means 10% work per day). Now every 3rd day B And C performs -ve work i.e, Work of every 3rd day = 1/20- 1/30-1/60 = 0/60 = 0. Days 1 2 3 4 5 6 7 8 9 10 11 12 13 14. Work% 10+10+0+10+10+0+10+10+0+10+10+0+10+10 = 100%. Hence work will complete in 14 days. Note: You can consider day 15 because in day 15 the work done will 0%. So more correct answer will 14 days. |

Satish Kumar said: (Jul 23, 2015) | |

How will make step 3 (1/10+1/10) explain? |

Varun said: (Aug 1, 2015) | |

How you can take total working days as 3 in last step? |

Narendra Patidar said: (Aug 10, 2015) | |

Please clear explain. |

Manisha said: (Aug 12, 2015) | |

Please tell me answer in LCM method. |

Lakhan said: (Aug 13, 2015) | |

Why 3 is multiplied by 5 in first method? |

Mareeswari said: (Aug 20, 2015) | |

Sir I can't understand the A is working two days please explain. |

Manju Verma said: (Aug 25, 2015) | |

Thank you for the solution its really easy way to understand. |

Jibinlal said: (Sep 2, 2015) | |

1/5 work is done in 3 days. Remaining work is 4/5. 1/5+1/5+1/5+1/5 = 3+3+3+3 = 12 days. Total = 12 days (4/5th work)+3 days (1/5th work). So 15 days. |

Tarun said: (Sep 3, 2015) | |

Anyone explain this question with LCM method. |

Manju said: (Sep 4, 2015) | |

LCM OF 20, 30, 60 is 60. Work out LCM method that means 10*2*3=60. |

Abhishek said: (Sep 5, 2015) | |

Let A does work in x days. Let B does work in y days. y - x = 60. 3x = y. 3x-x = 60. x = 30. y = 90. Hence, both will do work in (1/30+1/90) = 6/135. No of days = 135/6. 22.5 = 45/2. Answer = 22.1/2. |

Chittu said: (Sep 16, 2015) | |

B and C can complete this work in 20 days. B and C one day work will be = (1/20). Instead of assuming every third day B and C will join A in work, let us reduce the efficiency of B and C by 3 so this makes B and C one day as 1/60. So now doing the calculations. (1/20)+(1/60) = 1/15. Hence they can complete the work in 15 days. Please let me know in case of any clarifications needed. |

Chirag Gupta said: (Sep 18, 2015) | |

I read this but understood nothing please explain in easiest way. |

Shivam said: (Sep 20, 2015) | |

Let n be the no. of total days required; A's one day work = 1/10. B's = 1/30, C's = 1/60. So n(1/10)+n/3(1/30+1/60) = 1. From here, we get n = 15. |

Siddhu said: (Sep 25, 2015) | |

Here a work can be completed in by A in 20 days and on third day B and C assist that means A should work for 14 days and remaining 6 days work is assisted by B and C. So help me solve in this way. |

Akash said: (Sep 25, 2015) | |

Hey why in last we multiplied 1/5 with 3? |

Priya said: (Sep 29, 2015) | |

How to solve (1/20+1/30+1/60)? |

Parth said: (Oct 4, 2015) | |

What is the Mean assisted in this question? |

Praveen said: (Oct 14, 2015) | |

1/5x100-20 percent work done in 3 days therefore for 100 cross multiply 15 days. |

Nitin said: (Oct 25, 2015) | |

A = 20 = 1 day work of A = 3 LCM = 60. B = 30 = 1 day work of B = 2. C = 60 = 1 day work of C = 1. Day 1 2 3 1 2 3. Work 3 3 (A+B+C = 3+2+1). Work of 3 days = 2(A) + (A+B+C). 2*3+(3+2+1) = 12. 3 days = 12 unit work (which number we multiply both side that do not greater than the total unit work). 3*5 days = 12*5 unit work. 15 days = 60 unit. |

Onkar said: (Oct 30, 2015) | |

A can complete a work in 20 days, B can complete the same work in 30 days and with the help of C together if they complete the work in 60 days, so assume total work to be done is 60. (LCM of 20, 30, 60). Persons A B C per 3 day. Days- 20 30 60 12 = 6+6. Work/day- 3 2 1 5. ----------------------------------------------------------------. Work - 60 60 60 60. Per 3 day ----> 12. How many ----> 60 = (60*3)/12 = 15 days required. Days required. |

Junaid said: (Oct 30, 2015) | |

A do 1/20 = 0.05 work daily and in 4 days he do 0.05+0.05+0.05+0.05 = 0.2 work. But with the help of B and C the 4 days work is done in 3 days in this way. = 0.05+0.05+(0.05+1/30+1/60) = 0.2. Then every 4 days work is done in 3 days up-to 20 days. So 3+3+3+3+3 = 15. |

Venkat said: (Nov 4, 2015) | |

3 = 1/5. So, our complete work done at 1. So, 3*5 = 1/5*5. 15 = 1. |

Venkat said: (Nov 7, 2015) | |

Please tell me how 3 is coming in the last step? |

Harry Matharoo said: (Nov 8, 2015) | |

Now, 1/5 work is done in 3 days. Whole work will be done in (3 x 5) = 15 days. Where from 15 come from why we multiplied 3 with 5. |

Vicky said: (Nov 11, 2015) | |

Please anyone give clear explanation I confused. |

Vani said: (Nov 23, 2015) | |

In last step why did they multiplied with 3? |

Teja said: (Dec 17, 2015) | |

This is question which is completely related to alternate days. In alternate days question whenever work done is not given you have to take it as '1'. Here we got that 1/5 of the work is done in 3 days which means 1/5 work=3 days. 1=how many days? Cross multiply, you get days=15. This concept works for all of the alternate days question. When ever work is not given take it as '1'. |

Jitesh said: (Jan 13, 2016) | |

A = 20. B = 30. C = 60. LCM is 60 so a do the work in 1 day is 3 unit B is doing work in 1 day is 2 unit and C is 1 unit. 1st (A) day = 3. 2nd (A) day = 3. 3rd (A+B+C) day = 3+2+1 = 6. Total work done in 3 days is 12 unit. So for 60 unit (60/12 = 5). 5*3 days = 15 days. |

