Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 23)
23.
A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?
Answer: Option
Explanation:
| 2(A + B + C)'s 1 day's work = |  |
1 | + | 1 | + | 1 |  |
= | 15 | = | 1 | . |
| 30 | 24 | 20 | 120 | 8 |
| Therefore, (A + B + C)'s 1 day's work = | 1 | = | 1 | . |
| 2 x 8 | 16 |
| Work done by A, B, C in 10 days = | 10 | = | 5 | . |
| 16 | 8 |
| Remaining work = | |
1 - | 5 | |
= | 3 | . |
| 8 | 8 |
| A's 1 day's work = | |
1 | - | 1 | |
= | 1 | . |
| 16 | 24 | 48 |
| Now, | 1 | work is done by A in 1 day. |
| 48 |
| So, | 3 | work will be done by A in | |
48 x | 3 | |
= 18 days. |
| 8 | 8 |
Discussion:
61 comments Page 1 of 7.
Vikas kumar said:
3 years ago
LCM = 120.
A=2.5,
B=3.5,
C=1.5.
A + B + C = 7.5 for 10days all works togather then 7.5*10=75,
Remaining=120 - 75 = 45.
For A = 45/2.5 = 18days.
A=2.5,
B=3.5,
C=1.5.
A + B + C = 7.5 for 10days all works togather then 7.5*10=75,
Remaining=120 - 75 = 45.
For A = 45/2.5 = 18days.
(18)
Nikhil said:
1 decade ago
A+B = 30 days.
B+C = 24 days.
C+A = 20 days.
--------------------
divide by taking lcm (30,24,20) = 120.
A+B = 4 units.
B+C = 5 units.
C+A = 6 units.
-------------------
2( A+B+C) = 15 units.
A+B+C =7.5 units.
They work together 10 days, means in 10 days they have completed 75 units out of 120 units. now,
Remaining units are (120-75) = 45.
A+B+C = 7.5 (put B+C= 5 as given),
A = 2.5.
So for completing 45 units, A need (45/2.5) = 18 days.
B+C = 24 days.
C+A = 20 days.
--------------------
divide by taking lcm (30,24,20) = 120.
A+B = 4 units.
B+C = 5 units.
C+A = 6 units.
-------------------
2( A+B+C) = 15 units.
A+B+C =7.5 units.
They work together 10 days, means in 10 days they have completed 75 units out of 120 units. now,
Remaining units are (120-75) = 45.
A+B+C = 7.5 (put B+C= 5 as given),
A = 2.5.
So for completing 45 units, A need (45/2.5) = 18 days.
(18)
KEYA said:
4 years ago
(A+B+C) = 1/16.
(B+C) = 1/24.
Hence,
A = 1/16 - 1/24 = 1/48.
(B+C) = 1/24.
Hence,
A = 1/16 - 1/24 = 1/48.
(16)
Subash said:
7 years ago
By approx too we can solve this;
a+b+c=1/10,
b+c=1/24,
a=1/10-1/24,
a=17.13,
Near by value 18.
a+b+c=1/10,
b+c=1/24,
a=1/10-1/24,
a=17.13,
Near by value 18.
(5)
Bikash Choudhury said:
1 decade ago
LCM of 30, 24, 20 is 120.
A+B B+C C+A A+B+C.
4 5 6 2 (a+b+c) = 15.
=> a+b+c = 15/2.
A+B+C do for 10 means they work together 10*15/2 = 75 units of work.
Work remains 120-75 = 45 unit.
A 's one day work = (A+B+C)-(B+C) = 15/2-5 = 5/2.
Hence A can finish the remaining work = 45*2/5 = 18.
A+B B+C C+A A+B+C.
4 5 6 2 (a+b+c) = 15.
=> a+b+c = 15/2.
A+B+C do for 10 means they work together 10*15/2 = 75 units of work.
Work remains 120-75 = 45 unit.
A 's one day work = (A+B+C)-(B+C) = 15/2-5 = 5/2.
Hence A can finish the remaining work = 45*2/5 = 18.
(3)
S. Kalyan said:
5 years ago
A+B ----> (5/30),
B+C ----> (5/24),
C+A ----> (5/20).
Numerator (A+B+C) worked together for 10 days that means ' A ' Worked 5 days with ' B ' and 5 days with C . Similarly ' B ' Worked 5 days with ' A ' and 5 days with C. And Similarly ' C ' Worked 5 days with ' A ' and 5 days with B.
Now Denominator is their 1-day work :
(5/30) + (5/24) + (5/20) + A * (x) = 1.
Here ' A* (x) ' Means A Completed the Remaining Work (Given in Question) but We Don't Know How many days he Completed the work that is why I have taken (x). Here Represents the days ' A ' Has Worked to Complete that Job.
Now Equation Becomes :
(1/6) + (5/24) + (1/4) + (x * A) = 1.
(5/8) + (x * A) = 1 = > x * A = 1 - (5/8).
=> x * A = (3/8).
To Find ' A's ' 1 Day Work = Total Work Done By all of them - ( Eliminate 2 Variables to Get Value ) (i.e B and C )
So Equation Become,
A's 1 Day Work = (A + B + C)'s 1 day's work - (B + C)'s 1 day's work.
A's 1 Day Work = (1/16) - (1/24) = (1/48).
Now we Got A's 1 Day Work Substitute Above i.e in (x * A).
=> (x/ 48) = (3/8).
Cross Multiplying we get;
=> 8x = 48 * 3.
=> x = 18 days.
Hope You Understood.
B+C ----> (5/24),
C+A ----> (5/20).
Numerator (A+B+C) worked together for 10 days that means ' A ' Worked 5 days with ' B ' and 5 days with C . Similarly ' B ' Worked 5 days with ' A ' and 5 days with C. And Similarly ' C ' Worked 5 days with ' A ' and 5 days with B.
Now Denominator is their 1-day work :
(5/30) + (5/24) + (5/20) + A * (x) = 1.
Here ' A* (x) ' Means A Completed the Remaining Work (Given in Question) but We Don't Know How many days he Completed the work that is why I have taken (x). Here Represents the days ' A ' Has Worked to Complete that Job.
Now Equation Becomes :
(1/6) + (5/24) + (1/4) + (x * A) = 1.
(5/8) + (x * A) = 1 = > x * A = 1 - (5/8).
=> x * A = (3/8).
To Find ' A's ' 1 Day Work = Total Work Done By all of them - ( Eliminate 2 Variables to Get Value ) (i.e B and C )
So Equation Become,
A's 1 Day Work = (A + B + C)'s 1 day's work - (B + C)'s 1 day's work.
A's 1 Day Work = (1/16) - (1/24) = (1/48).
Now we Got A's 1 Day Work Substitute Above i.e in (x * A).
=> (x/ 48) = (3/8).
Cross Multiplying we get;
=> 8x = 48 * 3.
=> x = 18 days.
Hope You Understood.
(2)
Vinod said:
7 years ago
Thank you @Nihkil.
(2)
Anandhi said:
1 decade ago
A+B+C one day work = 1/16.
Given B+C = 1/24.
Therefore A+B+C = 1/16.
A+1/24 = 1/16.
A = 1/16-1/24.
A = 1/48.
Given B+C = 1/24.
Therefore A+B+C = 1/16.
A+1/24 = 1/16.
A = 1/16-1/24.
A = 1/48.
(2)
Nandhakumar said:
1 decade ago
how to find A's 1's day work
(1)
Sikandar said:
1 decade ago
As 1 day work = (A+B+C)-(B+C).
= 1/16-1/24 = 1/48.
= 1/16-1/24 = 1/48.
(1)
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