NIMCET 2026 Previous Year Questions (PYQs) – Page 3 of 12

The main idea of the passage is that astronomers confirmed the accelerating expansion of the universe.

The passage also says that this acceleration is driven by "dark energy".

The study used "weak gravitational lensing" and involved nearly $446000$ galaxies, $1000$ hours of observation, and $600$ Earth orbits by the Hubble Space Telescope.

Option $1$ only talks about the study details but misses the main finding.

Option $2$ is incorrect because the passage does not say that dark matter was noticed for the first time.

Option $3$ wrongly says that weak gravitational lensing is the best technique to study dark matter, which is not the main idea.

Option $4$ correctly captures the main idea and includes the important method and observation details.

The word corroborate means to support or confirm something with evidence.

Here, the sentence means that the new findings from the flight data recorder support or confirm the theory of pilot suicide.

So, the correct word is corroborate.

We know that
$A \Delta B = (A \cap B^c) \cup (A^c \cap B)$

Now,
$(A \Delta B) \cap C = [(A \cap B^c) \cup (A^c \cap B)] \cap C$

Using distributive law,
$(A \Delta B) \cap C = (A \cap B^c \cap C) \cup (A^c \cap B \cap C)$

Option $1$, option $2$ and option $4$ represent the same set.

But option $3$ is
$(A \cap B)^c \cap C$

This means all elements of $C$ except the elements common in $A$ and $B$.
It also includes elements which are neither in $A$ nor in $B$, but present in $C$.

So, it is not equal to $ (A \Delta B) \cap C $.

Given relation:
$a \sim b$ if $a - 2b$ is divisible by $3$

This means
$a - 2b \equiv 0 \pmod 3$

So,
$a \equiv 2b \pmod 3$

Checking reflexive:
For reflexive relation, $a \sim a$ must be true for every $a$.

Now,
$a - 2a = -a$

For all values of $a$, $-a$ is not divisible by $3$.
So, relation is not reflexive.

Checking symmetric:
If $a \sim b$, then
$a \equiv 2b \pmod 3$

Multiplying both sides by $2$,
$2a \equiv 4b \pmod 3$

Since $4 \equiv 1 \pmod 3$,
$2a \equiv b \pmod 3$

So,
$b - 2a \equiv 0 \pmod 3$

Hence, $b \sim a$.
So, relation is symmetric.

Checking transitive:
Let $a \sim b$ and $b \sim c$.

Then,
$a \equiv 2b \pmod 3$

and
$b \equiv 2c \pmod 3$

So,
$a \equiv 2(2c) \pmod 3$

$a \equiv 4c \pmod 3$

Since $4 \equiv 1 \pmod 3$,
$a \equiv c \pmod 3$

But for $a \sim c$, we need
$a \equiv 2c \pmod 3$

This is not always true.
So, relation is not transitive.

Therefore, $\sim$ is symmetric but neither transitive nor reflexive.

Even elements in the domain are $2$ and $4$.
Even elements in the codomain are $2,4,6,8$.

The two even elements of the domain must be mapped to even elements of the codomain injectively.

Number of ways:
${}^4P_2=4\times 3=12$

Now, $2$ elements of the codomain are already used, so $6$ elements are left.

The odd elements of the domain are $1$ and $3$.
They can be mapped injectively to the remaining $6$ elements.

Number of ways:
${}^6P_2=6\times 5=30$

So,
$n=12\times 30=360$

Now,
$360=2^3\times 3^2\times 5^1$

So,
$a=3,\ b=2,\ c=1$

Therefore,
$a+b+c=3+2+1=6$

For $n$ sets, the maximum number of regions in a Venn diagram is:

$2^n$

Here,
$n=5$

So, number of regions is:

$2^5=32$

Since
$R\subset \mathbb{N}\times \mathbb{N}$,

the relation $R$ can have at most countably infinite elements.

Now, suppose for some $a\in \mathbb{N}$, the set $R_a$ is infinite.

Since $R_a\subseteq \mathbb{N}$, it is countably infinite.

Also, for every element $b\in R_a$, the ordered pair $(a,b)$ belongs to $R$.

So, $R$ is also infinite. Since $R\subset \mathbb{N}\times \mathbb{N}$, $R$ is countably infinite.

Therefore, both $R_a$ and $R$ have the same cardinality.

We check the possible membership of each element in $A,B,C$.

Condition given is:

$(A\cap B)\subseteq C\subseteq (A\cup B)$

For one element:

If the element is in neither $A$ nor $B$, then it cannot be in $C$.
Number of choices $=1$

If the element is only in $A$, then it may or may not be in $C$.
Number of choices $=2$

If the element is only in $B$, then it may or may not be in $C$.
Number of choices $=2$

If the element is in both $A$ and $B$, then it must be in $C$.
Number of choices $=1$

Total choices for one element:

$1+2+2+1=6$

Since there are $n$ elements, total number of triples is:

$6^n$

There are three variables $p,q,r$.

So, total number of ordered tuples is:

$2^3=8$

The implication $A\Rightarrow r$ is false only when $A$ is true and $r$ is false.

Here,
$A=(\neg p\vee q)$

Now, $A=(\neg p\vee q)$ is false only when both $\neg p$ and $q$ are false.

That means:

$p$ is true and $q$ is false.

So, $A$ is false in only $1$ case out of $4$ possible cases of $(p,q)$.

Therefore, $A$ is true in $3$ cases.

For implication to be false, $A$ must be true and $r$ must be false.

So, false cases $=3$

Hence, true cases:

$8-3=5$

Given,
$P(X\text{ passes})=\frac{2}{5}$

So,
$P(X\text{ does not pass})=1-\frac{2}{5}=\frac{3}{5}$

Also,
$P(Y\text{ passes})=\frac{3}{4}$

So,
$P(Y\text{ does not pass})=1-\frac{3}{4}=\frac{1}{4}$

Since both events are independent,

$P(\text{neither }X\text{ nor }Y\text{ passes})=\frac{3}{5}\times \frac{1}{4}$

$=\frac{3}{20}$