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’ .

The correct answer is 260.

Step-by-step Explanation:

Let's analyze the given conditions step-by-step to find the values of the variables:

1. Identify the possible values of $c$, $d$, and $b$:
- From note (i), $c$ is a single-digit prime number greater than 2. The prime numbers meeting this condition are:
$c\in \{3,5,7\}$
- From note (ii), $d$ and $b$ are two-digit prime numbers less than 20. The prime numbers meeting this condition are:
$\{11,13,17,19\}$
- From note (iii), $d$ is greater than 11. Therefore, $d$ can be:
$d\in \{13,17,19\}$
- From note (iv), $b$ is greater than $d$ ($b>d$). This implies $d$ cannot be 19. Thus, $d$ can be 13 or 17, and $b$ can be 17 or 19.

2. Apply the condition $3c>b$:
- From note (viii), we have $3c>b$.
- Since the maximum possible value of $c$ is 7, we have $3c\le 3\times 7=21$.
- For $3c>b$ to hold, let's test the possible values:
If $b=19$, then $3c>19$. The only prime value for $c$ that satisfies this is $c=7$ (since $3\times 7=21>19$).
- If $b=19$, then since $d<b$ and $d\in \{13,17\}$, $d$ can be 13 or 17.

3. Use the quadratic equations and their roots to narrow down:
- Note (vii) states that $e=b+1$.
- If $b=19$, then:
$e=19+1=20$
- Let's test the value of $d=13$:
Equation 2 becomes: $d{y}^2+ey+c=0$
$13y^2+20y+7=0$
- Solving this quadratic equation:
$13y^2+13y+7y+7=0$
$13y(y+1)+7(y+1)=0$
$(13y+7)(y+1)=0$
The roots are $y=-1$ and $y=-\frac{7}{13}$.
Since $-1<-\frac{7}{13}$, the smallest root of Equation 2 is $-1$.

- According to note (v), the smallest roots of both equations are the same, so $x=-1$ must be the smallest root of Equation 1:
$a{x}^2+bx+c=0$
Substitute $x=-1$, $b=19$, and $c=7$:
$a{(-1)}^2+19(-1)+7=0$
$a-19+7=0$
$a=12$
- Let's check Equation 1 with $a=12$:
$12x^2+19x+7=0$
$12x^2+12x+7x+7=0$
$12x(x+1)+7(x+1)=0$
$(12x+7)(x+1)=0$
The roots are $x=-1$ and $x=-\frac{7}{12}$.
The smallest root is indeed $-1$, and neither root is irrational, satisfying all constraints.

4. Calculate the value of $d\times e$:
Using $d=13$ and $e=20$:
$d\times e=13\times 20=260$