1. Write the given frequency table

Ages and frequencies are:

$$\begin{array}{c|cccccc} \text{Age} & 15 & 16 & 17 & 18 & 19 & 20 \\ \hline \text{Frequency} & 5 & 8 & 5 & 12 & x & y \end{array}$$

Total students $=40$, so

$$5+8+5+12+x+y=40$$ $$30+x+y=40$$ $$x+y=10 \qquad ...(1)$$

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2. Find the median

Since there are $40$ students, the median is the average of the $20^{\text{th}}$ and $21^{\text{st}}$ observations.

Now compute cumulative frequencies:

• Age $15$: positions $1$ to $5$
• Age $16$: positions $6$ to $13$
• Age $17$: positions $14$ to $18$
• Age $18$: positions $19$ to $30$

Thus both the $20^{\text{th}}$ and $21^{\text{st}}$ observations are $18$.

So the median is

$$M=18$$

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3. Use mean deviation about the median

Mean deviation about median is given by

$$\text{M.D. about median} = \frac{\sum f|x_i-M|}{N}$$

Given this value is $1.25$ and $N=40$, so

$$\frac{\sum f|x_i-18|}{40}=1.25$$ $$\sum f|x_i-18|=50$$

Now calculate:

• For age $15$: $|15-18|=3$, contribution $=5\cdot 3=15$
• For age $16$: $|16-18|=2$, contribution $=8\cdot 2=16$
• For age $17$: $|17-18|=1$, contribution $=5\cdot 1=5$
• For age $18$: $|18-18|=0$, contribution $=12\cdot 0=0$
• For age $19$: $|19-18|=1$, contribution $=x\cdot 1=x$
• For age $20$: $|20-18|=2$, contribution $=y\cdot 2=2y$

Therefore,

$$15+16+5+0+x+2y=50$$ $$36+x+2y=50$$ $$x+2y=14 \qquad ...(2)$$

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4. Solve the equations

From (1):

$$x+y=10$$

From (2):

$$x+2y=14$$

Subtracting,

$$y=4$$

Then

$$x=10-4=6$$

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5. Compute $4x+5y$

$$4x+5y = 4(6)+5(4)=24+20=44$$

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6. Check with options

The value is

$$\boxed{44}$$

So the correct option is C.