Question

A and B together can do a piece of work in 6 days. if A's one day work is one and a half times the one day's work of B, find in how many days each alone can finish the work?​

Answer 1

Hint: To solve this question, we first assign the unknown values with a variable. We should find how many days will A and B take to finish the work so we will consider them as x and y. Then we will check their one day work and compare. We also know that A’s work is one and a half times of B. Hence, we will get two quadratic equations. By solving them simultaneously we will get the answer.

Complete step-by-step answer:
Let us consider that A needs $ x $ days to finish the work alone and B requires $ y $ days to finish the work .
Then , we see that
One day work of A is $ \dfrac{1}{x} $
One day work of B is $ \dfrac{1}{y} $
Total one day work of A and B is $ \dfrac{1}{x} + \dfrac{1}{y} $ which is equal to $ \dfrac{1}{6} $ .
Then, \[\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{6} - (i)\]
It is also mentioned in the question that, A's one day work is one and a half times the one day's work of B.
Hence, $ \dfrac{1}{x} = \dfrac{3}{2} \times \dfrac{1}{y} - (ii) $
Now comparing (i) and (ii) equation we get,
\[
  \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{6} \\
  \dfrac{3}{{2y}} + \dfrac{1}{y} = \dfrac{1}{6} \\
  \dfrac{1}{y}\left( {\dfrac{3}{2} + 1} \right) = \dfrac{1}{6} \\
  \dfrac{1}{y}\left( {\dfrac{5}{2}} \right) = \dfrac{1}{6} \\
  y = \dfrac{{6 \times 5}}{2} \\
  y = 15 \;
 \]
Now,
 $
  \dfrac{1}{x} = \dfrac{3}{2} \times \dfrac{1}{y} \\
  \dfrac{1}{x} = \dfrac{3}{2} \times \dfrac{1}{{15}} \\
  x = \dfrac{{15 \times 2}}{3} \\
  x = 10 \;
  $
Hence , A requires 10 days to finish the work alone and B requires 15 days to finish the work. $ $
So, the correct answer is “10 days”.

Note: We see that the number of workers is inversely proportional to the number of days required to finish the work. If we have more workers, then the work will get over in less number of days. Number of days will also increase if the workers are less.