Question : If $\frac{x}{y}=\frac{4}{5}$, then the value of $(\frac{4}{7}+\frac{2y–x}{2y+x})$ is:
Option 1: $\frac{3}{7}$
Option 2: $1\frac{1}{7}$
Option 3: $1$
Option 4: $2$
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Correct Answer: $1$
Solution : Given: $\frac{x}{y}=\frac{4}{5}$ Divide $y$ in both numerator and denominator of the fraction, $\frac{2y – x}{2y+x}$, we get, $\frac{2–\frac{x}{y}}{2+\frac{x}{y}}$ Substitute the value of $\frac{x}{y}=\frac{4}{5}$ in the above equation, we get, = $\frac{2–\frac{4}{5}}{2+\frac{4}{5}}$ = $\frac{\frac{10–4}{5}}{\frac{10+4}{5}}=\frac{6}{14}=\frac{3}{7}$ The value of the expression $(\frac{4}{7}+\frac{2y–x}{2y+x})$ is given as, $(\frac{4}{7}+\frac{3}{7})=(\frac{4+3}{7})=\frac{7}{7}=1$ Hence, the correct answer is $1$.
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Question : If $X$ is 20% less than $Y$, then find the values of$\frac{Y–X}{Y}$ and $\frac{X}{X–Y}$.
Option 1: $\frac{1}{5}$ and $-4$
Option 2: $5$ and $-\frac{1}{4}$
Option 3: $\frac{2}{5}$ and $-\frac{5}{2}$
Option 4: $\frac{3}{5}$ and $-\frac{5}{3}$
Question : If $x+\frac{2}{x}=1$, then the value of $\frac{x^2+7x+2}{x^2+13x+2}$ is:
Option 1: $\frac{5}{7}$
Option 2: $\frac{3}{7}$
Option 3: $\frac{4}{7}$
Option 4: $\frac{2}{7}$
Question : If $x=5–\sqrt{21}$, the value of $\frac{\sqrt{x}}{\sqrt{32–2x}–\sqrt{21}}$ is:
Option 1: $\frac{1}{\sqrt2}(\sqrt{3}–\sqrt{7})$
Option 2: $\frac{1}{\sqrt2}(\sqrt{7}–\sqrt{3})$
Option 3: $\frac{1}{\sqrt2}(\sqrt{7}+\sqrt{3})$
Option 4: $\frac{1}{\sqrt2}(7–\sqrt{3})$
Question : If $\left(3 y+\frac{3}{y}\right)=8$, then find the value of $\left(y^2+\frac{1}{y^2}\right)$.
Option 1: $5\frac{1}{9}$
Option 2: $4\frac{5}{6}$
Option 3: $7\frac{1}{9}$
Option 4: $9\frac{1}{9}$
Question : If $x+\frac{1}{x}=3$, then the value of $\frac{3x^{2}-4x+3}{x^{2}-x+1}$ is:
Option 1: $\frac{4}{3}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{5}{3}$
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