Pat, Kate, and Marck charged a total of 162 hours of a certain project. If Pat charged twice as much time to the project as Kate and 1/3 as much times as Marck. How many more hours did Marck charge to the project than Kate?
A. 18
B. 36
C. 72
D. 90
E. 108
The OA is D.
I think that it can be solved as follow,
If Pat, Kate, and Marck charged o total of 162 hours, it can be written as
$$P+K+M=162\ \ (1)$$
Pat charged twice as much time as Kate, it can be written as
$$P=2K\ \ (2)$$
Pat charged 1/3 as much times as Mark, that's mean
$$P=\frac{1}{3}M\ \ (3)$$
We have 3 equations and 3 incognitas, solving it we get,
$$From\ \left(2\right)\ \Rightarrow K=\frac{P}{2}\ $$
$$From\ \left(3\right)\ \Rightarrow M=3P\ $$
Then, substituting them into (1), we get
$$P+\frac{P}{2}+3P=162\ \Rightarrow P=36$$
Hence
$$K=18,\ \ M=108$$
Finally
$$M-K=108-18=90$$
Is there another strategic approach to solve this PS question? Can any experts help, please?
A. 18
B. 36
C. 72
D. 90
E. 108
The OA is D.
I think that it can be solved as follow,
If Pat, Kate, and Marck charged o total of 162 hours, it can be written as
$$P+K+M=162\ \ (1)$$
Pat charged twice as much time as Kate, it can be written as
$$P=2K\ \ (2)$$
Pat charged 1/3 as much times as Mark, that's mean
$$P=\frac{1}{3}M\ \ (3)$$
We have 3 equations and 3 incognitas, solving it we get,
$$From\ \left(2\right)\ \Rightarrow K=\frac{P}{2}\ $$
$$From\ \left(3\right)\ \Rightarrow M=3P\ $$
Then, substituting them into (1), we get
$$P+\frac{P}{2}+3P=162\ \Rightarrow P=36$$
Hence
$$K=18,\ \ M=108$$
Finally
$$M-K=108-18=90$$
Is there another strategic approach to solve this PS question? Can any experts help, please?
