Help with two basic word problems

[sbart]

I need help with the following:
Question 1:
Sixteen construction workers finished half a road in ten days. If four more workers are added to the crew, how long will it take them to finish the road?

I get the answer of 7.5 days, however, the solution states 8 days.


Question 2:
What is the probability of getting a pair if you are dealt two cards from a standard deck.

According to the answer key, the answer is 1/7

Please help if you can!

 

[soroban]

Hello, sbart!

1. Sixteen construction workers finished half a road in ten days.
If four more workers are added to the crew, how long will it take them to finish the road?

I get the answer of 7.5 days, however, the solution states 8 days.

Of course, we assume that all workers work at the same constant rate.


\(\displaystyle \begin{array}{cccc}\text{Workers} & \text{Days} & \text{Road} \\ \hline
16 & 10 & \frac{1}{2} \\ 16 & 1 & \frac{1}{20} \\ 1 & 1 & \frac{1}{320} \end{array}\)


With 20 workers, we have:

. . \(\displaystyle \begin{array}{cccc}\text{Workers} & \text{Days} & \text{Road} \\ \hline 20 & 1 & \frac{1}{16} \\ 20 & 8 & \frac{1}{2} \end{array}\)


With 20 workers, it will take 8 more days to finish the other half of the road.

2. What is the probability of getting a pair if you are dealt two cards from a standard deck?
According to the answer key, the answer is 1/7 . 1/17 ?

The first card can be any card: .\(\displaystyle \frac{52}{52} \,=\,1\)

The second card must match the value of the first card: .\(\displaystyle \frac{3}{51} \,=\,\frac{1}{17}\)

. . \(\displaystyle P(\text{Pair}) \:=\:1\cdot\dfrac{1}{17} \:=\:\dfrac{1}{17}\)

 

[sbart]

A little more help

Thanks I also got the solution of 1/17 for the card problem using the same method and for the life of me can not see any way to get the 1/7. On the road problem how do you get the 20 workers 1 day and 1/16 of the road? To me it looks as if you have a 25% increase in manpower and therefore they would complete the road in 7.5 days.

Hello, sbart!


Of course, we assume that all workers work at the same constant rate.


\(\displaystyle \begin{array}{cccc}\text{Workers} & \text{Days} & \text{Road} \\ \hline
16 & 10 & \frac{1}{2} \\ 16 & 1 & \frac{1}{20} \\ 1 & 1 & \frac{1}{320} \end{array}\)


With 20 workers, we have:

. . \(\displaystyle \begin{array}{cccc}\text{Workers} & \text{Days} & \text{Road} \\ \hline 20 & 1 & \frac{1}{16} \\ 20 & 8 & \frac{1}{2} \end{array}\)


With 20 workers, it will take 8 more days to finish the other half of the road.




The first card can be any card: .\(\displaystyle \frac{52}{52} \,=\,1\)

The second card must match the value of the first card: .\(\displaystyle \frac{3}{51} \,=\,\frac{1}{17}\)

. . \(\displaystyle P(\text{Pair}) \:=\:1\cdot\dfrac{1}{17} \:=\:\dfrac{1}{17}\)