Question

Mr. Sawyer drove his car from his home to New York at the rate of 45 mph and returned over the same road at the rate of 40 mph. If his time returning exceeded his time going by 30 min, find his time going and his time returning.

Answer 1

Answer: Time taken by him in going = 8 hours

Time taken by him in returning =  9 hours

Step-by-step explanation:

Let the total distance from home to New York is x miles,

$\text{ Since, Time} = \frac{\text{Distance}}{\text{Speed}}$

Also, he drove his car from his home to New York at the rate of 45 mph,

⇒ $\text{ Time taken by him in going } = \frac{x}{45}\text{ hours}$

And, returned over the same road at the rate of 40 mph.

⇒  $\text{ Time taken by him in returning } = \frac{x}{40}\text{ hours}$

According to the question,

Time taken by him in returning - Time taken by him in going = 30 minutes = 1/2 hours,    ( 1 hours = 60 minutes )

⇒ $\frac{x}{40}-\frac{x}{45}=\frac{1}{2}$

⇒ $\frac{9x}{360}-\frac{8x}{360}=\frac{1}{2}$

⇒ $\frac{x}{360} = \farc{1}{2}$

⇒ $2x=720$

⇒ $x=360\text{ miles}$

Hence, the total distance from home to New York = x miles = 360 miles

⇒ $\text{ Time taken by him in going } = \frac{x}{45}\text{ hours}$

$=\frac{360}{45}=8\text{ hours}$

⇒  $\text{ Time taken by him in returning } = \frac{x}{40}\text{ hours}$

$=\frac{360}{40}=9\text{ hours}$