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 2 of 36.
Piyush chandana said:
4 years ago
Mem ------> time ---- efficiency ----work
a------> 20 ---- 3 ---- 60
b------> 30 ---- 2 ---- 60
c------> 60 ---- 1 ---- 60
Now we make a series like :-
Day1 total unit of work will be -> 3 units.
Day2 total unit of work will be-> 3 units.
Day3 total unit of work will be -> 3+1+2 = 6(because b and c helped him).
Day1 Day2 Day3
3 3 6.
Work will be-> 12 units = 3 days.
The total units of work is 60 units then divide 60 with 12(60/12=5).
Now the total units are 5 times the 3 days of work than days will also multiply by 5 which is 3*5="15 days" (Ans).
a------> 20 ---- 3 ---- 60
b------> 30 ---- 2 ---- 60
c------> 60 ---- 1 ---- 60
Now we make a series like :-
Day1 total unit of work will be -> 3 units.
Day2 total unit of work will be-> 3 units.
Day3 total unit of work will be -> 3+1+2 = 6(because b and c helped him).
Day1 Day2 Day3
3 3 6.
Work will be-> 12 units = 3 days.
The total units of work is 60 units then divide 60 with 12(60/12=5).
Now the total units are 5 times the 3 days of work than days will also multiply by 5 which is 3*5="15 days" (Ans).
(18)
Balalakshmi said:
1 decade ago
Answer explained:
A is working for 2 days alone so A's two day work is first calculated as
A's 2 day work is =1/20*2 = 1/10
Now B and C work with A for only one day(ie 3rd day) and A also working along with B and C in the same 3rd day so A,B and C one day work(ie 3rd day work) is calculated as=(1/20+1/30+1/60)=1/5
Now A's 2 day work+ ABC one day work(ie 3rd day work is calculated)= (1/10+1/10)=1/5
Therefore in 3 days only 1/5 of work is completed
Thus to complete full work 3*5=15 days
(3days=1/5 of full wok completed
3*5 days=full work completed
15 days = full work completed)
A is working for 2 days alone so A's two day work is first calculated as
A's 2 day work is =1/20*2 = 1/10
Now B and C work with A for only one day(ie 3rd day) and A also working along with B and C in the same 3rd day so A,B and C one day work(ie 3rd day work) is calculated as=(1/20+1/30+1/60)=1/5
Now A's 2 day work+ ABC one day work(ie 3rd day work is calculated)= (1/10+1/10)=1/5
Therefore in 3 days only 1/5 of work is completed
Thus to complete full work 3*5=15 days
(3days=1/5 of full wok completed
3*5 days=full work completed
15 days = full work completed)
Magnesh said:
8 years ago
Simple, 3rd day B and C will hep A.
So first 2 days A will be working his work for two days is 2*(1/20) = 1/10.
Now 3rd day he is helped by B and C .
So A's work + B's work + C's work for 1 day is( 1/20 +1/30 +1/60) take LCM and solve we get 1/10.
Now A's 2days of work + (A+B+C's )work for 1 day.
1/10 +1/10 = 1/5.
So 1/5 work is completed which includes A's 2 days of work and Band C assistance for A (previous step same thing)
now
1/5 work is done in --- 3days
1( indicates complete work ) in ---- ? days
Ans is 15 days.
So first 2 days A will be working his work for two days is 2*(1/20) = 1/10.
Now 3rd day he is helped by B and C .
So A's work + B's work + C's work for 1 day is( 1/20 +1/30 +1/60) take LCM and solve we get 1/10.
Now A's 2days of work + (A+B+C's )work for 1 day.
1/10 +1/10 = 1/5.
So 1/5 work is completed which includes A's 2 days of work and Band C assistance for A (previous step same thing)
now
1/5 work is done in --- 3days
1( indicates complete work ) in ---- ? days
Ans is 15 days.
Akshay said:
8 years ago
Take LCM ot 20,30 and 60 you will get total units of work i.e 360 units.
Now find the 1 day work by using UNITS/DAYS.
