Rachel has two crayons. The fat crayon burns down to nothing after 15 hours. Her skinny crayons wear down to nothing ofter 6 hours. At 10:30 she began coloring with both crayons of the same height, one fat one skinny. At what time will the skinny crayon be half the height of the fat crayon?
To solve this problem, let's break it down step by step.
1. Determine the rate at which each crayon wears down:
- The fat crayon burns down to nothing after 15 hours, so it burns at a rate of 1/15 of its height per hour.
- The skinny crayon wears down to nothing after 6 hours, so it wears down at a rate of 1/6 of its height per hour.
2. We know that at 10:30, Rachel starts coloring with both crayons, and they are of the same height. Let's assume that the initial height of each crayon is 'h' units.
3. We need to find the time at which the skinny crayon will be half the height of the fat crayon. Let's call this time 't' hours.
4. Since the fat crayon burns down to nothing after 15 hours, its height at time 't' is given by: h - (1/15) * t.
5. Similarly, since the skinny crayon wears down to nothing after 6 hours, its height at time 't' is given by: h - (1/6) * t.
6. We want the skinny crayon to be half the height of the fat crayon. So, we can set up the following equation: (1/2) * (h - (1/15) * t) = h - (1/6) * t.
7. Now, solve the equation for 't':
(1/2) * h - (1/30) * t = h - (1/6) * t.
(1/30) * t = (1/2) * h - (1/6) * t.
(7/30) * t = (1/2) * h.
t = (30/7) * (1/2) * h.
8. Since we know the initial height of each crayon is 'h', we substitute this value into the equation to obtain the time 't'.
9. Finally, add the time 't' to 10:30 to find the exact time at which the skinny crayon will be half the height of the fat crayon.
With these steps, you should be able to calculate the exact time.