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In a network of car dealerships, a group of d sales director  [#permalink]

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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

Originally posted by skamal7 on 22 Apr 2013, 21:57.
Last edited by Bunuel on 18 Sep 2013, 00:23, edited 1 time in total.
Edited the question and the OA.
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Re: In a network of car dealerships, a group of d sales director  [#permalink]

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1
4
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?

The total # of directors = d;
The total # of associates = ad;

Total # of cars sold = 10d + 20ad.

(1) The total number of cars sold was 270 --> 10d + 20ad = 270 --> d(1 + 2a) = 27. Four cases are possible:
d = 1 and a = 13,
d = 3 and a = 4,
d = 9 and a = 1,
d = 27 and a = 0.

Not sufficient.

(2) a > d > 2. Clearly insufficient.

(1)+(2) Since from (2) a > d > 2, then only one case remains from (1): d = 3 and a = 4. Sufficient.

Hope it's clear.
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Re: n a network of car dealerships, a group of d sales  [#permalink]

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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
$$a=13,$$ $$d=1,$$ $$20*13+10*1=270$$ or $$a=12,$$ $$d=3,$$ $$20*12+10*3=270$$ Not sufficient

(2) a > d > 2
Clearly not sufficient

1+2
$$a=10, d=7, 10*20+7*10=270$$ or $$a=11,d=5,11*20+5*10=270$$. Not sufficient
E
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Re: n a network of car dealerships, a group of d sales  [#permalink]

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statement 2 says:
2) a > d > 2
but in test case u have taken " a=5,d=10 .So a<d which contradicts the question stem
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Re: n a network of car dealerships, a group of d sales  [#permalink]

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Ya now its correct and thanks for editing ur posts Senior Manager  Status: Verbal Forum Moderator
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A network of car dealerships, employs d sales directors, who  [#permalink]

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A network of car dealerships, employs d sales directors, who each supervise a sales associates. If each director sells 10 cars and each sales associate sells 20 cars, and no other cars were sold by the dealership, how many people total sold cars?

(1) The total number of cars sold was 270

(2) a>d>2

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked
Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient to answer the question asked
Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed

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Re: n a network of car dealerships, a group of d sales  [#permalink]

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Zarrolou 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
$$a=13,$$ $$d=1,$$ $$20*13+10*1=270$$ or $$a=12,$$ $$d=3,$$ $$20*12+10*3=270$$ Not sufficient

(2) a > d > 2
Clearly not sufficient

1+2
$$a=10, d=7, 10*20+7*10=270$$ or $$a=11,d=5,11*20+5*10=270$$. Not sufficient
E

dx10 + ax20 = 270
here a=d

d=9
2d total number pf person who sold the car = 2 X9 = 18 A is the correct answer
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Re: n a network of car dealerships, a group of d sales  [#permalink]

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2
Zarrolou 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
$$a=13,$$ $$d=1,$$ $$20*13+10*1=270$$ or $$a=12,$$ $$d=3,$$ $$20*12+10*3=270$$ Not sufficient

(2) a > d > 2
Clearly not sufficient

1+2
$$a=10, d=7, 10*20+7*10=270$$ or $$a=11,d=5,11*20+5*10=270$$. Not sufficient
E

Hi Zarrolou,

Isn't the answer suppose to be C

Below is my explanation

St 1 and 2 are clearly not sufficient so we rule out option A, D, and B.

Combining we get

10d+ 20 a= 270 and that a>d>2

Possible cases are

let us say d=1, a =13 so Total no . of people =14
d=3, a=12, Total= 15
d=9,a=9, Total 18
d=5, a=11, Total 16 and so one
We see that as D increases A will decrease and hence the given condition that each sales director handle "a" sales associates will not hold.

From the above condition there is only 1 case possible which d=3, a=12 and T=15. Note that each Sales director handles 4 Sales associates.

Where am I wrong on this one???
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Re: A network of car dealerships, employs d sales directors, who  [#permalink]

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1
honchos wrote:
A network of car dealerships, employs d sales directors, who each supervise a sales associates. If each director sells 10 cars and each sales associate sells 20 cars, and no other cars were sold by the dealership, how many people total sold cars?

