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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET 2012 PYQ

MCA NIMCET PYQ 2012
If $H$ is the harmonic mean between $P$ and $Q$, then $\dfrac{H}{P} + \dfrac{H}{Q}$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Harmonic mean of $P$ and $Q$ is $H = \dfrac{2PQ}{P+Q}$
 
 Now, 
$\dfrac{H}{P} + \dfrac{H}{Q} = H\left(\dfrac{1}{P} + \dfrac{1}{Q}\right) = H \cdot \dfrac{P+Q}{PQ}$ 

Substitute 

$H = \dfrac{2PQ}{P+Q}$: 

$\dfrac{H}{P} + \dfrac{H}{Q} = \dfrac{2PQ}{P+Q} \cdot \dfrac{P+Q}{PQ} = 2$
 
So the correct Answer is $2$.
MCA NIMCET PYQ 2012
The number of values of $k$ for which the system of equations $(k+1)x + 8y = 4k$ and $kx + (k+3)y = 3k-1$ has infinitely many solutions, is





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Solution

$ (k+1)x + 8y = 4k $ 
$ kx + (k+3)y = 3k - 1 $ 
Since the system has infinitely many solutions, $ \dfrac{k+1}{k} = \dfrac{8}{k+3} = \dfrac{4k}{3k-1} $ 
Taking 1st and 3rd ratio: $ (k+1)(3k-1) = 4k^2 $ 
$ 3k^2 + 2k - 1 = 4k^2 $ 
$ k^2 - 2k + 1 = 0 $ 
$ (k-1)^2 = 0 $ 
$ \therefore k = 1 $
MCA NIMCET PYQ 2012
The sum $^{20}C_8 + ^{20}C_9 + ^{21}C_{10} + ^{22}C_{11} - ^{23}C_{11}$





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Use Pascal identity: $^{n}C_r + ^{n}C_{r+1} = ^{n+1}C_{r+1}$ 
So, $^{20}C_8 + ^{20}C_9 = ^{21}C_9$ 
Then, $^{21}C_9 + ^{21}C_{10} = ^{22}C_{10}$
Next, $^{22}C_{10} + ^{22}C_{11} = ^{23}C_{11}$ 
Thus the expression becomes: 
$^{23}C_{11} - ^{23}C_{11} = 0$ 
$\therefore$ the answer is 0.
MCA NIMCET PYQ 2012
The value of $\cot^{-1}(21) + \cot^{-1}(13) + \cot^{-1}(-8)$ is:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Use identity for inverse cotangent: 
$\cot^{-1}a + \cot^{-1}b = \cot^{-1}\left(\dfrac{ab - 1}{a + b}\right)$ 
Apply to the first two terms: 
$\cot^{-1}(21) + \cot^{-1}(13)$ 
$= \cot^{-1}\left(\dfrac{21 \cdot 13 - 1}{21 + 13}\right)$ 
$= \cot^{-1}\left(\dfrac{273 - 1}{34}\right)$ 
$= \cot^{-1}\left(\dfrac{272}{34}\right)$ 
$= \cot^{-1}(8)$ 
Now the expression becomes: 
$\cot^{-1}(8) + \cot^{-1}(-8)$ 
But we know: $\cot^{-1}(x) + \cot^{-1}(-x) = \pi$ 
So, $\cot^{-1}(8) + \cot^{-1}(-8) = \pi$
$\therefore$ The value is $\pi$.
MCA NIMCET PYQ 2012
Normal to the curve $y = x^3 - 3x + 2$ at the point $(2,4)$ is:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Differentiate: 
$\dfrac{dy}{dx} = 3x^2 - 3$ 
At $x = 2$: 
$\dfrac{dy}{dx} = 3(4) - 3 = 12 - 3 = 9$ 
So slope of tangent $m_t = 9$ 
Slope of normal: $m_n = -\dfrac{1}{9}$ 
Equation of normal at $(2,4)$: 
$y - 4 = -\dfrac{1}{9}(x - 2)$ 
Multiply by 9: 
$9y - 36 = -(x - 2)$ 
$9y - 36 = -x + 2$ 
$ x + 9y - 38 = 0$ 
Answer: (c) $x + 9y - 38 = 0$
MCA NIMCET PYQ 2012
The value of $\displaystyle \lim_{n\to\infty} \frac{\pi}{n}\left[\sin\frac{\pi}{n}+\sin\frac{2\pi}{n}+\cdots+\sin\frac{(n-1)\pi}{n}\right]$ is:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

