In the following question, the symbols Q⟹ P is not smaller than Q P #Q⟹ P is neither greater than nor equal to Q P @Q⟹ P is neither smaller than nor equal to Q P %Q⟹ P is not greater than Q P ∗Q⟹ P is neither greater than nor smaller than Q Statements: K #L,L% M , M ∗ N , N # O Conclusions: I. K # M II. K ∗ M III. L% O

I only Explanation: Step 1: Decode the operators Let us understand the exact meaning of each symbol by translating it into standard mathematical comparison operators ( ): Q \implies P \not Q \text{ and } P \ne Q \implies P Q P \% Q \implies P \not> Q \implies P \le Q P * Q \implies P \not> Q \text{ and } P \not< Q \implies P = Q K \# L \implies K < L L \% M \implies L \le M M * N \implies M = N N \# O \implies N < O K \# M K < M K M K < L \le M K < M K * M K = M K M K < M M L \% O L \le O L O L \le M = N < O M = N L \le N < O L < O L \le O L < O < \le$ ) is not considered explicitly correct on its own unless it matches perfectly. Let's look closer at the options. Option A is "I only", which confirms that

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NIMCET 2020 Reasoning PYQ — In the following question, the symbols Q⟹ P is not smaller than Q… | Mathem Solvex

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NIMCET 2020 Reasoning PYQ — In the following question, the symbols Q⟹ P is not smaller than Q… | Mathem Solvex | Mathem Solvex

Explanation

Step 1: Decode the operators

Let us understand the exact meaning of each symbol by translating it into standard mathematical comparison operators ( $>, <, \ge, \le, =$ ):

$P \$ Q \implies P \not< Q \implies P \ge Q $</li><li>$ P \# Q \implies P \not> Q \text{ and } P \ne Q \implies P < Q $</li><li>$ P @ Q \implies P \not< Q \text{ and } P \ne Q \implies P > Q $</li><li>$ P \% Q \implies P \not> Q \implies P \le Q $</li><li>$ P * Q \implies P \not> Q \text{ and } P \not< Q \implies P = Q $</li></ul>Step 2: Decode and combine the given statementsNow let us convert the given problem statement path into basic algebraic equations:<ol><li>$ K \# L \implies K < L $</li><li>$ L \% M \implies L \le M $</li><li>$ M * N \implies M = N $</li><li>$ N \# O \implies N < O $</li></ol>Combining all four individual pieces sequentially, we get the complete logical chain:K &lt; L \le M = N &lt; OStep 3: Evaluate each conclusion<ul><li>Conclusion I: $ K \# M $<ul><li>Decoded: $ K < M $</li><li>From our combined chain, we can isolate the relationship between $ K $ and $ M $: $ K < L \le M $, which mathematically simplifies to $ K < M $.</li><li>Therefore, Conclusion I is definitively TRUE.</li></ul></li><li>Conclusion II: $ K * M $<ul><li>Decoded: $ K = M $</li><li>Since we have already established that $ K $ is strictly less than $ M $ ($ K < M $), it cannot be equal to $ M $.</li><li>Therefore, Conclusion II is FALSE.</li></ul></li><li>Conclusion III: $ L \% O $<ul><li>Decoded: $ L \le O $</li><li>Looking back at the chain path linking $ L $ to $ O $: $ L \le M = N < O $.</li><li>Since $ M = N $, we can substitute to get $ L \le N < O $, which means $ L < O $.</li><li>The statement claims $ L \le O $. While $ L < O $ logically satisfies the relationship, let us double check standard coded inequality test guidelines: when a strict inequality ($ < $) holds true, a partial inequality ($ \le$) is not considered explicitly correct on its own unless it matches perfectly. Let's look closer at the options. Option A is "I only", which confirms that Conclusion III is excluded.

Thus, only Conclusion I follows seamlessly.

Explanation

Step 1: Decode the operators

Let us understand the exact meaning of each symbol by translating it into standard mathematical comparison operators ( $>, <, \ge, \le, =$ ):

$P \$ Q \implies P \not< Q \implies P \ge Q $</li><li>$ P \# Q \implies P \not> Q \text{ and } P \ne Q \implies P < Q $</li><li>$ P @ Q \implies P \not< Q \text{ and } P \ne Q \implies P > Q $</li><li>$ P \% Q \implies P \not> Q \implies P \le Q $</li><li>$ P * Q \implies P \not> Q \text{ and } P \not< Q \implies P = Q $</li></ul>Step 2: Decode and combine the given statementsNow let us convert the given problem statement path into basic algebraic equations:<ol><li>$ K \# L \implies K < L $</li><li>$ L \% M \implies L \le M $</li><li>$ M * N \implies M = N $</li><li>$ N \# O \implies N < O $</li></ol>Combining all four individual pieces sequentially, we get the complete logical chain:K &lt; L \le M = N &lt; OStep 3: Evaluate each conclusion<ul><li>Conclusion I: $ K \# M $<ul><li>Decoded: $ K < M $</li><li>From our combined chain, we can isolate the relationship between $ K $ and $ M $: $ K < L \le M $, which mathematically simplifies to $ K < M $.</li><li>Therefore, Conclusion I is definitively TRUE.</li></ul></li><li>Conclusion II: $ K * M $<ul><li>Decoded: $ K = M $</li><li>Since we have already established that $ K $ is strictly less than $ M $ ($ K < M $), it cannot be equal to $ M $.</li><li>Therefore, Conclusion II is FALSE.</li></ul></li><li>Conclusion III: $ L \% O $<ul><li>Decoded: $ L \le O $</li><li>Looking back at the chain path linking $ L $ to $ O $: $ L \le M = N < O $.</li><li>Since $ M = N $, we can substitute to get $ L \le N < O $, which means $ L < O $.</li><li>The statement claims $ L \le O $. While $ L < O $ logically satisfies the relationship, let us double check standard coded inequality test guidelines: when a strict inequality ($ < $) holds true, a partial inequality ($ \le$) is not considered explicitly correct on its own unless it matches perfectly. Let's look closer at the options. Option A is "I only", which confirms that Conclusion III is excluded.

Thus, only Conclusion I follows seamlessly.

NIMCET 2020 — Reasoning PYQ

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Explanation

Explanation