Ganesh said: (Jan 22, 2016) | |

Total work = 60 units (lcm of 20, 30 and 60). A completes work in 20 days. It means 3 units a day. B completes work In 30 days. It means 2 units a day. C completes work in 60 days. It means 1 unit a day. A is working 2 days alone. It means 3+3 = 6 units is completed. On third day B and C helping A. It means on third day 3 +2+1 = 6 units of work is completed. It means for 3 days 12 units of work is completed. For 60 units? 60*3/12 = 15. |

Samrat said: (Jan 27, 2016) | |

Note: A alone can do its work in 20 days. So A can do the work in 1 day is 1/20. 2nd day A can do work = (1/20)2 = 1/10. 3rd day A, B and C work together. Therefore A's 1 day work = 1/20, B's 1 day work = 1/30, C's 1 day work = 1/60. Together A, B and C do work in 1 day = 1/20 + 1/30 + 1/60. = (3+2+1)/60. = 6/60 = 1/10. So after 3 days A's done = 1/10 + 1/10. = 2/10 = 1/5. = (1/5)X days = 1. X = 5. 1/5 work = 3 days. (1/5)X days = (3)5 days = 15 days. |

Sudhansu Sekhar said: (Feb 3, 2016) | |

Why applied 1/10 + 1/10? |

Shridhar said: (Mar 1, 2016) | |

A as done the work for 2 days, he is asserted to B and C in 3rd, he as not done the work at 3rd day. Then why 1/20 + 1/30 + 1/60? |

Shridhar said: (Mar 2, 2016) | |

Got it on 3 day B & C is helping A. |

Ompal. said: (Mar 22, 2016) | |

A, B, and C person can a do work 20, 40 and 60 days respectivily. When they works alternative. They compeleted work how many days. |

Shruthi said: (Mar 26, 2016) | |

Please explain the 2nd step => (A + B + C) 's 1 day's work. How it becomes 6/60. |

Jemish said: (Mar 29, 2016) | |

My new approach to explaining the answer. => A is working continuously. => B and C join at every 3rd day. A's efficiency is used as it is, But B and C's efficiency is used 1/3rd part considering a task. thus, A--> 1/20 B--> 1/3 * (1/30) c--> 1/3 * (1/60) Now simple. To find out days required = 1/20 + 1/90 + 1/180 = 1/15. Answer: 15 days required. |

Jenni said: (Mar 29, 2016) | |

How to multiply and solve this equation? 1/20 + 1/30 + 1/60 |

Preethu said: (Apr 3, 2016) | |

Why do we multiply it by 5? How we get the answer as 15? Please help me. |

Aariz said: (Apr 17, 2016) | |

Hey, guys. I have some confusion in this question. Is there anyone from Delhi, please guide me to solve this question. |

Rayhan said: (Apr 23, 2016) | |

Good explanation. I really understand this. Thankyou all. |

Praveen said: (May 3, 2016) | |

(A + B + C) = 1/20 + 1/30 + 1/60. Given 180/1800. For 20, 30 & 60 the common multiple is 1800 (LCM). Then you'll get the sum of 180/1800. This is how I got 1/10 for the work done by A, B & C in 1 day. |

Bhavnesh said: (May 17, 2016) | |

Hi friends. I just want to know why we are taking 3 days work for 1/10 + 1/10. |

Ismail said: (May 18, 2016) | |

Why didn't we take first B + C one day work is 1/30 + 1/60 = 1/20? Please explain me. |

Aaron said: (Jun 1, 2016) | |

At the last step 1/5 * 3 we must do reciprocal right? Or just multiple and get 15 for the sake of the answer in the options. Please tell me. |

Srinivas said: (Jun 6, 2016) | |

How A's 2 day's work = 1/10? Please explain the step. |

Anu said: (Jun 6, 2016) | |

A = 20, B = 30 &C = 60. Total work (LCM of 20, 30, 60) = 60. A's 1day work= 60/20 = 3units. B's 1day work= 60/30 = 2. C's 1day work= 60/60 = 1. 1st day A will do the work =3 unit. 2nd day again A will do the work = 3. In 3rd-day, A + B + C will do the work = 3 + 2 + 1 = 6. After 3 days total work = 12 (3 + 3 + 6). After 6 days total work = 24. Similarly; After 12 days total work = 48. 15 days total work = 60. So total no days to complete the work (60) is 15 days. |

Sreelekshmi said: (Jun 6, 2016) | |

3 day's work is 1/5. So, 1 day's work is = 1/15. Total no of day's to complete the work is (reciprocal of 1day's work) = 15 day's. |

Emiley said: (Jun 14, 2016) | |

A can do apiece of work in 30 days. B in 50 days and C in 40 days. If A is assisted by B on one day and by C on the next day alternately work will be completed in? |

Sangeetha said: (Aug 2, 2016) | |

@Nikitha. Superb explanation. |

Pavankushoba said: (Aug 5, 2016) | |

Hi everyone here is my solution. According to problem (a + b + c) 1 day work = 1/10. As per the problem A is assisted by B and C every third day so first let us find, Work done by A in 2 days = 1/20 * 2 = 1/10. As mentioned in step 3 B and C are assisted on the third day. (So A's 2-day work, And one important point A also work along with B and C on third day = 1/10 + (1/20 + 1/30 + 1/60) = 1/5. Its 3 days work done by A B and C = 1/5. 1-day work = x (assume). By cross multiplication. We get 3x = 1/5. x = 1/15. So its 15 days thank you. |

Subash said: (Sep 3, 2016) | |

In 3 days they work is 1/5; Then remaining work is 1 - 1/5 = 4/5, So using proportions 1/5 : 45 :: 3 : x. Get 12days. + 3days = 15days. |

Parshu said: (Sep 3, 2016) | |

3 values there 3persons working to gather assisted A's B + C we calculate A + B + C/2 we get, A+ B + C = 1/20 + 1/30 + 1/60. LCM denomineor = 1/10 = 10 * 3 total 30/2 = 15-> 3persons work. |

Sid said: (Sep 15, 2016) | |

According to me, the answer is [1/10 + 1/10 + 1/20] = 1/4 for 3 days the whole days is 4 * 3 = 12. |

Harsha said: (Sep 16, 2016) | |

According to the question the A has been assisted by B and C after the third day. It means the A has to work for three days and the third day is followed by B and C. |

