Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 2)
2.
A can lay railway track between two given stations in 16 days and B can do the same job in 12 days. With help of C, they did the job in 4 days only. Then, C alone can do the job in:
Answer: Option
Explanation:
| (A + B + C)'s 1 day's work = | 1 | , |
| 4 |
| A's 1 day's work = | 1 | , |
| 16 |
| B's 1 day's work = | 1 | . |
| 12 |
 C's 1 day's work = |
1 | - |  |
1 | + | 1 |  |
= | |
1 | - | 7 | |
= | 5 | . |
| 4 | 16 | 12 | 4 | 48 | 48 |
| So, C alone can do the work in | 48 | = 9 | 3 | days. |
| 5 | 5 |
Discussion:
224 comments Page 17 of 23.
Pavani said:
7 years ago
(A+B+C)'s efficiency - (A+B)'s efficiency = C's efficiency.
(1/4) - (1/16+1/12) = C's efficiency.
5/48 = C's efficiency.
So, the time taken, to complete the total work, by C = 48/5= 9 3/5.
(1/4) - (1/16+1/12) = C's efficiency.
5/48 = C's efficiency.
So, the time taken, to complete the total work, by C = 48/5= 9 3/5.
Justine said:
7 years ago
How come that 48 gets retain? when you cross multiply 48/1 * 1/16 = 3 and 48/1 * 1/12 = 4?
Please explain me.
Please explain me.
Imran said:
7 years ago
Can anyone explain me this question by LCM method?
TRUPTIMAYEE BAHINIPATI said:
7 years ago
The solution in LCM method is;
Solution :
SUPPOSE TOTAL WORK=1UNIT.
A=15d L.C.M. of 15,20=60.
B=20d.
Then A=4.
B=3.
A+B=7unit=60.
4 unit=60/7*1/4=15/7day.
1day work of A+B=7/15unit.
Remaining work=1-7/15=8/15 unit left (Ans.)
Solution :
SUPPOSE TOTAL WORK=1UNIT.
A=15d L.C.M. of 15,20=60.
B=20d.
Then A=4.
B=3.
A+B=7unit=60.
4 unit=60/7*1/4=15/7day.
1day work of A+B=7/15unit.
Remaining work=1-7/15=8/15 unit left (Ans.)
Siba siddquie said:
7 years ago
A can lay railway track between two given stations in 16 days and B can do the same job in 12 days. With help of C, they did the job in 4 days only. Then, C alone can do the job in:
LCm of A,B,A+B+C (16,12,4)=total work=48...;;
A' 1 day work=48/16=3;
same B=48/12=4..A+B+C=48/4=12,
A+B+C=12,
3+4+C=12,
C 'one day work =5,
total work is =48.
C completes the work in (days) = total work/one day work..(formula) = 48/5=9(3/5) => answer.
LCm of A,B,A+B+C (16,12,4)=total work=48...;;
A' 1 day work=48/16=3;
same B=48/12=4..A+B+C=48/4=12,
A+B+C=12,
3+4+C=12,
C 'one day work =5,
total work is =48.
C completes the work in (days) = total work/one day work..(formula) = 48/5=9(3/5) => answer.
Jaga said:
7 years ago
Thanks @Aanurag.
Pooja said:
7 years ago
As per formula: If A completes 1/n work in 1 day then the days required by him to complete that 1 work alone will be 1 ÷ 1/n = n days.
Similarly, C can complete 5/48 part of the work in a day. To complete 1 work alone he will take 1 ÷ 5/48 days to 48/5.
Similarly, C can complete 5/48 part of the work in a day. To complete 1 work alone he will take 1 ÷ 5/48 days to 48/5.
Suchi said:
7 years ago
Thank you @Siba.
Aruva said:
7 years ago
Thank you @Soujanya.
Sunshine said:
7 years ago
A work = 16 days.
B work = 12 days.
Total work was done by A and B = 48 unit (LCM Of A and B).
So work done by A in one day = 48/16= 3 unit.
And work was done by B in one day = 48/12 = 4 unit.
So work done by A and B in 1 day = 3+4 = 7 unit.
Work was done by A and B in 4 days = 7*4 = 28 unit.
So remaining work = 48-28 = 20 unit.
So work done by C in one day= 20/4 = 5 unit.
So, the total work was done by C in days = 48/5 days =9*3/5 days.
B work = 12 days.
Total work was done by A and B = 48 unit (LCM Of A and B).
So work done by A in one day = 48/16= 3 unit.
And work was done by B in one day = 48/12 = 4 unit.
So work done by A and B in 1 day = 3+4 = 7 unit.
Work was done by A and B in 4 days = 7*4 = 28 unit.
So remaining work = 48-28 = 20 unit.
So work done by C in one day= 20/4 = 5 unit.
So, the total work was done by C in days = 48/5 days =9*3/5 days.
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