fiza gupta wrote:If a price was increased by x% and then decreased by Y%, is the new price is higher than the original?
(I) x > y
(II) x = 1.2y
Notice: Statement 1 and Statement 2 can be tested at the same time, since as long as x > y > 0, then if x = 1.2y, then x > y.
Statement 1:
Since x > y, the new price could be greater than the old price as x could be any number and y could be 0.
So the key to assessing Statement 1 is determining whether the new price could be lower than the old price.
Notice the following.
x is a percentage of a lower base.
y is a percentage of a higher base.
So if x = y, the decrease would be greater than the increase.
Example:
x = y = 20
Original Price: P
Price After Increase: 1.2 P
New Price: 1.2P * .80 = .96P
So even though x = y, New Price < Original Price
New Price can also be less than Original Price when x > y.
Try x = 60 and y = 50. 60 = 1.2 * 50. So these numbers will work for Statement 2 also.
Original Price: P
Price After Increase: 1.60P
New Price: 1.60P * 0.50 = .80P
New Price < Old Price
Two different answers.
Insufficient.
Statement 2:
If x and y are both relatively low, we may get a result different from the one we get when y ≥ 50.
Try x = 12 and y = 10.
Original Price: P
Price After Increase: 1.12P
New Price: 1.12P * .9 = 1.008P
New Price > Original Price
Since x = 50 and y = 60 and x = 12 and y = 10 fit both statements, we have two different answers for Statement 1, two for Statement 2 and two for the statements combined.
Insufficient.
The correct answer is
E.
