MCQ

MCQ 11 Mark

For the given five values $15,24,18,33,42$, the three years moving averages are

A
$19,25,33$
✓
$19,25,31$
C
$19,30,31$
D
$19,22,33$

Answer

Correct option: B.

$19,25,31$

(B) 19, 25, 31
Explanation: 3-years moving average are
$\frac{15+24+18}{3}, \frac{24+18+33}{3}, \frac{18+33+42}{3}$
i.e. $\frac{57}{3}, \frac{75}{3}, \frac{93}{3}$ i.e. $19,25,31$

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MCQ 21 Mark

If the marginal revenue function of a commodity is $\mathrm{MR}=2 \mathrm{x}-9 \mathrm{x}^{2}$, then he revenue function is

A
2-18x
✓
$x^{2}-3 x^{3}$
C
$2 x^{2}-9 x^{3}$
D
$18+x^{2}-3 x^{3}$

Answer

Correct option: B.

$x^{2}-3 x^{3}$

(B) $x^{2}-3 x^{3}$
Explanation: Given MR $=2 \mathrm{x}-9 \mathrm{x}^{2}$
$\therefore \mathrm{R}(x)=\int\left(2 x-9 x^{2}\right) d x$
$\Rightarrow \mathrm{R}(\mathrm{x})=\mathrm{x}^{2}-3 \mathrm{x}^{3}+\mathrm{k}$
We know that when $\mathrm{x}=0, \mathrm{R}(\mathrm{x})=0$
$\Rightarrow 0-0+\mathrm{k}=0 \Rightarrow \mathrm{k}=0$
$\therefore \mathrm{R}(\mathrm{x})=\mathrm{x}^{2}-3 \mathrm{x}^{3}$

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MCQ 31 Mark

The assumed hypothesis which is tested for rejection considering it to be true is called

A
true hypothesis
B
simple hypothesis
✓
null hypothesis
D
alternative hypothesis

Answer

Correct option: C.

null hypothesis

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MCQ 41 Mark

The solution set of system of linear inequalities $2(x+1) \leq x+5,3(x+2)>2-x, x \in R$ is

A
$[-1,3)$
B
$(-1,3)$
C
$[-1,3]$
✓
$(-1,3]$

Answer

Correct option: D.

$(-1,3]$

(D) $(-1,3]$
Explanation: (-1, 3]

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MCQ 51 Mark

Corner points of the feasible region determined by the system of linear constraints $(0,3),(1,1)$ and $(3,0)$. Let $z$$=p x+q y$, where $p, q>0$. Condition on $p$ and $q$ so that the minimum of $z$ occurs at $(3,0)$ and $(1,1)$ is

A
$p=3 q$
B
$p=2 q$
C
$p=q$
✓
$2 \mathrm{p}=\mathrm{q}$

Answer

Correct option: D.

$2 \mathrm{p}=\mathrm{q}$

(D) $2 \mathrm{p}=\mathrm{q}$
Explanation: We have $\mathrm{Z}=\mathrm{px}+\mathrm{qy}$, $\operatorname{At}(3,0) \mathrm{Z}=3 \mathrm{p}$...(i)
At $(1,1) \mathrm{Z}=\mathrm{p}+\mathrm{q} . .$. (ii)
Therefore, from (i) and (ii):
We have: $\mathrm{p}=\frac{q}{2}$
$2 \mathrm{p}=\mathrm{q}$

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MCQ 61 Mark

In a 100 m race $A$ and $B$ are two participants. If $A$ runs at 5 kilometer per hour and $A$ gives B a start of 8 m and still beats him by 8 seconds, then the speed of $B$ is:

A
$5.15 \mathrm{~km} / \mathrm{hr}$
B
$4.4 \mathrm{~km} / \mathrm{hr}$
✓
$4.14 \mathrm{~km} / \mathrm{hr}$
D
$4.25 \mathrm{~km} / \mathrm{hr}$

Answer

Correct option: C.

