Directions (46-47): Read the following quadric equation carefully and answer the questions given below.

Equation 1: $a{x}^2+bx+c=0$
Equation 2: $d{y}^2+ey+c=0$

Note:
(i) c is a single digit prime number greater than 2
(ii) d and b are two digits prime number less than 20
(iii) d is greater than 11
(iv) b is greater than d
(v) Smallest roots of both the equation are same
(vi) No root is irrational
(vii) e = b + 1
(viii) 3c is greater than b

Find the value of ‘a’ .

The correct answer is 12.

Let us analyze the given information step-by-step to find the values of the variables and solve for $a$:

Step 1: Determine the value of $c$
According to note (i), $c$ is a single-digit prime number greater than 2. The single-digit prime numbers are 2, 3, 5, and 7. Since $c>2$, the possible values for $c$ are:
$c\in \{3,5,7\}$

Step 2: Determine the values of $d$ and $b$
According to note (ii), $d$ and $b$ are two-digit prime numbers less than 20. The two-digit prime numbers less than 20 are 11, 13, 17, and 19.
According to note (iii), $d>11$. Therefore, the possible values for $d$ are 13, 17, or 19.
According to note (iv), $b>d$. This means that $b$ must be strictly greater than $d$ from the remaining primes. This leaves the following possible pairs for $(d,b)$:
1) $d=13,b=17$
2) $d=13,b=19$
3) $d=17,b=19$

Step 3: Apply the condition $3c>b$
According to note (viii), $3c>b$.
If $b=17$, then $3c>17\Rightarrow c>5.67$. Thus, the only possible value for $c$ is 7.
If $b=19$, then $3c>19\Rightarrow c>6.33$. Thus, the only possible value for $c$ is 7.
In both cases, we must have $c=7$.

Step 4: Check the roots of Equation 2 to find a valid pair
Equation 2 is given by:
$d{y}^2+ey+c=0$
According to note (vii), $e=b+1$. Let's test the potential pairs:

Case A: $d=13$, $b=17$
Here, $e=17+1=18$ and $c=7$.
Equation 2 becomes:
$13y^2+18y+7=0$
The discriminant of this quadratic equation is:
$D=e^2-4dc={18}^2-4\left(13\right)\left(7\right)=324-364=-40$
Since the discriminant is negative, the roots are imaginary, which violates note (vi) (roots must be rational/real). Thus, this case is invalid.

Case B: $d=13$, $b=19$
Here, $e=19+1=20$ and $c=7$.
Equation 2 becomes:
$13y^2+20y+7=0$
The discriminant is:
$D={20}^2-4\left(13\right)\left(7\right)=400-364=36$
Since 36 is a perfect square, the roots are rational, which satisfies note (vi). The roots are:
$y=\frac{-20\pm \sqrt{36}}{2\left(13\right)}=\frac{-20\pm 6}{26}$
The two roots are:
$y_1=\frac{-20-6}{26}=-1$
$y_2=\frac{-20+6}{26}=-\frac{7}{13}$
Comparing the two values, the smallest root is $-1$ (since $-1<-\frac{7}{13}$).

Step 5: Find the value of $a$
According to note (v), the smallest roots of both equations are the same. Therefore, the smallest root of Equation 1 must also be $-1$.
Equation 1 is:
$a{x}^2+bx+c=0$
Substituting $b=19$, $c=7$, and the root $x=-1$ into Equation 1:
$a{(-1)}^2+19(-1)+7=0$
$a-19+7=0$
$a-12=0$
$a=12$

Thus, the value of $a$ is 12.