# Problem of the week #74 - August 25th, 2013

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#### Jameson

Staff member
Here is a word problem aimed at the upper-high school level that you might see on a college entrance exam in the US. Welcome back to school everyone!

Problem:
On a partly cloudy day, Derek decides to walk back from work. When it is sunny, he walks at a speed of s miles/hr (s is an integer) and when it gets cloudy, he increases his speed to (s + 1) miles/hr. If his average speed for the entire distance is 2.8 miles/hr, what fraction of the total distance did he cover while the sun was shining on him?
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#### Jameson

Staff member
Congratulations to the following members for their correct solutions:

1) MarkFL
2) eddybob123

Solution (from MarkFL):
Let $d$ be the total distance walked, with $t_s$, the time during the walk it was sunny and $t_c$ the time it was cloudy. We may then state:

(1) $$\displaystyle d=st_s+(s+1)t_c$$

Dividing through by $d$, and rearranging, we obtain:

(2) $$\displaystyle \frac{st_s}{d}=1-\frac{(s+1)t_c}{d}$$

The expression on the left is what we are trying to find, as it represents the fraction of the total distance covered while the sun was shining. Using the information regarding the average speed, we may write:

(3) $$\displaystyle d=2.8\left(t_s+t_c \right)$$

Using (1) and (3), we find:

$$\displaystyle d=st_s+(s+1)t_c=2.8\left(t_s+t_c \right)$$

Solving this for $t_c$, we find:

(4) $$\displaystyle t_c=(2.8-s)\left(t_s+t_c \right)$$

Using (3) and (4), we may now write (2) as:

$$\displaystyle \frac{st_s}{d}=1-\frac{(s+1)(2.8-s)\left(t_s+t_c \right)}{2.8\left(t_s+t_c \right)}$$

$$\displaystyle \frac{st_s}{d}=1-\frac{(s+1)(2.8-s)}{2.8}$$

Now, since the average speed was non-integral, we know we must have:

$$\displaystyle 0<\frac{st_s}{d}<1$$

$$\displaystyle 0<1-\frac{(s+1)(2.8-s)}{2.8}<1$$

Multiply through by 2.8:

$$\displaystyle 0<2.8-(s+1)(2.8-s)<2.8$$

$$\displaystyle 0<2.8-2.8s+s^2-2.8+s<2.8$$

$$\displaystyle 0<s^2-1.8s<2.8$$

We find $s=2$ is the only integral solution which satisfies the inequality, hence:

$$\displaystyle \frac{st_s}{d}=1-\frac{(2+1)(2.8-2)}{2.8}=1-\frac{3(0.8)}{2.8}=1-\frac{6}{7}=\frac{1}{7}$$

Thus, we find Derek covered 1/7 of the total distance while the sun was shining on him.

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