This is a Riemann sum. 
 Rewrite the expression: 
 $\displaystyle \frac{\pi}{n}\sum_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)$ 
 Let $x_k = \frac{k\pi}{n}$ 
Then spacing $\Delta x = \frac{\pi}{n}$ 
 As $n\to\infty$, this becomes: 
 $\displaystyle \int_{0}^{\pi} \sin x , dx$ 
 Now evaluate: $\displaystyle \int_{0}^{\pi} \sin x  dx = [-\cos x]_{0}^{\pi}$ 
 $= (-\cos\pi) - (-\cos 0)$ 
 $= -(-1) - (-1)$ 
 $= 1 + 1 = 2$ 
 $\therefore$ the value of the limit is 2. 

MCA NIMCET PYQ 2012
The point on the curve $y = 6x - x^2$ where the tangent is parallel to the x-axis is:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Tangent ∥ x-axis ⇒ slope = 0 
Differentiate: $\dfrac{dy}{dx} = 6 - 2x$ 
Set derivative equal to zero: 
$6 - 2x = 0$ $2x = 6$ 
$x = 3$ 
Now find $y$: 
$y = 6(3) - (3)^2 = 18 - 9 = 9$ 
Therefore, the required point is $(3,9)$.
Answer: $(3,9)$
MCA NIMCET PYQ 2012
If $ I_1 = \displaystyle \int_{0}^{1} 2^{x^2},dx,\quad I_2 = \displaystyle \int_{0}^{1} 2^{x^3},dx,\quad I_3 = \displaystyle \int_{1}^{2} 2^{x^2},dx,\quad I_4 = \displaystyle \int_{1}^{2} 2^{x^3},dx,$ then





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

On the interval $0 \le x \le 1$: 
$ x^3 < x^2 \Rightarrow 2^{x^3} < 2^{x^2} $ 
So, $ I_2 = \displaystyle \int_0^1 2^{x^3},dx < \int_0^1 2^{x^2}dx = I_1 $ 
Thus, $ I_1 > I_2 $ 

 On the interval $1 \le x \le 2$: 
$ x^3 > x^2 \Rightarrow 2^{x^3} > 2^{x^2} $ 
So, $ I_4 = \displaystyle \int_1^2 2^{x^3}dx > \int_1^2 2^{x^2}dx = I_3 $ 
Thus, $ I_4 > I_3 $
MCA NIMCET PYQ 2012
The value of integral $\displaystyle \int_{0}^{\pi/2} \log \tan x dx$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Let $I = \displaystyle \int_{0}^{\pi/2} \log \tan x  dx$  
Using property $\displaystyle \int_{0}^{\pi/2} f(x)dx = \int_{0}^{\pi/2} f\left(\dfrac{\pi}{2}-x\right)dx$ 
$I = \displaystyle \int_{0}^{\pi/2} \log \tan\left(\dfrac{\pi}{2}-x\right) dx$ 
$= \displaystyle \int_{0}^{\pi/2} \log \cot x  dx$ 
$= \displaystyle \int_{0}^{\pi/2} \log\left(\dfrac{1}{\tan x}\right) dx$ 
$= \displaystyle \int_{0}^{\pi/2} [-\log \tan x]  dx$ 
$I= -I$ 
$\Rightarrow I + I = 0 \Rightarrow 2I = 0 \Rightarrow I = 0$ 
Answer: $0$
MCA NIMCET PYQ 2012
A determinant is chosen at random from the set of all determinants of matrices of order 2 with elements 0 and 1 only. The probability that the determinant chosen is non-zero is:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Total $2\times2$ matrices with entries $0$ or $1$: $2^4 = 16$ 
Determinant: $ad - bc$ 
Non-zero determinant occurs in two cases. 

