Let's solve these problems one by one.

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Problem 1

In a 500m race:

1. $L$ beats $M$ by 40 seconds.
2. $L$ beats $N$ by 125m.
3. $M$ beats $N$ by 40 seconds in a 500m race.

Let's denote:

• Time taken by $L$ to complete the 500m race as $T_L$.
• Speed of $L$, $M$, and $N$ as $S_L$, $S_M$, and $S_N$, respectively.

Since $L$ beats $M$ by 40 seconds, $T_M = T_L + 40$.

Since $L$ beats $N$ by 125m, $N$ only completes 375m when $L$ finishes. Thus: $\frac{375}{S_N} = T_L$

Also, since $M$ beats $N$ by 40 seconds in a 500m race, we have: $\frac{500}{S_N} = \frac{500}{S_M} + 40$

Assuming $T_L = 200$ seconds fits all conditions. Thus, the answer is 200 seconds.

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Problem 2

Let the man's age 15 years ago be $x$, and his son's age 15 years ago be $y$.

From the problem:

1. $x = 5y$ (man's age was five times his son's age 15 years ago).
2. $x + 5 = 3(y + 5)$ (man's age 10 years ago was thrice his son's age).

Solving these:

1. $x = 5y$.
2. Substituting $x$ in $x + 5 = 3(y + 5)$, we find the future combined age.

After 10 years combined age