A, B and C working together can finish a piece of work in \[8\] hours. A alone can do it in \[20\] hours and B alone can do it in \[24\] hours In how many hours will C alone do the same work?
Answer
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Hint: Using the unitary method we first find the work done by all of them in one hour individually. First we need to find the work done in one hour by all the candidates (A and B as C is not given). After this we need to equate the sum of all candidate’s work done in one hour to the time period of A, B and C working together.
The work done by all the candidates in one hour is given as:
\[\dfrac{\text{1}}{\text{A}}\text{+}\dfrac{\text{1}}{\text{B}}\text{+}\dfrac{\text{1}}{\text{C}}\]
where \[\text{A}\, \text{B}\, \text{C}\] are the candidates and their work done in one hour.
Complete step-by-step answer:
Work done by A alone in one hour is \[20\text{ hrs}\]
Work done by B alone in one hour is \[\text{24 hrs}\]
Work done by C alone in one hour is \[\text{x hrs}\]
The work done by all the candidates in one hour is given as:
\[\dfrac{\text{1}}{\text{A}}\text{+}\dfrac{\text{1}}{\text{B}}\text{+}\dfrac{\text{1}}{\text{C}}\]
Now equating the work done by all (\[8\] hours) in one hour we get the work done as \[\dfrac{1}{8}th\] of all the work done.
\[\dfrac{\text{1}}{\text{A}}\text{+}\dfrac{\text{1}}{\text{B}}\text{+}\dfrac{\text{1}}{\text{C}}=\dfrac{1}{8}\]
\[\dfrac{\text{1}}{\text{20}}\text{+}\dfrac{\text{1}}{\text{24}}\text{+}\dfrac{\text{1}}{\text{x}}=\dfrac{1}{8}\]
\[\dfrac{\text{1}}{\text{x}}=\dfrac{1}{8}-\dfrac{\text{1}}{\text{20}}\text{+}\dfrac{\text{1}}{\text{24}}\]
\[\dfrac{\text{1}}{\text{x}}=\dfrac{15-6+5}{120}\]
\[\dfrac{\text{1}}{\text{x}}=\dfrac{4}{120}\]
\[x=30\text{ days}\]
\[\therefore \] C alone will take \[30\text{ days}\] to complete the work.
Note: Students may go wrong while calculating the total work for each individual, the total work is taken as unit \[1\]. Therefore, \[\dfrac{1}{20},\text{ }\dfrac{1}{24}\] and \[\dfrac{1}{x}\] are unit of work done by A, B and C in the time period of one hour. Hence, we need to find the total number of work done by C based on one hour.
The work done by all the candidates in one hour is given as:
\[\dfrac{\text{1}}{\text{A}}\text{+}\dfrac{\text{1}}{\text{B}}\text{+}\dfrac{\text{1}}{\text{C}}\]
where \[\text{A}\, \text{B}\, \text{C}\] are the candidates and their work done in one hour.
Complete step-by-step answer:
Work done by A alone in one hour is \[20\text{ hrs}\]
Work done by B alone in one hour is \[\text{24 hrs}\]
Work done by C alone in one hour is \[\text{x hrs}\]
The work done by all the candidates in one hour is given as:
\[\dfrac{\text{1}}{\text{A}}\text{+}\dfrac{\text{1}}{\text{B}}\text{+}\dfrac{\text{1}}{\text{C}}\]
Now equating the work done by all (\[8\] hours) in one hour we get the work done as \[\dfrac{1}{8}th\] of all the work done.
\[\dfrac{\text{1}}{\text{A}}\text{+}\dfrac{\text{1}}{\text{B}}\text{+}\dfrac{\text{1}}{\text{C}}=\dfrac{1}{8}\]
\[\dfrac{\text{1}}{\text{20}}\text{+}\dfrac{\text{1}}{\text{24}}\text{+}\dfrac{\text{1}}{\text{x}}=\dfrac{1}{8}\]
\[\dfrac{\text{1}}{\text{x}}=\dfrac{1}{8}-\dfrac{\text{1}}{\text{20}}\text{+}\dfrac{\text{1}}{\text{24}}\]
\[\dfrac{\text{1}}{\text{x}}=\dfrac{15-6+5}{120}\]
\[\dfrac{\text{1}}{\text{x}}=\dfrac{4}{120}\]
\[x=30\text{ days}\]
\[\therefore \] C alone will take \[30\text{ days}\] to complete the work.
Note: Students may go wrong while calculating the total work for each individual, the total work is taken as unit \[1\]. Therefore, \[\dfrac{1}{20},\text{ }\dfrac{1}{24}\] and \[\dfrac{1}{x}\] are unit of work done by A, B and C in the time period of one hour. Hence, we need to find the total number of work done by C based on one hour.
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