Question 1181303
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Carlo can type a manuscript for 10 hours. He started working then Bobby joined him 2 hours later. 
they worked together for 4 hrs until Carlo decided to stop and then Bobby had to finish the manuscript alone. 
if Bobby finished the remaining portion in 6 more hours, how long can he type the whole manuscript working alone?
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        To this problem, I will give two solutions:  first  Arithmetic and the second  Algebra.



<pre>
                <U>Arithmetic solution</U>


Carlo's rate of work is 1/10 of the job per hour.

Carlo typed 2 hours + 4 hours = 6 hours and made {{{2/10}}} +{{{4/10}}} = {{{6/10}}}  of the job in 6 hours.

So, the remaining job,  which was 1 - {{{6/10}}} = {{{4/10}}} = {{{2/5}}},  was done by Bobby.


Bobby worked 4 hours + 6 hours = 10 hours and completed  {{{2/5}}}  of the job.

Hence, Bob's rate of work was  {{{((2/5))/10}}} = {{{2/(5*10)}}} = {{{2/50}}} = {{{1/25}}}.


It means that Bobby takes  25 hours to print the entire manuscript alone.    <U>ANSWER</U>



                <U>Algebra solution</U>


Carlo's rate of work is 1/10 of the job per hour.

Let 'b' be the Bobby's rate of work.  It is unknown value now, for which we should set uo an equation and find it.


In 2 hours, Carlo made  {{{2/10}}} = {{{1/5}}}  of the job.

In the next 4 hours, Carlo made  {{{4/10}}} = {{{2/5}}} of the job.

During these 4 hours, Bobby made 4b  parts of the job.

In next 6 hours, Bobby completed the job.


So, we write an equation for the whole job by summing these parts

    {{{1/5}}} + {{{2/5}}} + 4b + 6b = 1.


Simplify and find 'b'

    {{{3/5}}} + 10b = 1,

    10b = 1 - {{{3/5}}} = {{{2/5}}},

      b = {{{((2/5))/10}}} = {{{2/50}}} = {{{1/25}}}.


Hence, Bobby needs 25 hours to make the whole job working alone.

We get the same answer.
</pre>

Solved in simple manner in two different ways, for your better understanding.


This my solution teaches you on how to solve joint work problems using a conception of the rate of job.


It is MUCH SIMPLER than to follow the @mananth's approach.