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For doing some work, Ganesh takes $10$ days more than John. If both work together, they will complete the work in $12$ days. Find the number of days if Ganesh works alone.
A) $30$ days
B) $30$ weeks
C) $13$ days
D) $13$ weeks

Answer
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Hint:Let the number of days work by John and evaluate the total work by multiplying the number of days for both people. Make the equation using that they complete the work in $12$ days.

Complete step-by-step answer:
We are given that for doing some work, Ganesh takes $10$ days more than John.
Let the number of days for John if he works alone is $x$ days.
Therefore, the number of days for Ganesh if he works alone will be $x + 10$days.
Let the total work is the product of both the number of days if they work alone.
Therefore, total work will be $x(x + 10)$.
Now, we evaluate work for individuals for $1$ day.
To evaluate the work done in $1$ day, divide the total work by the total number of days.
Therefore, Work done by John in $1$ day is:
$\dfrac{{x(x + 10)}}{x} = x + 10$
Work done by Ganesh in $1$ day is:
$\dfrac{{x(x + 10)}}{{x + 10}} = x$
Therefore, total work done by both in one day is $x + x + 10 = 2x + 10$
Total days to complete the work by both will be:
$\dfrac{{x(x + 10)}}{{2x + 10}}$
According to the question they will complete the work in $12$ days.
Therefore, $\dfrac{{x(x + 10)}}{{2x + 10}} = 12$
Cross multiply and solve.
$
  {x^2} + 10x = 24x + 120 \\
   \Rightarrow {x^2} - 14x - 120 = 0 \\
 $
Evaluate the roots of the quadratic equation.
$
  {x^2} - 20x + 6x - 120 = 0 \\
   \Rightarrow x(x - 20) + 6(x - 20) = 0 \\
   \Rightarrow (x + 6)(x - 20) = 0 \\
   \Rightarrow x = - 6,20 \\
 $
Number of days can not be negative, therefore, $x = 20$
Number of days Ganesh takes to work alone will be $x + 10 = 20 + 10 = 30$ days.
Hence, option (A) is correct.

Note:In Time and Work questions when the numbers are given, let the total work is L.C.M. of those numbers in which they are completing the work and in the case of the polynomial, for total work take the product of the number of days.
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