For a knapsack problem, $P$ and $W$ are profits and weights for a set of $7$ items as given below:

$P={9,5,2,7,6,16,9}$

$W={2,5,6,11,1,9,3}$

What is the maximum value of items you can carry if maximum weight allowed is $25$ with a fractional knapsack?

Take items in decreasing profit by weight ratio:

Item 5: weight $1$, profit $6$

Item 1: weight $2$, profit $9$

Item 7: weight $3$, profit $9$

Item 6: weight $9$, profit $16$

Item 2: weight $5$, profit $5$

Total weight used:

$1+2+3+9+5=20$

Total profit:

$6+9+9+16+5=45$

Remaining capacity:

$25-20=5$

Next best item is item 4 with profit $7$ and weight $11$.

Fractional profit:

$5\times \frac{7}{11}=\frac{35}{11}=3.18$

Maximum profit:

$45+3.18=48.18\approx 48.1$

Correct option is:

$\boxed{(a)\ 48.1}$