MCQ

MCQ 11 Mark

The best-fitted trend line is one for which sum of squares of residuals or errors is:

A
Maximum
✓
Minimum
C
Positive
D
Negative

Answer

Correct option: B.

Minimum

(b) Minimum
Explanation
The line is termed as the line of best fir from which the sum of squares of distances from the points is minimized.

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MCQ 21 Mark

$\int \frac{(\log x)^5}{x}$ is equal to

A
$\frac{\log x^6}{6}+C$
B
$\frac{(\log x)^6}{3 x^2}+C$
✓
$\frac{(\log x)^6}{6}+C$
D
$\frac{\log x^6}{3 x^2}+C$

Answer

Correct option: C.

$\frac{(\log x)^6}{6}+C$

(c) $\frac{(\log x)^6}{6}+C$
Explanation
Put $\log x = t \Rightarrow \frac{1}{x} dx = dt$
$
\therefore \int \frac{(\log x)^5}{x} d x=\int t^5 d t=\frac{t^6}{6}+C=\frac{(\log x)^6}{6}+C
$

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MCQ 31 Mark

A simple random sample consists of four observations 1, 3, 5, 7. What is the point estimate of population standard deviation?

A
3.1
B
2.3
C
2.87
✓
2.58

Answer

Correct option: D.

2.58

(d) 2.58

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MCQ 41 Mark

How many of the following points satisfy the inequality 2x - 3y > - 5?
(1, 1), (-1, 1), (1, -1 ), (-1, - 1), (- 2, 1), (2, -1 ), (-1, 2) and (-2, -1).

✓
5
B
3
C
4
D
6

Answer

Correct option: A.

(a) 5
Explanation:
(1, 1), (1, -1), (-1 , -1), (2, -1) and (-2, -1) satisfy the inequality 2x - 3y > - 5.

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MCQ 51 Mark

Comer points of the feasible region for an LPP are : (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let z = 4x + 6y be the objective function. Then, Max. z - Min. z =

A
48
B
42
✓
60
D
18

Answer

Correct option: C.

Corner points	Z=4x+6y
(0,2)	12 (Min.)
(3,0)	12 (Min.)
(6,0)	24
(6,8)	72 (Max.)
(0,5)	30

Maximum of F - Minimum of F = 72 - 12 = 60

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MCQ 61 Mark

A man can row a boat in still water at 15 km/hr and speed of water current is 5 km/hr. The distance covered by the boat downstream in 24 minutes is

A
4 km
B
6 km
✓
8 km
D
16 km

Answer

Correct option: C.

8 km

(c) 8 km
Explanation:
The speed of boat in still water = 15 km/hr
Speed of water current = 5 km/hr
$\therefore$ Speed in down stream $=15+5=20 km / hr$
Time given $=24 min=\frac{24}{60} hr =\frac{2}{5} hr$
$\therefore$ Distance travelled $=$ speed $\times$ times
$=20 \times \frac{2}{5}=8 km$

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MCQ 71 Mark

If $\frac{|x-2|}{x-2} \geq 0$, then

A
$x \in(-\infty, 2)$
B
$x \in[2, \infty)$
✓
$x \in(2, \infty)$
D
$x \in(-\infty, 2]$

Answer

Correct option: C.

$x \in(2, \infty)$

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MCQ 81 Mark

If x is the least non-negative integer satisfying 218 $\equiv x(\bmod 7)$, then $x^2+1$ is equal to

A
50
B
1
✓
2
D
5

Answer

Correct option: C.

(c) 2
Explanation:
From the definition: $a \equiv b$ (mod m)
a is said to be congruent to b modulo m, if m divides (a - b) or (a - b) is divisble by m.
$
\begin{array}{l}
\Rightarrow 218 \equiv x(\bmod 7) \\
\Rightarrow \frac{(218-x)}{7}
\end{array}
$
for this to be hold true, $x$ must be 1 .
$
\Rightarrow x^2+1=1^2+1=2
$

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MCQ 91 Mark

The trace of the matrix $A =\left[\begin{array}{rrr}1 & -5 & 7 \\ 0 & 7 & 9 \\ 11 & 8 & 9\end{array}\right]$ is

✓
17
B
25
C
3
D
12

Answer

Correct option: A.

(a) 17
Explanation
As the trace of a matrix is the sum of on – diagonal elements,
So, 1 + 7 + 9 = 17
Trace = 17

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MCQ 101 Mark

Two pipes A and B can fill a cistern in 10 minutes and 15 minutes respectively. Both the pipes are opened together, but after 3 minutes pipe B in turned off. How much time will the cistern take to be full?

✓
8 minutes
B
6 minutes
C
11 minutes
D
12 minutes

Answer

Correct option: A.

8 minutes

(a) 8 minutes
Explanation
In one min, (A + B) fill the cistern
$
=\frac{1}{10}+\frac{1}{15}=\frac{1}{6} \text { th }
$
In 3 mins. $( A + B )$ fill the cistern
$
=\frac{3}{6}=\frac{1}{2} \text { th }
$
Remaining part $=1-\frac{1}{2}=\frac{1}{2}$
$\because \frac{1}{10}$ th part in filled by A in one min.
$\therefore \frac{1}{2}$ nd part is filled by A in $10 \times \frac{1}{2}=5 min$
Total time $=3+5=8 min$

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MCQ 111 Mark

The order of the differential equation of all circles of given radius a is

A
1
B
4
C
3
✓
2

Answer

Correct option: D.

(d) 2
Explanation
Equation of all the circles of radius a is
(x - h)² + (y - k)² = a2
where h, k are arbitrary constants.
So, the order of differential equation is 2.

