PGDBA 2019 Question Paper

Let's tabulate the information in the graph ,we get 

From the table, it is evident that there are 20 members who got at least 320

Hence C is the correct answer.

Let's tabulate the information in the graph ,we get 

 No. of labourers who receive a wage less than 250 will be at least 10 but not more than 40 in which case all the people in the range 200-240 will be receiving a wage less than 250.

So if all the people in the range  >= 240 - <280 are included then to a maximum there will be 40 labourers to whom the wage will be less than 250.

Hence B is the correct answer.

Let's tabulate the information in the graph ,we get 

The total number of labourer = 100

50% of the total labourers is 50 who will fall in the range >=280 - <320

$$\therefore$$ The maximum wage such that at least 50% of the labourers definitely earn more than that is 280

Hence B is the correct answer.

Let's solve the options one by one,

In 2017

Option A:The percentage increase of Foreign Tourists in FEB = 570-500/500 = 14% 

Option B:The percentage increase of Foreign Tourists in JUL = 660-600/600 = 10%

Option C:The percentage increase of Foreign Tourists in SEP = 875-700/700 = 25%

Option D:The percentage increase of Foreign Tourists in DEC = 1090-1000/1000 = 9%

From the above values it is clear that the percentage increase of Foreign Tourists in DEC is minimum.

Hence D is the correct answer

Let's solve the options one by one,

In 2017

Option A:The percentage increase of Foreign Tourists in MAR= 720-570/570 = 26.32% 

Option B:The percentage increase of Foreign Tourists in SEP= 875-696/696 = 25.72% 

Option C:The percentage increase of Foreign Tourists in OCT= 984-875/875 = 12.46% 

Option D:The percentage increase of Foreign Tourists in NOV= 1170-984/984 = 18.90% 

It is clear that the percentage increase of Foreign Tourists in MAR is maximum

Hence A is the correct answer.

Logarithmic function is defined only for positive values 

So $${\frac{2 - x}{9 - x^2}}$$ >0 and $$9-x^{2}$$ should not equal 0.

x $$\in$$ (-3,2)$$\cup$$ (3,$$\infty$$)

D is the correct answer.

Let's represent the system of linear equations
$$2x+ y+7z = a$$
$$6x-2y+11z = b$$
$$2x-y+3z = c$$ in the Matrix Equation form

$$\begin{bmatrix}2 & 1 &7 & a  \\6 & -2 & 11 & b &\\2 & -1 &3 & c \end{bmatrix}$$

Order of the matrix = 3

Now represent the matrix in echelon form 

$$\begin{bmatrix}2 & 1 &7 & a  \\0 & -5 & -10 & b-3a &\\0 & -2 &-4 & c-a \end{bmatrix}$$

$$\begin{bmatrix}2 & 1 &7 & a  \\6 & -2 & 11 & b &\\0 & 0 &0 & 5c-5a-2b+6a \end{bmatrix}$$

Rank of the matrix = No of Non-zero rows after transforming the matrix into echelon form

The system of linear equations will have infinite solutions only if Rank of the matrix is not equal to Order of the Matrix

$$\therefore$$ 5c-5a-2b+6a=0

a-2b+5c=0

D is the correct answer.

$$\frac{1}{\alpha\ +1}+\frac{1}{\beta\ +1}=\frac{\left(\alpha\ +\beta\ +2\right)}{\alpha\ \beta\ +\alpha\ +\beta\ +1}$$

Since $$\alpha\ ,\beta\ $$ are roots of the given equation:

$$\alpha+\beta\ =-3$$

$$\alpha\times\beta\ =-3$$

Then we have: $$\frac{\left(\alpha\ +\beta\ +2\right)}{\alpha\ \beta\ +\alpha\ +\beta\ +1}=\frac{\left(-3+2\right)}{-3-3+1}=\frac{1}{5}$$

Let $$e^x+e^{-x}=P$$

$$P^3+3P^2+3P-7 = 0$$

$$\left(P-1\right)(P^2+4P+7)=0$$

P = 1 and the other two roots are imaginary roots 

$$e^x + e^{-x} = 1$$ but the minimum value of $$e^x + e^{-x} is 2$$, as $$e^x$$ is a positive number and sum of a positive number and its reciprocal is at least 2.

No. of real roots is 0.

Hence A is the correct answer.

$$e^k = \sum_{n = 0}^\infty \frac{k^n}{n!} = 1 + k + \frac{k^2}{2!} + \frac{k^3}{3!}+ ....$$

We are given,

 $$x = -\frac{1}{1!}\cdot\frac{3}{4} + \frac{1}{2!}\cdot(\frac{3}{4})^2 -\frac{1}{3!}\cdot(\frac{3}{4})^3 +....$$

This can be written as,

$$e^{-\frac{3}{4}}\ =\ 1\ -\ \frac{1}{1!}\left(\frac{3}{4}\right)\ +\ \frac{1}{2!}\left(\frac{3}{4}\right)^2-\frac{1}{3!}\left(\frac{3}{4}\right)^3....$$

$$x\ =\ -\ \frac{1}{1!}\left(\frac{3}{4}\right)\ +\ \frac{1}{2!}\left(\frac{3}{4}\right)^2-\frac{1}{3!}\left(\frac{3}{4}\right)^3....\ \ =\ e^{-\frac{3}{4}}\ -\ 1$$

We also know that,

$$\log\left(1\ +\ k\right)\ =\ k\ -\ \frac{k^2}{2}\ +\ \frac{k^3}{3}+\ ....$$

We can also write,

$$y=x-\frac{x^2}{2}+\frac{x^3}{3}-...\ =\ \log\left(1\ +\ x\right)$$

Substituting the value of x, we get,

$$y\ =\ \log\left(1\ +\ e^{-\frac{3}{4}}\ -\ 1\right)\ =\ \log\left(e^{-\frac{3}{4}}\right)\ =\ -\frac{3}{4}$$

Hence, A is the correct answer.