Case 1: $ad = 1$ and $bc = 0$ 
$a = 1, d = 1$ $(b,c)$ can be $(0,0), (0,1), (1,0)$ 
→ 3 matrices 

Case 2: $bc = 1$ and $ad = 0$ 
$b = 1,c = 1$ $(a,d)$ can be $(0,0), (0,1), (1,0)$ 
→ 3 matrices 
Total non-zero determinants = $3 + 3 = 6$
$P(\text{non-zero determinants})=\dfrac{6}{16}=\dfrac{3}{8}$
MCA NIMCET PYQ 2012
If $\sin^2 x = 1 - \sin x$, then $\cos^4 x + \cos^2 x$ is equal to:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

From the given equation: 
$\sin^2 x = 1 - \sin x$ 
So: $\cos^2 x = 1 - (1 - \sin x)$ 
$\cos^2 x = \sin x$ 

$\cos^4x+\cos^2x$
=$\sin^2 x + \sin x$
$ = 1$ 
MCA NIMCET PYQ 2012
The equation of the plane passing through the point $(1,2,3)$ and having the normal vector $N = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ is:





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Solution

Equation of plane with normal $(a,b,c)$ passing through $(x_1, y_1, z_1)$ is: 
$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$ 

Here: $a = 3; b = -1; c = 2$ 
$(x_1, y_1, z_1) = (1,2,3)$ 

So: $3(x - 1) - 1(y - 2) + 2(z - 3) = 0$ 
Expand: $3x - 3 - y + 2 + 2z - 6 = 0$ 
Combine constants: $3x - y + 2z - 7 = 0$ 
$3x - y + 2z = 7$ 
Thus the correct answer is: $3x - y + 2z = 7$
MCA NIMCET PYQ 2012
The value of $\displaystyle \int_{0}^{\sin^2 x} \sin^{-1}\sqrt{t} dt + \int_{0}^{\cos^2 x} \cos^{-1}\sqrt{t} dt$ is:





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Solution

Use the identity:
 $\sin^{-1}y+\cos^{-1}y=\dfrac{\pi}{2}$ 
So: $\cos^{-1}\sqrt{t}=\dfrac{\pi}{2}-\sin^{-1}\sqrt{t}$ 
Now substitute: $I=\displaystyle\int_0^{\sin^2 x}\sin^{-1}\sqrt{t}  dt + \int_0^{\cos^2 x} \left(\dfrac{\pi}{2}-\sin^{-1}\sqrt{t}\right) dt$ 
$I=\dfrac{\pi}{2}\cos^2 x + \int_0^{\sin^2 x}\sin^{-1}\sqrt{t}dt - \int_0^{\cos^2 x}\sin^{-1}\sqrt{t} dt$ 
Combine integrals: $I=\dfrac{\pi}{2}\cos^2 x + \int_{\cos^2 x}^{\sin^2 x}\sin^{-1}\sqrt{t} dt$ 
But: $\sin^2 x + \cos^2 x = 1$ 
Limits become from $1$ to $0$: 
$I=\dfrac{\pi}{2}(1 - \sin^2 x) + \int_{1}^{0}\sin^{-1}\sqrt{t} dt$ 

$I=\dfrac{\pi}{2} - \int_0^{1}\sin^{-1}\sqrt{t} dt$ 

Let $u=\sqrt{t}$, 
$dt=2udu$: 

$\displaystyle \int_0^{1}\sin^{-1}\sqrt{t} dt = 2\int_0^{1} u\sin^{-1}udu$ 

Standard result: $\displaystyle 2\int_0^{1} u\sin^{-1}u du = \dfrac{\pi}{4}$ 

Thus: $I = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$
MCA NIMCET PYQ 2012
Coefficients a, b, c of $ax^2 + bx + c = 0$ are chosen by tossing 3 fair coins. Head means 1, Tail means 2. Find the probability that the roots are imaginary





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Since each of $a,b,c$ can be $1$ or $2$: 
Total possible triples $(a,b,c)$: $2^3 = 8$. 
Roots are imaginary if: $b^2 - 4ac < 0$ 
So we check all $8$ cases. 

 List all possibilities 
 $(1,1,1)$: $1^2 - 4(1)(1) = 1 - 4 = -3 < 0$ ✔ 
 $(1,1,2)$: $1 - 8 = -7 < 0$ ✔ 
 $(1,2,1)$: $4 - 4 = 0$ ✘ (not imaginary) 
 $(1,2,2)$: $4 - 8 = -4 < 0$ ✔ 
 $(2,1,1)$: $1 - 8 = -7 < 0$ ✔ 
 $(2,1,2)$: $1 - 16 = -15 < 0$ ✔ 
 $(2,2,1)$: $4 - 8 = -4 < 0$ ✔ 
 $(2,2,2)$: $4 - 16 = -12 < 0$ ✔ 
 Count imaginary root cases 
Total imaginary cases = $7$ out of $8$. 
So probability: $\displaystyle P = \frac{7}{8}$ 
MCA NIMCET PYQ 2012
In a class of 100 students, 55 passed in Mathematics and 67 passed in Physics.The number of students who passed in Physics only is:





