A man left town am at 10:00 AM and traveled by car to town N at an average speed of .
He spent 2 hours for a meeting and returned through town M by bus at an average speed of .
f the distance covered by the bus was 2 km longer than that of the car and he arrived at town M at 1
55 PM .
alculate distance from M to N .

Ask by Li Mcguire. in Ghana

Mar 29,2025

Question

A man left town am at 10:00 AM and traveled by car to town N at an average speed of .
He spent 2 hours for a meeting and returned through town M by bus at an average speed of .
f the distance covered by the bus was 2 km longer than that of the car and he arrived at town M at 1
55 PM .
alculate distance from M to N .

Ask by Li Mcguire. in Ghana

Mar 29,2025

UpStudy AI · Accepted Answer

Let $$ d_c $$ be the distance traveled by car from the starting town (A) to town $$ N $$, and let $$ d_b $$ be the distance traveled by bus from $$ N $$ to town $$ M $$.

According to the problem, the bus distance is 2 km longer than the car distance:
$$
d_b = d_c + 2.
$$

The timeline is as follows:
- The car trip from A to $$ N $$ starts at 10:00 AM and takes 
  $$
\frac{d_c}{72} \text{ hours}.
  $$
- He spends 2 hours in a meeting.
- The bus trip from $$ N $$ to $$ M $$ takes 
  $$
\frac{d_b}{40} \text{ hours}.
  $$
- He arrives at $$ M $$ at 1:55 PM, so the total time elapsed is 3 hours and 55 minutes. Converting 55 minutes to hours:
  $$
3 \text{ hours } 55 \text{ minutes} = 3 + \frac{55}{60} = \frac{235}{60} \text{ hours}.
  $$

Thus, the sum of the travel times is
$$
\frac{d_c}{72} + 2 + \frac{d_b}{40} = \frac{235}{60}.
$$

Substitute $$ d_b = d_c + 2 $$ into the equation:
$$
\frac{d_c}{72} + 2 + \frac{d_c + 2}{40} = \frac{235}{60}.
$$

To simplify the calculations, we convert the time equation into minutes. Converting each part:
- Car time: 
  $$
\frac{d_c}{72} \text{ hours} = \frac{60d_c}{72} = \frac{5d_c}{6} \text{ minutes}.
  $$
- Meeting time: $$ 2 $$ hours $$ = 120 $$ minutes.
- Bus time:
  $$
\frac{d_c + 2}{40} \text{ hours} = \frac{60(d_c+2)}{40} = \frac{3(d_c+2)}{2} \text{ minutes}.
  $$
- Total time: 1:55 PM minus 10:00 AM $$ = 3 $$ hours 55 minutes $$ = 235 $$ minutes.

Now, we have:
$$
\frac{5d_c}{6} + 120 + \frac{3(d_c+2)}{2} = 235.
$$

Multiply the entire equation by 6 to eliminate denominators:
$$
5d_c + 720 + 9(d_c+2) = 1410.
$$

Distribute and combine like terms:
$$
5d_c + 720 + 9d_c + 18 = 1410,
$$
$$
14d_c + 738 = 1410.
$$

Solve for $$ d_c $$:
$$
14d_c = 1410 - 738 = 672,
$$
$$
d_c = \frac{672}{14} = 48 \text{ km}.
$$

Now, find the bus distance:
$$
d_b = d_c + 2 = 48 + 2 = 50 \text{ km}.
$$

Since the bus journey was from town $$ N $$ to town $$ M $$, the distance from $$ M $$ to $$ N $$ is:
$$
\boxed{50 \text{ km}}.
$$
The distance from M to N is 50 km.

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