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Set \(P\) consists of five distinct positive integers \(\{a, b, c, d, e\}.\) The median of the set is \(M\) and its stan

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by Gmat_mission » Fri Dec 18, 2020 12:53 am

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Set \(P\) consists of five distinct positive integers \(\{a, b, c, d, e\}.\) The median of the set is \(M\) and its standard deviation is an integer \(D\) such that \(6\le D \le 9.\) When \(M\) is divided by \(D,\) the remainder is \(5.\) When a positive integer \(x\) is added to all the numbers of the set \(P,\) the new median of the set \(P\) is \(M’\) and the new standard deviation is \(D’.\) If \(x\) leaves a remainder not less than \(8\) when divided by \(D’,\) and \(M’\) leaves a remainder \(r\) when divided by \(D’,\) by how much is \(r\) less than or greater than the remainder obtained when \(M\) was divided by \(D?\)

A. \(20\%\) less

B. \(20\%\) greater

C. \(25\%\) less

D. \(60\%\) greater

E. \(160\%\) greater

Answer: A

Source: e-GMAT
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Re: Set \(P\) consists of five distinct positive integers \(\{a, b, c, d, e\}.\) The median of the set is \(M\) and its

by deloitte247 » Sun Dec 27, 2020 8:15 am
$$Set\ P\ =\ \left\{a,\ b,\ c,\ d,\ e\right\}$$
$$median\ of\ P\ =\ m\ and\ SD\ =\ D\ such\ that\ 6\le D\le9$$
$$when\ \frac{M}{D\ }remainder\ =\ 5\ hence,\ M=KD+5.........eqn\ 1$$
$$when\ x\ is\ added\ to\ all\ the\ number\ of\ set\ P,\ median\ =\ M^1$$
$$M^1=M+X........eqn\ 2$$
$$new\ SD=D^1\ so\ D^1=D..........eqn\ 3$$
$$X\ leaves\ a\ remainder\ and\ not\ less\ than\ 8$$
$$X=yD+R\ and\ R\ is\ \ge8\ so\ D>8$$
$$from\ 6\le D\le9;\ D\ can\ only\ take\ the\ value\ of\ 9\ and\ R=8$$
$$M^1\ leaves\ a\ remainder\ r\ when\ divided\ by\ D^1$$
$$\sin ce\ D^1=D\ and\ D=$$
$$M^1=tD+r$$
$$M^1=9t+r........eqn\ 4$$
$$from\ eqn\ 2\ M^1=M+X\ where\ M=kD+5$$
$$and\ X=yD+R$$
$$M^1=kD+5+yD+R$$
$$M^1=\left(k+y\right)9+5+8$$
$$M^1=\left(k+y\right)9+13$$
$$M^1=\left(k+y+1\right)9+4......eqn\ 5$$
$$\exp res\sin g\ eqn\ 5\ in\ terms\ of\ eqn\ 4$$
$$M^1=9t+r.......eqn4$$
$$M^1=\left(k+y+1\right)+4........eqn5$$
$$9t=\left(k+y+1\right)and\ r=4$$
$$so\ the\ remainder\ when\ M\ is\ divided\ by\ D=5$$
$$\%\ that\ r<remainder\ from\ \frac{M}{D}=\left(\frac{5-4}{5}\right)\cdot\frac{100}{1}$$
$$=\frac{1}{5}\cdot100$$
$$=20\%$$
$$Answer\ =A$$
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Re: Set \(P\) consists of five distinct positive integers \(\{a, b, c, d, e\}.\) The median of the set is \(M\) and its

by deloitte247 » Sun Dec 27, 2020 8:15 am
$$Set\ P\ =\ \left\{a,\ b,\ c,\ d,\ e\right\}$$
$$median\ of\ P\ =\ m\ and\ SD\ =\ D\ such\ that\ 6\le D\le9$$
$$when\ \frac{M}{D\ }remainder\ =\ 5\ hence,\ M=KD+5.........eqn\ 1$$
$$when\ x\ is\ added\ to\ all\ the\ number\ of\ set\ P,\ median\ =\ M^1$$
$$M^1=M+X........eqn\ 2$$
$$new\ SD=D^1\ so\ D^1=D..........eqn\ 3$$
$$X\ leaves\ a\ remainder\ and\ not\ less\ than\ 8$$
$$X=yD+R\ and\ R\ is\ \ge8\ so\ D>8$$
$$from\ 6\le D\le9;\ D\ can\ only\ take\ the\ value\ of\ 9\ and\ R=8$$
$$M^1\ leaves\ a\ remainder\ r\ when\ divided\ by\ D^1$$
$$\sin ce\ D^1=D\ and\ D=$$
$$M^1=tD+r$$
$$M^1=9t+r........eqn\ 4$$
$$from\ eqn\ 2\ M^1=M+X\ where\ M=kD+5$$
$$and\ X=yD+R$$
$$M^1=kD+5+yD+R$$
$$M^1=\left(k+y\right)9+5+8$$
$$M^1=\left(k+y\right)9+13$$
$$M^1=\left(k+y+1\right)9+4......eqn\ 5$$
$$\exp res\sin g\ eqn\ 5\ in\ terms\ of\ eqn\ 4$$
$$M^1=9t+r.......eqn4$$
$$M^1=\left(k+y+1\right)+4........eqn5$$
$$9t=\left(k+y+1\right)and\ r=4$$
$$so\ the\ remainder\ when\ M\ is\ divided\ by\ D=5$$
$$\%\ that\ r<remainder\ from\ \frac{M}{D}=\left(\frac{5-4}{5}\right)\cdot\frac{100}{1}$$
$$=\frac{1}{5}\cdot100$$
$$=20\%$$
$$Answer\ =A$$
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