Latest Accenture Aptitude Question SOLUTION: 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? A. 12 days B. 15 days C.

• Placement Papers
• /
• Accenture
• /
• Numerical Ability
• /
• Time and Work

Accenture Question Papers: Accenture Interview Experience(7) ACCENTURE Company Profile

(#M40004058) ACCENTURE Numerical Ability question accenture latest puzzle Keep an EYE

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?

A. 12 days
B. 15 days
C. 16 days
D. 18 days

Asked In Accenture (12 years ago)

Unsolved Read Solution (22)

Is this Puzzle helpful?   (229)   (50) Submit Your Solution Time and Work

Read Solution (22) :

Jyoti  (12 years ago)

Work done in 3 days=(3/20)+(1/30)+(1/60)=1/5
Remaining work=4/5
If 1/5 work is done in 3 days then,
4/5 work is done in (3*5*4)/5=12days
Total no. of days will become 12+3=15days for the complete work to be done.:)

Is this solution Helpfull? Yes (54) | No (17)  

J.Durga  (11 years ago)

A's 2 day's work =2 * (1/20)= 1/10
(A + B + C)'s 1 day's work =((1/20)+(1/30)+(1/60))= 1/10
Work done in 3 days =(1/10)+(1/10)= 1/5
Now, (1/5) work is done in 3 days.
Whole work will be done in (3 x 5) = 15 days.

Is this solution Helpfull? Yes (45) | No (9)  

anil kumar  (11 years ago)

The work done by A +B +C in 3 Days is 3/20+1/20+1/60=12/60 i.e 1/5
hence 1/5 work done in 3 days
so 1 work done in 3*5=15 days
Option :B

Is this solution Helpfull? Yes (26) | No (8)  

Subhalakshmi.M  (8 years ago)

A can do 1/20 units of work per day.
B can do 1/30 units of work per day.
C can do 1/60 units of work per day.

Here,B and C can able to assist A only on every 3rd day.
So,we can express as:
1/20+1/3(1/30+1/60)
=1/20+1/3(3/60)
=1/20+3/180
=1/20+1/60
=4/60
=1/15

So the work can be finished in 15 days if A is being assisted by B and C on every 3rd day.

Is this solution Helpfull? Yes (17) | No   

saif alam  (9 years ago)

assume total work=lcm of(20,30,60=60) so total work=60 units
a--20 60/20=3unit/day
b--30 60/30=2 unit/day
c--60 60/60=1 unit/day
.............................
a+b+c =6 units/day
work done by a in 2 days=6 unit and next day assisted by b & c so total work on 3rd day=6 unit..thus in first 3 days work completed=6+6=12 units
if 12 unit................3 days
60 unit..............3*60/12=15days

Is this solution Helpfull? Yes (16) | No (3)  

Darshit Pathak  (9 years ago)

A completes 5% of total work in a day(100/20)...B->3.3% C->1.6%
Work done by (B+C) in a day is 5%(approx)
A does work alone for 2 days n completes 10% in 2 days.
on 3rd day A is assisted by B and C so total work done on 3rd day is (5+5)=10%

work done in 3 days=20%
days required to complete 100% work=15 days.

Is this solution Helpfull? Yes (3) | No (3)  

Deepak Kumar  (7 years ago)

work done in 24/120= 1/5;
then remaining work =4/5
if 1/5 work is done in 3 days.
4/5 work is done in (3*5*4)/5=12 days
total no. of days will beccome 12+3= 15 days....
so Ans is = 15 days..

Is this solution Helpfull? Yes (2) | No   

Gajula Hari Chandana  (7 years ago)

A can do the work in 20 days
B can do the work in 30 days
C can do the work in 60 days
total work is 60(L.C.M of 20,30,60=60),
so A`s 1 day work 3
B`s 1 day work 2
C`s 1 day work 1
A`s 2 day work is 6
on 3rd day B & C will work with A,then (A+B+C)`s total work in 3rd day is 6(3+2+1)
then at total 3 day work is 12
then 60 work can be done in 15 days..

