NIMCET 2026 Previous Year Questions (PYQs) – Page 10 of 12

Total number of students is:

$\text{Rank from top}+\text{Rank from bottom}-1$

$=35+32-1$

$=67-1$

$=66$

Given initial values:

$X=2,\quad Y=4$

Rule:

$X=\frac{XY}{2}$ and $Y=Y+1$

Now update step by step:

First iteration:

$X=\frac{2\times 4}{2}=4,\quad Y=5$

Second iteration:

$X=\frac{4\times 5}{2}=10,\quad Y=6$

Third iteration:

$X=\frac{10\times 6}{2}=30,\quad Y=7$

Fourth iteration:

$X=\frac{30\times 7}{2}=105,\quad Y=8$

Fifth iteration:

$X=\frac{105\times 8}{2}=420,\quad Y=9$

Sixth iteration:

$X=\frac{420\times 9}{2}=1890,\quad Y=10$

Seventh iteration:

$X=\frac{1890\times 10}{2}=9450$

Now $X=9450$, which is greater than $3000$.

So, the game stops.

Therefore, the final value of $X$ is $9450$.

Simple interest is given by:

$SI=\frac{PRT}{100}$

Here, rate is same and interest is also same.

So, principal is inversely proportional to time.

Times are:

$2$ years, $3$ years and $4$ years.

Therefore, ratio of amounts invested is:

$\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$

LCM of $2,3,4$ is $12$.

Multiplying each term by $12$:

$6:4:3$

$P6$ is the son-in-law of $P3$.

So, $P6$ is married to the daughter of $P3$.

Also, $P2$ is the mother of $P6$'s wife.

So, $P6$'s wife is the daughter of $P2$ and $P3$.

Hence, $P1$ can be taken as the wife of $P6$.

Now, $P5$ and $P7$ are the grandsons of $P3$.

Also, $P4$ is the sister of $P7$.

So, $P4$ is the daughter of $P1$ and $P6$.

Therefore, the relationship between $P1$ and $P4$ is Mother-Daughter.

Given,

$T$ is at the bottom.

$P$ is just above $T$.

So, from bottom:

$T,\ P$

Now, $R$ is above $S$ but below $Q$.

So, their order from bottom to top is:

$S,\ R,\ Q$

Therefore, complete order from bottom to top is:

$T,\ P,\ S,\ R,\ Q$

Hence, the box at the top is $Q$.

Arun says:

"My father's only son"

This refers to Arun himself.

So, the woman is the daughter of Arun.

Therefore, the woman is Arun's daughter.

Let the present age of $B$ be $x$ years.

Then present age of $A$ is:

$2x$

Five years ago,

Age of $A=2x-5$

Age of $B=x-5$

According to the question:

$2x-5=3(x-5)$

$2x-5=3x-15$

$3x-2x=15-5$

$x=10$

So, present age of $A$ is:

$2x=2(10)=20$

A train crosses Park Street every $45$ minutes.

One train crossed Park Street $15$ minutes ago.

So, the next train will cross after:

$45-15=30$ minutes

The next train will cross at $9:45$ am.

Therefore, the current time is:

$9:45\text{ am}-30\text{ minutes}=9:15\text{ am}$

Let the total bill be Rs $x$.

Amount paid by $9$ people:

$9\times 400=3600$

So, amount paid by the $10^{\text{th}}$ person is:

$x-3600$

Group average is:

$\frac{x}{10}$

According to the question,

$x-3600=\frac{x}{10}+900$

$x-\frac{x}{10}=4500$

$\frac{9x}{10}=4500$

$x=5000$

So, amount paid by the $10^{\text{th}}$ person is:

$x-3600=5000-3600$

$=1400$

In container $A$, alcohol : water $=5:3$.

So, alcohol fraction in $A$ is:

$\frac{5}{8}$

In container $B$, alcohol : water $=1:3$.

So, alcohol fraction in $B$ is:

$\frac{1}{4}$

Required total mixture is $2.1$ litres, and alcohol and water are equal.

So, required alcohol quantity is:

$\frac{2.1}{2}=1.05$

Let $x$ litres be drawn from container $A$.

Then liquid drawn from container $B$ will be:

$2.1-x$

Now,

$\frac{5x}{8}+\frac{1}{4}(2.1-x)=1.05$

Multiply by $8$:

$5x+2(2.1-x)=8.4$

$5x+4.2-2x=8.4$

$3x=4.2$

$x=1.4$

Therefore, $1.4$ litres should be drawn from container $A$.