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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.
Financial data: Profit from operations $59,800; Finance costs $12,000; Profit for year $47,800; Ordinary share capital $700,000; Retained earnings $72,500; 10% Debentures $120,000. What was the ROCE?
Option C (6.70%) is correct.
Capital Employed = ($700,000 + $72,500) + $120,000 = $892,500
ROCE = $59,800 ÷ $892,500 × 100 = 6.70%
What mass of sucrose (C$_{12}$H$_{22}$O$_{11}$, molar mass = 342 g mol$^{-1}$) is required to prepare 2 L of a 0.1 M aqueous solution?
Moles needed $= M \times V = 0.1\ \text{mol/L} \times 2\ \text{L} = 0.2\ \text{mol}$
Mass $= 0.2 \times 342 = \mathbf{68.4\ \text{g}}$
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