In a certain town, 25% of the families own a phone and 15% own a car; 65% families own neither a phone nor a car and 2000 families own both a car and a phone. Consider the following three statements:
(i) 5% families own both a car and a phone.
(ii) 35% families own either a car or a phone.
(iii) 40000 families live in the town.
Then,

Let us denote the following events for the families of the town:

$$P:\text{ the family owns a phone}$$ $$C:\text{ the family owns a car}$$

We are told that

$$P(P)=25\% = 0.25,\qquad P(C)=15\% = 0.15,\qquad P(\text{neither }P\text{ nor }C)=65\% = 0.65.$$

Because a family must belong either to the union of the two ownership groups or to the “neither” group, we have

$$P(P\cup C) = 1 - P(\text{neither }P\text{ nor }C).$$

Substituting the given value,

$$P(P\cup C)=1-0.65=0.35=35\%.$$

Thus, 35% of the families own either a phone or a car (or both). This immediately verifies statement (ii).

Now we want the percentage of families that own both a phone and a car. We use the well-known principle of inclusion-exclusion, which in probability form states

$$P(P\cup C)=P(P)+P(C)-P(P\cap C).$$

Solving this relation for $$P(P\cap C)$$, we have

$$P(P\cap C)=P(P)+P(C)-P(P\cup C).$$

Substituting the numerical percentages,

$$P(P\cap C)=0.25+0.15-0.35=0.40-0.35=0.05=5\%.$$

Therefore 5% of the families own both items, confirming statement (i).

We are also told that exactly 2000 families are in this “both” category. Let the total number of families in the town be $$N$$. Then

$$0.05N=2000.$$

Dividing by $$0.05$$ (which is $$\tfrac5{100}$$) gives

$$N=\frac{2000}{0.05}=2000\times20=40000.$$

Hence the town contains 40000 families, validating statement (iii).

We have now shown that statements (i), (ii) and (iii) are all correct. The option that lists all three statements as correct is Option C.

Hence, the correct answer is Option C.