8Number 8ExplanationIf x=56x = \frac{5}{6}x=65 and y=37y = \frac{3}{7}y=73. Which of the following relationships is correct between quantities PPP and QQQ based on the information provided? PPPQQQ1x1+2y\frac{\frac{1}{x}}{1 + \frac{2}{y}}1+y2x11x+y\frac{1}{x + y}x+y1P<QP < QP<QP>QP > QP>QP=QP = QP=QP=2QP = 2QP=2QThe information provided is not sufficient to decide on one of the four options aboveExplanationGiven x=56x = \frac{5}{6}x=65 and y=37y = \frac{3}{7}y=73. Calculating the value of PPP P=1x1+2yP = \frac{\frac{1}{x}}{1 + \frac{2}{y}}P=1+y2x1=15/61+23/7= \frac{\frac{1}{5/6}}{1 + \frac{2}{3/7}}=1+3/725/61=651+2⋅73= \frac{\frac{6}{5}}{1 + 2 \cdot \frac{7}{3}}=1+2⋅3756=651+143= \frac{\frac{6}{5}}{1 + \frac{14}{3}}=1+31456=6533+143= \frac{\frac{6}{5}}{\frac{3}{3} + \frac{14}{3}}=33+31456=65173= \frac{\frac{6}{5}}{\frac{17}{3}}=31756=65⋅317= \frac{6}{5} \cdot \frac{3}{17}=56⋅173=1885= \frac{18}{85}=8518 The value of P=1885P = \frac{18}{85}P=8518 (less than 111, approx 0.210.210.21). Calculating the value of QQQ Q=1x+yQ = \frac{1}{x + y}Q=x+y1=156+37= \frac{1}{\frac{5}{6} + \frac{3}{7}}=65+731=13542+1842= \frac{1}{\frac{35}{42} + \frac{18}{42}}=4235+42181=15342= \frac{1}{\frac{53}{42}}=42531=4253= \frac{42}{53}=5342 The value of Q=4253Q = \frac{42}{53}Q=5342 (less than 111, approx 0.790.790.79). Comparing PPP and QQQ P=1885≈0.21P = \frac{18}{85} \approx 0.21P=8518≈0.21Q=4253≈0.79Q = \frac{42}{53} \approx 0.79Q=5342≈0.79 It is clear that 0.21<0.790.21 < 0.790.21<0.79. Therefore, P<QP < QP<Q.
