Latest IBM Aptitude Question SOLUTION: A can do in 6 day and B can do in 8 and C can do in 12 days but b left after working for 6 days for how many no of day A and C shluld work ?

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A can do in 6 day and B can do in 8 and C can do in 12 days but b left after working for 6 days for how many no of day
A and C shluld work ?

Asked In IBM Debdulal Mandal (9 years ago)

Solved sruthi Read Solution (17)

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Read Solution (17) :

sruthi  (9 years ago) Best Solution

a=1/6
b=1/8
c=1/12
b left after 6 days so 6*1/8=3/4
work remaining =1-3/4=1/4
a+c=1/6+1/12=18/72=1/4 ( a &c take 4 days to complete the whole work ) 1 work=4 days
no of days for 1/4th of work = 1 day

Is this solution Helpfull? Yes (38) | No (3)  

koppolu naga sravani  (9 years ago)

B for one day work is 1/8
he worked for 6 days, so 6*1/8=3/4 work will be completed.
remaining work is 1-3/4=1/4
A and C can complete the work in 4 days
but there is only 1/4 work that is to be completed, so A and C can complete 1/4 work in 1 day..
answer: 1 day

Is this solution Helpfull? Yes (32) | No (1)  

Bhaskar Balu Pawar  (9 years ago)

try to find out their efficiencies and LCM first
so we get
LCM of 6,8,12=48 nothing but a total work to do
efficiencies
A=48/6=8
B=48/8=6
C=48/12=4
therefore B finishes his work in 6 days 6*6=36
remaining work=48-36=12
that can be finished by A & C in
12/(8+4)=12/12
1 must be the answer

Is this solution Helpfull? Yes (24) | No (13)  

Sri Reddy  (9 years ago)

let x be the no. of days..
6/8+x/6+x/12=1
solving we get x=1 ...so 1 day is the answer

Is this solution Helpfull? Yes (18) | No (2)  

karthik  (9 years ago)

6(a+b+c)+x(a+c)=1
9/4+x/4=1
x=5days,remaining days will be 5days

Is this solution Helpfull? Yes (6) | No (1)  

pavan repalle  (9 years ago)

this type of problems try like this. lets assume total work = total no of chocolates.
a= 6 days
b= 8 days
c= 12 days
find out LCM of a,b, and c = 48
So total chocolates = 48 (assume total work = total chocolates).
a can able to eat all chocolates in 6 days. So in one day he can able to eat = 8 choclates
b can able to eat all chocolates in 8 days. So in one day he can able to eat = 6 choclates
c can able to eat all chocolates in 12 days. So in one day he can able to eat = 4 choclates.


Now look at the question b worked for 6 days, so he can eat 36 choclates in 6 days( 1 day = he can eat 6 chocs)
remaining chocolates = 48-36 = 12.
A and C can combinely eat 12 chocolates per day. So they will take 1 days to complete the remaining work

Is this solution Helpfull? Yes (6) | No (3)  

rakesh  (9 years ago)

a,b,c worked for 6days their work is 6*(1/6+1/8+1/12)= 9/4 remaning work is 1-9/4 = 5/4(work cannot be -ve) and now a and c has to complete the remaining work =efficiency /time ... so a+c eff= 1/6+1/12=1/4... time= 1/4*4/5=ie. .......5days............guys let me know if i am wrong ........

Is this solution Helpfull? Yes (4) | No (3)  

senthilkumar sivraman  (9 years ago)

A-1/6
B-1/8
C-1/12
B(6days)--->>6*1/8--->>3/4
Remaining work--->>1/4
A+C(1day)--->>>1/6+1/12--->>1/4
SO they complete it in 1 day

Is this solution Helpfull? Yes (3) | No   

tirupathi reddy  (9 years ago)

No need to work because A lonely can do whole work in 6 days.
They can do double work in 6 days.

A do in 6 days, b Do in 8 days, c do in 12 days.
B left after working 6 days,
The work done by three of them is ( in 6 days)
6(1/6+1/8+1/12) = 9/4

Is this solution Helpfull? Yes (2) | No (9)  

Balaji  (9 years ago)

Guys plz Try Understand the question...
There is NO A+B+C work..
Simply A do a WOrk in 6 days
B do in 8 days
c do in 12 days..
do not combine three ...
just consider B work only, If B continues to work it complete in 4 days,
but due to some problem it only works for 6 days, remaing 1/4 work should b completed in 4 days...

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

Harsha Vardhan  (7 years ago)

Take an LCM of (6,8,12)=24..
So A can complete in 6days compare 6 wid 24==>6*4===>4.
So B can complete in 8days compare 8 wid 24==>8*3===>3.
So C can complete in 12days compare 12 wid 24==>12*2===>2.
B left after 6days means they worked for 6days and then B left..
So 6(A+B+C)===> 6(4+3+2)= =>6*9=54.
A+B+C= 54-24===> i.e., 30.
So after 6days B left.. A+C= 30/4+2
A+C= 30/6====> A+C= 5days..

Is this solution Helpfull? Yes (1) | No (3)  

chandravardhan  (9 years ago)

FOR 6 days
6(1/6+1/8+1/12)

a&c can do it in

={1/6+1/12} - 6{1/6+1/8+1/12}
= simplify this

Is this solution Helpfull? Yes | No (7)  

Praveen Kumar Vishwakarma  (9 years ago)

A and C working together only 1 day

Is this solution Helpfull? Yes | No (1)  

Prachi  (9 years ago)

A+B+C = (1/6+1/8+1/12) = 9/24

1 day work = 9/24*6 = 9/4

The work will be done by three of them will be 9/4
Remaining work done = 1-9/4 =5/4.
multiply it with B days = 5/4*8 = 10 days.ans

Is this solution Helpfull? Yes | No (3)  

shaik sharina  (9 years ago)

ans:1 day
chacolate method:
lcm of a,b,c=144
so work = 144chacolates
a can eat 144/6=24 chacolates per day
b can eat 144/8=18 c/day
c can eat 144/12=12 c/d
b can do awork in 6 days=6*18=108
a+c eat remaining chacolates=12+24=36chacolates/day
so ans is 1day

Is this solution Helpfull? Yes | No (1)  

shaik sharina  (9 years ago)

ans:1 day
chacolate method:
lcm of a,b,c=144
so work = 144chacolates
a can eat 144/6=24 chacolates per day
b can eat 144/8=18 c/day
c can eat 144/12=12 c/d
b can do awork in 6 days=6*18=108
a+c eat remaining chacolates=12+24=36chacolates/day
so ans is 1day

Is this solution Helpfull? Yes | No (1)  

saket biharee  (9 years ago)

3 women=2 man
then 21 women =14 man
then apply formula ( m 1 *d 1* h 1)/w 1=(m 2*d 2* h 2)/w 2
15*8*21=14*6*d 2

d 2= (15*8*21)/(14*6) =30

Is this solution Helpfull? Yes | No (3)  

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