PGDBA 2016 Question Paper

The number of students scoring less than 60 = 5+20+30=55

The number of students scoring less than 80 marks but not less than 40 marks = 30+15 = 45

There is a discrepancy in this question that appeared in PGDBA 2016. Assuming that the average in each range lies exactly in the middle, we can get the answer to be 51.25. But it has not been mentioned and hence, the question is incorrect.

let all the students in the group 40-60, score atleast 50,

Then the number off students who score atleast 50 = 30+15+10= 55

Percentage = (55/80)*100 = 68.75%

let all the students in the group 40-60, score less than 50,

Then the number off students who score atleast 50 = 15+10= 25

Percentage = (25/80)*100 = 31.25%

C is the correct answer.

Here we have to find the extreme cases

Let us consider all the students indicated by the graph 20-40 scored more than 30, 

Then the number of students scored less than 30 = 5

Percentage of students scored less than 30 = $$\frac{5}{80}$$ = 6.25%

Let us consider all the students indicated by the graph 20-40 scored less than 30,

Then the number of students scored less than 30 = 25

Percentage of students scored less than 30 = $$\frac{25}{80}$$ = 31.25%

B is the correct answer.

Let the six-letter word be ABCDEF

Number of possibilities for A = 11, B = 10, C = 5, D=5, E = 9, F = 8

Required number of ways = 11*10*5*5*9*8 = 198000

For the number to be divisible by 5, the last digit must be either 0 or 5

If 0 is the last digit then the first digit must also be 0, then it will be a 7 digit number.

So the last digit can only be 5.

If the last digit is 5, then the first digit is also 5.

We have select values of second, third, fourth digits only because fifth, sixth and seventh will be same as second, third, fourth digits.

Second, third and fourth digits can have 10 possibilities.

The required number of ways = 1*10*10*10 = 1000

A is the correct answer.

$$x^2 + 4 \mid x \mid - 4 = 0$$

Case 1:

x > 0

$$x^2 + 4  x - 4 = 0$$

x = -2+2$$\sqrt{2}$$, -2-2$$\sqrt{2}$$

Since x > 0, x = -2+2$$\sqrt{2}$$

Case 2:

x < 0

$$x^2 - 4 x - 4 = 0$$

x = 2+2$$\sqrt{2}$$, 2-2$$\sqrt{2}$$

Since x < 0, x = 2-2$$\sqrt{2}$$

Product of values of x

= (-2 + 2$$\sqrt{2}$$) * (2 - 2$$\sqrt{2}$$)

=  -12 + 8$$\sqrt{2}$$

= $$-4(\sqrt2 - 1)^2$$

If m = 10 => $$10_{C_6}\left(x^3\right)^4\left(1\right)^6=\ 210x^{12}$$

Hence $$10_{C_5}\left(x^3\right)^5\left(1\right)^5=\ 252x^{15}$$

$$\frac{m^2}{n^2}$$ = $$\frac{\frac{m*(2\alpha+ (m-1)d)}{2}}{\frac{n*(2\alpha+ (n-1)d)}{2}}$$

$$\frac{m}{n}$$ = $$\frac{(2\alpha+ (m-1)d)}{(2\alpha+ (n-1)d)}$$

$$m(2\alpha+ (n-1)d) = n(2\alpha+ (m-1)d))$$

$$(n-m)d=(n-m)2\alpha\ $$

d = 2$$\alpha$$

$$50^{th}$$ term = $$\alpha$$+49*2$$\alpha$$ = 99$$\alpha$$