$4.14 \mathrm{~km} / \mathrm{hr}$

(C) $4.14 \mathrm{~km} / \mathrm{hr}$
Explanation: A's Speed $=\frac{\text { Distance }}{\text { Time Travelled }}$
$\Rightarrow$ A's Speed $=5 \mathrm{kmph}=\frac{100 \mathrm{~m}}{\text { Time Travelled }}$
$\Rightarrow$ Total time taken by A to complete $100 \mathrm{~m}=\frac{100}{\left(\frac{5 \times 1000}{3600}\right)}$ seconds $=72$ seconds
$\Rightarrow$ B's Speed $=\frac{\text { Distance Travelled by B }}{\text { Time Taken byB }}=\frac{\frac{(100-8)}{\frac{(7+-8)}{3600}}}{\frac{(7 m p h}{}=\frac{92 \times 36}{800} \mathrm{kmph}=4.14 \mathrm{kmph},{ }^{2} \mathrm{k}}$

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MCQ 71 Mark

$x$ and $b$ are real numbers. If $b>0$ and $|x|>b$, then

A
$x \in(-b, b)$
B
$x \in[-\infty$, b)
C
$x \in(-b, \infty)$
✓
$x \in(-\infty,-b) \cup(b, \infty)$

Answer

Correct option: D.

$x \in(-\infty,-b) \cup(b, \infty)$

(D) $x \in(-\infty,-b) \cup(b, \infty)$
Explanation: $x \in(-\infty,-b) \cup(b, \infty)$

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MCQ 81 Mark

In what ratio must rice at ₹ 29.30 per kg be mixed with rice at ₹ 30.80 per kg so that the mixture be worth ₹ 30 per kg?

A
$7: 8$
B
$3: 8$
C
$8: 3$
✓
$8: 7$

Answer

Correct option: D.

$8: 7$

(D) $8: 7$
Explanation: 8:7

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MCQ 91 Mark

Solution of the differential equation $x \frac{d y}{d x}+2 y=\mathrm{x}^{2}$ is

A
$y=\frac{x^{2}}{4}+\mathrm{C}$
✓
$y=\frac{x^{4}+\mathrm{C}}{4 x^{2}}$
C
$y=\frac{x^{2}+C}{4 x^{2}}$
D
$y=\frac{x^{2}+C}{x^{2}}$

Answer

Correct option: B.

$y=\frac{x^{4}+\mathrm{C}}{4 x^{2}}$

(B) $y=\frac{x^{4}+\mathrm{C}}{4 x^{2}}$
Explanation: $\frac{d y}{d x}+\frac{2}{x} y=x \Rightarrow$ I.F. $=e^{\int \frac{2}{x} d x}=e^{2 \log x}=\mathrm{x}^{2}$
$\therefore$ Solution is $\mathrm{y} \cdot \mathrm{x}^{2}=\int \mathrm{x} \cdot \mathrm{x}^{2} \mathrm{dx}+\mathrm{C}_{1}$
$\mathrm{y} \cdot \mathrm{x}^{2}=\frac{x^{4}}{4}+\mathrm{C}_{1} \Rightarrow \mathrm{y}=\frac{x^{4}+C}{4 x^{2}}$

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MCQ 101 Mark

A pipe A can fill a tank in 25 minutes and pipe B can empty the full tank in 50 minutes. The time taken by two pipes to fill the tank is:

A
20 minutes
B
30 minutes
✓
50 minutes
D
10 minutes

Answer

Correct option: C.

50 minutes

(C) 50 minutes
Explanation: Part of tank filled by A and B in 1 minute $=\frac{1}{25}-\frac{1}{50}$
$=\frac{2-1}{50}$
$=\frac{1}{50}$
$\because \frac{1}{50}$ part of tank is filled in 1 minute
$\therefore 1$ part of tank is filled in 50 minute
Hence, time taken by two pipe to fill the tank $=50$ minute

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MCQ 111 Mark

The degree of the differential equation $\frac{d^{2} y}{d x^{2}}+3\left(\frac{d y}{d x}\right)^{2}=x^{2} \log \left(\frac{d^{2} y}{d x^{2}}\right)$ is

A
1
B
3
C
2
✓
not defined

Answer

Correct option: D.

not defined

(D) not defined
Explanation: As the term $\log \left(\frac{d^{2} y}{d x^{2}}\right)$ is not a polynomial in $\frac{d^{2} y}{d x^{2}}$. So, the degree of the given differential equation is not defined.

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MCQ 121 Mark

A random variable ' X ' has the following probability distribution:

X	0	1	2	3	4	5	6	7
P(X)	0	2k	2k	3k	$k ^2$	2$k ^2$	7$k ^2$	2k

The value of $k$ is

A
-1
B
1
C
$-\frac{1}{10}$
✓
$\frac{1}{10}$

Answer

Correct option: D.

