Question

A person increases a number by $x\%$ and then decreases it by $y\%$. Increases it again by $x\%$ and then again decreased by $y\%$. The resultant number is $z\%$ more than the original number. If $x,y$ and $z$ are all positive numbers then which of the following are not possible?

• A. $x=y$
• B. $x>y$
• C. $x<y$
• D. $x\ge y$

Answer 1

Answer: (A), (C), (D)

$x=y$
$x<y$
$x\ge y$

Let the no. be $n$

After increase by $x\%$ it becomes $(100+x)\%\times n=n(1+\frac{x}{100})$

After decrease by $y\%$ it becomes $(100-y)(100+x)\%\times n=n(1+\frac{x}{100})(1-\frac{y}{100})$

Increasing again by $x\%$ and decreasing by $y\%$ we get new number $n_0=n(1+\frac{x}{100}{)}^2(1-\frac{y}{100}{)}^2$

Given that $n_0=(100+z)\%\times n=n(1+\frac{z}{100})$

$⟹(1+\frac{z}{100})=(1+\frac{x}{100}{)}^2(1-\frac{y}{100}{)}^2$

$⟹(1+\frac{z}{100})=\left[\right(1+\frac{x}{100}\left)\right(1-\frac{y}{100}){]}^2$

$⟹\left(\frac{z}{100}\right)=\left[\right(1+\frac{x-y}{100})-(\frac{xy}{100}){]}^2-1$

$⟹\left(\frac{z}{100}\right)=\left[\right(\frac{x-y}{100})-(\frac{xy}{100}\left)\right][2+\frac{x-y}{100}-\frac{xy}{1000}]$

We know that $0\le x\le 100$ and $0\le y\le 100$

$⟹$ $z$ will be negative.

Hence $x=y$ is not possible.

Similarly $x<y$ is not possible.

Hence $x<y,x=y,x\ge y$ is not possible.