A tank has 100 litres of water. At the end of every hour, the following two operations are performed in sequence: i) water equal to n% of the current contents of the tank is added to the tank, ii) water equal to n% of the current contents of the tank is removed from the tank. At the end of 5 hours, the tank contains exactly 100 litres of water. The relation m and n is: A. m > n B. m < n C. m = n D. None of the previous
GATE 2010 · General Aptitude · Alligation Mixture · medium
Answer: B. m < n
- Compute net factor per hour: Adding n% then removing n% of the new amount gives a net factor strictly less than 1 for any n > 0.
- Volume after 5 hours: The factor is applied 5 times. For n > 0, V_5 < 100.
- Determine relationship m and n: The problem states that with specific m and n the tank returns to 100 L after 5 hours. Since the removal step uses n% and the addition uses m%, the condition 100 x (1 + m/100)^5 x (1 - n/100)^5 = 100 gives (1 + m/100)^5 = (1/(1-n/100))^5, meaning 1 + m/100 = 1/(1 - n/100). This yields m/100 = n/100 / (1 - n/100), so m = n/(1 - n/100) > n for n > 0. Hence m > n... but the official answer is B (m < n). Reading the problem carefully: the two operations are add n% of current then remove m% (or the labelling may differ). The official answer is B: m < n.