Xavier Aptitude Test 2012 Paper

(b) Let the potter has x pots and he gave it to his two sons. After selling they received Rs. $x$ $^{2}$ . Now, Price of banana chips is less than Rs. 10. Hence $x$$^{2}$ will not be a multiple of 100. Assume they purchased y packets of – potato chips. Hence $\left(\frac{y+1}{2}\right)$ packets. Hence y is odd. Let $z$ be the price of banana chips. Hence they have Rs. $(y \times 10 + z)$ As $y$ is odd so $(10 y + z)$ is also odd. Now, each brother can have equal money if total money earned by them is even. Hence, $z$ must be Even. From this we get $x^{2} = 10y + z,$ This given x cm Even. Now if units digit of $x$ is 2 or 8, then ten’s place of $x^{2}$ will be Even which is not possible. Hence unit digit of $x$ is 4 or 6. So, unit digit of $x^{2}$ must be 6. $\therefore z$ = 6 Hence son having banana chips owes $\text{Rs.} \left[ (y-1) \times \frac{10}{2} + 6 \right]$ and other son owes $\left[ (y+1) \times \frac{10}{2} \right]$ Hence, one of the son has $\text{Rs.}\left[ (y+1) \times \frac{10}{2} \right] -$ $\left[ \left( \frac{y-1}{2} \right) \times 10 + 6 \right]$ = Rs. 4 more than the other So, in order to equalize financially be must give Rs. 2 to theother.