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If \(x \ne -1,\) which is greater, \(\dfrac1{x+1}\) or \(\dfrac{x}2 ?\)

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Source: — Data Sufficiency |
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Re: If \(x \ne -1,\) which is greater, \(\dfrac1{x+1}\) or \(\dfrac{x}2 ?\)

by deloitte247 » Sun Sep 06, 2020 12:55 am

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$$Statement\ 1\ =>\ x\ge0$$
$$if\ x\ =\ 0\ then\ \frac{1}{x+1}=\frac{1}{1}=1$$
$$and\ \frac{x}{2}=\frac{0}{2}=0$$
$$\frac{1}{\left(x+1\right)}>\frac{x}{2}$$
$$if\ x=1\ then\ \frac{1}{x+1}=\frac{1}{2\ }and\ \frac{x}{2}=\frac{1}{2}$$ $$if\ x=1\ then\ \frac{1}{x+1}=\frac{1}{2\ }and\ \frac{x}{2}=\frac{1}{2}$$
$$\frac{1}{\left(x+1\right)}=\frac{x}{2}$$
$$if\ x=2\ then\ \frac{1}{x+1}=\frac{1}{3}and\ \frac{x}{2}=\frac{2}{2}=1$$
$$\frac{1}{\left(x+1\right)}<\frac{x}{2}$$
The available information is not enough to arrive at a definite answer. Statement 1 is NOT SUFFICIENT

$$Statement\ 2\ =>\ x<3$$
$$if\ x=2,\ \frac{1}{\left(x+1\right)}<\frac{x}{2}$$
$$if\ x=1,\ \frac{1}{\left(x+1\right)}=\frac{x}{2}$$
$$if\ x=0,\ \frac{1}{\left(x+1\right)}>\frac{x}{2}$$
The available information does not provide a definite answer. Statement 2 is NOT SUFFICIENT

$$Combining\ both\ statements\ together\ =>$$
$$from\ statement\ 1\ =>\ x\ge0$$
$$from\ statement\ 2\ =>\ x<3$$
$$0\le x<3$$
The possible values of x include 0, 1, and 2. Substituting these values of x will not provide a definite answer as they all yield different results. Both statements together ARE NOT SUFFICIENT

$$Answer\ =\ E$$
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