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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.

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  1. cinar
    • 3 years ago
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    1/a +1/b =1/6 a=b+6

  2. hartnn
    • 3 years ago
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    wouldn't it be a=b+5 ?

  3. cinar
    • 3 years ago
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    yes (:

  4. namonminejasisyt
    • 3 years ago
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    oh wait2. my bad, it's all about work problems. not word problems.

  5. namonminejasisyt
    • 3 years ago
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    the work formula is Q=RT

  6. namonminejasisyt
    • 3 years ago
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    Can you please give me the whole answer?

  7. cinar
    • 3 years ago
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    1/(b+5) +1/b=1/6

  8. cinar
    • 3 years ago
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    b^2-7b-30=0

  9. namonminejasisyt
    • 3 years ago
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    okay thanks.

  10. cinar
    • 3 years ago
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    b=10

  11. cinar
    • 3 years ago
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    b+5=15

  12. amriju
    • 3 years ago
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    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
    • 3 years ago
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    it's all about work problems. anyone give me the right answer? the whole correct answer with the formula on.

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