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:
359 comments Page 10 of 36.
Abhishek said:
2 decades ago
@chirag:
Step 3 gives work done in first 3 day i.e. work done in 1st day + work done in 2nd day +work done in 3rd day = 1/20 (by A alone) + 1/20 (by B alone) + 1/10 (done by A, B and C together)
Similarly they continue and as per the question on every 3rd day A is joined by B and C.
Step 3 gives work done in first 3 day i.e. work done in 1st day + work done in 2nd day +work done in 3rd day = 1/20 (by A alone) + 1/20 (by B alone) + 1/10 (done by A, B and C together)
Similarly they continue and as per the question on every 3rd day A is joined by B and C.
Vishal grade said:
1 decade ago
A's 1 day work = 5%.
B's 1 day work = 3.33%.
C's 1 day work = 1.66%.
A, B and C' 1 day work = 5+3.3+1.6 = 10%.
On 1st 2 days total work = 5+5 = 10%.
On 3rd day total work(by all 3)=10%.
In 3 day total work done=10+10=20%.
20% work done in ---- 3 days.
100% work done in --- 15 days.
B's 1 day work = 3.33%.
C's 1 day work = 1.66%.
A, B and C' 1 day work = 5+3.3+1.6 = 10%.
On 1st 2 days total work = 5+5 = 10%.
On 3rd day total work(by all 3)=10%.
In 3 day total work done=10+10=20%.
20% work done in ---- 3 days.
100% work done in --- 15 days.
Kanishk said:
5 years ago
@Keerthi.
We are not directly multiplying.
We just use the unitary method. Here the total work done is 1. In fact, in all these sums total work is 1.
Now,
Since 1/5th of the work is done in 3 days,
Therefore, 1 of the work( whole work) is done in 3*5 days.
Hope you understand.
We are not directly multiplying.
We just use the unitary method. Here the total work done is 1. In fact, in all these sums total work is 1.
Now,
Since 1/5th of the work is done in 3 days,
Therefore, 1 of the work( whole work) is done in 3*5 days.
Hope you understand.
Tuhin said:
1 decade ago
EASY WAY FOR THIS QUESTION :
Let total no. of days be= N
Work done by A,B,C respectively per day are :1/20,1/30,1/60
As B and C join every third day there work done by them :(N/3)/30 and (N/3)/60 respectively.
And for A : N/20
Total work=
N/20+(N/3)/30+(N/3)/60=1
We get N as 15
Let total no. of days be= N
Work done by A,B,C respectively per day are :1/20,1/30,1/60
As B and C join every third day there work done by them :(N/3)/30 and (N/3)/60 respectively.
And for A : N/20
Total work=
N/20+(N/3)/30+(N/3)/60=1
We get N as 15
Riya said:
8 years ago
If we use the LCM method.
We get the 1 day's work of A= 3; B= 2; c= 1. & total work done by a,b,c together on 3rd day will be 6 , therefore; 60/6 = 10 &
2 days work by A= 3*2=6 , 60/6= 10.
Therefore 3 days work = 10 + 10 =20.
Please explain the solution after this.
We get the 1 day's work of A= 3; B= 2; c= 1. & total work done by a,b,c together on 3rd day will be 6 , therefore; 60/6 = 10 &
2 days work by A= 3*2=6 , 60/6= 10.
Therefore 3 days work = 10 + 10 =20.
Please explain the solution after this.
Jitesh said:
1 decade ago
A = 20.
B = 30.
C = 60.
LCM is 60 so a do the work in 1 day is 3 unit B is doing work in 1 day is 2 unit and C is 1 unit.
1st (A) day = 3.
2nd (A) day = 3.
3rd (A+B+C) day = 3+2+1 = 6.
Total work done in 3 days is 12 unit. So for 60 unit (60/12 = 5).
5*3 days = 15 days.
B = 30.
C = 60.
LCM is 60 so a do the work in 1 day is 3 unit B is doing work in 1 day is 2 unit and C is 1 unit.
1st (A) day = 3.
2nd (A) day = 3.
3rd (A+B+C) day = 3+2+1 = 6.
Total work done in 3 days is 12 unit. So for 60 unit (60/12 = 5).
5*3 days = 15 days.
Ganbrave said:
1 decade ago
@Poornima
A works for 2 days and B,C joins him on 3rd day.
so total combination of them is now 3days
Based on these 3days we can furthur calculate..ok??
A's 2 days work=1/10 and on 3rd day A,B,C together work=1/10
So sum of them gives u total 3days of work done by all of them.
A works for 2 days and B,C joins him on 3rd day.
so total combination of them is now 3days
Based on these 3days we can furthur calculate..ok??
A's 2 days work=1/10 and on 3rd day A,B,C together work=1/10
So sum of them gives u total 3days of work done by all of them.
Raj said:
1 decade ago
Question is on every 3rd day, it means from 20 days B and C will assist him on every 3rd day. So count every 3rd day till 20 days.
So we get 6 days. It means B and C will assist him 6 days from 20 days i.e. on 3rd, 6th, 9th, 12th, 15th and 18th day. I think the answer is 16.
So we get 6 days. It means B and C will assist him 6 days from 20 days i.e. on 3rd, 6th, 9th, 12th, 15th and 18th day. I think the answer is 16.
Kapil dev said:
8 years ago
day-1, day-2, day-3.
A. A. A+B+C.
Every 3rd-day b & c joining.
So 3(3A+B+C)=3(1/20)+1/30+1/60,
=9+2+1/60,
=1/5,
3A+B+C=1/3*5.
=1/15.
15 days work is complited.
A. A. A+B+C.
Every 3rd-day b & c joining.
So 3(3A+B+C)=3(1/20)+1/30+1/60,
=9+2+1/60,
=1/5,
3A+B+C=1/3*5.
=1/15.
15 days work is complited.
Rudra said:
9 years ago
Here, first of all, we have to find the total work.
So for this, L.C.M of 20 30 60 = 60,
a's b's c's one day work =3 2 1 (one day work)
Now a's 2days work = 6.
b's, c,s one day's woks = 2.
+1= 3.
So three days work 3 + 3 + 3 + 3 = 12.
3 days = 12
3*5 =15 days, 12 * 5 = 60.
So for this, L.C.M of 20 30 60 = 60,
a's b's c's one day work =3 2 1 (one day work)
Now a's 2days work = 6.
b's, c,s one day's woks = 2.
+1= 3.
So three days work 3 + 3 + 3 + 3 = 12.
3 days = 12
3*5 =15 days, 12 * 5 = 60.
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