$\frac{1}{10}$

(D) $\frac{1}{10}$
Explanation: $0+2 \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+\mathrm{k}^{2}+2 \mathrm{k}^{2}+7 \mathrm{k}^{2}+2 \mathrm{k}=1$
$\Rightarrow 10 \mathrm{k}^{2}+9 \mathrm{k}-1=0 \Rightarrow(10 \mathrm{k}-1)(\mathrm{k}+1)=0$
$\Rightarrow k=\frac{1}{10},-1$ but $\mathrm{k} \neq-1$
$\Rightarrow k=\frac{1}{10}$

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MCQ 131 Mark

In a binomial distribution, the probability of getting success is $\frac{1}{4}$ and standard deviation is 3 . Then, its mean is

A
10
B
6
C
8
✓
12

Answer

Correct option: D.

(D) 12
Explanation: $p=\frac{1}{4}, \sqrt{n p q}=3$
$\Rightarrow q=\frac{3}{4}, \mathrm{npq}=9$
$\Rightarrow$ Mean $=\mathrm{np}=\frac{9}{q}$
$\Rightarrow$ Mean $=9 \times \frac{4}{3}=12$

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MCQ 141 Mark

For the curve $\sqrt{x}+\sqrt{y}=1, \frac{d y}{d x}$ at $\left(\frac{1}{4}, \frac{1}{4}\right)$ is

A
2
✓
-1
C
-2
D
1

Answer

Correct option: B.

-1

(B) -1
Explanation: $\sqrt{x}+\sqrt{y}=1$
Differentiating with respect to x ,
$\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0$
$\frac{d y}{d x}=-\sqrt{\frac{y}{x}}$
$\frac{d y}{d x}\left(\frac{1}{4^{-1} \frac{1}{4}}\right)=-\sqrt{\frac{\frac{1}{4}}{\frac{1}{4}}}=-1$

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MCQ 151 Mark

Any feasible solution which maximizes or minimizes the objective function is called:

A
An objective feasible solution
B
A reasonable feasible solution
✓
An optimal feasible solution
D
A regional feasible solution

Answer

Correct option: C.

An optimal feasible solution

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MCQ 161 Mark

A certain sum of money amounts to ₹ 5832 in 2 years at $8 \%$ p.a. compound interest. The sum invested is

A
₹ 5280
B
₹ 5400
C
₹ 5200
✓
₹ 5000

Answer

Correct option: D.

₹ 5000

(D) ₹ 5000
Explanation: Let sum invested be ₹ $x$, rate $=8 \%$, time $=2$ years
Amount = ₹ 5832
$\therefore 5832=x\left(1+\frac{8}{100}\right)^{2}$
$\Rightarrow 5832-\mathrm{x} \times\left(\frac{27}{25}\right)^{2}$
$\Rightarrow x=\frac{5832 \times 25 \times 25}{27 \times 27}=5000$
$\therefore$ Sum invested =₹$ 5000$

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MCQ 171 Mark

Which of the following is an assumption underlying the use of the t-distribution?

✓
The sample size are drawn from a normally distributed population.
B
Sample standard deviation is an unbiased estimate of the population variance.
C
All of these
D
The variance of the population is known.

Answer

Correct option: A.

The sample size are drawn from a normally distributed population.

(A) The sample size are drawn from a normally distributed population.
Explanation: The sample size are drawn from a normally distributed population.

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MCQ 181 Mark

If the matrix $\left[\begin{array}{ccc}0 & -1 & 3 x \\ 1 & y & -5 \\ -6 & 5 & 0\end{array}\right]$ is skew-symmetric, then

✓
$x=2, y=0$
B
$x=2, y=-1$
C
$x=-2, y=0$
D
$x=-2, y=1$

Answer

Correct option: A.

$x=2, y=0$

(A) $x=2, y=0$
Explanation: Let $\mathrm{A}=\left[\begin{array}{rrr}0 & -1 & 3 x \\ 1 & y & -5 \\ -6 & 5 & 0\end{array}\right]$, then $\mathrm{A}^{\prime}=-\mathrm{A}$
$\Rightarrow\left[\begin{array}{rrr}0 & 1 & -6 \\ -1 & y & 5 \\ 3 x & -5 & 0\end{array}\right]=\left[\begin{array}{rrr}0 & 1 & -3 x \\ -1 & -y & 5 \\ 6 & -5 & 0\end{array}\right]$
$\Rightarrow-3 x=-6 \Rightarrow x=2, y=-y \Rightarrow 2 y=0 \Rightarrow y=0$
$\therefore \mathrm{x}=2, \mathrm{y}=0$
$\therefore$ Option ( $\mathrm{x}=2, \mathrm{y}=0$ ) is the correct answer.

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Applied Maths MCQ

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