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Solution

Use the formula: 
$\text{Physics only} = n(P) - n(M \cap P)$ 
First find $n(M \cap P$) using: 
$n(M) + n(P) - n(M \cap P) = 100$ 
So: $55 + 67 - n(M \cap P) = 100$ 
$122 - n(M \cap P) = 100$ 
$n(M \cap P) = 22$ 
Now: $\text{Physics only} = 67 - 22 = 45$
MCA NIMCET PYQ 2012
If $(4,-3)$ and $(-9,7)$ are two vertices of a triangle and $(1,4)$ is its centroid, find the area of the triangle.





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Let the third vertex be $(x,y)$. Centroid formula: $\displaystyle \left(\frac{4 + (-9) + x}{3}, \frac{-3 + 7 + y}{3}\right) = (1,4)$

From x–coordinate: $\dfrac{4 - 9 + x}{3} = 1$ 
$x - 5 = 3$ 
$x = 8$

From y–coordinate: $\dfrac{-3 + 7 + y}{3} = 4$ 
$y + 4 = 12$ 
$y = 8$ 
So third vertex is: $(8,8)$

Vertices: $A(4,-3), B(-9,7), C(8,8)$ 

Area: $\displaystyle \text{Area} = \frac{1}{2}\left| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right|$ Substitute: 
$\displaystyle = \frac{1}{2}\left|4(7-8) + (-9)(8+3) + 8(-3-7) \right|$ 
$\displaystyle = \frac{1}{2}\left| 4(-1) - 9(11) + 8(-10) \right|$ 
$\displaystyle = \frac{1}{2}\left| -4 - 99 - 80 \right|$ 
$\displaystyle = \frac{1}{2} \times 183$ $\displaystyle = \frac{183}{2}$
MCA NIMCET PYQ 2012
The equation of ellipse with major axis along the x–axis and passes through the point $(4,3)$ and $(-1,4)$.





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Ellipse form: 
$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\quad a^2 > b^2$ 

Using points: $\displaystyle \frac{16}{a^2} + \frac{9}{b^2} = 1$ 
$\displaystyle \frac{1}{a^2} + \frac{16}{b^2} = 1$ 

Solve gives: 
$\displaystyle 7x^2 + 15y^2 = 247$ 
MCA NIMCET PYQ 2012
If the circles $ x^2 + y^2 + 2x + 2ky + 6 = 0$ and $x^2 + y^2 + 2ky + k = 0$ intersect orthogonally, then $k$ is:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Two circles: $C_1:\; x^2 + y^2 + 2x + 2ky + 6 = 0$ 
 $C_2:\; x^2 + y^2 + 2ky + k = 0$ 

 Orthogonality condition for circles 
 $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ 
 and 
 $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ is: 
 $2(g_1g_2 + f_1f_2) = c_1 + c_2$ 

 From $C_1$: $g_1 = 1,\; f_1 = k,\; c_1 = 6$ 
 From $C_2$: $g_2 = 0,\; f_2 = k,\; c_2 = k$ 

 Apply formula: 
 $2(1\cdot 0 + k\cdot k) = 6 + k$ 
 $2k^2 = 6 + k$ 
 $2k^2 - k - 6 = 0$ 
 $(2k + 3)(k - 2) = 0$ 
 So: $k = 2 \;\text{or}\; -\dfrac{3}{2}$
MCA NIMCET PYQ 2012
Focus of the parabola $x^2 + y^2 - 2xy - 4(x + y - 1) = 0$ is:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Given equation: 
 $x^2 + y^2 - 2xy - 4x - 4y + 4 = 0$ 