Sarathy said: (Sep 22, 2016) | |

A's total work = 1. A's 1 day work = 1/20 = 0.05. B's 1 day work = 1/30 = 0.03. c's 1 day work = 1/60 = 0.01. B + C combined work together = 0.03 + 0.01 = 0.04. A's 3 day work = 0.05 + 0.05 + 0.05 + 0.04 = 0.19. If answer is 15 means = 5 * (0.19) = 0.95 remaining 0.05. So, the answer is wrong. In 16th day = (15days work + A's one day work). = (0.95 + 0.05), = 1. So, 16 days is the correct answer. |

Puspanjalisahoo said: (Sep 29, 2016) | |

I am not understanding A's 2 day's work. Please explain. |

Priya said: (Oct 21, 2016) | |

@All. Please explain the last step. |

Rudra said: (Oct 28, 2016) | |

Here, first of all, we have to find the total work. So for this, L.C.M of 20 30 60 = 60, a's b's c's one day work =3 2 1 (one day work) Now a's 2days work = 6. b's, c,s one day's woks = 2. +1= 3. So three days work 3 + 3 + 3 + 3 = 12. 3 days = 12 3*5 =15 days, 12 * 5 = 60. |

Anand J said: (Nov 4, 2016) | |

I'm finding difficulty in understanding the question. How to analyze the question? Please help me out. |

Mohammed said: (Nov 22, 2016) | |

Why here they take 2? |

Vikash said: (Dec 11, 2016) | |

Here first we take LCM of 20,30,60 that is 60. After that , A work done = 60/20 = 3 Similar, B =60/30 = 2 and C = 60/60 = 1. According to questions, A is assisted by B and C on every 3rd day. i.e : 3 persons are Working. Total work = A + A +(A + B + C) (Here a ,B and come together on every third day). Total work = 3 + 3 + (3 + 2 + 1) = 12. Now work = 60 is divided by 12 rem is 0. Means 60/12 = 5 . We got 5 and 3 persons are working. So that, 5 *3 = 15 days. |

Aiswarya said: (Dec 15, 2016) | |

How did we get that 1/10+1/10 ? Can anyone help me? |

Prashant Hatwar said: (Dec 20, 2016) | |

Here A's 2 days work = 1/20 * 2 = 1/10, And every third day A, B and C working together, Means A, B, C's 1 day work is A + B + C = 1/20 + 1/30 + 1/60 = 3 + 2 + 1/60 = 6/60 = 1/10 So here complete 3 days work is 1/10 + 1/10 = 2/10 =1/5. It means 1/5 work done in 3 days. So work remains is 1 - 1/5 = 4/5. So here, 1/5 : 4/5 :: 3 : x 1/5 * x = 4/5 * 3 x =12. It means 4/5 work done in 12 days, so complete work done in, 12 + 3 = 15days. Therefore, the answer in 15. |

Rah said: (Feb 4, 2017) | |

Suppose B and C are helping a, = b 1/30 + c 1/60 + a 1/20. But B and C are helping only on third day so( b+c / 3)+a = (1/30+1/60 =3/60 = 1/20/3=1/60)+ 1/20= 1/20+1/60=4/60=1/15 = 15. |

Subash Patra said: (Feb 24, 2017) | |

Take whole work is 1.B and C assists A only 3rd day. So A"s two day work is 1/10. On 3rd day total work is done is 1/5. That means 1/5 work done taken in 3rd day. So whole work(1) is done in 15 days. |

Lkesh R said: (Mar 9, 2017) | |

For whole work take lcm of 20,30 and 60 ie 60 so whole work is 60. A alone work = 60/20 = 3w/d. B=2w/d. C=1w/d. A+b+c= 6w/d. A work 3 /day so 60/3=10 days. It means A alone work 10 days and ie, 10*3(A per day work)= 30. Remaining work 60-30=30. So remaining 30 work done by; A+B+C= 30/6(pr day work)= 5 days, So total days 10(alone A)+ 5(A+B+C)=15 days. |

Jpv said: (Mar 21, 2017) | |

Hi, I'm not getting the below logic in quotes " Complete 3 days work is 1/10 + 1/10 = 2/10 =1/5" Shouldn't that be [ (A's ist day)+ (A's 2nd day) + (A+B+C combined on 3rd day)]? ie = 1/20 + 1/10 + 1/10? Please correct me if I am wrong. |

Jpv said: (Mar 21, 2017) | |

Works that done till 3rd day: [ (A's ist day)+ (A's 2nd day) + (A+B+C combined on 3rd day)]. ie = 1/20+1/10+1/10, = 1/4. ie remaining pending work = 3/4. So, the no of days Al required to complete the work= 20*(3/4), = 15 days. |

Amit Yadav said: (Apr 11, 2017) | |

Explain the Step 1. A's 2 day's work = 1/20*2= 1/10. Explain it clearly. |

Akash Kumar said: (Apr 12, 2017) | |

Here 3'day's work is 1/5 so next 3 day work is also 1/5. 1/5+ 1/5+ 1/5+ 1/5+ 1/5 i.e its happen 5 times so 3x5= 15 days. |

Shubham said: (May 20, 2017) | |

Please explain the last step ( 1/10 + 1/10) = 1/5. |

B Sundara said: (Jun 3, 2017) | |

Since B and C assisting A on every 3rd day, B & C takes 30x3=90 days and 60x3=180 days respectively to do the work. Therefore 1/20+1/90+1/180 = 12/180 = 1/15 or 15 days. |

Abhishek Rajput said: (Jun 6, 2017) | |

solve :: x * (1/20)+x/3*(1/30+1/60) = 1 for x, where x is number of days A worked. Please, anyone solve this. |

Bhaskar I said: (Jun 26, 2017) | |

A - 20. B - 30. C - 60. Total work is 60. Because LCM of ( 20,30,60 ) is 60. Then A one day work is 60/20 = 3, B one day work is. 60/30 = 2, C one day work is 60 /60 = 1, 1st day work is only A = 3. 2nd day work is only A =3. 3rd day work is A+B+C (3+2+1) = 6. So., 3 day work 12. 6 day work 24. 9 day work 36. 12 day work 48. 15 day work 60. Total 60 work is completed in 15 day. It's simply |

Bhuvana said: (Jul 1, 2017) | |

It is solved by chocholate method 20 30 60 lcm of those 60. 3 2 1 are their 1 day works. a a a+b+c a a a+b+c a a a+b+c a a a+b+c a a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 3 +3 + 7 +3 +3 +7 +3 +3 +7 +3 +3 +7 +3 +3 +7=60 days. |

