Find the number of days for John to complete the work alone in the following case:For doing some work, John takes 10 days more than Lee. If they both work together the work will complete in 12 days.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

It is given that for doing some work John takes 10 days more than Lee. If both work together the work will complete in 12 days.

Let the rate at which John completes the work be J and the rate at which Lee completes the work be L

We know that J + L = 1/12

1/L - 1/J = 10

=> L - J = 10 L*J

Substitute L = 1/12 - J

=> 1/12 - J - J = 10*(1/12 - J)*J

=> (1 - 24J)/12 = 10J/12 - 120J^2/12

=> 1 - 24J = 10J - 120J^2

=> 120J^2 - 34J + 1 = 0

=> 120J^2 - 30J - 4J + 1 = 0

=> 30*J(4J - 1) - 1(4J - 1) = 0

=> (30J - 1)(4J - 1) = 0

=> J = 1/30 or J = 1/4

The rate at which John works is 1/30

The number of days taken by John alone to finish the work is 30.

shrinath's profile pic

shrinath | Student, Undergraduate | (Level 1) Honors

Posted on

Let Lee do the work in      --> 'x' days

each day he does(rate)    --> 1/x

then John does in work in --> (x+10) days

each day he does(rate)    --> 1/(x+10)

(John and Lee) combined do work in --> 12 days

rate of doing work combined --> 1/12

So, 1/x + 1/(x+10)=1/12

(2x+10)/(x^2+10x) = 1/12

or    (x^2+10x)/(2x+10) = 12 --> x^2 -14x -120=0

implies x=20..

i.e lee does work in 20 days

----->john alone does work in 20+10 =30 days.

Answer is 30 days...

 

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