Aptitude Discussion

Q. |
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 sheep 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. |
$x > y$ |

✖ B. |
$y > x$ |

✖ C. |
$x = y$ |

✖ D. |
Cannot be determined |

**Solution:**

Option(**A**) is correct

Let us assume the value of $x$ to be $10\%$.

Therefore, the number of sheep in the herd at the beginning of year 2001 (end of 2000) will be 1 million + $10\%$ of 1 million = 1.1 million

In 2001, the numbers decrease by $y\%$ and at the end of the year the number sheep in the herd = 1 million.

i.e., 0.1 million sheep have died in 2001.

In terms of the percentage of the number of sheep alive at the beginning of 2001,

it will be $(0.1/1.1)\times 100 \% = 9.09\%$.

From the above illustration it is clear that $x > y$

**Kedari Gowri**

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Its easy to understand with this assumption