ABC26GN3019 · Ratio and Proportion
Subject: General Aptitude · Chapter: Ratio and Proportion · Exam: · Marks: · Difficulty:
If $(x+y):(x-y)=4: 1$, then $\left(x^{2}+y^{2}\right):\left(x^{2}-y^{2}\right)$ = ?
(a)8 : 17
(b)17 : 8
(c)16 : 1
(d)25 : 9
Answer
Explanation
$$\begin{aligned} & \frac{(x+y)}{(x-y)}=4 \Rightarrow x+y=4 x-4 y \Rightarrow 3 x=5 y \\ & \Rightarrow \frac{x}{y}=\frac{5}{3} \\ & \Rightarrow \frac{x^{2}}{y^{2}}=\frac{25}{9} \Rightarrow \frac{x^{2}+y^{2}}{x^{2}-y^{2}}=\frac{\frac{x^{2}}{y^{2}}+1}{\frac{x^{2}}{y^{2}}-1}=\frac{\frac{25}{9}+1}{\frac{25}{9}-1}=\frac{34}{9} \times \frac{9}{16}=\frac{17}{8} . \\ & \therefore\left(x^{2}+y^{2}\right):\left(x^{2}-y^{2}\right)=17: 8 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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