Is this solution Helpfull? Yes (2) | No (2)  

JUHI RAI  (9 years ago)

answer is 15days

Is this solution Helpfull? Yes (1) | No (4)  

B.Naga Sivani  (8 years ago)

15 days is ans

Is this solution Helpfull? Yes (1) | No (2)  

C.G.Kavya sree  (8 years ago)

LCM(20,30,60)=120
A does 6 units/day
B does 4 units/day
C does 2 units/days

so, 6+6+12,6+6+12,6+6+12,6+6+12,6+6+12
24*5=120
5*3=15 days

Is this solution Helpfull? Yes (1) | No (4)  

B.Naga Sivani  (8 years ago)

15 days is ans

Is this solution Helpfull? Yes | No (1)  

DHIVYA  (8 years ago)

work done by A in one day =1/20
work done by A in two days=2/20
On third day A,B,C are working together, so
work done by three of them in one day=(1/10)+(1/30)+(1/60)=(1/10)
work completed in third day=(1/5)
Remaining work to be completed=1-(1/5)=4/5
Three days are required to complete 1/5th work, So numbr of days required to complete (4/5)th work is
=4*3
=12 days

Is this solution Helpfull? Yes | No   

Neeraj  (8 years ago)

b.
A's 1 day work = 1/20
(A+B+C) 1 day work 2/20
total work completed on 15th day

Is this solution Helpfull? Yes | No   

alpa sharma  (7 years ago)

AVG=SUM/8
AVG+2=SUM-(W1+W2)+80/8
W1+W2/2=32

Is this solution Helpfull? Yes | No   

PRIYA DEV  (7 years ago)

A's 2days work= 2*1/20= 1/10
Third day i.e (A+B+C)'s 1days work= 1/20+1/30+1/60= 1/10
Work done in 3 days= 1/10+1/10= 1/5
Now, 1/5 work done in 3 days= 3*5= 15days

Is this solution Helpfull? Yes | No   

smriti  (7 years ago)

total work= 60
a=60/20=3
b=60/30=2
c=60/60=1
3+3+3+2+1=12
60/12*3=15

Is this solution Helpfull? Yes | No   

c.vinay kumar  (7 years ago)

LCM of 20,30,60 is 60, MEANS TOTAL WORK IS 60
A does it work for - 60/20=3 units/day
B does it work for -60/30=2 units /day
c does it work for - 60/60=1 unit /day
first day work done=3
second day work done =3
third day A+B+C=6
FOR OVER ALL THREE DAY WORK DONE IS=12
THEN, 3 DAYS*5 = WORK DONE FOR 3 DAYS IS 12*5
15 DAYS = TOTAL WORK 60

ANSWER IS 15 DAYS

Is this solution Helpfull? Yes | No   

nishant raj  (7 years ago)

As the question says,
A, B, C can do the work in 20, 30 and 60 days respectively,
so taking a common multiple of these three numbers, say 60 (LCM can go fine with this too). And 60 is the total work to do in units.
now we need to calculate efficiency, A can complete work in 20 days, so each day he needs to do (60/20)=3 unit work,
similarly, B=2 unit/day
C=1 unit/day
now as question says,on third day both B and C joins A for his work;
A+A+(A+B+C) this is for first three days
on calculating we get,3+3+(3+2+1) ==12 units in three days.
so multiplying 5 both sides,
we get 60 units in 15 days,ANS

Is this solution Helpfull? Yes | No   

Nirnoy Nimesh Behera  (6 years ago)

Total units of work to be completed= LCM of 20,30,60 = 60 units
So in 1 day A does=60/20= 3 units
B= 2 units
C= 1 unit
& A is assisted by B and C on every third day
So, 1st day 3 units (by A)
2nd day 3 units (by A)
3rd days 6 units (by A+B+C)
In 3 days 12 units are completed, therefore in (60/12)*3 days = 15 days, the work can be completed.

Is this solution Helpfull? Yes | No   

Kumari sanjana  (6 years ago)

A=5%
B=3.33%
C=1.66%
(Work done by them in a day)
In 2 days A will do 10% work and on 3rd day A+B+C= 10%+10% = 20% work is done
20% work = 3 day
100% work = 100*3/20=15 days
So, 15 day is the answer

Is this solution Helpfull? Yes | No   

saiprakash  (6 years ago)

A=W/20 per day work
B=W/30 C=W/60
3x(w/20)+x(w/30+w/60)=w
3x/20+3x/60=1
12x/60=1
x=5
so 3(5)=15 days
so Option B

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.