A survey shows that 63% of the people in a city read newspaper A whereas 76% read news paper B. If $$x$$% of the people read both the newspapers, then a possible value of $$x$$ can be:

Let us denote by $$A$$ the set of people who read newspaper A and by $$B$$ the set of people who read newspaper B.

We are told that

$$\text{Percentage in }A = 63\%,$$

$$\text{Percentage in }B = 76\%,$$

and we wish to find a possible value of

$$x\% = \text{Percentage in }A\cap B,$$

that is, the percentage of people who read both newspapers.

We begin with the well-known principle of inclusion-exclusion for two sets. It states

$$|A\cup B| = |A| + |B| - |A\cap B|.$$

Replacing absolute values by percentages, we can write

$$\underbrace{($$ percentage that read at least one of A or B $$)}_{\le 100\%} = 63\% + 76\% - x\%.$$

Because a percentage cannot exceed $$100\%,$$ the left side is at most $$100\%.$$ Hence we obtain the inequality

$$63 + 76 - x \le 100.$$

Simplifying, we have

$$139 - x \le 100,$$

so

$$-x \le 100 - 139 = -39,$$

and multiplying both sides by $$-1$$ (which reverses the inequality sign) gives

$$x \ge 39.$$

Therefore the minimum possible value of $$x$$ is $$39\%.$$ Any smaller value would force the union $$A\cup B$$ above $$100\%,$$ which is impossible.

Next, observe that the intersection cannot be larger than the smaller of the two individual percentages, because one cannot have more people in both sets than are present in a single set. Formally,

$$x \le \min(63,\,76) = 63.$$

Putting the two bounds together, the percentage $$x$$ has to satisfy

$$39 \le x \le 63.$$

Now we test the four answer choices:

A. $$29\%$$    (which is < 39)  &rightarrow;  impossible, reject.

B. $$37\%$$    (which is < 39)  &rightarrow;  impossible, reject.

C. $$65\%$$    (which is > 63)  &rightarrow;  impossible, reject.

D. $$55\%$$    (which lies between 39 and 63)  &rightarrow;  possible, accept.

Hence, the correct answer is Option D.