## namonminejasisyt 3 years ago Mar and Ros working together can both encode a maniscript in 6 hours. If mar requires 5 hours longer to do the encoding than ros..how many hours can each do the encoding alone? it's all about problem solving.

1. cinar

1/a +1/b =1/6 a=b+6

2. hartnn

wouldn't it be a=b+5 ?

3. cinar

yes (:

4. namonminejasisyt

oh wait2. my bad, it's all about work problems. not word problems.

5. namonminejasisyt

the work formula is Q=RT

6. namonminejasisyt

7. cinar

1/(b+5) +1/b=1/6

8. cinar

b^2-7b-30=0

9. namonminejasisyt

okay thanks.

10. cinar

b=10

11. cinar

b+5=15

12. amriju

lets try simple algebra....mar( i hope she's pretty) takes 5 hours more than ros...now lets assume ros takes 'x' hours to do that crap alone...so mar takes x+5 hours..clearly. Now how much can ros do in one hour....she can encode(1/x) part...rite..? tAnd mar can do (1/x+5) part....now they r united...in one hour they together do ((1/x)+(1/x+5)) part.... now that much part in 1 hour...and the total work is ofcourse 1 part...so u use unitary method to find out the time taken in terms of x... and equate it with the accual time reqd thats 6 hours....find x..thats time for ros...the eqn is.."x^2-7x-30=0 " and x=10..

13. namonminejasisyt

it's all about work problems. anyone give me the right answer? the whole correct answer with the formula on.