 Group terms: 
 $(x - y)^2 - 4(x + y) + 4 = 0$ 
 Let: $u = x - y,\;\; v = x + y$ 
 Then equation becomes: 
 $u^2 - 4v + 4 = 0$ 
 $u^2 = 4(v - 1)$ 
 This is the standard parabola: 
 $u^2 = 4p(v - 1)$ 
 Comparing gives: 
 $4p = 4 \Rightarrow p = 1$ 
 Vertex in $(u,v)$: $(0,1)$ 
 Focus in $(u,v)$: $(0, 1 + p) = (0,2)$ 
 Convert to $(x,y)$: 
 $x - y = 0$ 
 $x + y = 2$ 
 Solving: $x = 1,\; y = 1$ 
 Therefore, the focus is: $\boxed{(1,1)}$
MCA NIMCET PYQ 2012
If $\mathbf{a},\; \mathbf{b},\; \mathbf{c}$ are unit vectors such that $\mathbf{a} + \mathbf{b} + \mathbf{c} = 0$, then the value of $\mathbf{a}\cdot \mathbf{b} + \mathbf{b}\cdot \mathbf{c} + \mathbf{c}\cdot \mathbf{a}$ is:





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Solution

Given: $\mathbf{a} + \mathbf{b} + \mathbf{c} = 0$ 
 Square both sides: 
 $(\mathbf{a} + \mathbf{b} + \mathbf{c})\cdot(\mathbf{a} + \mathbf{b} + \mathbf{c}) = 0$ 
 $\mathbf{a}\cdot\mathbf{a} + \mathbf{b}\cdot\mathbf{b} + \mathbf{c}\cdot\mathbf{c} + 2(\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a}) = 0$ 
 Since each of $\mathbf{a},\;\mathbf{b},\;\mathbf{c}$ is a unit vector: $\mathbf{a}\cdot\mathbf{a} = \mathbf{b}\cdot\mathbf{b} = \mathbf{c}\cdot\mathbf{c} = 1$ 
 Thus: $3 + 2(\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a}) = 0$ 
 $2(\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a}) = -3$
 $\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a} = -\dfrac{3}{2}$ 
 Therefore: ${-\dfrac{3}{2}}$
MCA NIMCET PYQ 2012
If $ \cos(\alpha+\beta)=\frac{4}{5}$ and  $\sin(\alpha-\beta)=\frac{5}{13},\quad 0<\alpha,\beta<\frac{\pi}{4},$ then $ \tan(2\alpha)$ is equal to





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Given: 
$\cos(\alpha+\beta)=\frac{4}{5}$ 
Since $0<\alpha,\beta<\frac{\pi}{4}$, 
therefore $\alpha+\beta$ is an acute angle. 

$\sin(\alpha+\beta)=\sqrt{1-\cos^2(\alpha+\beta)}$ 
$=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$ 

Therefore, $\tan(\alpha+\beta)=\frac{3}{4}$ 

Also, $\sin(\alpha-\beta)=\frac{5}{13}$ 

Since the sine value is positive, $\alpha-\beta>0$. 

Therefore, $\cos(\alpha-\beta)=\sqrt{1-\sin^2(\alpha-\beta)}$ 

$=\sqrt{1-\frac{25}{169}}=\frac{12}{13}$ 

Hence, $\tan(\alpha-\beta)=\frac{5}{12}$ 

Now, $2\alpha=(\alpha+\beta)+(\alpha-\beta)$ 

Using the tangent addition formula: 

$\tan(2\alpha)=\frac{\tan(\alpha+\beta)+\tan(\alpha-\beta)}{1-\tan(\alpha+\beta)\tan(\alpha-\beta)}$ 

$=\frac{\frac{3}{4}+\frac{5}{12}}{1-\left(\frac{3}{4}\right)\left(\frac{5}{12}\right)}$ 

$=\frac{\frac{14}{12}}{1-\frac{15}{48}}$ 

$=\frac{\frac{7}{6}}{\frac{33}{48}}$ 

$=\frac{7}{6}\times\frac{48}{33}$ 

$=\frac{56}{33}$ 

Therefore, the correct answer is: $\frac{56}{33}$
MCA NIMCET PYQ 2012
The number of words that can be formed by using the letters of the word 'MATHEMATICS' that start as well as end with T is





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Solution

Letters and frequencies: 
$M=2,\ A=2,\ T=2,\ H=1,\ E=1,\ I=1,\ C=1,\ S=1$ (total $11$ letters) 
Condition: Starts with T and ends with T 
Fix $T$ at first and last position. 
Remaining letters = $9$ letters: $M=2,\ A=2,\ H=1,\ E=1,\ I=1,\ C=1,\ S=1$ 
Number of distinct arrangements of these $9$ letters: 
Using division method $\displaystyle \frac{9!}{2!,2!}$ 
Final answer: ${\frac{9!}{2!,2!} = 90720}$
MCA NIMCET PYQ 2012
If $A-B=\frac{\pi}{4}$, then $(1+\tan A)(1-\tan B)$ is equal to





