Answer

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**Hint:**We first have to express the given equation as an equation with terms A, B, C and D.

We have to write the work done by A, B, C, and D with the given conditions by taking the reciprocal of the time taken. We then have to convert \[15\dfrac{3}{4}\] into a fraction. Therefore, the work done by them can be found by adding all the obtained equations. Add the RHS taking the LCM. Divide the whole equation by 3 and cancel out 3 from the numerator and denominator of the LHS. We get the value of A+B+C+D which is the work done by them. We need to take the reciprocal of work done to get the time taken by A, B, C, and D to reap the field.

**Complete step by step solution:**

According to the question, we are asked to find the total number of days it takes to reap together by A, B, C and D.

We have been given that A, B and C can reap a field in \[15\dfrac{3}{4}\] days.

Let us convert the statement into an equation, we get

Time taken by A, B and C to do the work are \[15\dfrac{3}{4}\] days.

We need to convert \[15\dfrac{3}{4}\] into a fraction for further calculation.

To convert a whole number \[a\dfrac{b}{c}\] into a fraction, we use the formula \[a\dfrac{b}{c}=\dfrac{ac+b}{c}\].

Here a=15, b=3 and c=4.

Therefore, \[15\dfrac{3}{4}=\dfrac{15\times 4+3}{4}\]

\[\Rightarrow 15\dfrac{3}{4}=\dfrac{60+3}{4}\]

Hence, we get \[15\dfrac{3}{4}=\dfrac{63}{4}\].

Time taken to complete the work by A, B and C is \[\dfrac{63}{4}\] days.

Therefore, the work done by A, B and C in one day is the reciprocal of the number of days.

Let us convert the statement into an equation, we get

\[A+B+C=\dfrac{4}{63}\] ---------------(1)

Also, we know that the time taken by B, C and D to reap is 14 days.

Therefore, the work done by B, C and D is

\[B+C+D=\dfrac{1}{14}\] ---------------(2)

Then, we have been given that the time taken by C, D and A to reap is 18 days.

Therefore, the work done by C, D and A is

\[C+D+A=\dfrac{1}{18}\] ---------------(3)

Similarly, we know that the time taken by D, A and B to reap is 21 days.

Therefore, the work done by D, A and B is

\[D+A+B=\dfrac{1}{21}\] ---------------(4)

Let us now add all the four equations (1), (2), (3) and (4).

We get the work done is

\[A+B+C+B+C+D+C+D+A+D+A+B=\dfrac{4}{63}+\dfrac{1}{14}+\dfrac{1}{18}+\dfrac{1}{21}\]

Let us now group all the similar terms.

\[\Rightarrow \left( A+A+A \right)+\left( B+B+B \right)+\left( C+C+C \right)+\left( D+D+D \right)=\dfrac{4}{63}+\dfrac{1}{14}+\dfrac{1}{18}+\dfrac{1}{21}\]

On further simplifications, we get

\[\Rightarrow 3A+3B+3C+3D=\dfrac{4}{63}+\dfrac{1}{14}+\dfrac{1}{18}+\dfrac{1}{21}\]

We find that 3 are common in the LHS of the equation. On taking 3 common from the equation, we get

\[3\left( A+B+C+D \right)=\dfrac{4}{63}+\dfrac{1}{14}+\dfrac{1}{18}+\dfrac{1}{21}\] ---------------(5)

Now, we have to solve the RHS of equation (5).

Let us take the LCM OF 63, 14, 18 and 21.

\[3\left| \!{\underline {\,

63,14,18,21 \,}} \right. \]

\[3\left| \!{\underline {\,

21,14,6,7 \,}} \right. \]

\[2\left| \!{\underline {\,

7,14,2,7 \,}} \right. \]

\[7\left| \!{\underline {\,

7,7,1,7 \,}} \right. \]

\[1\left| \!{\underline {\,

1,1,1,1 \,}} \right. \]

Therefore, LCM= \[3\times 3\times 2\times 7\]

LCM=126.

In equation (5), we get

\[3\left( A+B+C+D \right)=\dfrac{4\times 2+9+7+6}{126}\]

\[\Rightarrow 3\left( A+B+C+D \right)=\dfrac{8+9+7+6}{126}\]

\[\Rightarrow 3\left( A+B+C+D \right)=\dfrac{30}{126}\]

We can write the above equation as

\[3\left( A+B+C+D \right)=\dfrac{3\times 10}{3\times 42}\]

Since 3 are common in both the numerator and denominator of RHS, we can cancel 3.

\[\Rightarrow 3\left( A+B+C+D \right)=\dfrac{10}{42}\]

Now, on further simplification, we get

\[3\left( A+B+C+D \right)=\dfrac{2\times 5}{2\times 21}\]

Since 2 are common in both the numerator and denominator of RHS, we can cancel 2.

\[\Rightarrow 3\left( A+B+C+D \right)=\dfrac{5}{21}\]

Now, let us divide the whole equation by 3. We get

\[\dfrac{3\left( A+B+C+D \right)}{3}=\dfrac{5}{21\times 3}\]

\[\Rightarrow \dfrac{3\left( A+B+C+D \right)}{3}=\dfrac{5}{63}\]

We find that 3 are common in both the numerator and denominator of the LHS.

Let us cancel 3, we get

\[A+B+C+D=\dfrac{5}{63}\]

Therefore, the work done by A, B, C and D is \[\dfrac{5}{63}\].

The time taken by A, B, C and D is \[\dfrac{1}{\dfrac{5}{63}}=\dfrac{63}{5}\].

**Hence, the time taken by A, B, C and D together reap the field is \[\dfrac{63}{5}\] days.**

**Note:**We can further simplify the obtained answer by converting it into a mixed fraction.

We have to divide 63 by 5, that is the divisor is 5 and the dividend is 63.

\[5\overset{12}{\overline{\left){\begin{align}

& 63 \\

& \dfrac{5}{\begin{align}

& 13 \\

& \dfrac{10}{3} \\

\end{align}} \\

\end{align}}\right.}}\]

Here, the quotient is 12 and the remainder is 3.

We have to write the mixed fraction as \[quotient\dfrac{remainder}{divisor}\].

Therefore, \[\dfrac{63}{5}=12\dfrac{3}{5}\].

Hence, the time taken by A, B, C and D together to reap the field is \[12\dfrac{3}{5}\] days.

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