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 30 of 36.
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.
Sai Kumar Reddy Nossam said:
7 years ago
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.
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:
7 years ago
How did (1/10+1/10) = 1/5 came?
Veeresh said:
7 years ago
@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.
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:
6 years ago
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.
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:
6 years ago
Why 3 is multiplied by 5 in the last step? Please tell me.
Nikhil patil said:
6 years ago
Please explain the 3rd step.
Himanshu said:
6 years ago
How it is 3*5 at last?
Raju Kamalla said:
6 years ago
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.
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:
6 years ago
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.
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.
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