**Work Problems**

The formula for work problems is *Work* = *Rate* Ã— *Time*, or *W = R Ã— T*. The amount of work done is usually 1 unit. Hence, the formula becomes 1 = *R Ã— T*. Solving this for R gives *R* = $\frac{1}{T}$ .

**Example 1: **

If Johnny can mow the lawn in 30 minutes and with the help of his brother, Bobby, they can mow the lawn 20 minutes, how long would it take Bobby working alone to mow the lawn?

(A) 1/2 hour

(B) 3/4 hour

(C) 1 hour

(D) 3/2 hours

(E) 2 hours

Let r = 1/t be BobbyÂ’â€™s rate. Now, the rate at which they work together is merely the sum of their rates:

*Total Rate = JohnnyÂ’â€™s Rate + BobbyÂ’â€™s Rate*

$\frac{1}{20}$ = $\frac{1}{30}$ + $\frac{1}{t}$

$\frac{1}{20}$ - $\frac{1}{30}$ = $\frac{1}{t}$

$\frac{\mathrm{30\; -\; 20}}{\mathrm{30\; \xc2\xb7\; 20}}$ = $\frac{1}{t}$

$\frac{1}{60}$ = $\frac{1}{t}$

t = 60

Hence, working alone, Bobby can do the job in 1 hour.

The answer is (C).

**Example 2: **

A tank is being drained at a constant rate. If it takes 3 hours to drain 6/7 of its capacity, how much longer will it take to drain the tank completely?

(A) 1/2 hour

(B) 3/4 hour

(C) 1 hour

(D) 3/2 hours

(E) 2 hours

Since 6/7 of the tankÂ’â€™s capacity was drained in 3 hours, the formula W = R Ã— T becomes $\frac{6}{7}$ = R Ã— 3. Solving for R gives R = 2/7. Now, since 6/7 of the work has been completed, 1/7 of the work remains. Plugging this information into the formula W = R Ã— T gives $\frac{1}{7}$ = $\frac{2}{7}$ Ã— T . Solving for T gives T = 1/2.

The answer is (A).