Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 3)
3.
A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
Answer: Option
Explanation:
| A's 2 day's work = |  |
1 | x 2 |  |
= | 1 | . |
| 20 | 10 |
| (A + B + C)'s 1 day's work = | |
1 | + | 1 | + | 1 | |
= | 6 | = | 1 | . |
| 20 | 30 | 60 | 60 | 10 |
| Work done in 3 days = | |
1 | + | 1 | |
= | 1 | . |
| 10 | 10 | 5 |
| Now, | 1 | work is done in 3 days. |
| 5 |
Whole work will be done in (3 x 5) = 15 days.
Discussion:
357 comments Page 7 of 36.
Sameer said:
5 years ago
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.
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.
T baba said:
7 years ago
@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.
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.
Ranju said:
1 decade ago
(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.
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.
Jemish said:
9 years ago
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.
=> 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.
Lkesh R said:
8 years ago
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.
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.
Nirmala Rai said:
5 years ago
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.
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.
Nitin said:
10 years ago
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.
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.
Raghunath Kisan said:
4 years ago
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.
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.
(124)
Sruthi said:
7 years ago
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.
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.
Harry said:
1 decade ago
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
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
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers