Question

Working individually A, B and C can finish a piece of work in 16 days, 20 days and 30 days respectively. In how many days can A, B and C together complete a work which is \[3\dfrac{1}{2}\] times the previous work.

Answer 1

Hint: Here in this question we have determined the number of days taken by A, B and C together to complete the work. So first we determine how much work is done by an individual each day. Then we determine the total work done by all three members in a day. By taking the reciprocal of total work done by 3 members in a day will represent the number of days of previous work. Then on multiplying the previous work and \[3\dfrac{1}{2}\] , we can determine the solution for the given question.

Complete step by step solution:
Work is defined as something which has an effect or outcome; often the one desired or expected. The basic concept of Time and Work is similar to that across all Arithmetic topics, i.e. the concept of Proportionality. Efficiency is inversely proportional to the Time taken when the amount of work done is constant.
Now we consider the question,
The number of days A can complete the work = 16 days.
The number of days B can complete the work = 20 days.
The number of days C can complete the work = 30 days.
Now we calculate for one how much work will be done by A, B and C
The work done by A for one day = \[\dfrac{1}{{16}}\]
 The work done by B for one day = \[\dfrac{1}{{20}}\]
The work done by C for one day = \[\dfrac{1}{{30}}\]
The total work done by A, B and C for one day = \[\dfrac{1}{{16}} + \dfrac{1}{{20}} + \dfrac{1}{{30}}\]
On adding these three fractions, we should take LCM for 16, 20 and 30. So we have
 \[
  2\left| \!{\underline {\,
  {16,20,30} \,}} \right. \\
  2\left| \!{\underline {\,
  {8,10,15} \,}} \right. \\
  2\left| \!{\underline {\,
  {4,5,15} \,}} \right. \\
  2\left| \!{\underline {\,
  {2,5,15} \,}} \right. \\
  5\left| \!{\underline {\,
  {1,5,15} \,}} \right. \\
  3\left| \!{\underline {\,
  {1,1,3} \,}} \right. \\
  \,\,\,\,1,1,1 \\
 \]
By taking LCM we have
  \[ \Rightarrow \dfrac{{\dfrac{1}{{16}} \times 240 + \dfrac{1}{{20}} \times 240 + \dfrac{1}{{30}} \times 240}}{{240}}\]
On simplifying we have
 \[ \Rightarrow \dfrac{{15 + 12 + 8}}{{240}}\]
 \[ \Rightarrow \dfrac{{35}}{{240}}\]
Therefore the total work done by the A, B and C for one day = \[\dfrac{{35}}{{240}}\]
In one day \[\dfrac{{35}}{{240}}\] part of the work done by A, B and C.
Therefore to complete work by A, B and C together they will take \[\dfrac{{240}}{{35}}\] days.
The \[\dfrac{{240}}{{35}}\] days. will represent the previous work.
Now we have to determine the days when A, B and C together complete a work which is \[3\dfrac{1}{2}\] times the previous work.
Therefore we have
 \[ \Rightarrow 3\dfrac{1}{2} \times \dfrac{{240}}{{35}}\] days
Convert the mixed fraction \[3\dfrac{1}{2}\] into an improper fraction.
 \[ \Rightarrow \dfrac{7}{2} \times \dfrac{{240}}{{35}}\] days
By using the tables of multiplication we simplify it. So we have
 \[ \Rightarrow 24\] days.
Therefore in 24 days A, B and C together can complete the work
So, the correct answer is “24 days”.

Note: While adding the fractions where the value of denominators differ, we should take LCM for the denominators. To find the LCM we use the prime factorisation method. So we have to divide the numbers by a common number, if any one of the numbers is not divisible by the number which we had considered we write the number as it is. For example: If we want to find the LCM for the numbers 2 and 9
 \[
  2\left| \!{\underline {\,
  {2,9} \,}} \right. \\
  \,\,\,\,1,9 \;
 \]
The 9 is not divisible by 2, so we are writing as it is.