(1) The total number of cars sold was 270

(2) a>d>2

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked
Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient to answer the question asked
Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed

Merging similar topics.
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Re: In a network of car dealerships, a group of d sales  [#permalink]

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(1) --> d*10+ad*20 =270; d(10+2a)=270
Many values could solve this equation, Not Sufficient

(2) --> a>d>2
Clearly insufficient

(1)+(2),
The first value of d and a is 3 and 4 (a and d must be integer), then
d(10+2a)=270 --> 3(10+2*4)=270 --> so d must be 4 and a must be 3.
If we increase d to 4 and a to 5, then the value is to big to get 270. Sufficient, answer C
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Re: In a network of car dealerships, a group of d sales director  [#permalink]

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Bunuel 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?

The total # of directors = d;
The total # of associates = ad;

Total # of cars sold = 10d + 20ad.

(1) The total number of cars sold was 270 --> 10d + 20ad = 270 --> d(1 + 2a) = 27. Four cases are possible:
d = 1 and a = 13,
d = 3 and a = 4,
d = 9 and a = 1,
d = 27 and a = 0.

Not sufficient.

(2) a > d > 2. Clearly insufficient.

(1)+(2) Since from (2) a > d > 2, then only one case remains from (1): d = 3 and a = 4. Sufficient.

Hope it's clear.

Hi Bunuel,

For AC A, how can we figure out the possible possible valid combinations for 'a' and 'd' quickly? it would take significant time starting from 1 and tracking back from 27. Thank you.
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Re: In a network of car dealerships, a group of d sales director  [#permalink]

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BobbyAssassinCross wrote:
Bunuel 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?

The total # of directors = d;
The total # of associates = ad;

Total # of cars sold = 10d + 20ad.

(1) The total number of cars sold was 270 --> 10d + 20ad = 270 --> d(1 + 2a) = 27. Four cases are possible:
d = 1 and a = 13,
d = 3 and a = 4,
d = 9 and a = 1,
d = 27 and a = 0.

Not sufficient.

(2) a > d > 2. Clearly insufficient.

(1)+(2) Since from (2) a > d > 2, then only one case remains from (1): d = 3 and a = 4. Sufficient.

Hope it's clear.

Hi Bunuel,

For AC A, how can we figure out the possible possible valid combinations for 'a' and 'd' quickly? it would take significant time starting from 1 and tracking back from 27. Thank you.

27 can be broken into the product of two positive integers only in 4 ways:
1*27
3*9
9*3
27*1.

From here it should not take much time to get the values of a and d.
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Re: In a network of car dealerships, a group of d sales director  [#permalink]

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skamal7 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

# of people who sold cars = d + d*a
Each director has a associates.

We should acknowledge (2) first because it is a simple statement giving us a constraint rather than a concrete data about question and it is easy to prove that insufficient.

(2) a > d > 2
Insufficient.

(1) The total number of cars sold was 270

i.e. 10d + 20d*a = 270
=> d + 2d*a = 27

1 Equation, 2 Variables. Don't be too hasty to jump on to (C) for this situation. There have been a lot of DS questions where such situations have yielded a unique answer because of the presence of "invisible" constraint - a & d are Positive Integers .

d*(1 + 2a) = 27

Notice that (1+2a) and d will always be Odd because their multiple is Odd.

Therefore,

If d = 1; 1+2a = 27 => Odd => a = 13
d = 3; 1+2a = 9 => Odd => a = 4

2 Solutions Possible. Statement (1) - Not Sufficient.

(1) + (2)..

a > d > 2 & d(1+2a) = 27

We do not have to check for every Odd value of d. We can find the factors of 27 and then move forward.

27 = $$3^3$$

Therefore,
Factors of 27 - 1, 3, 9, 27.

Therefore, If d != 1 (Constraint in Statement (2))
If d = 3; a = 4 - Valid Solution.
For d = 9; a != 1 (Constraint in Statement (2))
d = 27; a != 0 (Constraint in Statement (2))

Hence, C is the Answer.
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