Three persons $$A,B,C$$ participate in a running race of $$1\text{ km}$$ distance. When $$A$$ and $$B$$ run, $$A$$ wins by $$60$$ seconds. When $$A$$ and $$C$$ run, $$A$$ wins by $$375\text{ m}$$. When $$B$$ and $$C$$ run, $$B$$ wins by $$30$$ seconds. If the time taken by $$B$$ to run $$1\text{ km}$$ is $$x$$ minutes and $$30$$ seconds, then $$x$$ is

Let the times taken by $$A,$$ $$B,$$ and $$C$$ to complete the race be $$T_A,$$ $$T_B,$$ and $$T_C$$ seconds, respectively.

Since $$B$$ finishes $$60$$ seconds after $$A$$,

$$T_B=T_A+60.$$

Also, $$C$$ finishes $$30$$ seconds after $$B$$, so

$$T_C=T_B+30=T_A+90.$$

When $$A$$ finishes the $$1000\text{ m}$$ race, $$C$$ has covered only

$$1000-375=625\text{ m}.$$

Since both run for the same amount of time until $$A$$ finishes,

$$\dfrac{1000}{s_A}=\dfrac{625}{s_C},$$

where $$s_A$$ and $$s_C$$ are the speeds of $$A$$ and $$C$$.

Thus,

$$s_C=\dfrac{625}{1000}s_A=\dfrac{5}{8}s_A.$$

Now,

$$T_A=\dfrac{1000}{s_A}\quad \text{and}\quad T_C=\dfrac{1000}{s_C}.$$

Therefore,

$$\dfrac{T_A}{T_C}=\dfrac{\dfrac{1000}{s_A}}{\dfrac{1000}{s_C}}=\dfrac{s_C}{s_A}=\dfrac{5}{8}.$$

Hence,

$$T_C=\dfrac{8}{5}T_A.$$

But,

$$T_C=T_A+90.$$

So,

$$T_A+90=\dfrac{8}{5}T_A.$$

Multiplying both sides by $$5$$,

$$5T_A+450=8T_A.$$

$$3T_A=450.$$

$$T_A=150\text{ seconds}.$$

Therefore,

$$T_B=150+60=210\text{ seconds}.$$

Since $$210$$ seconds is equal to $$3$$ minutes $$30$$ seconds.

Thus, the value of $$ x = 3$$.