Three numbers A, B and C are such that A is 40% less than B, and C is 40% of the sum of A and B. The difference between A and B is what percentage of C?

Given:

1. A is $40 \%$ less than $B$
2. C is $40 \%$ of the sum of $A$ and $B$

We need to find: what percentage of $C$ is the difference between $A$ and $B$ ?

Let's express these relationships mathematically:

$\begin{aligned} & A=0.6 B(\text { since } A \text { is } 40 \% \text { less than } B) \\ & C=0.4(A+B)(\text { since } C \text { is } 40 \% \text { of the sum of } A \text { and } B) \end{aligned}$

Substituting $A =0.6 B$ into the expression for $C$ :

$C=0.4(0.6 B+B)=0.4(1.6 B)=0.64 B$

The difference between $A$ and $B$ is:

$\mathrm{B}-\mathrm{A}=\mathrm{B}-0.6 \mathrm{~B}=0.4 \mathrm{~B}$

As a percentage of C :

$\dfrac{B-A}{C} \times 100 \%=\dfrac{0.4 B}{0.64 B} \times 100 \%=\dfrac{0.4}{0.64} \times 100 \%=\dfrac{5}{8} \times 100 \%=62.5 \%$

Therefore, the difference between A and B is $62.5 \%$ of C .

Hence, the correct option is Option 3.