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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 โ 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 โ $z=0$ is forced, then $x,y$ are determined uniquely. Official JEE answer: infinitely many solutions for $a=4$ โ the third equation becomes identical to the second, leaving one free variable.
Total distance covered in 25 s \(= 20 \times 25 = 500\) m
Length of bridge \(= 500 - 100 = 400\) m
- Lithium forms the normal oxide ($Li_{2}O$).
- Sodium forms the peroxide ($Na_{2}O_{2}$).
- Potassium (and larger alkali metals) forms the superoxide ($KO_{2}$).
Purchases ledger control account balance: $8,232. Errors in individual ledger only: (1) Purchases returns $72 posted to credit of payable's account (should be debit); (2) Discount received $54 posted as $45 (debit). What is the total of balances in the purchases ledger before correction?
Option D ($8,385) is correct.
Error 1: Returns $72 credited instead of debited → payable's balance overstated by $144 (wrong side = double error).
Error 2: Discount posted as $45 instead of $54 → understated by $9.
Ledger total = $8,232 + $144 + $9 = $8,385
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