Question : If $a^3+b^3=9$ and $a+b=3$, then the value of $\frac{1}a+\frac{1}b$ is:
Option 1: $\frac{1}2$
Option 2: $\frac{3}2$
Option 3: $\frac{5}2$
Option 4: $–1$
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Correct Answer: $\frac{3}2$
Solution : Given: That $a+b=3$ and $a^{3} + b^{3} = 9$ We know that $(a+ b)^{3}=a^3 + b^3 + 3ab(a + b)$ So, $a^3 + b^3 + 3ab(a + b) = 27$ ⇒ $9 + 3ab(a + b) = 27$ ⇒ $3ab(3) = 27 - 9$ ⇒ $9ab=18$ ⇒ $ab=2$ So, $\frac{1}a+\frac{1}b= \frac{(a+b)}{ab}$ = $\frac{3}{2}$ Hence, the correct answer is $\frac{3}{2}$.
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Question : The arrangement of the fractions $\frac{4}{3}, -\frac{2}{9}, -\frac{7}{8}, \frac{5}{12}$ in ascending order is _____.
Option 1: $–\frac{7}{8}, –\frac{2}{9}, \frac{5}{12}, \frac{4}{3}$
Option 2: $–\frac{7}{8}, –\frac{2}{9}, \frac{4}{3}, \frac{5}{12}$
Option 3: $–\frac{2}{9}, –\frac{7}{8}, \frac{5}{12}, \frac{4}{3}$
Option 4: $–\frac{2}{9}, –\frac{7}{8}, \frac{4}{3}, \frac{5}{12}$
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 $(9-3x)-(17x-10)=1$, then the value of $x$ is:
Option 1: $1$
Option 2: $–1$
Option 3: $\frac{9}{10}$
Option 4: $–\frac{9}{10}$
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 $\frac{1}{a}–\frac{1}{b}=\frac{1}{a–b}$, then the value of $a^{3}+b^{3}$ is:
Option 1: 0
Option 2: –1
Option 3: 1
Option 4: 2
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