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 35 of 36.
Azhar Khan said:
4 years ago
Easy way to answer is:
LCM of 20 30 60 = 60.
A = 20 * 3 = 60,
B = 30 * 2 = 60,
C = 60 * 1 = 60,.
A + B + C = 3 + 2 + 1 = 6 ( together)
A's 2 days work = 2 * 3 = 6.
Total work done by them = 6 + 6 = 12 unit
Total work is 60 unit.
So, 60 ÷ 15 = 5 units ( In one day by them )
Then in 3 days,
3*5 = 15 days
LCM of 20 30 60 = 60.
A = 20 * 3 = 60,
B = 30 * 2 = 60,
C = 60 * 1 = 60,.
A + B + C = 3 + 2 + 1 = 6 ( together)
A's 2 days work = 2 * 3 = 6.
Total work done by them = 6 + 6 = 12 unit
Total work is 60 unit.
So, 60 ÷ 15 = 5 units ( In one day by them )
Then in 3 days,
3*5 = 15 days
(29)
Raghunath Kisan said:
4 years ago
LCM of 20,30,60 = 60.
Efficiency of One day work of;
A = 60/20 = 3,
B = 60/30 = 2,
C = 60/60 = 1.
A + B + C = 3 + 2 + 1 = 6 ( together),
A's 2 days work = 3 * 2 = 6.
Total work done by them = 6 + 6 = 12 unit (in 3 days),
Then,
(Total work)/(Total work done in 3 days),
That is, 60/12 = 5,
So,
1 = 3, then 5 = 5 * 3 = 15 Days.
Efficiency of One day work of;
A = 60/20 = 3,
B = 60/30 = 2,
C = 60/60 = 1.
A + B + C = 3 + 2 + 1 = 6 ( together),
A's 2 days work = 3 * 2 = 6.
Total work done by them = 6 + 6 = 12 unit (in 3 days),
Then,
(Total work)/(Total work done in 3 days),
That is, 60/12 = 5,
So,
1 = 3, then 5 = 5 * 3 = 15 Days.
(124)
Suresh said:
4 years ago
@Monoj.
Best explanation.
Best explanation.
(24)
Sawaira ikram said:
4 years ago
Work done by A = 1/20.
Work done by B= 1/30.
Work done by C= 1/60.
Now, work done by A in 2 days is= (1/20) * 2 = 1/10.
And work done by A,B,C in one day= 1/20 + 1/30 + 1/60 = (3+2+1)/60 = 6/60 = 1/10.
And the work done in first 3 days = 1/10 + 1/10 = 1/5.
So we can say that 1/5 part of work is done= in 3 days.
And 1 complete work is done in=. 3/(1/5) = 3*5 = 15 days.
Work done by B= 1/30.
Work done by C= 1/60.
Now, work done by A in 2 days is= (1/20) * 2 = 1/10.
And work done by A,B,C in one day= 1/20 + 1/30 + 1/60 = (3+2+1)/60 = 6/60 = 1/10.
And the work done in first 3 days = 1/10 + 1/10 = 1/5.
So we can say that 1/5 part of work is done= in 3 days.
And 1 complete work is done in=. 3/(1/5) = 3*5 = 15 days.
(126)
Prem Prasidda said:
3 years ago
A = 1/20.
B = 1/30.
C = 1/60.
Now, 1/20+1/30+1/60=5/60 = 1/12.
Here, 1 days efficiency of A is 3, B is 2 and C is 1.
Now,
A can do the work in first day= 3,
" in the second day=3+3=6,
" in third day = 6+6=12,
So, in three days, A completed the work 12.
Now,
12 work = 3 days.
60 work = 3/12*60= 15 days.
B = 1/30.
C = 1/60.
Now, 1/20+1/30+1/60=5/60 = 1/12.
Here, 1 days efficiency of A is 3, B is 2 and C is 1.
Now,
A can do the work in first day= 3,
" in the second day=3+3=6,
" in third day = 6+6=12,
So, in three days, A completed the work 12.
Now,
12 work = 3 days.
60 work = 3/12*60= 15 days.
(102)
Harshal sonwane said:
3 years ago
1/5 work is done in 3 days.
The total work 1 done in x days.
Simply cross multiply;
x*1/5 = 3 * 1.
x/5 = 3,
x = 15.
The total work 1 done in x days.
Simply cross multiply;
x*1/5 = 3 * 1.
x/5 = 3,
x = 15.
(103)
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)
Aya adel said:
2 years ago
a=1/20
b=1/30
c=1/60.
So a + b + c = 1/10.
Work done in 3 days =1/20 + 1/20 + 1/10 = 1/5.
The second 3 days will be 2/5,
Third 3 days 3/5,
Fourth 3 days 4/5,
Fifth 3 days 5/5,
So, A needs 3*5 = 15 days.
b=1/30
c=1/60.
So a + b + c = 1/10.
Work done in 3 days =1/20 + 1/20 + 1/10 = 1/5.
The second 3 days will be 2/5,
Third 3 days 3/5,
Fourth 3 days 4/5,
Fifth 3 days 5/5,
So, A needs 3*5 = 15 days.
(132)
Naren kumar B said:
2 years ago
A = 20,
B = 30,
C = 60.
Find 1/A + ((1/B + 1/C)/3).
= 15 => Answer.
B = 30,
C = 60.
Find 1/A + ((1/B + 1/C)/3).
= 15 => Answer.
(193)
Subham said:
8 months ago
A can do 20.
B can do 30.
C can do 60.
Total work (LCM of Total time)= 60
Now one day work
A's One day work 3.
B's One day work 2.
C's One day work 1.
Now A 2 days work 3 × 2 = 6.
Now B and c join after 3 days (2+1)×3 = 9
Now 6 + 9 = 15 (Ans).
B can do 30.
C can do 60.
Total work (LCM of Total time)= 60
Now one day work
A's One day work 3.
B's One day work 2.
C's One day work 1.
Now A 2 days work 3 × 2 = 6.
Now B and c join after 3 days (2+1)×3 = 9
Now 6 + 9 = 15 (Ans).
(83)
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