Step 1: Define VariablesNbsp;
Let Anil's speed be \(v_A\) km/h.Nbsp;
Let Sunil's speed be \(v_S\) km/h.Nbsp;
They meet after 1 hour and 30 minutes, which is \(1.5\) hours. At this point, the total distance they covered combined is \(45 \,km.\)
Step 2: Relate the Distances Traveled
In the \(1.5\) hours before they met:
Anil covered a distance of \(v_A \times 1.5\).
Sunil covered a distance of \(v_S \times 1.5\).
Since they met after covering the total distance:
\[v_A \times 1.5 + v_S \times 1.5 = 45.\]
Simplifying this equation gives:
\[1.5(v_A + v_S) = 45,\]
\[v_A + v_S = 30. \quad \text{(Equation 1)}\]
Step 3: Relation Between Times After Meeting
After meeting, Anil takes 1 hour and 15 minutes (\(1.25\) hours) longer than Sunil to reach their respective destinations.
The remaining distance for Anil is \(45 - 1.5v_A\).
The remaining distance for Sunil is \(45 - 1.5v_S\).
Using the formula \(\text{time} = \frac{\text{distance}}{\text{speed}}\), we can express the time taken after meeting:
Anil's time after meeting: \(\frac{45 - 1.5v_A}{v_A}\).
Sunil's time after meeting: \(\frac{45 - 1.5v_S}{v_S}\).
The problem states Anil takes \(1.25\) hours more than Sunil after meeting, so:
\[\frac{45 - 1.5v_A}{v_A} = \frac{45 - 1.5v_S}{v_S} + 1.25. \quad \text{(Equation 2)}\]
Step 4: Solve the Equations
From Equation 1, we can express \(v_S\) as:
\[v_S = 30 - v_A.\]
Substitute this expression for \(v_S\) into Equation 2:
\[\frac{45 - 1.5v_A}{v_A} = \frac{45 - 1.5(30 - v_A)}{30 - v_A} + 1.25.\]
Simplify the numerator of the second term: \(45 - 1.5(30 - v_A) = 45 - 45 + 1.5v_A = 1.5v_A\).
So, the equation becomes:
\[\frac{45 - 1.5v_A}{v_A} = \frac{1.5v_A}{30 - v_A} + 1.25.\]
To eliminate the denominators, multiply both sides by \(v_A(30 - v_A)\):
\[(45 - 1.5v_A)(30 - v_A) = 1.5v_A^2 + 1.25v_A(30 - v_A).\]
Expand both sides of the equation:
\[1350 - 45v_A - 45v_A + 1.5v_A^2 = 1.5v_A^2 + 37.5v_A - 1.25v_A^2.\]
Simplify and combine like terms:
\[1350 - 90v_A + 1.5v_A^2 = 0.25v_A^2 + 37.5v_A.\]
Rearrange into a standard quadratic form \(ax^2 + bx + c = 0\):
\[1.25v_A^2 - 127.5v_A + 1350 = 0.\]
Divide the entire equation by 1.25 to simplify:
\[v_A^2 - 102v_A + 1080 = 0.\]
Use the quadratic formula \(v_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1\), \(b=-102\), and \(c=1080\):
\[v_A = \frac{-(-102) \pm \sqrt{(-102)^2 - 4(1)(1080)}}{2(1)}.\]
\[v_A = \frac{102 \pm \sqrt{10404 - 4320}}{2},\]
\[v_A = \frac{102 \pm \sqrt{6084}}{2},\]
\[v_A = \frac{102 \pm 78}{2}.\]
This gives two possible solutions for \(v_A\):
\[v_A = \frac{102 + 78}{2} = \frac{180}{2} = 90,\]
\[v_A = \frac{102 - 78}{2} = \frac{24}{2} = 12.\]
A speed of \(90 \, \text{km/h}\) for Anil is generally considered unrealistic in such problems. Therefore, we select the more reasonable speed:
\[v_A = 12 \, \text{km/h}.\]
Final Answer
Anil's speed is:
\[\boxed{12 \, \text{km/h}}.\]