Linear Equations - Integral Solutions

a(a + b + 1) = 14 ...... (1)

b(a + b + 1) = 28 ...... (2)

$$\frac{a}{b}=\frac{1}{2}$$

b = 2a

Substituting in (1), we get

a(3a + 1) = 14

$$3a^2+a-14=0$$

$$3a^2-6a+7a-14=0$$

$$3a\left(a-2\right)+7\left(a-2\right)=0$$

Given, a and b are natural numbers.

Therefore, a = 2 and b = 4

2a + b = 2(2) + 4 = 8

The answer is option A.

It is given, y(x + z) = 19

y cannot be 19. 

If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.

Therefore, y = 1 and x + z = 19

It is given, z(x + y) = 51

z can take values 3 and 17

Case 1:

If z = 3, y = 1 and x = 16

xyz = 3*1*16 = 48

Case 2:

If z = 17, y = 1 and x = 2

xyz = 17*1*2 = 34

Minimum value xyz can take is 34.

Let the number of questions attempted be x+y out of which x are correct and y are incorrect and the number of questions unattempted be z.

It is given,

x + y + z = 75 ...... (1)

3x - y + z = 97 ...... (2)

(2)-(1) -> x - y = 11

(1)+(2) -> 2x + z = 86

z > x + y

z > 75 - z

z > 37.5

Minimum possible value of z is 38

2x + 38 = 86

2x = 48

x = 24

The maximum number of correct answers is 24.