Yash said: (Jul 2, 2017) | |

Why are we multiplying by 5 for whole work? Explain. |

Shruthi said: (Jul 8, 2017) | |

From the explaination, we got to know that in 3 day's 1/5 of the work completed. If A's one day's work is 1/20 then he can complete in 20 days. Same way we should find out 1 day work for the above problem. i.e., If 3 day's work=1/5. then one day work=(1/5)/3 =>1/15, =>so 1 day work=1/15 then he can complete in 15 days. |

Deepa said: (Jul 8, 2017) | |

Can you explain me how it is 1/5 and 3*15? I'm not getting this concept. |

Pooja Kulkarni said: (Jul 12, 2017) | |

It can be done in an easier way than the given solution: Let the number of days required = x. So basically, B will contribute (1/3)rd of its work in 30 days and C will contribute (1/3)rd of its work in 60 days. Work by A in one day: 1/20. Work by B in one day: 1/30. Work by C in one day: 1/60. So total work can be easily calculated as: (x/20) + (1/3)*(x/30) + (1/3)*(x/60) = 1. So, 12x = 180, And thus x = 15 days. |

Surya said: (Jul 19, 2017) | |

Usually, we are taking 20 days means 1/20. Then why there finding 2 days work is multiplied by 2? My thoughts are 1/20*1/2. Please explain me. |

Veena said: (Aug 10, 2017) | |

Why we add 1/10 and 1/10 here? Please explain. |

Dhivyabharathi said: (Aug 10, 2017) | |

Nice explanation @Nikita. |

Salman said: (Aug 14, 2017) | |

How 6/60 come? |

Vishnu Vardhan said: (Aug 19, 2017) | |

According to me, the answer is 12. |

Magnesh said: (Aug 31, 2017) | |

Simple, 3rd day B and C will hep A. So first 2 days A will be working his work for two days is 2*(1/20) = 1/10. Now 3rd day he is helped by B and C . So A's work + B's work + C's work for 1 day is( 1/20 +1/30 +1/60) take LCM and solve we get 1/10. Now A's 2days of work + (A+B+C's )work for 1 day. 1/10 +1/10 = 1/5. So 1/5 work is completed which includes A's 2 days of work and Band C assistance for A (previous step same thing) now 1/5 work is done in --- 3days 1( indicates complete work ) in ---- ? days Ans is 15 days. |

Shanmugapriya said: (Sep 29, 2017) | |

Nice explanation, Thanks @Vijay. |

Nike said: (Oct 13, 2017) | |

We have A work 2 days alone third-day B&C are added that means A+A+ (A+B+C) THIS is 3days work. So A can do 2days work=1/10, and A+B+C 1day work=1/10, so add 1/10+1/10=1/5then total day work by A is 3*5=15. |

Anuj said: (Oct 28, 2017) | |

Nice explanation, Thanks. |

Divya said: (Nov 27, 2017) | |

Simply; A's 3 days works ----- 3/20. B and C join on 3rd day so; (A+B+C)'s 3 days work will be ---- 3/20 +1/30+ 1/60 = 12/60 = 1/5. 3 days work -- 1/5 then; 1 day work -- 1/15 part of 1 work (i.e 1/15 work in 1 days), 15 days = 1 work. |

Akshay said: (Dec 16, 2017) | |

Take LCM ot 20,30 and 60 you will get total units of work i.e 360 units. Now find the 1 day work by using UNITS/DAYS. A=18 units B=12 units C=6 units Now A+B+C 1 day work is 18+12+6=36 units. A+B+C will complete the work in 10 days(360/36). Now, B and C joins every 3rd day which means A will work for 2 days alone and A+B+C will work on the 3rd day. A 1day work = 18, A 2 day work= 18*2=36 units, A+B+C 1 day work is 36 units, So 3 days of work =36+36= 72 units, So 1-day work will be 72/3. Now total work was 360 units. Total days= 360*3/72 = 15 Days. |

Rizwan said: (Dec 19, 2017) | |

How this 2 step comes 1/60? explain please. |

Kapil Dev said: (Jan 5, 2018) | |

day-1, day-2, day-3. A. A. A+B+C. Every 3rd-day b & c joining. So 3(3A+B+C)=3(1/20)+1/30+1/60, =9+2+1/60, =1/5, 3A+B+C=1/3*5. =1/15. 15 days work is complited. |

Abdullah Shaikh said: (Jan 14, 2018) | |

LCM method: The LCM of A,B,C is 60 so consider total work to be 60 unit. Since A take 20 days, he does 3 unit per day. Similarly, B does 2 unit per day. Similarly, C does 1 unit per day. So total unit of work done by 3 combined is 6 unit. So by Addition of each day of Unit of work done until reaching total work. So the addition of A alone and A+B+C can be summarized as. A doing work for first 2 days will be 2*3, total Work 6 done. On 3rd day work done is 6, total work done 12. Again A working alone for 4th and 5th is 2*3 =6, Total work 18 On 6th day work done (A+B+C) 6, Total work 24 On 7th and 8th A is 6, total work 30. On 9th day again (A+B+C) work done 6, TOTAL 36. On 10 and 11th A alone work was done 6, Total work 42. On 12th (A+B+C) work done 6, Total work 48. On 13 and 24th A alone is 6, Total work 54. On 15th A+B+C is 6, Total work was done 60 unit, work completed. |

Vaibhav More said: (Feb 13, 2018) | |

As A does work in 20 days. B does in 30 days. C does in 60 days. Take the lcm and you will get total work = 60 parts. No of parts A does work = 60/20 = 3 part / day. Similarly B = 2 parts /day and c= 1 parts/day. If A does work for 2 days he covers 3+3 parts = 6 parts. And every third day B & C joins and then A + B +C = 6 parts on 3rd day. Total work from first day to third day = 6 + 6 parts= 12 parts. Total work = 60 parts. Therefore no of days require = 5 * 3 = 15 days. |

Sangi said: (Feb 18, 2018) | |

Why 3 is multiplied with 5 at last? Please explain. |

Anu said: (Mar 8, 2018) | |

I really didn't get the last step where they said whole work will be done in 15 days. Can anyone explain how it comes? |

