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18 Dec 2012, 06:47
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D
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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 ?
(1) The increased price per share of Stock X was equal to the original price per share of Stock Y.
(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) The increased price per share of Stock X was equal to the original price per share of Stock Y.
(2) The increase in the price per share of Stock X was 10/11 the decrease in the price per share of Stock Y.
18 Dec 2012, 06:51
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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.
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.
25 Nov 2015, 17:42
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Given
1.1x
0.9y
(0.9y/x)*100 = ¿?
THE KEY: This is a COMBO QUESTION --> Knowing y/x (or knowing that you have sufficient information to know y/x) is sufficient.
Statement 1
1.1x = y
If you manipulate the equation, you can get y/x
Statement 2
0.1x = (10/11) 0.1y
If you manipulate the equation, you can get y/x
Note the difference between:
- the increase IN THE price of x -->0.1x
- the decrease IN THE price of y --> 0.1y
- the INCREASED price in x --> 1.1x
- the DECREASED price in y -->0.9y
1.1x
0.9y
(0.9y/x)*100 = ¿?
THE KEY: This is a COMBO QUESTION --> Knowing y/x (or knowing that you have sufficient information to know y/x) is sufficient.
Statement 1
1.1x = y
If you manipulate the equation, you can get y/x
Statement 2
0.1x = (10/11) 0.1y
If you manipulate the equation, you can get y/x
Note the difference between:
- the increase IN THE price of x -->0.1x
- the decrease IN THE price of y --> 0.1y
- the INCREASED price in x --> 1.1x
- the DECREASED price in y -->0.9y
