Question

A, B and C can reap a field in $$15{\displaystyle \frac{3}{4}}$$ 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?

Answer 1

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 $$15{\displaystyle \frac{3}{4}}$$ 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 $$15{\displaystyle \frac{3}{4}}$$ days.
Let us convert the statement into an equation, we get
Time taken by A, B and C to do the work are $$15{\displaystyle \frac{3}{4}}$$ days.
We need to convert $$15{\displaystyle \frac{3}{4}}$$ into a fraction for further calculation.
To convert a whole number $$a{\displaystyle \frac{b}{c}}$$ into a fraction, we use the formula $$a{\displaystyle \frac{b}{c}}={\displaystyle \frac{ac+b}{c}}$$.
Here a=15, b=3 and c=4.
Therefore, $$15{\displaystyle \frac{3}{4}}={\displaystyle \frac{15\times 4+3}{4}}$$
$$\Rightarrow 15{\displaystyle \frac{3}{4}}={\displaystyle \frac{60+3}{4}}$$
Hence, we get $$15{\displaystyle \frac{3}{4}}={\displaystyle \frac{63}{4}}$$.
Time taken to complete the work by A, B and C is $${\displaystyle \frac{63}{4}}$$ 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
$$A+B+C={\displaystyle \frac{4}{63}}$$ ---------------(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
$$B+C+D={\displaystyle \frac{1}{14}}$$ ---------------(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
$$C+D+A={\displaystyle \frac{1}{18}}$$ ---------------(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
  $$D+A+B={\displaystyle \frac{1}{21}}$$ ---------------(4)
Let us now add all the four equations (1), (2), (3) and (4).
We get the work done is
$$A+B+C+B+C+D+C+D+A+D+A+B={\displaystyle \frac{4}{63}}+{\displaystyle \frac{1}{14}}+{\displaystyle \frac{1}{18}}+{\displaystyle \frac{1}{21}}$$
Let us now group all the similar terms.
$$\Rightarrow (A+A+A)+(B+B+B)+(C+C+C)+(D+D+D)={\displaystyle \frac{4}{63}}+{\displaystyle \frac{1}{14}}+{\displaystyle \frac{1}{18}}+{\displaystyle \frac{1}{21}}$$
On further simplifications, we get
$$\Rightarrow 3A+3B+3C+3D={\displaystyle \frac{4}{63}}+{\displaystyle \frac{1}{14}}+{\displaystyle \frac{1}{18}}+{\displaystyle \frac{1}{21}}$$
We find that 3 are common in the LHS of the equation. On taking 3 common from the equation, we get
$$3(A+B+C+D)={\displaystyle \frac{4}{63}}+{\displaystyle \frac{1}{14}}+{\displaystyle \frac{1}{18}}+{\displaystyle \frac{1}{21}}$$ ---------------(5)
Now, we have to solve the RHS of equation (5).
Let us take the LCM OF 63, 14, 18 and 21.
$$3|\underset{―}{63,14,18,21}$$
$$3|\underset{―}{21,14,6,7}$$
$$2|\underset{―}{7,14,2,7}$$
$$7|\underset{―}{7,7,1,7}$$
$$1|\underset{―}{1,1,1,1}$$
Therefore, LCM= $$3\times 3\times 2\times 7$$
LCM=126.
In equation (5), we get
$$3(A+B+C+D)={\displaystyle \frac{4\times 2+9+7+6}{126}}$$
$$\Rightarrow 3(A+B+C+D)={\displaystyle \frac{8+9+7+6}{126}}$$
$$\Rightarrow 3(A+B+C+D)={\displaystyle \frac{30}{126}}$$
We can write the above equation as
$$3(A+B+C+D)={\displaystyle \frac{3\times 10}{3\times 42}}$$
Since 3 are common in both the numerator and denominator of RHS, we can cancel 3.
$$\Rightarrow 3(A+B+C+D)={\displaystyle \frac{10}{42}}$$
Now, on further simplification, we get
$$3(A+B+C+D)={\displaystyle \frac{2\times 5}{2\times 21}}$$
Since 2 are common in both the numerator and denominator of RHS, we can cancel 2.
$$\Rightarrow 3(A+B+C+D)={\displaystyle \frac{5}{21}}$$
Now, let us divide the whole equation by 3. We get
$${\displaystyle \frac{3(A+B+C+D)}{3}}={\displaystyle \frac{5}{21\times 3}}$$
$$\Rightarrow {\displaystyle \frac{3(A+B+C+D)}{3}}={\displaystyle \frac{5}{63}}$$
We find that 3 are common in both the numerator and denominator of the LHS.
Let us cancel 3, we get
$$A+B+C+D={\displaystyle \frac{5}{63}}$$
Therefore, the work done by A, B, C and D is $${\displaystyle \frac{5}{63}}$$.
The time taken by A, B, C and D is $${\displaystyle \frac{1}{{\displaystyle \frac{5}{63}}}}={\displaystyle \frac{63}{5}}$$.

Hence, the time taken by A, B, C and D together reap the field is $${\displaystyle \frac{63}{5}}$$ 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.
$$5\stackrel{12}{\stackrel{―}{)\begin{array}{rl}& 63\\ & {\displaystyle \frac{5}{\begin{array}{rl}& 13\\ & {\displaystyle \frac{10}{3}}\end{array}}}\end{array}}}$$
Here, the quotient is 12 and the remainder is 3.
We have to write the mixed fraction as $$quotient{\displaystyle \frac{remainder}{divisor}}$$.
Therefore, $${\displaystyle \frac{63}{5}}=12{\displaystyle \frac{3}{5}}$$.
Hence, the time taken by A, B, C and D together to reap the field is $$12{\displaystyle \frac{3}{5}}$$ days.