Manisha said: (Mar 20, 2018) | |

A work in one day=1/20, B work in one day=1/30, C work in one day=1/60. as A work for 2 days and then on third day B and C do work. total work done in three days = A 3 days work+B 1 day work+ C one day work, =3/20+1/30+1/60, =9/60+2/60+1/60 =12/60. Total work done in three days =1/5. 3=1/5part. Let x days required to complete work=1 part. 3/x=1/5/1. x=3*5=15 days. |

Roshan said: (Mar 30, 2018) | |

@ALL. In simple method. 1/5 work is done in 3 days, 2/5 work is done in 6 days, 3/5 work is done in 9 days, 4/5 work is done in 12 days, 5/5 work is done in 15 days. |

Kiran said: (Apr 27, 2018) | |

My answer is 17. . LCM = 60, Efficiency of A = 3/day, Efficiency of B = 2/day, Efficiency of C = 1/day. . 3+3+6+3+3+6+3+3+6+3+3+6+3+36+3+3= 60, Total 17 days. |

Tejeswar Sai said: (May 31, 2018) | |

Thank you @Vikas, very good explanation. |

Dhara Jawale said: (Jun 13, 2018) | |

Please explain the second step in detail. |

Shashank Tripathi said: (Jun 28, 2018) | |

A-> 20 days, B->30 days, C->60 days, L.C.M of (20,30,60)=60. Work = 60 unit. so the capacity of A = 60/20 = 3 unit /day. so the capacity of B = 60/30 = 2 unit /day. so the capacity of C = 60/60 = 1 unit /day. day 1 : A -> 3 unit/day. day 2 : A -> 3 unit/day. day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C) Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days. 12 unit in 3 days -> equation (1) so for 60 unit work multiply by 5 in equation (1) 60 unit work in 15 days. So A can do in 15 days. |

Shashank Tripathi said: (Jun 28, 2018) | |

A-> 20 days, B->30 days, C->60 days, L.C.M of (20,30,60)=60. Work = 60 unit. so the capacity of A = 60/20 = 3 unit /day. so the capacity of B = 60/30 = 2 unit /day. so the capacity of C = 60/60 = 1 unit /day. day 1 : A -> 3 unit/day. day 2 : A -> 3 unit/day. day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C) Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days. 12 unit in 3 days -> equation (1) so for 60 unit work multiply by 5 in equation (1) 60 unit work in 15 days. So A can do in 15 days. |

Shashank Tripathi said: (Jun 28, 2018) | |

A-> 20 days, B->30 days, C->60 days, L.C.M of (20,30,60)=60. Work = 60 unit. so the capacity of A = 60/20 = 3 unit /day. so the capacity of B = 60/30 = 2 unit /day. so the capacity of C = 60/60 = 1 unit /day. day 1 : A -> 3 unit/day. day 2 : A -> 3 unit/day. day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C) Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days. 12 unit in 3 days -> equation (1) so for 60 unit work multiply by 5 in equation (1) 60 unit work in 15 days. So A can do in 15 days. |

Riya said: (Jul 3, 2018) | |

If we use the LCM method. We get the 1 day's work of A= 3; B= 2; c= 1. & total work done by a,b,c together on 3rd day will be 6 , therefore; 60/6 = 10 & 2 days work by A= 3*2=6 , 60/6= 10. Therefore 3 days work = 10 + 10 =20. Please explain the solution after this. |

Shashikumar G said: (Jul 22, 2018) | |

Nice explanation, Thanks @Shashank Tripathi. |

T Baba said: (Jul 31, 2018) | |

@ALL. As per my calculation. If it takes A 1/20 a day then two days= 1/20+1/20=2/20=1/10 If B and C Joins A it will be 1/20+1/30+1/60=6/60=1/10 So for 3 days, it will require A two days work plus A+B+C work= 1/10+1/10 as shown above=1/5. If 1/5 of the total work is 3 days, then 1 divided by 1/5( bearing in mind that 1/5 is 3 days)=5 then 5 *3=15 days. |

Shambhawi said: (Aug 2, 2018) | |

I have a very simple solution. Here it is -> Let the total number of days in which the work is finished is x. A will work for x days, B and C will work for x/3 days. Thus, (A's x days of work) + (B's x/3 days of work) + (C's x/3 days of work) = 1. Thus -> x/20 + x/(3*30) + x/(3*60) = 1, Thus-> x = 15. |

Pakiil said: (Aug 9, 2018) | |

The work done in 3 days is 1/5. So the work remaining is 4/5. 1/5=3 days, 2/5=3 days, 3/5=3 days, 4/5= 3 days, 5/5= 3 days, The total days=15. |

Mamun said: (Aug 9, 2018) | |

Thanks all. |

Manish Raj said: (Aug 12, 2018) | |

A's work = 20, A+B+C = 10, Together {A+(A+B+C)}/2 = (30+20)/2 = 15. |

Lavanya said: (Aug 22, 2018) | |

The total work done for three days means. There should be 1/10+1/10+1/10 should come right. Why they used it for 2 days only? |

Md Toufique said: (Aug 28, 2018) | |

Let us suppose total work to be done is 60 unit, then A, B and C's one day work is 3, 2, 1 respectively, Then in every 3 days they will finish 12 unit of work (3, 3, 6), Then, 12 + 12 + 12 + 12+ 12 =60 work done so total days = 3+3+3+3+3=15. |

Rishi said: (Sep 9, 2018) | |

How can you say the total work is 1? |

Sha said: (Sep 13, 2018) | |

How does the last step is (3*5=15)? How come 3 here? |

Dipankar said: (Sep 15, 2018) | |

12-60. 1-5, So 3=5*3=15. |

Prem K said: (Sep 28, 2018) | |

A's 1 day's work is 1/20. A's 3 day's work is 3/20. (B+C)'s 1 day's work is 1/30+1/60 = 1/20. B and C will work on every third day means A will work three days and B and C will work for 1 day. (A+B+C)'s 3 day's work = A's 3 day's work + (B+C)'s 1 day's work, = 3/20 + 1/20, = 1/5. 3 day's work is 1/5 then, whole work will be done in (3*5) = 15 days. |

Gayatri Dhokne said: (Oct 12, 2018) | |

Take the LCM of A, B, C which is 60. For LCM we have to multiply A, B, C by 3,2,1 respectively, this is work done per day by each of them. For the first 2 days, only A works so, 2*3=6. On the 3rd day A is assisted by B&C so,total work done in one day is 6units, Therefore, the work they have done in 3days is 6+6=12, Total work is done =60. Days require to complete the work by A,B, C =60 ÷ 12=5days. Days required by A to complete work = 5*3 = 15days. |

