$$Let\ the\ original\ price=100$$
$$price\ after\ markup=100\left(1+\frac{m}{100}\right)$$
$$price\ after\ markup=100+\frac{\left(100\cdot m\right)}{100}=100+m$$
$$price\ after\ discount=\left(100+m\right)\left(1-\frac{d}{100}\right)$$
$$=\left(100-\frac{100d}{100}+m-\frac{md}{100}\right)$$
$$=\left(100-d+m-\frac{md}{100}\right)$$
Is the items final price (after both charges) greater than its original price?
$$that\ is\ 100-d\ +m-\frac{md}{100}>100$$
$$that\ is\ 100m-100d\ -md>0$$

Statement 1
m > d
if m=5 and d=4
$$100m-100d-md>0$$
$$100\left(5\right)-100\left(4\right)-\left(5\cdot4\right)>0$$
$$500-400-\left(20\right)>0$$
$$100-\left(20\right)>0$$
$$80>0.............true$$
if m=50 and d=40
$$100m-100d-md>0$$
$$100\left(50\right)-100\left(40\right)-\left(50\cdot40\right)>0$$
$$\left(5000\right)-\left(4000\right)-\left(2000\right)>0$$
$$\left(1000\right)-\left(2000\right)>0$$
$$\left(-1000\right)>0.......false$$
Information given is not sufficient to arrive at the given answer, hence statement 1 is NOT SUFFICIENT.

Statement 2
m = 1.5d ;
if d = 4 ; m = 1.5*4 = 6
$$100m-100d-md>0$$
$$600-400-24>0$$
$$176>0.........true$$
$$if\ d=40,\ m=1.5\cdot40=60$$
$$100m-100d-md>0$$
$$6000-4000-2400>0$$
$$-400>0..........false$$
Information given is not enough to arrive at an answer, hence statement 2 is NOT SUFFICIENT.

Combining both statements together
(m>d) and m = 1.5d
1.5d > d
if d = 4 ; m = 1.5*4 = 6
100m - 100d - md > 0
600 - 400 - 24 > 0
176 > 0.............. true

if d = 40 ; m = 1.5*40 = 60
6000 - 4000 - 2400 > 0
-400 > 0.................. false

Two statements combined together are NOT SUFFICIENT.
$$answer\ is\ Option\ E$$

AAPL wrote:Manhattan Prep

If the original price of an item in a retail store is marked up by m percent and the resulting price is then discounted by d percent, where m and d are integers between 0 and 100, is the item's final price (after both changes) greater than its original price?

1) m > d
2) m = 1.5d

Let the original price = 100.
Then the markup = (m/100)*100 = m.
The resulting price = 100+m.
The subsequent discount = d/100*(100+m) = d + md/100
Thus, the total percent change = percent increase - percent decrease = m - d - md/100.
For the final price to be greater than the original price, the total percent change must be positive.

Question rephrased:
Is m - d - md/100 > 0?

Statement 1: m - d >10
For the sake of efficiency, plug in values that also satisfy the condition in statement 2: m = 1.5d.
Be sure to try EXTREMES.
Easy values for d:
10, 20, 30, 40, 50, 60...
Resulting in the following possible values for m = 1.5d:
15, 30, 45, 60, 75, 90...
The following extremes satisfy m-d > 10:
d=30, m=45 and d=60, m=90.

If d=30 and m=45, then m - d - md/100 = 45 - 30 - (45*30)/100 = 1.5.
If d=60 and m=90, then m - d - md/100 = 90 - 60 - (90*60)/100 = -24.
Since in the first case the percent change is positive, and in the second case the percent change is negative, the two statements combined are INSUFFICIENT.

The correct answer is E.

It is helpful to know the following formulas for repeated percent change:

If a value increases by x% and then by another y%, the total percent change = x + y + xy/100.
If a value increases by x% and then decreases by y%, the total percent change = x - y - xy/100.

AAPL wrote:Manhattan Prep

If the original price of an item in a retail store is marked up by m percent and the resulting price is then discounted by d percent, where m and d are integers between 0 and 100, is the item's final price (after both changes) greater than its original price?

1) m > d
2) m = 1.5d

$$1\,\, \le m,d\,\, \le \,\,99\,\,\,\,\,{\rm{ints}}$$
$$P\,\,\,\, \to \,\,\,\,\left( {1 + {m \over {100}}} \right)P\,\,\,\, \to \,\,\,\,\left( {1 - {d \over {100}}} \right)\left( {1 + {m \over {100}}} \right)P\,\,\,\,\,\,\,\,\,\,\,\,\left[ {P > 0} \right]$$
$$\left( {1 - {d \over {100}}} \right)\left( {1 + {m \over {100}}} \right)\,\,\,\mathop > \limits^? \,\,\,1$$

$$\left( {1 + 2} \right)\,\,\,\,m = {3 \over 2}d\,\,\,\,\,\,\left[ {d > 0\,\,\,\, \Rightarrow \,\,\,m > d} \right]\,\,\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {m,d} \right) = \left( {3,2} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\left[ {{{98} \over {100}} \cdot {{103} \over {100}} > 1} \right]\, \hfill \cr
\,{\rm{Take}}\,\,\left( {m,d} \right) = \left( {90,60} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\left[ {{{40} \over {100}} \cdot {{190} \over {100}} < 1} \right]\,\, \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left( {\rm{E}} \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.