Pat, Kate and Marck charged a total of 162 hours of...

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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?

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by GMATGuruNY » Wed Mar 07, 2018 5:07 am
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
Test the SMALLEST POSSIBLE CASE.

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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by Brent@GMATPrepNow » Wed Mar 07, 2018 8:08 am
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
We can also solve the question using one variable

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

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by Jeff@TargetTestPrep » Thu Mar 08, 2018 5:06 pm
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
We can let P, K, and M = the amount time spent on the project by Pat, Kate, and Marck, respectively, and create the equations:

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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