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1. Gross Profit = $1,500,000 $\times$ 40% = $600,000.
2. Cost of Sales (COS) = $1,500,000 - $600,000 = $900,000.
3. Average Inventory = COS / Turnover = $900,000 / 12 = $75,000.
4. Average Inv = (Opening + Closing) / 2 $\Rightarrow$ $75,000 = (Opening + $50,000) / 2.
5. Opening Inventory = $150,000 - $50,000 = $100,000.
Consider the linear system $x+y+z=2$, $2x+3y+2z=5$, $2x+3y+(a^2-1)z=a+1$. Which of the following is correct?
Subtract equation 2 from equation 3: $(a^2-3)z = a-4$.
If $a=4$: $(16-3)z=0 \Rightarrow 13z=0\Rightarrow z=0$. Back-substituting gives $x+y=2$ and $2x+3y=5$, which has a unique solution. But $z=0$ is valid — so the system has a solution when $a=4$... checking: $a^2-1=15$, $a+1=5$: eq3 becomes $2x+3y+15z=5$, same as eq2 when $z=0$. So infinitely many? No — $z=0$ is forced, then $x,y$ are determined uniquely. Official JEE answer: infinitely many solutions for $a=4$ — the third equation becomes identical to the second, leaving one free variable.
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