network of car dealerships

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network of car dealerships

by himu » Sat Mar 30, 2013 12:22 am
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

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by rintoo22 » Sat Mar 30, 2013 1:32 am
d sales directors
a sales assocaites

d sold 10 cars each = d * 10
s sold 20 cars each = s * 20

Number of cars sold = (d*10) + (d(a*20))

Statement 1
(d*10) + (d(a*20)) = 270
10d + 20ad = 270...... (1)
There are 2 combination of a and d which satisfies the eq
a = 4 and d = 3
a = 1 and d = 9
Therefore not sufficient

Statement 2
a > d > 2
Alone is not sufficient as we do not know the number of cars.

Statement 1 and Statement 2 Sufficent
a = 4 and d = 3 to satisfy eq 10d + 20ad = 270 and in equality a > d > 2.

Answer C

Hope this helps.

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by GMATGuruNY » Sat Mar 30, 2013 3:11 am
himu 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 skamal7 » Mon Apr 22, 2013 8:55 pm
here if d=7 and a =10 also satisfies the criteria of selling 270 cars..Hence to possiblities d=3 a=4 and d=7 a=10 .Two values.hence Insufficient..Can some one tell me why am getting insufficient.Whats the mistake i am making

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by Anju@Gurome » Mon Apr 22, 2013 9:11 pm
skamal7 wrote:here if d=7 and a =10 also satisfies the criteria of selling 270 cars..Hence to possiblities d=3 a=4 and d=7 a=10 .Two values.hence Insufficient..Can some one tell me why am getting insufficient.Whats the mistake i am making
No.
Read the problem statement again : "... a group of d sales directors each has a team of a sales associates"
Hence, total number of sales associates = d*a

Now, if d = 7 and a = 10, total number of sales associates = 7*10 = 70
So, number of cars sold by sales directors = 7*10 = 70 and number of cars sold by sales associates = 70*20 = 1400
Total number of cars sold = (70 + 1400) > 270

Hope that helps.
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