Question:

Is $x + y - z + t$ even?
Statement I
I. $x + y + t$ is even.
Statement II
II. $t$ and $z$ are od(d)

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In parity (even/odd) problems, use addition and subtraction rules: even ± even = even, odd ± odd = even, even ± odd = od(d)
Updated On: Aug 6, 2025
  • The question can be answered with the help of statement I alone.
  • The question can be answered with the help of statement II alone.
  • Both statement I and statement II are needed to answer the question.
  • The question cannot be answered even with the help of both the statements.
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The Correct Option is C

Solution and Explanation

Step 1: From Statement I If $x + y + t$ is even, we still don’t know about $z$, so cannot conclude about $x + y - z + t$. Step 2: From Statement II $t$ odd and $z$ odd → $-z + t$ is even (odd − odd = even). Still no info about $x + y$. Step 3: Combining Statements From I: $x + y + t$ even, and from II: $t$ odd → $x + y$ must be odd (odd + odd = even). Now, $x + y$ odd and $-z + t$ even → odd + even = odd → expression is od(d) Step 4: Conclusion Both statements are required to deduce the parity.
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