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Solution

We know, $\tan(A-B)=\tan\frac{\pi}{4}=1$ 

Using the formula, 
$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ 
Therefore, $\frac{\tan A-\tan B}{1+\tan A\tan B}=1$ 
$\tan A-\tan B=1+\tan A\tan B$ 
Now, $(1+\tan A)(1-\tan B)$ 
$=1+\tan A-\tan B-\tan A\tan B$ 
$=1+(\tan A-\tan B)-\tan A\tan B$ 
$=1+(1+\tan A\tan B)-\tan A\tan B$ 
$=2$ 
Therefore, the correct answer is: 2
MCA NIMCET PYQ 2012
Let $P(E)$ denote the probability of event $E$. Given $P(A) = 1$, $P(B) = \frac{1}{2}$, the values of $P(A \mid B)$ and $P(B \mid A)$ respectively are





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Solution

Since $P(A)=1$, event $A$ is a sure event. 
Therefore for any event $B$, 
 $P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(B)}{P(B)}=1$ 
 Also, $P(B \mid A)=\frac{P(A \cap B)}{P(A)}=\frac{P(B)}{1}=\frac{1}{2}$ 
 Thus the correct pair is: 
 $P(A \mid B)=1$ and $P(B \mid A)=\frac{1}{2}$
MCA NIMCET PYQ 2012
Which of the following is correct?





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Solution

Here, $\sin 1^\circ$ means the angle is measured in degrees, while $\sin 1$ means the angle is measured in radians. Convert $1$ radian into degrees: $1\text{ radian}=\frac{180^\circ}{\pi}\approx57.3^\circ$ 

Therefore, $1^\circ<1\text{ radian}<90^\circ$ 
Since the sine function is increasing in the first quadrant, $\sin 1^\circ<\sin 1$ 

Therefore, the correct answer is: $\sin 1^\circ<\sin 1$
MCA NIMCET PYQ 2012
If $\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$, then $\sin 2\theta$ is equal to





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Given, $\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$ 

Using, $\cos x=\sin\left(\frac{\pi}{2}-x\right)$ 

Therefore, $\sin(\pi\cos\theta)=\sin\left(\frac{\pi}{2}-\pi\sin\theta\right)$ 

This gives the following two possible cases. 

Case 1: $\pi\cos\theta=\frac{\pi}{2}-\pi\sin\theta$ 
$\cos\theta+\sin\theta=\frac{1}{2}$ 
Squaring both sides, $(\cos\theta+\sin\theta)^2=\frac{1}{4}$ 

$\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta=\frac{1}{4}$ 

$1+\sin 2\theta=\frac{1}{4}$ 
$\sin 2\theta=-\frac{3}{4}$ 

Case 2: 
$\pi\cos\theta=\pi-\left(\frac{\pi}{2}-\pi\sin\theta\right)$ 
$\cos\theta-\sin\theta=\frac{1}{2}$ 
Squaring both sides, 
$(\cos\theta-\sin\theta)^2=\frac{1}{4}$ 
$1-\sin 2\theta=\frac{1}{4}$ 
$\sin 2\theta=\frac{3}{4}$ 
Hence, $\sin 2\theta=\pm\frac{3}{4}$ 
 Therefore, the correct answer is: $\boxed{\text{(a) }\pm\frac{3}{4}}$
MCA NIMCET PYQ 2012
The equation $(\cos p-1)x^2+(\cos p)x+\sin p=0$ where $x$ is a variable, has real roots. Then, the interval of $p$ is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

For the quadratic equation to have real roots, its discriminant must be non-negative. 

Here, $a=\cos p-1,\qquad b=\cos p,\qquad c=\sin p$ 

Therefore, $\Delta=b^2-4ac$ 

$\Delta=\cos^2p-4(\cos p-1)\sin p$ 

$\Delta=\cos^2p+4(1-\cos p)\sin p$ 

For 0<p<0$\pi$, $\sin p >0$ and $1-\cos p>0$ 

Therefore, $\cos^2p+4(1-\cos p)\sin p>0$ 

Hence, the equation has two real roots for $p\in(0,\pi)$.
MCA NIMCET PYQ 2012
Pointing towards a man in the photograph, lady said "the father of his brother is the only son of my mother". How is the man related to lady?





