Jamia Millia Islamia MCA Mathematical Reasoning Previous Year Questions (PYQs)
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Which of the following is correct?
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Solution
Since $1^\circ = \dfrac{\pi}{180}$ radians and $\sin x$ increases for small $x$ in radians,
$\sin(1^\circ) = \sin\left(\dfrac{\pi}{180}\right) \approx 0.01745 < \sin(1 \text{ rad}) \approx 0.841$.
Hence, $\sin 1^\circ < \sin 1$.
If $P,Q,R$ have truth values $T,T,F$, then the truth values of
$\Big(P\to(Q\to R)\Big)\to\Big((P\to Q)\to(P\to R)\Big)$ and $P\to(Q\vee R)$ are:
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Solution
$Q\to R = T\to F = F$, so $P\to(Q\to R)=T\to F=F$.
$P\to Q = T\to T = T$, $P\to R = T\to F = F$, hence $(P\to Q)\to(P\to R)=T\to F=F$.
Therefore $(F)\to(F)=T$.
Also $Q\vee R = T\vee F = T$, so $P\to(Q\vee R)=T\to T=T$.
The contrapositive of $p \to (\neg q \to \neg r)$ is —
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Solution
Given statement: $p \to (\neg q \to \neg r)$
Contrapositive = $\neg(\neg q \to \neg r) \to \neg p$
Now, $\neg(\neg q \to \neg r) \equiv (\neg q \land r)$
Hence, contrapositive = $(\neg q \land r) \to \neg p.$
Consider the following statements:
S1: There exists infinite sets A, B, C such that A ∪ C = B ∪ C is finite.
S2: There exist two irrational numbers x and y such that (x+y) is rational.
Which of the following is true about S1 and S2?
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Solution
Consider the statements:
S1: There exists infinite sets A,B,C such that A C = B C is finite.
If A and C are infinite, then A C must be infinite. So this is impossible.
S2: There exist two irrational numbers x and y such that (x+y) is rational.
The statement $p \to (q \to p)$ is equivalent to...
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Solution
$q \to p$ is true whenever $p$ is true.
Hence, $p \to (q \to p)$ is always true.
At any time, the total number of people on the earth who have shaken hands an odd number of times is:
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Solution
By the Handshake Lemma, in any graph, the number of vertices with odd degree is always even.
Ram secures $100$ marks in maths, then he will get a smartphone.” The converse is:
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Solution
Converse of “$P\Rightarrow Q$” is “$Q\Rightarrow P$”.
$\boxed{\text{If Ram will get a smartphone, then he secures 100 marks in maths.}}$
Negation of $,q \vee \sim (p \wedge r),$ is
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Solution
$\neg\big(q \vee \neg(p\wedge r)\big)=\neg q \wedge \neg\neg(p\wedge r)=\neg q \wedge (p\wedge r)$.
$\boxed{,\sim q \wedge (p \wedge r),}$
$(p\wedge \neg q)\wedge(\neg p\wedge q)$ is
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Solution
Expression forces $p\wedge\neg p$ and $q\wedge\neg q$ simultaneously → impossible.
$\boxed{\text{Contradiction}}$
The compound statement with ‘And’ is false if ______ of its component statements are.
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Solution
In logic, a conjunction $p \land q$ is true only if both $p$ and $q$ are true.
Hence, it is false if any one is false.
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