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 9 of 36.
Deepa said:
3 months ago
LCM of A, B, C's works = 60,
And the work done by A in 1 day = 3 (60/20),
so in 2 days work done by A is = 3*2= 6,
And work done by A+B+C in 1 day = 6 [(60/20) + (60/30) + (60/60)],
So total work done for 3 days = 12,
So the total number of days required to complete 60 units of work = 60 * 3/12 = 15.
And the work done by A in 1 day = 3 (60/20),
so in 2 days work done by A is = 3*2= 6,
And work done by A+B+C in 1 day = 6 [(60/20) + (60/30) + (60/60)],
So total work done for 3 days = 12,
So the total number of days required to complete 60 units of work = 60 * 3/12 = 15.
(7)
Sravya Suryadevara said:
7 years ago
Here in question, it is mentioned that A alone worked for 2 days, on the third day B and C assisted him.
So,
A's 2 days work= 2(1/20) = 1/10.
On third day three together worked.
i.e.,
3rd day work = (1/20)+(1/30)+(1/60) = 1/10.
Work done in 3 days= (1/10)+(1/10) = 1/5.
Total work is 3*5 = 15 days.
So,
A's 2 days work= 2(1/20) = 1/10.
On third day three together worked.
i.e.,
3rd day work = (1/20)+(1/30)+(1/60) = 1/10.
Work done in 3 days= (1/10)+(1/10) = 1/5.
Total work is 3*5 = 15 days.
Mayur S said:
2 years ago
As LCM(20, 30, 60) = 60.
A's rate = 3, B's rate = 2 and C' rate = 1.
As work = 60 units
A will work for 2 days = 2 x 3 = 6.
B and C will join A on third day = (2+3+1) x 1 = 6
Work done in 3 days = 12 units.
Work done each day = 4.
The time required for 60 units of work at a rate of 4 is 15 days.
A's rate = 3, B's rate = 2 and C' rate = 1.
As work = 60 units
A will work for 2 days = 2 x 3 = 6.
B and C will join A on third day = (2+3+1) x 1 = 6
Work done in 3 days = 12 units.
Work done each day = 4.
The time required for 60 units of work at a rate of 4 is 15 days.
(386)
Naresh said:
1 decade ago
If we add the work of b and c for 1 day we get 1/30 + 1/60=1/20.
First day = 1/20.
Second day =2/20.
Third day =4/20 // work of b & c is added 3/20 + 1/20.
Every 3 days 4/20.
For 6 days we get 8/20.
For 9 days we get 12/20.
For 12 days we get 16/20.
For 15 days we get 20/20 = 1. Job over.
First day = 1/20.
Second day =2/20.
Third day =4/20 // work of b & c is added 3/20 + 1/20.
Every 3 days 4/20.
For 6 days we get 8/20.
For 9 days we get 12/20.
For 12 days we get 16/20.
For 15 days we get 20/20 = 1. Job over.
Suribabu said:
1 decade ago
In this problem first we understand the work done till 3rd days.
A can do his work for 2 days (2*(1/20)) and the 3rd day. A can do work with B & C i.e. 1/20+1/30+1/60 = 1/10 so up-to 3 days.
A can do 1/10+1/10 = 1/5 work (3 days work) then total work done in 3*5 = 15 days.
Hope understand.
A can do his work for 2 days (2*(1/20)) and the 3rd day. A can do work with B & C i.e. 1/20+1/30+1/60 = 1/10 so up-to 3 days.
A can do 1/10+1/10 = 1/5 work (3 days work) then total work done in 3*5 = 15 days.
Hope understand.
Bhuvana said:
8 years ago
It is solved by chocholate method
20 30 60 lcm of those 60.
3 2 1 are their 1 day works.
a a a+b+c a a a+b+c a a a+b+c a a a+b+c a a
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
3 +3 + 7 +3 +3 +7 +3 +3 +7 +3 +3 +7 +3 +3 +7=60 days.
20 30 60 lcm of those 60.
3 2 1 are their 1 day works.
a a a+b+c a a a+b+c a a a+b+c a a a+b+c a a
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
3 +3 + 7 +3 +3 +7 +3 +3 +7 +3 +3 +7 +3 +3 +7=60 days.
Xyz said:
1 decade ago
@Yashraj
When we want to do (1/20)+(1/30)+(1/60) generally, we take the help of L.C.M, But in above case Highest number is 60 which is Divisible by all the remaining number i.e. 20 and 30.
Hence we take L.C.M. as 60 i.e. (3+2+1)/60 becomes 6/60.
Build your Basic maths Strong My friend.
When we want to do (1/20)+(1/30)+(1/60) generally, we take the help of L.C.M, But in above case Highest number is 60 which is Divisible by all the remaining number i.e. 20 and 30.
Hence we take L.C.M. as 60 i.e. (3+2+1)/60 becomes 6/60.
Build your Basic maths Strong My friend.
Vinay vidhani said:
1 decade ago
Please tell me some one where I'm doing mistake.
Hear A can do a work in 20 days B-30 days and C doing in 60 days.
A's 3 day work is 1/20*3.
B's work will be on 3rd day is 1/90 and C's - 1/180 they work together then answer is coming 6 days please tell about that's solution soon please.
Hear A can do a work in 20 days B-30 days and C doing in 60 days.
A's 3 day work is 1/20*3.
B's work will be on 3rd day is 1/90 and C's - 1/180 they work together then answer is coming 6 days please tell about that's solution soon please.
JPV said:
8 years ago
Works that done till 3rd day:
[ (A's ist day)+ (A's 2nd day) + (A+B+C combined on 3rd day)].
ie = 1/20+1/10+1/10,
= 1/4.
ie remaining pending work = 3/4.
So, the no of days Al required to complete the work= 20*(3/4),
= 15 days.
[ (A's ist day)+ (A's 2nd day) + (A+B+C combined on 3rd day)].
ie = 1/20+1/10+1/10,
= 1/4.
ie remaining pending work = 3/4.
So, the no of days Al required to complete the work= 20*(3/4),
= 15 days.
Abhishek said:
1 decade 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.
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