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MCQ 121 Mark

If X has a Poisson distribution such that P(X = 1) = P(X = 2) and e^-2 = 0.1353, then P(X = 4) is

A
0.0213
✓
0.0902
C
0.9098
D
0.9787

Answer

Correct option: B.

0.0902

(b) 0.0902
Explanation:
Given $P ( X =1)= P ( X =2)$
$
\begin{array}{l}
\Rightarrow \lambda e^{-\lambda}=\frac{\lambda^2 e^{-\lambda}}{2!} \\
\Rightarrow \lambda^2-2 \lambda=0 \Rightarrow \lambda=0,2 \\
\Rightarrow \lambda=2
\end{array}
$
Now, $P ( X =4)=\frac{2^4 \cdot e^{-2}}{4!}=\frac{16 \times 0.1353}{24}=0.0902$

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MCQ 131 Mark

In a series of three trials, the probability of two successes is 9 times the probability of three successes. Then, the probability of success in each trial is

A
$\frac{1}{3}$
B
$\frac{3}{4}$
C
$\frac{1}{2}$
✓
$\frac{1}{4}$

Answer

Correct option: D.

$\frac{1}{4}$

(d) $\frac{1}{4}$
Explanation
Given $n =3$ and $P ( X =2)=9 P ( X =3)$.
$
\begin{array}{l}
\text { So, }{ }^3 C_2 p^2 \cdot q=9 \times{ }^3 C_3 \cdot p^3 \\
\Rightarrow 3 p^2 q=9 p^3 \Rightarrow 3 p^2(q-3 p)=0 \\
\Rightarrow q=3 p \\
\because p+q=1 \Rightarrow p+3 p=1 \Rightarrow p \frac{1}{4}
\end{array}
$

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MCQ 141 Mark

If $S =\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, then $\operatorname{adj} A$ is

✓
$\left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right]$
B
$\left[\begin{array}{ll}d & c \\ b & a\end{array}\right]$
C
$\left[\begin{array}{rr}-d & -b \\ -c & a\end{array}\right]$
D
$\left[\begin{array}{ll}d & b \\ c & a\end{array}\right]$

Answer

Correct option: A.

$\left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right]$

(a) $\left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right]$
Explanation
$S =\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
$\begin{array}{l} M _{11}= d \Rightarrow A _{11}= d \\ M _{12}= c \Rightarrow A _{12}=- c \\ M _{21}= b \Rightarrow A _{21}=- b \\ M _{22}= a \Rightarrow A _{22}= a \\ \Rightarrow \operatorname{Adj}( A )=\left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right]\end{array}$

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MCQ 151 Mark

The solution set of the inequation 2 x + y > 5 is

A
whole xy-plane the points lying on the line 2x + y = 5
✓
open half-plane not containing the origin
C
whole xy-plane except the points lying on the line 2x + y = 5
D
half plane that contains the origin

Answer

Correct option: B.

open half-plane not containing the origin

(b) open half-plane not containing the origin
Explanation
Given inequation is 2x + y > 5
Now, we convert the inequation into an intercept line equation form, we can clearly see the intercepts of the inequation on x-axis and y-axis.
2x + y > 5
[dividing the whole inequation by 5]
$\begin{array}{l}\frac{2 x}{5}+\frac{y}{5}>\frac{5}{5} \\ \frac{x}{\frac{5}{2}}+\frac{y}{5}>1 \\ \frac{x}{2.5}+\frac{y}{5}>1\end{array}$
Therefore, from the above inequation, we can say that 2.5 and 5 are the intercepts of the x-axis and y-axis respectively. Now by plotting these on the graph, we can clearly see the graph of the inequation.

From the graph, it is clear that the solution set of the inequation.
2x + y > 5 is the open half-plane not containing the origin.

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MCQ 161 Mark

The value of a machine purchased two years ago depreciates at the annual rate of 10%. If its present value is ₹97,200, then its value after 3 years is

A
₹ 80,859 approx
✓
₹ 70,859 approx
C
₹ 88,509 approx
D
₹ 71,859 approx

Answer

Correct option: B.

₹ 70,859 approx

(b) ₹ 70,859 approx
Explanation
Given, P = ₹ 97,200, i = 10% p.a.
$
\Rightarrow i=\frac{10}{100}=0.1
$
So, value after 3 years
$
=97,200 \times(1-0.1)^3
$
$=97,200 \times 0.729$
= ₹ 70,858.80
= ₹ 70,859 (approx.)

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MCQ 171 Mark

A statement made about a population parameter for testing purpose is called

A
statistic
B
level of significance
✓
hypothesis
D
parameter

Answer

Correct option: C.

hypothesis

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MCQ 181 Mark

If $A=\left[\begin{array}{rr}2 & 3 \\ 5 & -2\end{array}\right]$ be such that $A^{-1}=k A$, then $k$ equals

A
19
✓
$\frac{1}{19}$
C
$-\frac{1}{19}$
D
-19

Answer

Correct option: B.

$\frac{1}{19}$

(b) $\frac{1}{19}$
Explanation:
A $=\left[\begin{array}{ll}2 & 3 \\ 5 & -2\end{array}\right]$
Using adjoint matrix
$
\begin{array}{l}
A^{-1}=\frac{-1}{19}\left[\begin{array}{ll}
-2 & -3 \\
-5 & 2
\end{array}\right] \\
A^{-1}=\frac{1}{19}\left[\begin{array}{cc}
2 & 3 \\
5 & -2
\end{array}\right]=\frac{1}{19} A \\
k=\frac{1}{19}
\end{array}
$

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Applied Maths MCQ

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