PGDBA 2017 Question Paper

From the table , 

Total population in 1901 = Rural population in 1901 + Urban population in 1901

=225+25 = 250

Total population in 1981 = Rural population in 1981 + Urban population in 1981

=450+150 = 600

Percentage increase in population from 1901 to 1981 is $$\frac{600-250}{250}$$ * 100

=350/250 *100

=140%

Hence C is the correct answer.

Since area is not known density of urban areas cannot be determined.

Hence D is the correct answer.

Let's solve the options one by one .

Option A : 1971-81 $$\frac{150-100}{100}$$ = 0.5

Option B : 1961-71 $$\frac{100-79}{79}$$ = 0.266

Option C : 1951-61 $$\frac{79-62}{62}$$ = 0.274

Option D : 1941-1951 $$\frac{62-44}{44}$$ = 0.41

Among the above values,the largest rate of increase in urban population occurred during 1971-81

Hence A is the correct answer.

from the table 

Rate of urban growth per year over the period 1901-81 = $$\frac{150}{25*9}$$

=$$\frac{150}{225}$$

=0.66 million 

Hence A is the correct answer.

Let's solve the options one by one ,

Option A :Rate of increase of urban population during 1951-61 = $$\frac{79-62}{62}$$ = 0.274

Option B :Rate of increase of urban population during 1941-51 = $$\frac{62-44}{44}$$ = 0.409

Option C :Rate of increase of urban population during 1931-41 = $$\frac{44-33}{33}$$ = 0.33

Option D :Rate of increase of urban population during 1921-31 = $$\frac{12-11}{11}$$ = 0.091

Among the above values smallest rate of increase was during 1951-61

Hence A is the correct answer.

Let '$$\alpha$$' be the common root which satisfies both the equations.
$$\alpha^{2}$$ + $$\alpha$$ + a = 0 and $$\alpha^{2}$$ + a$$\alpha$$ + 1 = 0
On subtracting the equations,
a($$\alpha$$ - 1) + 1- a=0
$$\alpha$$ = 1
Substitute the value of $$\alpha$$ in any of the equations, we get 
a=-2

C is the correct answer.

Centre and radius of a circle $$x^2 + y^2 +2gx +2fy + c = 0$$ is (-g,-h) and $$\sqrt{g^{2}+h^{2}-c}$$
Centre and the radius of $$C_1$$ is (2,2) and $$\sqrt{2}$$
Centre and the radius of $$C_2$$ is (5,5) and $$\sqrt{50-k}$$
Distance between the centres of the two circles = $$\sqrt{18}$$
According to the properties of the circle 
If two circles have two common tangents then ,
|Difference of the radii|  < Distance between the centres of the two circles < |Sum of the radii|
$$\sqrt{50-k}$$ - $$\sqrt{2}$$ <  $$\sqrt{18}$$ < $$\sqrt{50-k}$$ + $$\sqrt{2}$$
Let's look at the options one by one ,

Option A : let k= 1  |7-$$\sqrt{2}$$| < 3$$\sqrt{2}$$ < |7+$$\sqrt{2}$$|  ------------------->  Does not satisfy 
Option B : let k= 25 |5-$$\sqrt{2}$$| < 3$$\sqrt{2}$$ < |5+$$\sqrt{2}$$|  -------------------> Satisfied
Option C : let k= 49  |1-$$\sqrt{2}$$| < 3$$\sqrt{2}$$ < |1+$$\sqrt{2}$$|  ------------------->  Does not satisfy  

Hence B is the correct answer.

F(x)=

         $$\lim_{x \rightarrow -1^{-}}$$ (2x-1) = 2(-1)-1 = -3
         $$\lim_{x \rightarrow -1^{+}}$$ $$(x^{2} + 1)$$ = 1+1 = 2
Since the left-hand limit and right-hand limit are not equal,the function is discontinuous at -1.
           $$\lim_{x \rightarrow 1^{-}}$$ $$(x^{2} + 1)$$ = 1+1 = 2
          $$\lim_{x \rightarrow -1^{+}}$$ x+1 = 1+1 = 2
Since the left-hand limit and right-hand limit are equal, the function is continuous at 1.
So the function is continuous everywhere except at -1.

C is the correct answer.

$$3 + 7 + 13 + 21 + 31 + 43 +...$$
The difference of two consecutive terms is  4,6,8,10
Since the difference of two consecutive terms is in AP, the general term will be a quadratic expression in n
So the general term is t(n) = $$an^{2}+bn+c$$
t(1)= a+b+c = 3
t(2) = 4a+2b+c = 7
t(3) = 9a+3b+c = 13
we get the values of a ,b,c as 1,1,1 respectively 
General term = $$1n^{2}+n+1$$
$$\sum1n^{2}+n+1$$ = n($$\frac{n^{2}+3n+5}{3}$$) 
Substitute n = 50
=50*885

Hence D is the correct answer.