Bunuel wrote:

If a price was increased by \(x\)% and then decreased by \(y\)%, is the new price higher than the original?

(1) \(x \gt y\)

(2) \(x = 1.2y\)

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.