Sravya Suryadevara said: (Oct 27, 2018) | |

Here in question, it is mentioned that A alone worked for 2 days, on the third day B and C assisted him. So, A's 2 days work= 2(1/20) = 1/10. On third day three together worked. i.e., 3rd day work = (1/20)+(1/30)+(1/60) = 1/10. Work done in 3 days= (1/10)+(1/10) = 1/5. Total work is 3*5 = 15 days. |

Ajaykumarchevuri said: (Nov 21, 2018) | |

A-20days B-30days. C-60days LCM is 60 So, A one day work 3 units (20*3) B one day work 2 units(30*2) C one day work 1 unit (1*60) Then; 6 units ( both A+B+C one day work on every third day) So A 2 days work = 3units *2 = 6 units A+B+C on 3 Rd day = 6 units*1= 6 units So, weget 12 units. So 12/60 = 1/5 (A 2 days work +(A+B+C) work on a third day/total work) So 3days * 1/5 = 15 days. Here the answer. |

Sruthi said: (Dec 6, 2018) | |

LCM of A,B & C is 60. so the total work to be complete is 60. A->3units/day{60/20}. B->2units/day{60/30}. c->1unit/day(60/60}. A's 2days work-> 3*2 = 6units. 3rd day A is assisted by B & C -> A + B + C. A + B + C 1 day work->3 + 2 + 1 = 6. 2days A alone->6units. 3rd day->6 {A + B + C work}. 60/12 * 3 = 15. |

Sai Kumar Reddy Nossam said: (Dec 9, 2018) | |

Common multiple of A,B & C is 120. A's one day work =120/20 = 6 units, B's one day work =120/30 = 4 units, C' one day work = 120/60 = 2 units, work done in three days by A, B, C= 24 units. So, the total work = (120/24) * 3 = 15 days. |

Nazam said: (Dec 29, 2018) | |

How did (1/10+1/10) = 1/5 came? |

Veeresh said: (Jan 28, 2019) | |

@All. A is 1st-day work 1/20. A is 2nd-days work 1/2. 02/20. a+b+c together work is = 1/20 + 1/30 + 1/60. Add 2/20+1/20+1/30+1/60 = 1/5. For 3 days = 1/5 * 1/3 = 1/15 ie 15 days. |

Ashish Agrawal said: (Feb 21, 2019) | |

A=30 B=20 C=60 Now take the lcm that is 60 mean A will make 3 chairs for example in one day B will make 2 chairs in one day and will make c will make 1 chair ( chairs example is for understanding). Now A will make 3 chairs in one day and after 3 days it will make 9chair ( 3 *3) and at the third day ( B +c) will also make 3 chairs mean at the third day A +B +C WILL MAKE 12 chair in 3 days to make 60 chairs which our lcm it will require 15 days ( 12*5 = 60) that is 3*5 is 15 days. |

Simita said: (Mar 4, 2019) | |

Why 3 is multiplied by 5 in the last step? Please tell me. |

Nikhil Patil said: (Apr 27, 2019) | |

Please explain the 3rd step. |

Himanshu said: (Jun 6, 2019) | |

How it is 3*5 at last? |

Raju Kamalla said: (Jun 18, 2019) | |

A's 1 day work is 1/20, B's one day work is 1/30, C's one day work is 1/60. Given problem A is doing 2 days work alone and assisted B&C on the third day and repeats this cycle. So, the Total work done in 3 days= (A's 2days work) + (A+B+C's 1 day work) i.e., (1/20x2) +(1/20+1/30+1/60). (1/10) +( (3+2+1)/60) (hence 20,30,60 lcm is 60), => 1/10+(6/60), => 1/10+1/10 = 2/10=> 1/5. Here the work done in 3 days is 1/5; So total days required is 3x5=15. |

Akshara said: (Jul 10, 2019) | |

Efficiency total work-60. A - 20 days----> 3 B - 30 days----> 2 c - 60 days-----> 1 A+B+C work=6---->3 days work=12. The total work to done is 60 hence 12 * 5=60. Hence days are 5 * 3=15 is the right answer. |

Ravi said: (Aug 13, 2019) | |

First, we took lcm of all 3 that is A, B and C that is 60. On the basis of this A's efficiency is 3, B's efficiency is 2 and C's 1. So first 2 days A will do the work that is 3x2=6. Third-day work will be 6 as all three are doing the work. So 3 day's work will be 12. On the basis of the unitary method in every 3 day, they are doing 12 work so 3 day's efficiency will be 4unit. So, total work that is 60 divide by 4= 15. |

Aishwarya Subhash said: (Aug 30, 2019) | |

That is 1/15 work is done in 1 Day. Now 4/5 part of work that is (1-1/5) will be done in :15*4/5 =12 days. |

Gabriel said: (Aug 30, 2019) | |

One simple way : 1/20 + (1/90 + 1/180) = 1/15. |

Gourab said: (Sep 16, 2019) | |

In 3 days ,the total work = 3/20 + 1/30 + 1/60 = 1/5. 1/5 % done in 3 days. 1% done in 3/(1/5) = 3 * 5 = 15 days. |

Rashmi Maurya said: (Sep 20, 2019) | |

a b c 20 30 60 (LCM=120) 6 4 2 (a20/120=6, b30/720=4, c60/120=2). 6 + 4 + 2 = 12, 12/120 = 5 5 * 3 = 15. |

A.Venkatesh Reddy said: (Oct 10, 2019) | |

A=20 A Effi=3. B=30 LCM=60 B Effi=2. So,Work=60 => 3+3+6=12(3 days work)=> 12*5=60=>15days. C=60 C Effi=1. |

Sivapriya said: (Oct 25, 2019) | |

A complete in 20days. B complete in 30days. C complete in 60days. Lcm(20,30,60)= 60. So total work = 60. A's one day work = 60/20 = 3. B's one day work = 60/30 = 2. C's one day work = 60/60 = 1. A+B+C 's one day work = 3+2+1=6. First day A's complete = 3unit of work. Sec day A complete = 3unit of work. 3rd day A+B+C = 6unit of work. . . So on First 3day total work completed= 3+3+6=12 unit of work. So (total work / first schedule work) x schedule days. = (60/12) x 3. = 5 x 3 = 15days. |

