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16 Sep 2014, 00:58
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If a price was increased by \(x\%\) and then decreased by \(y\%\), where x and y are both non-zero, is the new price higher than the original?
(1) \(x \gt y\)
(2) \(x = 1.2y\)
(1) \(x \gt y\)
(2) \(x = 1.2y\)
16 Sep 2014, 00:58
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Official Solution:
If a price was increased by \(x\%\) and then decreased by \(y\%\), where x and y are both non-zero, is the new price higher than the original?
Let \(P\) represent the original price.
(1) \(x > y\).
If \(x\) is significantly larger than \(y\), the new price is higher than the original. However, if \(x\) is only slightly larger, the new price is lower. For example, if \(x = 20\) and \(y = 19\), the new price is \(P*1.20*0.81 < P\).
(2) \(x = 1.2y\)
Similar reasoning applies here. If \(y\) is large, the new price is smaller (if \(y = 100\), the new price is 0). If \(y\) is small, the new price is higher than the original (if \(y = 10\), the new price is \(P*1.12*0.9 = P*1.008 > P\)).
(1)+(2) Combining Statement 1 with Statement 2 provides no additional information. For example, if \(P = 100\), \(x = 120\), and \(y = 100\), the new price becomes 0, which is lower than the original price of 100. However, if \(P = 100\), \(x = 12\), and \(y = 10\), the new price becomes \(100 * 1.12 * 0.9 = 100 * 1.008\), which is higher than the original price of 100. Therefore, the given information is not sufficient to determine whether the new price is higher or lower than the original price.
Answer: E
If a price was increased by \(x\%\) and then decreased by \(y\%\), where x and y are both non-zero, is the new price higher than the original?
Let \(P\) represent the original price.
(1) \(x > y\).
If \(x\) is significantly larger than \(y\), the new price is higher than the original. However, if \(x\) is only slightly larger, the new price is lower. For example, if \(x = 20\) and \(y = 19\), the new price is \(P*1.20*0.81 < P\).
(2) \(x = 1.2y\)
Similar reasoning applies here. If \(y\) is large, the new price is smaller (if \(y = 100\), the new price is 0). If \(y\) is small, the new price is higher than the original (if \(y = 10\), the new price is \(P*1.12*0.9 = P*1.008 > P\)).
(1)+(2) Combining Statement 1 with Statement 2 provides no additional information. For example, if \(P = 100\), \(x = 120\), and \(y = 100\), the new price becomes 0, which is lower than the original price of 100. However, if \(P = 100\), \(x = 12\), and \(y = 10\), the new price becomes \(100 * 1.12 * 0.9 = 100 * 1.008\), which is higher than the original price of 100. Therefore, the given information is not sufficient to determine whether the new price is higher or lower than the original price.
Answer: E
10 Oct 2016, 12:37
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Bunuel
Lets assume that New price is higher than the old price and see if it always holds true.
\(P*\frac{100+X}{100}*\frac{100-Y}{100} - P > 0\)
Solving, 100*(X-Y) > XY
1. Given, \(x \gt y\)
Now when X is infinitesimally larger then y , then L.H.S of the equality tends to 0 and RHS is X squared or Y squared, which can be large depending on the value of the X
When X is much larger than Y, say X = 100000, Y = 0.0001, then LHS is much larger then RHS
Hence the inequality flips sign based on the values of X and Y so not sufficient.
2. Putting \(x = 1.2y\) in our derived inequality
20> 1.2 Y
hence again whether this always holds true will depend on the value of Y. Not sufficient.
The condition given in point 2 also satisfies point 1 hence together also they are not sufficient.