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

The father of 'his brother' means the father of the man in the photograph.
He is the only son of my mother means he is my Brother.
So, the father of the man in photograph is the  brother of the lady.
Hence, the man in the photograph is the nephew of that lady.
MCA NIMCET PYQ 2012
Three men A, B and C play cards. If one loses the game he has to give ₹3. If he wins the game he will gain ₹3 each from the other two losers. If A has won 3 games, B loses ₹3, C wins ₹12, then the total number of games played is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

B loses ₹3 and C wins ₹12.

Total balance must be zero:
\(A + (-3) + 12 = 0\)
\(A + 9 = 0\)
\(A = -9\)

So, A finally loses ₹9.

A has won 3 games.
In each winning game, A gains ₹3 from B and ₹3 from C.

Gain per winning game:
\(3 + 3 = 6\)

Total gain from 3 wins:
\(3 \times 6 = 18\)

But A finally loses ₹9, so total loss in losing games:
\(18 + 9 = 27\)

In each losing game, A gives ₹3.

Number of losing games:
\(\frac{27}{3} = 9\)

Total games played:
\(3 + 9 = 12\)

Therefore, the correct answer is \(12\).
MCA NIMCET PYQ 2012
A causes B or C, but not both
F occurs only if B occurs
D occurs if B or C occurs
E occurs only if C occurs
J occurs only if E or F occurs
D causes G, H or both
H occurs if E occurs
G occurs if F occurs





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution


MCA NIMCET PYQ 2012
A causes B or C, but not both
F occurs only if B occurs
D occurs if B or C occurs
E occurs only if C occurs
J occurs only if E or F occurs
D causes G, H or both
H occurs if E occurs
G occurs if F occurs





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution


MCA NIMCET PYQ 2012
A causes B or C, but not both
F occurs only if B occurs
D occurs if B or C occurs
E occurs only if C occurs
J occurs only if E or F occurs
D causes G, H or both
H occurs if E occurs
G occurs if F occurs
If J occurs which must have occurred





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution


MCA NIMCET PYQ 2012
If ‘ROAST’ is coded as ‘PQYUR’ in a certain language, then ‘SLOPPY’ is coded in that language as:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
If ‘lelibroon’ means ‘yellow hat’, ‘plekafroti’ means ‘flower graden’ and ‘frotimix’ means ‘garden salad’, then which word could mean ‘yellow flower’?





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
The ratio between a two-digit number and the sum of the digits of that number is 4 : 1. If the digit in the unit’s place is 3 more than the digit in ten’s place, then the number is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
Two positions of a dice are shown below. When number 1 is on the top, what number will be at the bottom?





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
A, B, C, D, E, F and G are sitting in a line facing East. C is immediate to the right of D. B is at one of the extreme ends and has E as his neighbour. G is between E and F. D is sitting third from the South end. Who is sitting third from North?





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
If A is brother of B, C is brother of B and A is brother of D, then which of the following must be true?





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012

Directions: Read the following passage carefully and answer the questions.

Five houses lettered A, B, C, D and E are built in a row next to each other. The houses are lined up in the order A, B, C, D and E. Each of the five houses have coloured roofs and chimneys. The roof and chimney of each house must be painted as follows.

  1. The roof must be painted either green, red or yellow.
  2. The chimney must be painted either white, black or red.
  3. No house may have the same colour chimney as the colour of roof.
  4. No house may use any of the same colours that adjacent house uses.
  5. House E has a green roof.
  6. House B has a red roof and a black chimney.
Which of the following is true?





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Fixed data:

B = Red roof + Black chimney

E = Green roof

Since A and C are adjacent to B, they cannot use Red or Black colour. Therefore, A and C must have White chimney, and their roof can be Green or Yellow.

Case House A House B House C House D House E
1 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Red roof, Black chimney Green roof, White chimney
2 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, White chimney
3 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, Red chimney
4 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, White chimney
5 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, Black chimney
6 Green roof, White chimney Red roof, Black chimney Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney
7 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Red roof, Black chimney Green roof, White chimney
8 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, White chimney
9 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, Red chimney
10 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, White chimney
11 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, Black chimney
12 Yellow roof, White chimney Red roof, Black chimney Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney
MCA NIMCET PYQ 2012

Directions: Read the following passage carefully and answer the questions.

