Question

A shoe repairman is working with his assistant, who takes twice as long to repair a pair of shoes. Together they can fix 16 pairs of shoes in an eight-hour day. How long does it take the repairman to fix one pair of shoes by himself?

Answer 1

$\boxed{\boxed{Answer: \tex{ } 45 minutes}}$

Hello!

The number of pairs of shoes each man can repair is its speed times the time.

Call x the repair speed of the assistant. Then, the repair speed of the repairman is 2x, and the joint speed is 2x + x = 3x.

Also, the joint speed is 16 pairs in 8 hours = 16 pairs/ 8 hours = 2 pairs/hour.

Thus, we can equate:

$3x = 2 pairs/hour, \text { or } x = (2/3) pairs / hour.$

That is the speed of the assistant. Then, the speed of the repairman is:

$2x = 2 (2/3) pairs/hour\\2x = (4/3) pairs/hour.$

That means that the repairman takes the inverse of 4/3 of an hour to repair one pair of shoes. That is 3/4 of an hour (to repair on pair)

• 3/4 of an hour is 45 minutes and this is the answer.

Answer 2

Answer:

3 hours.          

Step-by-step explanation:

Let x be the time taken by shoe repairman to repair one pair.

We have been given that his assistant, who takes twice as long to repair a pair of shoes. So time taken by his assistant to repair one pair of shoes would be 2x.

The number of pair of shoes repaired by repairman in one hour would be $\frac{1}{x}$.

The number of pair of shoes repaired by assistant in one hour would be $\frac{1}{2x}$.

We have been given that together they can fix 16 pairs of shoes in an eight-hour day. We can represent this information in an equation as:

$\frac{1}{x}+\frac{1}{2x}=\frac{8}{16}$

$\frac{1}{x}+\frac{1}{2x}=\frac{1}{2}$

Let us have a common denominator.

$\frac{2*1}{2*x}+\frac{1}{2x}=\frac{1}{2}$

$\frac{2}{2x}+\frac{1}{2x}=\frac{1}{2}$

$\frac{2+1}{2x}=\frac{1}{2}$

$\frac{3}{2x}=\frac{1}{2}$  

Upon cross multiplying our equation we will get,

$2x*1=3*2$

$2x=3*2$

$\frac{2x}{2}=\frac{3*2}{2}$

$x=3$

Therefore, it take 3 hours for the repairman to fix one pair of shoes by himself.