A=18 units
B=12 units
C=6 units
Now A+B+C 1 day work is 18+12+6=36 units.
A+B+C will complete the work in 10 days(360/36).
Now, B and C joins every 3rd day which means A will work for 2 days alone and A+B+C will work on the 3rd day.
A 1day work = 18,
A 2 day work= 18*2=36 units,
A+B+C 1 day work is 36 units,
So 3 days of work =36+36= 72 units,
So 1-day work will be 72/3.
Now total work was 360 units.
Total days= 360*3/72 = 15 Days.
Now find the 1 day work by using UNITS/DAYS.
A=18 units
B=12 units
C=6 units
Now A+B+C 1 day work is 18+12+6=36 units.
A+B+C will complete the work in 10 days(360/36).
Now, B and C joins every 3rd day which means A will work for 2 days alone and A+B+C will work on the 3rd day.
A 1day work = 18,
A 2 day work= 18*2=36 units,
A+B+C 1 day work is 36 units,
So 3 days of work =36+36= 72 units,
So 1-day work will be 72/3.
Now total work was 360 units.
Total days= 360*3/72 = 15 Days.
Shashank Tripathi said:
7 years ago
A-> 20 days,
B->30 days,
C->60 days,
L.C.M of (20,30,60)=60.
Work = 60 unit.
so the capacity of A = 60/20 = 3 unit /day.
so the capacity of B = 60/30 = 2 unit /day.
so the capacity of C = 60/60 = 1 unit /day.
day 1 : A -> 3 unit/day.
day 2 : A -> 3 unit/day.
day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C)
Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days.
12 unit in 3 days -> equation (1)
so for 60 unit work multiply by 5 in equation (1)
60 unit work in 15 days.
So A can do in 15 days.
B->30 days,
C->60 days,
L.C.M of (20,30,60)=60.
Work = 60 unit.
so the capacity of A = 60/20 = 3 unit /day.
so the capacity of B = 60/30 = 2 unit /day.
so the capacity of C = 60/60 = 1 unit /day.
day 1 : A -> 3 unit/day.
day 2 : A -> 3 unit/day.
day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C)
Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days.
12 unit in 3 days -> equation (1)
so for 60 unit work multiply by 5 in equation (1)
60 unit work in 15 days.
So A can do in 15 days.
Shashank Tripathi said:
7 years ago
A-> 20 days,
B->30 days,
C->60 days,
L.C.M of (20,30,60)=60.
Work = 60 unit.
so the capacity of A = 60/20 = 3 unit /day.
so the capacity of B = 60/30 = 2 unit /day.
so the capacity of C = 60/60 = 1 unit /day.
day 1 : A -> 3 unit/day.
day 2 : A -> 3 unit/day.
day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C)
Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days.
12 unit in 3 days -> equation (1)
so for 60 unit work multiply by 5 in equation (1)
60 unit work in 15 days.
So A can do in 15 days.
B->30 days,
C->60 days,
L.C.M of (20,30,60)=60.
Work = 60 unit.
so the capacity of A = 60/20 = 3 unit /day.
so the capacity of B = 60/30 = 2 unit /day.
so the capacity of C = 60/60 = 1 unit /day.
day 1 : A -> 3 unit/day.
day 2 : A -> 3 unit/day.
day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C)
Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days.
12 unit in 3 days -> equation (1)
so for 60 unit work multiply by 5 in equation (1)
60 unit work in 15 days.
So A can do in 15 days.
Shashank Tripathi said:
7 years ago
A-> 20 days,
B->30 days,
C->60 days,
L.C.M of (20,30,60)=60.
Work = 60 unit.
so the capacity of A = 60/20 = 3 unit /day.
so the capacity of B = 60/30 = 2 unit /day.
so the capacity of C = 60/60 = 1 unit /day.
day 1 : A -> 3 unit/day.
day 2 : A -> 3 unit/day.
day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C)
Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days.
12 unit in 3 days -> equation (1)
so for 60 unit work multiply by 5 in equation (1)
60 unit work in 15 days.
