Let price in 2020 = y
Price in 2011 = y * (1 - m/100)
Price in 2012 = price in 2011 + (1 + n/100)
$$=y\cdot\left(1-\frac{m}{100}\right)\cdot\left(1+\frac{n}{100}\right)$$
Target question=> Was the price of automobile lower in 2010 than in 2012?
$$i.e\ y<y\cdot\left(1-\frac{m}{100}\right)\cdot\left(1+\frac{n}{100}\right)$$
Divide through by y, we have
$$1<\left(1-\frac{m}{100}\right)\cdot\left(1+\frac{n}{100}\right)$$
$$1<\left(\frac{100-m}{100}\right)\cdot\left(\frac{100+n}{100}\right)$$
$$1<\left(\frac{\left(100-m\right)\cdot\left(100+n\right)}{100\cdot100}\right)$$
On cross multiplying, it gives
$$10000<\left(100-m\right)\cdot\left(100+n\right)$$
$$10000<10000+100n-100m-mn$$
$$10000-10000<+100n-100m-mn$$
$$0<+100n-100m-mn$$
$$mn<+100n-100m$$
$$So,\ is\ 100n-100m\ >\ mn?$$
Statement 1=> m<n
If m=2 and n=3
Then 100n - 100m > mn
300 - 200 > 6 [This is true and 100n - 100m > mn]
If m=19, and n=20, then, we have
2000 - 1900 > 380 [This is false and 100n - 100m < mn].
Therefore, the information available in statement 1 is NOT SUFFICIENT enough to arrive at a definite answer.
Statement 2=> mn < 100n - 100m
This directly answers the target question as 100n - 100m > mn can be rewritten as mn < 100n - 100m. Hence, statement 2 is SUFFICIENT.
Since statement 2 alone is SUFFICIENT, then the correct answer is option B.
