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?
Pat, Kate and Marck charged a total of 162 hours of...
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Test the SMALLEST POSSIBLE CASE.AAPL wrote: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
Let M = 6.
Since Pat charged 1/3 as much as Mark, P = (1/3)(6) = 2.
Since Pat charged twice as much as Kate, Kate's amount is 1/2 Pat's amount:
K = (1/2)(2) = 1.
Smallest possible total = 6+2+1 = 9.
Since (actual total)/(smallest possible total) = 162/9 = 18, the values for M, P and K must each be increased by a factor of 18:
M = 6*18 = 108, P = 2*18 = 36, K = 1*18 = 18, T = M+P+K = 108+36+18 = 162.
Thus:
M-K = 108-18 = 90.
The correct answer is D.
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We can also solve the question using one variableAAPL wrote: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
We can see that Kate charged the fewest hours, so...
Let x =the number of hours Kate charged
Pat charged twice as much time to the project as Kate
So, 2x = the number of hours Pat charged
Pat charged 1/3 as much times as Mark
In other words, Mark charged THREE TIMES as much time as Pat
So, 3(2x ) = the number of hours Mark charged
In other words, 6x = the number of hours Mark charged
Pat, Kate and Mark charged a total of 162 hours to a certain project.
We can write: x + 2x + 6x = 162
Simplify: 9x = 162
Solve: x = 18
So, Kate charged 18 hours
When we plug x = 18 into 6x, we see that Mark charged 108 hours
How many more hours did Mark charge to the project than Kate?
Answer = 108 - 18
= 90
= D
Cheers,
Brent
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- Jeff@TargetTestPrep
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We can let P, K, and M = the amount time spent on the project by Pat, Kate, and Marck, respectively, and create the equations:AAPL wrote: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
P + K + M = 162
and
P = 2K
P/2 = K
and
P = (M)(1/3)
3P = M
Substituting, we have:
P + P/2 + 3P = 162
Multiplying by 2, we have:
2P + P + 6P = 324
9P = 324
P = 36, so M = 108 and K = 18.
M - K = 108 - 18 = 90.
Answer: D
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