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 5 of 23.
Jansi said:
1 decade ago
Hello Mr. Khan,
You can refer June -19th tuesday Mr. Amit.J has explained good. He given clear explanation.
But Mr.Amit you had used this formula, (a+b+c) = a*b*c/ (ab+bc+ca). Always we can use this formula or any other restriction is there. Can you explain me?
You can refer June -19th tuesday Mr. Amit.J has explained good. He given clear explanation.
But Mr.Amit you had used this formula, (a+b+c) = a*b*c/ (ab+bc+ca). Always we can use this formula or any other restriction is there. Can you explain me?
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.
Sudha said:
1 decade ago
At the end of explanation,
C's 1 day work is 5/48 I calculated up to to this but how it comes work in 48/5 days.
Because acc. to formula if A's 1 day work is 1/n' then A can finish work in n' days.
Means only it comes 48. I am just confuse. please explain me.
C's 1 day work is 5/48 I calculated up to to this but how it comes work in 48/5 days.
Because acc. to formula if A's 1 day work is 1/n' then A can finish work in n' days.
Means only it comes 48. I am just confuse. please explain me.
Shiyamala said:
1 decade ago
A=1/16;
B=1/12;
A+B=[1/16+1/12]
=[16+12/192]{1/16*12/12=16/192;1/12*16/16=16/192)
=28/192
=7/48 {4*7=28; 48*4=192)
C-[A+B]=1/4-7/48 {1/4*12/12=12/48)
=12-7/48
=5/48
but c work alone.so,48/5=9*3/5
B=1/12;
A+B=[1/16+1/12]
=[16+12/192]{1/16*12/12=16/192;1/12*16/16=16/192)
=28/192
=7/48 {4*7=28; 48*4=192)
C-[A+B]=1/4-7/48 {1/4*12/12=12/48)
=12-7/48
=5/48
but c work alone.so,48/5=9*3/5
Happy Singh said:
1 decade ago
Easy method in step by step :
Given values :
A=1/16, b=1/12.
(A+B+C)'S 1 day's work = 1/4.
So we can write,
(A+B+c)=1/4.
(1/16+1/12+C)=1/4.
C=(1/4-(1/16+1/12)).
C=(1/4-7/48).
C=(5/48).
C's one day work is = 5/48,
so, C can do the work in 48/5 days.
Given values :
A=1/16, b=1/12.
(A+B+C)'S 1 day's work = 1/4.
So we can write,
(A+B+c)=1/4.
(1/16+1/12+C)=1/4.
C=(1/4-(1/16+1/12)).
C=(1/4-7/48).
C=(5/48).
C's one day work is = 5/48,
so, C can do the work in 48/5 days.
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.)
Ovii said:
1 decade ago
@aparna
Little help to the explanation of 1's days work.
If A's 1 day work = 1/n than the complete/whole work can be done in n*1 days.
It can be modify to above 1 day's work, e.g
if A's 5 days work = 1/25, than the complete/whole work = 5*25= 125.
Little help to the explanation of 1's days work.
If A's 1 day work = 1/n than the complete/whole work can be done in n*1 days.
It can be modify to above 1 day's work, e.g
if A's 5 days work = 1/25, than the complete/whole work = 5*25= 125.
Shivaji said:
2 years ago
Simple logic;
A-----> 16 days.
B ----> 12days.
with the help of C.
AB+C -----> 4 days.
Total work (LCM) = 48.
Efficiency = A--3, B--4, AB+C--12.
A+B = 7,
AB+C = 12
Then C = 12-7 = 5.
C can do work in = Total work/Efficiency = 48/5 = 9*3/5.
A-----> 16 days.
B ----> 12days.
with the help of C.
AB+C -----> 4 days.
Total work (LCM) = 48.
Efficiency = A--3, B--4, AB+C--12.
A+B = 7,
AB+C = 12
Then C = 12-7 = 5.
C can do work in = Total work/Efficiency = 48/5 = 9*3/5.
(54)
Abhishek Kumar said:
2 years ago
Simple logic;
A-----> 16 days.
B ----> 12days.
with the help of C.
AB+C -----> 4 days.
Total work (LCM) = 48.
Efficiency = A--3, B--4, AB+C--12.
A+B = 7,
AB+C = 12
Then C = 12-7 = 5.
C can do work in = Total work/Efficiency = 48/5 = 9*3/5.
A-----> 16 days.
B ----> 12days.
with the help of C.
AB+C -----> 4 days.
Total work (LCM) = 48.
Efficiency = A--3, B--4, AB+C--12.
A+B = 7,
AB+C = 12
Then C = 12-7 = 5.
C can do work in = Total work/Efficiency = 48/5 = 9*3/5.
(113)
Tharun A said:
5 months ago
A-----> 16 days.
B ----> 12days.
with the help of C.
AB + C -----> 4 days.
Total work (LCM) = 48.
Efficiency = A-3, B -4, AB + C - 12.
A + B = 7,
AB + C = 12
Then C = 12 - 7 = 5.
C can do work in = Total work/Efficiency = 48/5 = 9*3/5.
B ----> 12days.
with the help of C.
AB + C -----> 4 days.
Total work (LCM) = 48.
Efficiency = A-3, B -4, AB + C - 12.
A + B = 7,
AB + C = 12
Then C = 12 - 7 = 5.
C can do work in = Total work/Efficiency = 48/5 = 9*3/5.
(58)
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