Five houses lettered A, B, C, D and E are built in a row next to each other. The houses are lined up in the order A, B, C, D and E. Each of the five houses have coloured roofs and chimneys. The roof and chimney of each house must be painted as follows.

  1. The roof must be painted either green, red or yellow.
  2. The chimney must be painted either white, black or red.
  3. No house may have the same colour chimney as the colour of roof.
  4. No house may use any of the same colours that adjacent house uses.
  5. House E has a green roof.
  6. House B has a red roof and a black chimney.
If house C has a yellow roof, then which of the following must be true?





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Fixed data:

B = Red roof + Black chimney

E = Green roof

Since A and C are adjacent to B, they cannot use Red or Black colour. Therefore, A and C must have White chimney, and their roof can be Green or Yellow.

Case House A House B House C House D House E
1 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Red roof, Black chimney Green roof, White chimney
2 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, White chimney
3 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, Red chimney
4 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, White chimney
5 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, Black chimney
6 Green roof, White chimney Red roof, Black chimney Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney
7 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Red roof, Black chimney Green roof, White chimney
8 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, White chimney
9 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, Red chimney
10 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, White chimney
11 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, Black chimney
12 Yellow roof, White chimney Red roof, Black chimney Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney
MCA NIMCET PYQ 2012

Directions: Read the following passage carefully and answer the questions.

Five houses lettered A, B, C, D and E are built in a row next to each other. The houses are lined up in the order A, B, C, D and E. Each of the five houses have coloured roofs and chimneys. The roof and chimney of each house must be painted as follows.

  1. The roof must be painted either green, red or yellow.
  2. The chimney must be painted either white, black or red.
  3. No house may have the same colour chimney as the colour of roof.
  4. No house may use any of the same colours that adjacent house uses.
  5. House E has a green roof.
  6. House B has a red roof and a black chimney.
What is the maximum number of green roofs?





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Fixed data:

B = Red roof + Black chimney

E = Green roof

Since A and C are adjacent to B, they cannot use Red or Black colour. Therefore, A and C must have White chimney, and their roof can be Green or Yellow.

Case House A House B House C House D House E
1 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Red roof, Black chimney Green roof, White chimney
2 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, White chimney
3 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, Red chimney
4 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, White chimney
5 Green roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, Black chimney
6 Green roof, White chimney Red roof, Black chimney Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney
7 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Red roof, Black chimney Green roof, White chimney
8 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, White chimney
9 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Black chimney Green roof, Red chimney
10 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, White chimney
11 Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney Yellow roof, Red chimney Green roof, Black chimney
12 Yellow roof, White chimney Red roof, Black chimney Yellow roof, White chimney Red roof, Black chimney Green roof, White chimney
MCA NIMCET PYQ 2012
For the word, four spellings are given. Choose the correct one.





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
Choose the wrongly spelled word.





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
Choose the word or phrase that is most similar in meaning to the word POLEMIC.





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
Pick the synonym of the word 'Mearge'.





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
Choose the words that best express the meaning of the given idiom-Mud slinging.





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
Pick the antonym of the word 'Timid'.





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
Let x = 11111010 and y = 00001010 be two 8-bit 2’s complement numbers. Their product in 2’s complement notation is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
The range of numbers that can be stored in 8 bits, if negative numbers are stored in 2’s complement form is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
Primary storage is … as compared to secondary memory.





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
Which of the following units is used to supervise each instruction in the CPU?





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
(2FA0C)16 is equivalent to





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
The decimal equivalent to octal 111010 is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
The number of words that can be formed by using the letters of the word MATHEMATICS that start as well as end with T is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
If $A-B=\frac{\pi}{4}$, then (1 + tan A)(1 – tan B) is equal to





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
Let P(E) denote the probability of event E. Given P(A) = 1, P(B) =$\frac{1}{2}$ the value of P(A|B) and P(B|A) respectively are





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

MCA NIMCET PYQ 2012
The number of different license plates that can be formed in the format 3 English letters (A….Z) followed by 4 digits (0, 1, …9) with repetitions allowed in letters and digits is equal to





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution


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