mtk10 wrote:

Bunuel wrote:

The price per share of Stock X increased by 10 percent over the same time period that the price per share of Stock Y decreased by 10 percent. The reduced price per share of Stock Y was what percent of the original price per share of Stock X ?

Let the initial price per share of stock X be \(x\), so after increase by 10% it would become \(1.1x\);

Let the initial price per share of stock Y be \(y\), so after decrease by 10% it would become \(0.9y\).

Question: \(\frac{0.9y}{x}=\frac{9y}{10x}=?\)

(1) The increased price per share of Stock X was equal to the original price per share of Stock Y --> \(1.1x=y\) --> \(\frac{9y}{10x}=\frac{9*1.1x}{10x}=0.99\) or 99%. Sufficient.

(2) The increase in the price per share of Stock X was 10/11 the decrease in the price per share of Stock Y --> \(1.1x-x=\frac{10}{11}*(y-0.9y)\) --> \(0.1x=\frac{10}{11}*0.1y\) --> \(1.1x=y\) the same info as in (1). Sufficient

Answer: D.

Shouldn't the highlighted part be [ y- .1y] as its 10/11 of the decease and it is to be decreased by 10 %, correct me if I'm wrong.

Regards

It's should be the way it is written.

The decrease in the price per share of Stock Y is \(y - 0.9y\). 10/11 times that is \(\frac{10}{11}*(y-0.9y)\). I think you misunderstood the statement.

But the stem states that decrease in y is 10 %, Would we not write this question 1 -.1 =.9 . Sorry, i guess I am overlooking something. I am not understanding why are you subtracting .9y when question states that decrease is 10 % not 90 %.