A shepherd has 1 million sheep at the beginning of Year 2000. The numb

We can create the following equation.

(1,000,000)(1 + x/100)(1 - y/100) = 1,000,000

(1 + x/100)(1 - y/100) = 1

[(100 + x)/100][(100 - y)/100] = 1

[(100 + x)(100 - y)/10,000] = 1

(100 + x)(100 - y) = 10,000

10,000 + 100x - 100y - xy = 10,000

100x - 100y - xy = 0

100(x - y) = xy

Since 100, x, and y are positive quantities, x - y must be also positive. That is, x - y > 0, or x > y.

Alternate Solution:

We can solve this problem by simplifying the numbers and then evaluating it intuitively rather than mathematically.

First, let’s assume the shepherd starts the year 2000 with 100 sheep and this number increases by 10%. (The value of x is, therefore, 10.) Thus, by the end of 2000, there are 110 sheep in the flock.

In 2001, the shepherd loses a percentage of the flock such that there are 100 sheep on hand at the end of 2001. If the loss percentage, y%, were 10%, then the shepherd would have 110 – 11 = 99 sheep on hand at the end of 2001, which means that 10% for the value of y% is too large a loss percentage. Therefore, y% must be some value less than 10%, which means that y is less than x or, conversely, x is greater than y.

Answer: C