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

Question

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

(1) The total number of cars sold was 270

(2) a > d > 2

Bunuel · Accepted Answer

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

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.

Answer: C.

Hope it's clear.

Zarrolou · Answer

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

(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

skamal7 · Answer

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

skamal7 · Answer

Ya now its correct and thanks for editing ur posts :-D

honchos · Answer

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

WoundedTiger · Answer

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???

khairilthegreat · 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

OreoShake · Answer

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.

Bunuel · Answer

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.

umg · Answer

# 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.

tanishqgirotra · Answer

I think statement A is also alone sufficicient:
 we have the number of directors as 'd', and let's for now call all the number of associates under all the directors as 'a'. So, 10d + 20a = 270
Now, d cannot be 1 as the question says it is a group of d directors, d cannot be 2 as 250/20 is not an integer, d can be 3 leading to 'a = 12' and 12/3 = 4 associates under each director. D cannot be 9 as the question says team of 'a' associates and definitely d cannot be 27 as that would lead to a = 0. So, we are left with only 1 option that is 3 directiors and 4 associates under each director and 12 persons in total. Correct me if I am wrong anywhere. I legit eliminated the pairs beacuse of these reasonings.

bumpbot · Answer

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

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