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
\(a=13,\) \(d=1,\) \(20*13+10*1=270\) 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

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

Ya now its correct and thanks for editing ur posts

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

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

(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

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.

# 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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