Let the time taken by machine X to complete 1 job = x
Let the time taken by machine Y to complete 1 job = y
The work rate of machine X = 1/x and that of machine Y = 1/y
Target question => If both are working alone, then the time taken by machine Y is what percentage more/less than that of machine X?
$$If\ x\ is\ greater\ than\ y\ as\ a\ percentage\ then;$$
$$\left(\frac{x-y}{y}\right)\cdot\frac{100}{1}=>\ \left(\frac{x}{y}-1\right)\cdot\frac{100}{1}$$
$$if\ y\ is\ greater\ than\ x\ as\ a\ percentage\ then;$$
$$\left(\frac{y-x}{y}\right)\cdot\frac{100}{1}=>\ \left(1-\frac{x}{y}\right)\cdot\frac{100}{1}$$
$$so\ evaluating\ the\ ratio\ \frac{x}{y\ }answers\ the\ t\arg et\ question$$
Statement 1 => Machine X and Y, working together, complete a production order of the same size in two-thirds the time that machine Y, working alone does
$$The\ combined\ rate\ of\ both\ machines\ working\ together=>$$
$$\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}$$
$$time\ taken\ by\ both\ of\ them\ together\ =\ \frac{combined\ rate}{job}$$
$$Combined\ time\ =\ \frac{y+x}{xy}\div\frac{1}{1}$$
$$=\ \frac{y+x}{xy}\cdot\frac{1}{1}=>\frac{y+x}{xy}$$
$$Time\ taken\ by\ both\ machines\ together\ =>\ \frac{xy}{x+y}$$
$$\frac{xy}{x+y}=\frac{2}{3}\left(y\right)$$
$$\frac{xy}{x+y}\div\frac{y}{1}=\frac{2}{3}$$
$$\frac{xy}{x+y}\cdot\frac{1}{y}=\frac{2}{3}$$
$$\frac{x}{x+y}=\frac{2}{3}$$
$$3x=2\left(x+y\right)$$
$$3x=2x+2y$$
$$3x-2x=2y$$
$$\frac{x}{y}=\frac{2y}{y}=>\ \frac{x}{y}=\frac{2}{1}$$
$$\sin ce\ we\ have\ the\ ratio,\ the\ t\arg et\ question\ can\ be\ answered$$
$$Hence,\ statement\ 1\ is\ SUFFICIENT$$
Statement 2 => Machine Y, working alone, fills a production order of twice the size in 6 hrs
Statement 2 does not provide any information on machine X so statement 2 is NOT SUFFICIENT
Since only statement 1 is SUFFICIENT,
Answer = A
