Solve the following problems:

$Here, relative speed of Y in terms of X$

$=\frac{4x}{3}-x=\frac{4x-3x}{3}=\frac{x}{3} mph.$

$X will cover in (11.30 - 10) = 1.5 hours = ( x \times 1.5) miles = 1.5x miles$

$\therefore Required time taken by Y to catch X= \frac{1.5x}{{\displaystyle \frac{x}{3}}}=1.5x\times \frac{3}{x}=4.5x hours$

$\therefore Y will meet X at =11.30 am +4.5 hour = 4 pm$

$Alternative: Given that,$

$X\text{'} s speed is = x mph$

$\therefore Y\text{'}s speed is =\frac{4x}{3} mph$

$Let, X and Y will meet after p hours$

$$\begin{gathered} According to questions, \\ px = (p - 1.5) \times \frac{4x}{3} [As, d = time \times velocity] \end{gathered}$$

$$\begin{gathered} =px=\frac{4px}{3}-2x \\ =\frac{4px}{3}-px=2x \\ =\frac{4px-3px}{3}=2x \\ =px=6x \\ \therefore p=6 \\ \therefore Y will catch train X at = 10 am +6 hours = 4pm \left(answer\right) \end{gathered}$$

$Let, original \cos t price of an orange is x dollar.$

$At 40\% discount, the price of an orange is$

$=60\% of x =\frac{60x}{100}=\frac{3x}{5}Dollar$

$According to the question$

$$\begin{gathered} \frac{12}{{\displaystyle \frac{3x}{5}}}-\frac{12}{x}=4 \\ =12\times \frac{5}{3x}-\frac{12}{x}=4 \end{gathered}$$

$$\begin{gathered} =\frac{20}{x}-\frac{12}{x}=4 \\ =\frac{20-12}{x}=4 \\ =\frac{8}{x}=4 \end{gathered}$$

$\therefore x=\frac{8}{4}=2$

$\therefore Number of oranges that can be purchased by 24 dollars=\frac{24}{2}= 12 (answer$)