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# A, B and C can reap a field in $15\dfrac{3}{4}$ days; B, C and D in 14 days; C, D and A in 18 days; D, A and B in 21 days. In what time can A, B, C and D together reap it?

Last updated date: 15th Jun 2024
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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.