Time, Speed and Distance. Arithmetic question (Moderate Level) 
Ajay and Kajol start towards each other at the same time from Barabanki and Fatehpur for their destinations Fatehpur and Barabanki
  respectively which are 300 km apart. They meet each other 120 km away
  from Barabanki. Shahrukh starts from Barabanki to Fatehpur, 1 hour
  after Ajay starts. Shahrukh meets Kajol 1.5 hours after Shahrukh
  starts. If the speed of Shahrukh is at least 20 km/h faster than the
  speed of Kajol.
(a)Which of the following statements is true?
1)The minimum possible speed of Ajai is 45 km/h
2)The maximum possible speed of Ajai is 45 km/h
3)The minimum possible speed of Kajol is 60 km/h
4)The maximum possible speed of Kajol is 60 km/h
(b)What is the minimum speed of Shahrukh to overtake Ajay, before he meets Kajol?(use the data from the previous question if necessary)

I could solve the first question and got the maximum possible speed of Ajay to be 45km/h. The working is shown below. However, I could not solve the second question. Please help me regarding this.
My attempt:
Let Speed of Ajay be $S_A=2x$
Let Speed of Kajol be $S_K=3x$
Let Speed of Shahrukh be $S_S$
The Meeting time of Ajay and Kajol is constant(Same). So, the speed of Ajay and Kajol is proportional to the respective distances they covered before the meeting.

The note in the red ink is a personal note.
Solution in the book:

 A: Here's what I would do (almost the same as your way):
Let the speed of Ajay, Kajol, Shahrukh be $v_A;v_K;v_S$ respectively (in kilometers), assume that $v_A=2x$, then $v_K=3x$ and $v_S\ge 3x+20$.

Shahrukh starts from Barabanki to Fatehpur, $1$ hour after Ajay starts. Shahrukh meets Kajol $1.5$ hours after Shahrukh starts.

We have a formula
$$\dfrac{300-3x}{v_S+v_K}=1.5$$
where $300-3x$ is the distance between Ajay and Shahrukh when Sharukh starts in kilometers.
We will have $300-3x=1.5(v_S+v_K)\ge 1.5(3x+20+3x)$
$\Leftrightarrow 300-3x\ge 9x+30$
$\Leftrightarrow 12x\le 270\Leftrightarrow x\le 22.5$
The maximum speed of Ajay is $v_A=2x\le 45$km/h.
The maximum speed of Kajol is $v_A=3x\le 67.5$ km/h, so you have done task (a) correctly.

Ajay and Kajol start towards each other at the same time from Barabanki and Fatehpur for their destinations Fatehpur and Barabanki respectively which are $300$ km apart. They meet each other $120$ km away from Barabanki. 

Shahrukh needs to overtake Ajay, before he meets Kajol. This means Shahrukh must reach the "meeting point" $120$km away from his starting point before Ajay does so.
The maximum speed of Ajay is $45$km/h, so the minimum time taken for Ajay to reach the "meeting point" is $120\div 45=\dfrac{8}{3}$ hours or $2$ hours $40$ minutes, as $t_A=\dfrac{120}{v_A}\ge\dfrac{120}{45}=\dfrac{8}{3}$
Because Shahrukh starts one hour later, he needs to reach the $120$km "meeting point" in less than $\dfrac{5}{3}$ hours or $1$ hour $40$ minutes. This is a strict inequality: $v_S=\dfrac{120}{t_S}>120\div\dfrac{5}{3}=72$ km/h.
If we follow the task, saying the minimum speed of Shahrukh is $72$ km/h is not accurate, because it is a strict inequality, but just over $72$ km/h would be the answer.
