Two places A and B are 45 kms apart and connected by a straight road. Anil goes from A to B while Sunil goes from B to A. Starting at the same time, they cross each other in exactly 1 hour 30 minutes. If Anil reaches B exactly 1 hour 15 minutes after Sunil reaches A, the speed of Anil, in km per hour, is

The correct option is 12.

Let the speed of Anil be $a$ km/h and the speed of Sunil be $s$ km/h.

We are given that the distance between places A and B is 45 km. Since Anil starts from A toward B and Sunil starts from B toward A at the same time, they travel toward each other. Their relative speed is the sum of their individual speeds:
$a+s$
They cross each other in exactly 1 hour 30 minutes. Let us convert this time into hours:
$1\text{hour}30\text{minutes}=1.5\text{hours}=\frac{3}{2}\text{hours}$
Using the relation $\text{Distance}=\text{Speed}\times \text{Time}$, we can write the relative speed equation as:
$45=(a+s)\times 1.5$
Solving for the sum of their speeds:
$a+s=\frac{45}{1.5}=30\text{km/h}$
Thus, we can express Sunil's speed $s$ in terms of Anil's speed $a$:
$s=30-a$

Next, we calculate the time taken by each of them to complete the entire journey of 45 km:
Time taken by Anil to reach B, $T_a=\frac{45}{a}$ hours.
Time taken by Sunil to reach A, $T_s=\frac{45}{s}$ hours.

We are given that Anil reaches B exactly 1 hour 15 minutes after Sunil reaches A. Let us convert 1 hour 15 minutes into hours:
$1\text{hour}15\text{minutes}=1\frac{15}{60}\text{hours}=1\frac{1}{4}\text{hours}=\frac{5}{4}\text{hours}$
Therefore, the difference between their times is:
$T_a-T_s=\frac{5}{4}$
Substituting the expressions for $T_a$ and $T_s$ gives:
$\frac{45}{a}-\frac{45}{30-a}=\frac{5}{4}$

To solve this equation, first divide both sides by 5:
$\frac{9}{a}-\frac{9}{30-a}=\frac{1}{4}$
Now, find a common denominator on the left side:
$9\times \left(\frac{(30-a)-a}{a(30-a)}\right)=\frac{1}{4}$
Simplify the numerator:
$9\times \left(\frac{30-2a}{30a-a^2}\right)=\frac{1}{4}$
$\frac{270-18a}{30a-a^2}=\frac{1}{4}$
Cross-multiply to get a quadratic equation:
$4\times (270-18a)=30a-a^2$
$1080-72a=30a-a^2$
Rearrange all terms to one side:
$a^2-102a+1080=0$

We can solve this quadratic equation using the quadratic formula:
$a=\frac{-b\pm \sqrt{b^2-4AC}}{2A}$
Here, the coefficients are $A=1$, $b=-102$, and $C=1080$:
$a=\frac{102\pm \sqrt{{(-102)}^2-4\left(1\right)\left(1080\right)}}{2}$
Calculate the values inside the square root:
${(-102)}^2=10404$
$4\times 1\times 1080=4320$
$b^2-4AC=10404-4320=6084$
Taking the square root:
$\sqrt{6084}=78$
Now substitute this back to find the two possible roots:
$a=\frac{102\pm 78}{2}$
This gives the following options:
1) $a=\frac{102+78}{2}=\frac{180}{2}=90$
2) $a=\frac{102-78}{2}=\frac{24}{2}=12$

Since their combined speed is 30 km/h, the speed of Anil must be less than 30 km/h. Hence, $a=90$ is not a physically possible solution.
Therefore, the speed of Anil is 12 km/h.