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

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

by BTGmoderatorLU » Wed May 23, 2018 3:23 pm

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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

The OA is C.

I solved this PS question as follows,

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

I appreciate if anyone explains another way to solve this question. Thanks in advance!
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by [email protected] » Wed May 23, 2018 9:15 pm
Hi All,

We're told that A shepherd has 1 million sheep at the beginning of Year 2000. The numbers grow by X% (X > 0) during the year, then 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. We're asked which of the following is correct. This question can be solved in a number of different ways, including with logic or by TESTing VALUES.

From a logical standpoint: In 2000, the 'starting number' is 1 million, so after that number INCREASES (by X%), the total will be GREATER than 1 million. That larger number is what decreases by Y% - leading the total back to 1 million. The decrease from that larger number represents a SMALLER percent of the total (since the calculation begins with the larger starting number), so Y will ALWAYS be SMALLER than X.

Here's an example that proves the point by TESTing VALUES.
IF.... X = 25
Starting value = 1,000,000
increase by 25% = 1,250,000
To reduce this back to 1,000,000 we have to remove 250,000...
250,000/1,250,000 = 1/5 = 20%
So Y = 20
Thus X > Y

Final Answer: C

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Rich
Contact Rich at [email protected]
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by swerve » Thu May 24, 2018 9:37 am
Plugging numbers is a good help here,

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.

I hope this helps! Regards!
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by Scott@TargetTestPrep » Thu May 24, 2018 4:58 pm
BTGmoderatorLU wrote: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
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 (instead of 1 million) sheep and this number increases by 25%. (The value of x is, therefore, 25.) Thus, by the end of 2000, there are 125 sheep in the flock.

In 2001, the shepherd loses a percentage of the flock such that there are 100 sheep on hand at the beginning of 2002. Thus the percent decrease is (100 - 125)/125 = -25/125 = -1/5 or a 20% decrease. So y = 20 and therefore, x > y.

Answer: C

Scott Woodbury-Stewart
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