G.Prathyusha said: (Nov 2, 2019) | |

To complete 1/5th of work - 3 days. Hence to complete 5/5th of work - x days. By cross multiply, you will get x value as 15 days. |

Jayshri said: (Nov 6, 2019) | |

How got the 3rd step (1/10+1/10)? |

Common Man said: (Dec 10, 2019) | |

3days = 1/5. 6days = 2/5. 9days = 3/5. 12days = 4/5. 15days = 5/5 = 1 total work done is complete. |

16Cs1045 said: (Dec 13, 2019) | |

10 %for A in first 2 day 10%(5+3.33+1.66)5 for A in third day and 5 for B and C in third similarly for other 12 day work will complete. Therefore the total time required is 15 day. |

Samiksha Koli said: (Jan 7, 2020) | |

@All. I couldn't get this. Because A is assisted by B & C on the third day only. So, A B C will together work for first two days. And for last day A will work alone because he is assisted by B & C for third day. So total work of three days Will be A+B+C=1/10 for 2 days it will be 1/5 and A's last day's work will be 1/20. So, the total work of three days will be1/4. So total work has to be done in 3*4=12 days. Correct me, if I am wrong. |

Vivek said: (Jan 16, 2020) | |

I can't get 3rd step. How do 1/10+1/10 came and how it is divided by 5? |

Sameer said: (Mar 2, 2020) | |

Fraction of work done in one by A, B, and C is, 1/20, 1/30 and 1/60. In first 2 days, fraction of work completed by A is 2/20. Third day, B and C are assisting A then fraction of work completed on 3rd day will be 1/20 + 1/30+ 1/60 = 6/60 i.e 1/10. So total fraction of work done after 3 days. 2/20+1/10=4/20 i.e. 1/5. So complete work will be completed in 5 days. |

Rj Rishi Winston said: (Apr 2, 2020) | |

As per the question, A works all days but on every three days A gets assisted by B&C. Upon calculation through LCM method the total workdays =60 and also with the help of LCM it is possible to find A's & B+C each day work i.e. 60/20 =3, [60/30+ 60/60]. Thus A's efficiency for one day work is 3 and B+C =2+1 =3. On every 3days A finishes 9 work but as he is assisted by B&C on every 3rd day their work also gets added with A's work as a result of which in every 3rd day 9+3=12 work gets completed. Hence as on every 3rd day, 12 work gets completed the entire work of 60 would be completed by [60*3] /12 =15 days. |

Poonam said: (Jun 19, 2020) | |

Here is a solution: Efficiency A = 3 , B = 2, C = 1 and A+B+C = 6. Now, Work done in format ie A can do a work in 2day and next 1day with B&c . So. Eff * time = work. A = 3 * 2 = 6. A + B + C = 6 * 1 = 6. Means work done in 3 days is 6 + 6 = 12 work. Days -------- work 3 -------- 12 ? ------- 60 (total work) Days =(60*3) / 12. The answer is = 15 days. |

Billal Hossain said: (Jul 13, 2020) | |

Solution: A's 3 days work = 3/20. (B+C)'s 3rd day or 1 day work= 1/20. (A+B+C)'s 3 days work = (3/20+1/20). = 1/5. A's 1/5 part work in 3 days. 1 part ---> 15 days. A can do the wok in 15 days (Ans). |

Nirmala Rai said: (Jul 19, 2020) | |

3A----20 2B----30 1C----60 this come from L. C. M------60. Add:all efficiency A+B+C=3+2+1=6. Per day works of A+B+C=60/6=10. 3days work of A+B+C. 3 = (3+3per day work of A +6 per day work of a+b+c). 3 = 12(Tw=60 if we want 60in the table of 12 then we need to * ...12*5 and also *right side) 5*3 = 12 * 5. 15day = 60. A can do the work in 15 days. |

Santosh said: (Jul 30, 2020) | |

A =20 Days B=30 Days C=60 Days LCM = A,B,C LCM = 60 A = 60/20 =3. B = 60/30 = 2. C = 60/60 =1. AFTER 3 DAYS B, C JOINED. Total =3 + 3(3 + 1) = 15 DAYS. |

Hrishi said: (Aug 20, 2020) | |

Step 1: Work is done by A+B+C in one day can be calculated as 1/20 + 1/30 + 1/60 = 1/10. Step 2: For the first two days only A is doing work alone so the work is done by A in two days will be 2*1/20= 1/10. Step 3: Now total work is done in 3 days by A B & C will be = 1/10+1/10= 1/5. Step 4: But we need a complete work to be done so we have to convert 1/5 to 1 therefore if we do 1/5 work 5 times then work will be completed (1/5 x 5=1). Therefore 5 x 3 = 15. Hope you will understand this solution. |

Keerthi said: (Aug 23, 2020) | |

why are we multiplying 1/5 *3 in the last step? Please explain anyone. |

Kanishk said: (Aug 31, 2020) | |

@Keerthi. We are not directly multiplying. We just use the unitary method. Here the total work done is 1. In fact, in all these sums total work is 1. Now, Since 1/5th of the work is done in 3 days, Therefore, 1 of the work( whole work) is done in 3*5 days. Hope you understand. |

Pooja Pawatekar said: (Sep 7, 2020) | |

How can you take A's 2 days work? Can anyone explain this pls? |

Avinash Singh said: (Sep 19, 2020) | |

Here's an easy one: Let's assume that the total work is done in x days. Then x/3 days, A will require B and C's help and 2x/3 days A will be working alone. So the equation becomes: x/3(1/20) + 2x/3(1/20 + 1/30 + 1/60) = 1 [1 because 1 unit of work is done in x days]. Solve it and you'll get x = 15. Thanks! |

Britney said: (Oct 4, 2020) | |

How did we get 6/60? Anyone explain about it. |

Yoezer said: (Oct 10, 2020) | |

3 days 1/5. 6 days 2/5. 9 days 3/5. 12 days 4/5. 15 days 5/5( work completed). |

Sunil_ said: (Nov 3, 2020) | |

A work in one day = 1/20, B work in one day = 1/30, C work in one day = 1/60. As A work for 2 days and then on third day B and C do work. Total work done in three days = A 3 days work+B 1-day work+ C one day work, = 3/20 + 1/30 + 1/60, = 9/60 + 2/60 + 1/60 = 12/60. Total work is done in three days =1/5. 3 = 1/5part. Let x days required to complete work = 1 part. 3/x = 1/5/1. x = 3 * 5 = 15 days. |

