Question : If $x$ men can do a piece of work in $x$ days, then the number of days in which $y$ men can do the same work is:
Option 1: $xy$ days
Option 2: $\frac{y^{2}}{x}$ days
Option 3: $\frac{x^{2}}{y}$ days
Option 4: $x^{2}y$ days
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Correct Answer: $\frac{x^{2}}{y}$ days
Solution : Given: $x$ men can do a piece of work in $x$ days. We know the concept $M_1D_1=M_2D_2$, where $M_1$, $M_2$ are the persons and $D_1$, $D_2$ are the number of days. According to the question, ⇒ $x×x=y×D_2$ $\therefore D_2=\frac{x^{2}}{y}$ days Hence, the correct answer is $\frac{x^{2}}{y}$ days.
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Question : X can do a piece of work in $p$ days and Y can do the same work in $q$ days. Then the number of days in which X and Y can together do that work is:
Option 1: $\frac{p+q}{2}$
Option 2: $\frac{1}{p}$ + $\frac{1}{q}$
Option 3: $\frac{pq}{p+q}$
Option 4: $pq$
Question : If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$ is:
Option 1: $\frac{3}{4}$
Option 2: $\frac{4}{3}$
Option 3: $\frac{3}{5}$
Option 4: $\frac{5}{3}$
Question : If $xy(x+y)=m$, then the value of $(x^3+y^3+3m)$ is:
Option 1: $\frac{m^3}{xy}$
Option 2: $\frac{m^3}{(x+y)^3}$
Option 3: $\frac{m^3}{x^3y^3}$
Option 4: $mx^3y^3$
Question : If $x^4+y^4+x^2 y^2=17 \frac{1}{16}$ and $x^2-x y+y^2=5 \frac{1}{4}$, then one of the values of $(x-y)$ is:
Option 1: $\frac{5}{2}$
Option 2: $\frac{3}{4}$
Option 3: $\frac{5}{4}$
Option 4: $\frac{3}{2}$
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}$?
Option 1: $0$
Option 2: $\frac{1}{3}$
Option 3: $1$
Option 4: $3$
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