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

Question

A shepherd has 1 million sheep at the beginning of Year 2000. The numbers grow by x% (x > 0) during the year. A famine hits his village in the next year and many of his sheeps die. The sheep population decreases by y% during 2001 and at the beginning of 2002 the shepherd finds that he is left with 1 million sheep. Which of the following is correct?

(A) x^2 > y^3
(B) y^2 > x^3
(C) x > y
(D) y > x
(E) x = y

(A) x^2 > y^3 
(B) y^2 > x^3 
(C) x > y 
(D) y > x 
(E) x = y

jaffacake · Accepted Answer

Plugging numbers is a good help here imo.

Take 100 sheep and increase by 10%.
To decrease 110 sheep to their original count, one can obtain that X > Y is necessary as 10% of 110 = 11.

Rishirajchauhan11 · Answer

Since net change is 0 
 
Successive % change here can be described as : x - y - xy/100

Which is 0

So,

x-y -xy/100= 0

X- Y = XY/100

100(x-y) = xy /100

xy/ 100 is +ve , so x-y should also be +ve 
~ x>y

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emockus · Answer

I plugged in a value for x%...I used 10%:

1,000,000*10%=100,000
1,000,000+100,000=1,100,000

To return 1,100,000 back to the original 1,000,000; y% has to be less than 10% (1,100,000*10%=110,000 - will not equal 1,000,000). Therefore, x>y.

x>y Answer Choice "C"

ScottTargetTestPrep · Answer

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

= 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

PSKhore · Answer

Let's take the sheep as 100.
100 -> 120 which is 20% increase ...x%
120 -> 100 which is 20/120 ->1/6 -> 16.66...y%

x>y

MyGMATbuddy · Answer

Start: 100 sheep
After +x%: 100 + x
To return to 100 in year 2, we reduce (100 + x) by y%

So:
(100 + x) × (1 – y/100) = 100
→ Need smaller y% since it's applied on a larger base (100 + x)

So x > y

Answer: (C) x > y

Pankaj Jindal | GMAT Classic 730 | IIM Bangalore | Helping working professionals
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A shepherd has 1 million sheep at the beginning of Year 2000. The numb

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