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# Working hours problem

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Can you help me with this problem?

I can finish a job 2 hours longer than you. After I had been working for 1 hour,  you joined me and we completed the remaining portion of the job after 3 hours.

How long would it take me to do the job if I worked alone?

Sep 20, 2018

#1
+5226
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$$\text{Let's say I can finish the job in }n \text{ hours} \\ \text{thus I work at a rate of }\dfrac{1~job}{n~hr} \\ \text{You finish the job in (}n-2) \text{ hours}\\ \text{and thus work at a rate of }\dfrac{1~job}{(n-2)~hr}$$

$$\text{When working together the rates combine.}\\ \text{I work 1 hour at a rate of }\dfrac{1~job}{n~hr} \\ \text{and then work 3 hours at a rate of } \left(\dfrac{1}{n}+\dfrac{1}{n-2}\right) \dfrac {job}{hr} \text{completing the job}$$

$$1~job = \left(\dfrac {1~job}{n~hr} \cdot 1~hr+\left(\dfrac 1 n+\dfrac{1}{n-2} \right)\dfrac{job}{hr} \cdot 3~hr\right)$$

dropping units for a moment

$$1 = \dfrac 1 n +\dfrac{2n-2}{n(n-2)}\cdot 3$$

$$n^2 - 2n = 7n-8 \\ n^2 -9n+8 = 0 \\ (n-1)(n-8) = 0 \\ n=1,~8$$

$$\text{n can't be 1 as that would make your partner finish the job in (-1) hours}\\ \text{so it must be that }n=8~ hrs$$

.
Sep 20, 2018

#1
+5226
+2

$$\text{Let's say I can finish the job in }n \text{ hours} \\ \text{thus I work at a rate of }\dfrac{1~job}{n~hr} \\ \text{You finish the job in (}n-2) \text{ hours}\\ \text{and thus work at a rate of }\dfrac{1~job}{(n-2)~hr}$$

$$\text{When working together the rates combine.}\\ \text{I work 1 hour at a rate of }\dfrac{1~job}{n~hr} \\ \text{and then work 3 hours at a rate of } \left(\dfrac{1}{n}+\dfrac{1}{n-2}\right) \dfrac {job}{hr} \text{completing the job}$$

$$1~job = \left(\dfrac {1~job}{n~hr} \cdot 1~hr+\left(\dfrac 1 n+\dfrac{1}{n-2} \right)\dfrac{job}{hr} \cdot 3~hr\right)$$

dropping units for a moment

$$1 = \dfrac 1 n +\dfrac{2n-2}{n(n-2)}\cdot 3$$

$$n^2 - 2n = 7n-8 \\ n^2 -9n+8 = 0 \\ (n-1)(n-8) = 0 \\ n=1,~8$$

$$\text{n can't be 1 as that would make your partner finish the job in (-1) hours}\\ \text{so it must be that }n=8~ hrs$$

Rom Sep 20, 2018