So A can do in 15 days.
B->30 days,
C->60 days,
L.C.M of (20,30,60)=60.
Work = 60 unit.
so the capacity of A = 60/20 = 3 unit /day.
so the capacity of B = 60/30 = 2 unit /day.
so the capacity of C = 60/60 = 1 unit /day.
day 1 : A -> 3 unit/day.
day 2 : A -> 3 unit/day.
day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C)
Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days.
12 unit in 3 days -> equation (1)
so for 60 unit work multiply by 5 in equation (1)
60 unit work in 15 days.
So A can do in 15 days.
Pavankushoba said:
9 years ago
Hi everyone here is my solution.
According to problem (a + b + c) 1 day work = 1/10.
As per the problem A is assisted by B and C every third day so first let us find,
Work done by A in 2 days = 1/20 * 2 = 1/10.
As mentioned in step 3 B and C are assisted on the third day.
(So A's 2-day work, And one important point A also work along with B and C on third day = 1/10 + (1/20 + 1/30 + 1/60) = 1/5.
Its 3 days work done by A B and C = 1/5.
1-day work = x (assume).
By cross multiplication.
We get 3x = 1/5.
x = 1/15.
So its 15 days thank you.
According to problem (a + b + c) 1 day work = 1/10.
As per the problem A is assisted by B and C every third day so first let us find,
Work done by A in 2 days = 1/20 * 2 = 1/10.
As mentioned in step 3 B and C are assisted on the third day.
(So A's 2-day work, And one important point A also work along with B and C on third day = 1/10 + (1/20 + 1/30 + 1/60) = 1/5.
Its 3 days work done by A B and C = 1/5.
1-day work = x (assume).
By cross multiplication.
We get 3x = 1/5.
x = 1/15.
So its 15 days thank you.
Bhaskar I said:
8 years ago
A - 20.
B - 30.
C - 60.
Total work is 60.
Because LCM of ( 20,30,60 ) is 60.
Then A one day work is 60/20 = 3,
B one day work is. 60/30 = 2,
C one day work is 60 /60 = 1,
1st day work is only A = 3.
2nd day work is only A =3.
3rd day work is A+B+C (3+2+1) = 6.
So.,
3 day work 12.
6 day work 24.
9 day work 36.
12 day work 48.
15 day work 60.
Total 60 work is completed in 15 day.
It's simply
B - 30.
C - 60.
Total work is 60.
Because LCM of ( 20,30,60 ) is 60.
Then A one day work is 60/20 = 3,
B one day work is. 60/30 = 2,
C one day work is 60 /60 = 1,
1st day work is only A = 3.
2nd day work is only A =3.
3rd day work is A+B+C (3+2+1) = 6.
So.,
3 day work 12.
6 day work 24.
9 day work 36.
12 day work 48.
15 day work 60.
Total 60 work is completed in 15 day.
It's simply
Kiran said:
4 years ago
Hi, can anyone tell me what's wrong here?
Given,
A=20days.
B=30days.
C=60days.
I took as 60units of work by taking LCM of A, B, C.
A's one day work=3 units/day.
B's one day work=2 units/day.
A's two days work = 6 units.
B's one day work= 2 units.
C's one day work= 1 unit.
A+B+C's 3 days work = 9 units.
So, I can take A+B+C's one day work = 3 units =>9/3=3 units/day.
So, to complete 60 units of work they will take 20days =>20*3=60 units.
So, I am saying that answer is "20 days".
Correct me if I'm wrong.
Given,
A=20days.
B=30days.
C=60days.
I took as 60units of work by taking LCM of A, B, C.
A's one day work=3 units/day.
B's one day work=2 units/day.
A's two days work = 6 units.
B's one day work= 2 units.
C's one day work= 1 unit.
A+B+C's 3 days work = 9 units.
So, I can take A+B+C's one day work = 3 units =>9/3=3 units/day.
So, to complete 60 units of work they will take 20days =>20*3=60 units.
So, I am saying that answer is "20 days".
Correct me if I'm wrong.
(1)
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