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 ‘d * e’ .

To find the value of $d*e$, let us analyze the given conditions step-by-step:

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 greater than 2 are 3, 5, and 7.
So, $c\in \{3,5,7\}$.

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 prime numbers between 10 and 20 are 11, 13, 17, and 19.
According to Note (iii), $d$ is greater than 11. Therefore, $d\in \{13,17,19\}$.
According to Note (iv), $b$ is greater than $d$.
Since $b$ must also be a prime number less than 20, $d$ cannot be 19. 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: Analyze e and evaluate the target value (d * e)
According to Note (vii), we have $e=b+1$.
Let us calculate $d*e$ for each of the possible cases:
Case 1: If $d=13$ and $b=17$, then $e=17+1=18$.
$d*e=13*18=234$ (not in options).

Case 2: If $d=13$ and $b=19$, then $e=19+1=20$.
$d*e=13*20=260$ (matches Option 4).

Case 3: If $d=17$ and $b=19$, then $e=19+1=20$.
$d*e=17*20=340$ (not in options).

Thus, the correct pair must be $d=13$, $b=19$, and $e=20$.

Step 4: Verify with remaining conditions
According to Note (viii), $3c$ is greater than $b$.
Since $b=19$, we have:
$3c>19\Rightarrow c>6.33$
Since $c\in \{3,5,7\}$, the only value that satisfies this condition is $c=7$.

Now, let us substitute the values into Equation 2:
$d{y}^2+ey+c=0$
$13y^2+20y+7=0$
$13y^2+13y+7y+7=0$
$13y(y+1)+7(y+1)=0$
$(13y+7)(y+1)=0$

The roots are $y=-1$ and $y=-\frac{7}{13}$.
Both roots are rational, satisfying Note (vi). The smallest root is $-1$.

For Equation 1:
$a{x}^2+bx+c=0$
$a{x}^2+19x+7=0$

Since the smallest root of both equations is the same (Note v), $x=-1$ must be a root of Equation 1:
$a{(-1)}^2+19(-1)+7=0$
$a-19+7=0\Rightarrow a=12$

Substituting $a=12$ back to find the other root of Equation 1:
$12x^2+19x+7=0$
$12x^2+12x+7x+7=0$
$(12x+7)(x+1)=0$

The roots are $x=-1$ and $x=-\frac{7}{12}$.
Both roots are rational, and the smallest root is $-1$, which is consistent with all notes.

Therefore, the value of $d*e$ is indeed $13*20=260$.