NIMCET 2013 Previous Year Questions (PYQs) – Page 1 of 12
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If $A=\begin{bmatrix} 1 &0 &0 \\ 0& 1 &1 \\ 0&-2 & 4 \end{bmatrix}$ and $6A^{–1} = A^{2} + cA + dI$, where $A^{–1}$ is A- inverse, I is the identify matrix, then (c, d)
is
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Let $\vec{a}=\hat{j}-\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ . Then the vector $\vec{b}$ satisfying $(\vec{a} \times \vec{b})+ \vec{c} =0$ and $\vec{a} . \vec{b}=3$ is
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Find the number of elements in the union of 4 sets A, B, C and D having 150, 180, 210 and 240
elements respectively, given that each pair of sets has 15 elements in common. Each triple of sets has
3 elements in common and $A \cap B \cap C \cap D = \phi$
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Solution
Given: \(|A|=150,\ |B|=180,\ |C|=210,\ |D|=240\); every pair has 15 common elements; every triple has 3 common elements; and \(A\cap B\cap C\cap D=\varnothing\).
Use Inclusion–Exclusion for 4 sets:
\[ \begin{aligned} |A\cup B\cup C\cup D| &= \sum |A_i| \;-\; \sum |A_i\cap A_j| \;+\; \sum |A_i\cap A_j\cap A_k| \;-\; |A\cap B\cap C\cap D|\\ &= (150+180+210+240)\;-\; \binom{4}{2}\cdot 15 \;+\; \binom{4}{3}\cdot 3 \;-\; 0\\ &= 780 \;-\; 6\cdot 15 \;+\; 4\cdot 3\\ &= 780 - 90 + 12\\ &= \boxed{702}. \end{aligned} \]
Answer: 702
If the straight line ax + by + c = 0 always passes through (1, –2), then a, b, c are in
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A six faced die is a biased one. It is thrice more likely to show an odd number than to show an even
number. It is thrown twice. The probability that the sum of the numbers in the two throws is even is
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Solution
Let $P(\text{odd}) = 3k$ and $P(\text{even}) = k$.
Since $P(\text{odd}) + P(\text{even}) = 1$:
$3k + k = 1$
$4k = 1$
$k = \frac{1}{4}$
So,
$P(\text{odd}) = \frac{3}{4}$
$P(\text{even}) = \frac{1}{4}$
The sum of two numbers is even if:
1. both are odd, or
2. both are even.
Therefore,
$P(\text{even sum}) = P(\text{odd})^2 + P(\text{even})^2$
$= \left(\frac{3}{4}\right)^2 + \left(\frac{1}{4}\right)^2$
$= \frac{9}{16} + \frac{1}{16}$
$= \frac{10}{16}$
$= \frac{5}{8}$
So the probability that the sum is even is $\frac{5}{8}$.
If $I_n = \int_0^{\pi/4} tan^{n} \theta d\theta$ , then $I_8 + I_6$ equals
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Let $\Delta ABC$ be a triangle whose area is $10\sqrt{3}$ units with side lengths $|AB|= 8$ units and $|AC|=5$
units. Find possible values of the angle A
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Person A can hit a target 4 times in 5 attempts. Person B - 3 times in four attempts. Person C – 2
times in 3 attempts. They fire a volley. The probability that the target is hit at least two times. Is
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Solution
Person A can hit the target with probability $\frac{4}{5}$
Person B with probability $\frac{3}{4}$
Person C with probability $\frac{2}{3}$
They fire together.
We need:
Probability(target is hit at least two times)
This means: exactly 2 hits OR exactly 3 hits.
Let $p_A=\frac{4}{5}$, $p_B=\frac{3}{4}$, $p_C=\frac{2}{3}$
and miss probabilities:
$q_A=\frac{1}{5}$, $q_B=\frac{1}{4}$, $q_C=\frac{1}{3}$.
1.Exactly 2 hits:
-A and B hit, C misses: $p_A p_B q_C$
- A and C hit, B misses: $p_A q_B p_C$
- B and C hit, A misses: $q_A p_B p_C$
So,
$P(\text{exactly 2}) = p_A p_B q_C + p_A q_B p_C + q_A p_B p_C$
$= \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} \;+\; \frac{4}{5} \cdot \frac{1}{4} \cdot \frac{2}{3} \;+\; \frac{1}{5} \cdot \frac{3}{4} \cdot \frac{2}{3}$
$= \frac{1}{5} + \frac{2}{15} + \frac{2}{20}$
$= \frac{4}{20} + \frac{8}{60} + \frac{6}{60}$
$= \frac{12}{60} + \frac{14}{60}$
$= \frac{26}{60} = \frac{13}{30}$
2. Exactly 3 hits:
$P(\text{all hit}) = p_A p_B p_C$
$= \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3}$
$= \frac{2}{5}$
Therefore,
$P(\text{at least 2 hits}) = P(\text{exactly 2}) + P(\text{exactly 3})$
$= \frac{13}{30} + \frac{2}{5}$
$= \frac{13}{30} + \frac{12}{30}$
$= \frac{25}{30}$
$= \frac{5}{6}$
Final Answer: $\frac{5}{6}$
The value of the integral $\int _0^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$ is
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If $\omega$ is a cube root of unity, then find the value of determinant $\begin{vmatrix} 1+\omega &\omega^{2} &-\omega \\ 1+\omega^{2}&\omega &-\omega^{2} \\ \omega^{2}+\omega&\omega &-\omega^{2} \end{vmatrix}$
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Solution
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