Arun said: (Nov 8, 2020) | |

Worker - TotalDay - OneDay A - 20 - 1/20 B - 30 - 1/30 C - 60 - 1/60 1st Day - A. 2nd Day- B. 3rd Day - A, B, C. They said A lonely can work for 2 day. 2A = 2(1/20) =1/10. Then in third day A, B, C together; (A+B+C) = 1/20+1/30+1/60, = 1/10. 3 Days(not 3rd Day) work is; =2A + (A+B+C). = 1/10+1/10. 3Days =1/5 work, 6Days =2/5 work, 9Days =3/5 work, 12Days=4/5 work, 15Days= 5/5 =1 work completed. |

Abdul Moeed said: (Dec 9, 2020) | |

Use unity rule at the last step, 1/5 work is done in = 3 days. 1 work (whole work) is done in = 3/1/5 = 3*5= 15 days. |

Usman said: (Dec 11, 2020) | |

Solution: A do a piece of work in = 1/20 days. 'A' work alone for 2 days, so the work done by A on the second day is = 2/20 = 1/10 On the third day, A is assisted by B&C, then the work done by three of them is = 1/10 + 1/20 + 1/30 = 3/20. Total work done by A in 3 days is = 1/20 + 3/20 = 1/5. As we have to find number of days, so, 1/5 part of work done by A is in = 3 days. 1 part of the work done by A is in = 3x5 = 15 days. In 15 days, A can do the work if he is assisted by B& C on the third day. |

Vijay said: (Jan 6, 2021) | |

How we got only A done 3 days of work? Please explain it. |

T.S said: (Jan 18, 2021) | |

@Vijay We know that the whole work done is 1. here 3 days work is 1/5. So, if, 1/5 work=3days. then, 1 work= ? Cross multiply we get ?= 3 x 5= 15days. |

Amit Bhau said: (Feb 14, 2021) | |

3 days ---> work 1/5 x days ---> to get total work =1 Cross multiplication, then we get x = 15. |

Disha said: (Feb 21, 2021) | |

Every 3 days 1/5 work is done. Then in how many days 1 whole work is done. Mathematically = 1/5:3::1:x. 3*1/ (1/5) = 3÷1/5= 3*5/1= 3*5= 15. |

Kiran said: (Jun 16, 2021) | |

Hi, can anyone tell me what's wrong here? Given, A=20days. B=30days. C=60days. I took as 60units of work by taking LCM of A, B, C. A's one day work=3 units/day. B's one day work=2 units/day. A's two days work = 6 units. B's one day work= 2 units. C's one day work= 1 unit. A+B+C's 3 days work = 9 units. So, I can take A+B+C's one day work = 3 units =>9/3=3 units/day. So, to complete 60 units of work they will take 20days =>20*3=60 units. So, I am saying that answer is "20 days". Correct me if I'm wrong. |

Sairam said: (Jun 30, 2021) | |

@Kiran. You didn't account for A's third day of work. A does 6 units of work in two days and another 3 units on the third day. So in total, A has done 9 units of work in 3 days. |

Poojitha said: (Jul 4, 2021) | |

@Dhanam. Because on third day they are doing combined work it means that for 1st 2 days they have worked alone here in question. About A has asked so he is doing only A. |

Ananthu said: (Aug 10, 2021) | |

For the peopel who are confused with last step. We knew that 3 days work is 1/5.. the remaining work is 1-1/5=4/5.. ie already worked 3 days here. Then we are calculating that how many 3 days are required to complete the remaining work 4/5. Ie 4/5/1/5 = 4/5 * 5/1 = 4.. 4 (three days required). ie 4*3 = 12 we already worked 3 days then total days will be 12 + 3 = 15. |

Piyush Chandana said: (Aug 11, 2021) | |

Mem ------> time ---- efficiency ----work a------> 20 ---- 3 ---- 60 b------> 30 ---- 2 ---- 60 c------> 60 ---- 1 ---- 60 Now we make a series like :- Day1 total unit of work will be -> 3 units. Day2 total unit of work will be-> 3 units. Day3 total unit of work will be -> 3+1+2 = 6(because b and c helped him). Day1 Day2 Day3 3 3 6. Work will be-> 12 units = 3 days. The total units of work is 60 units then divide 60 with 12(60/12=5). Now the total units are 5 times the 3 days of work than days will also multiply by 5 which is 3*5="15 days" (Ans). |

Aayush Pandey said: (Aug 21, 2021) | |

A. B. C. Take => a=20, b=3, c=60 days respectively. The Lcm of 20, 30, 60 is 60. So the efficiency of a, b, c is :. Eff days. 3<---a---20---->. 2<---b---30----> 1<---c---60---->. And 60 (total work). Now work done by A in 2 days will be => 3*3 = 6. On the 3rd day a, b, c will work so Work done will be = 3+2+1 = 6. Work done in 3 days = 6+6 =12. So, (12 work) is done in 3 days, (1 work) will be done in = 3/12 days. (total work is 60) , will be done in = (3/12) *60 = 60/4 = 15 days. |

Azhar Khan said: (Aug 25, 2021) | |

Easy way to answer is: LCM of 20 30 60 = 60. A = 20 * 3 = 60, B = 30 * 2 = 60, C = 60 * 1 = 60,. A + B + C = 3 + 2 + 1 = 6 ( together) A's 2 days work = 2 * 3 = 6. Total work done by them = 6 + 6 = 12 unit Total work is 60 unit. So, 60 ÷ 15 = 5 units ( In one day by them ) Then in 3 days, 3*5 = 15 days |

Raghunath Kisan said: (Aug 30, 2021) | |

LCM of 20,30,60 = 60. Efficiency of One day work of; A = 60/20 = 3, B = 60/30 = 2, C = 60/60 = 1. A + B + C = 3 + 2 + 1 = 6 ( together), A's 2 days work = 3 * 2 = 6. Total work done by them = 6 + 6 = 12 unit (in 3 days), Then, (Total work)/(Total work done in 3 days), That is, 60/12 = 5, So, 1 = 3, then 5 = 5 * 3 = 15 Days. |

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