One grocery clerk can stock a shelf in 40 min, whereas a second clerk requires 20 min to stock the same shelf. How long would it take to stock the shelf if the two clerks worked together?
Is this how you do this:
t/40 +t/24 =1
t*3/40*3 + t*5/24*5 =1
3t/120 + 5t/120 =1
(8t/120)*120 = 1(120)
8t/8 = 120/8
=15
Yes, you are on the right track in setting up the equation to solve the problem. Let me explain the steps in more detail:
Let's assume that it takes a certain amount of time, "t", for the two clerks working together to stock the shelf.
The first clerk can stock the shelf in 40 minutes, which means that in one minute, the first clerk can complete 1/40th of the shelf. So, in "t" minutes, the first clerk can complete t/40th of the shelf.
Similarly, the second clerk can stock the shelf in 20 minutes, which means that in one minute, the second clerk can complete 1/20th of the shelf. So, in "t" minutes, the second clerk can complete t/20th of the shelf.
When the two clerks work together, they are completing the entire shelf, so the fractions for each clerk sum up to 1. Therefore, we have the equation:
t/40 + t/20 = 1
To solve for "t", we need to find a common denominator for the fractions. In this case, the common denominator is 40, which is a multiple of both 40 and 20. So we can rewrite the equation as:
t/40 + 2t/40 = 1
Combining like terms, we get:
3t/40 = 1
To isolate "t", we can multiply both sides of the equation by the reciprocal of 3/40, which is 40/3:
(3t/40) * (40/3) = 1 * (40/3)
The 40/3 cancel out on the left side, leaving us with:
t = 40/3
Now, to simplify the answer, we can divide 40 by 3, which gives us approximately 13.33. Therefore, it would take approximately 13.33 minutes, or 13 minutes and 20 seconds, for the two clerks to stock the shelf together.