In a network of car dealerships, a group of d sales director

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In a network of car dealerships, a group of d sales directors each has a team of a sales associates. In a given month, if the directors each sold 10 cars and each sales associate sold 20 cars, and all cars were sold by either a sales director or a sales associate, how many people total sold cars?

(1) The total number of cars sold was 270

(2) a > d > 2


OA: C
Last edited by RBBmba@2014 on Fri May 08, 2015 12:38 am, edited 1 time in total.

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by GMATGuruNY » Thu May 07, 2015 7:14 am
RBBmba@2014 wrote:In a network of car dealerships, a group of d sales directors each has a team of a sales associates. In a given month, if the directors each sold 10 cars and each sales associate sold 20 cars, and all cars were sold by either a sales director or a sales associate, how many people total sold cars?

(1) The total number of cars sold was 270

(2) a > d > 2
For each of the d directors, there are a associates.
Thus, the total number of associates = d*a.
Each director sells 10 cars, for a total of 10d cars.
Each associate sells 20 cars, for a total of 20ad cars.
Thus:
Total cars sold = 10d + 20ad.

Statement 1: The total number of cars sold was 270.
10d + 20ad = 270.
d + 2ad = 27.
d(1 + 2a) = 27.
Thus, d is a factor of 27: 1, 3, 9, or 27.
Since d can be different values, INSUFFICIENT.

Statement 2: a > d > 2
Since a and d can be different values, INSUFFICIENT.

Statements combined: d(2a+1) = 27 and a>d>2.
d is a factor of 27 greater than 2: 3, 9, or 27.
If d=3, we get:
3(2a+1) = 27
2a+1 = 9
2a=8
a=4.
If the value of d INCREASES, then the value of a must DECREASE, violating the constraint that a>d.
Thus, d=3 and a=4.
SUFFICIENT.

The correct answer is C.
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by RBBmba@2014 » Fri May 08, 2015 2:06 am
Thanks Mitch. I missed the catch "the total number of associates = d*a".

The approach I followed (obviously after getting the above missing link right, from your solution) is plugging in values ,although it took a lot of time to reach to answer. Whereas, your equation based approach is a faster one BUT I guess, it's a bit tricky to understand during test environment that this 'equation based approach' could be used in such a way.

Most critical part would be I think, to quickly understand this -

d(1 + 2a) = 27.
Thus, d is a factor of 27: 1, 3, 9, or 27.


So, my question is whether there is any other smarter approach to this 700+ level (I presume it's so) problem?

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by RBBmba@2014 » Tue May 12, 2015 2:34 am
Mitch - could you please provide a quick feedback on the above post ? Much thanks in advance!

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by GMATGuruNY » Tue May 12, 2015 6:16 am
RBBmba@2014 wrote:Thanks Mitch. I missed the catch "the total number of associates = d*a".

The approach I followed (obviously after getting the above missing link right, from your solution) is plugging in values ,although it took a lot of time to reach to answer. Whereas, your equation based approach is a faster one BUT I guess, it's a bit tricky to understand during test environment that this 'equation based approach' could be used in such a way.
The purpose of algebra is to show how the values in a problem are CONSTRAINED.

When we evaluate statement 1, we have the following information:
Each director sold 10 cars.
Each associate sold 20 cars.
The total number of cars sold was 270.

This information clearly implies an equation:
(number of cars sold by the directors) + (number of cars sold by the associates) = 270.

Since this equation is clearly implied, we should check how it constrains the unknowns in the problem.
Most critical part would be I think, to quickly understand this -

d(1 + 2a) = 27.
Thus, d is a factor of 27: 1, 3, 9, or 27.


So, my question is whether there is any other smarter approach to this 700+ level (I presume it's so) problem?
Equation implied by statement 1:
10d + 20ad = 270.

When given an equation, SIMPLIFY.
Here, we can simplify by dividing by 10 and by factoring out d:
d + 2ad = 27.
d(1 + 2a) = 27.

Now we can see how the values of a and d are constrained.
d and (1 + 2a) must both be factors of 27.

Please note that there is no such thing as a "smarter" approach.
Whenever it is suggested that a particular solution is the "best," be skeptical:
You must determine whether that approach truly is the best way for YOU.
The best approach for one test-taker might not be the best approach for another.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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I unlock the best way for YOU to solve problems.

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