kiran

• Oct 9th, 2005

 

A=600days,B=500days

Satya

• Jan 17th, 2006

 

Let, days taken by A = x days

Then, days taken by B = 6x days (since he is 6 times slower)

So, the difference = 6x - x = 5x days.

But, this difference = 100 days

Therefore, 5x = 100 implies x = 20 days

So, A takes 20 days to finish a job and B takes 120 days to finish a job.

if they work together, total time taken = 1/(1/20+1/120) = 120/7 = 17 1/7 days

abhinav

• Feb 20th, 2006

 

A completed work in 100 days before and A is 6 times faster than B..

 let total days to work is n.

A=6B;

A=(n-100) days taken to complete the work;

where as B=n days ;

A=(B-100)

therefore B=A+100;

A=6(A+100);

A=6A+600;

-600=6A-A

-600=5A

A=-600/5

A=300

B=400;

Ash

• Apr 30th, 2006

 

Since a is 6 times faster 6A=B

ie if b compeletes the work in n days B=n

& A=(n-100)

ie 6(n-100)=n

5n=600

n=120

ie B's work=120

A works 120-100=20

swathi

• Jul 8th, 2006

 

1/A=6/B , B=6A,A=B-100,solving 2 equations

A=20,B=120   ,(1/20)+(1/120)=7/120

so he takes 120/7days