A certain dealership has a number of cars to be sold by its

Yes.

A certain dealership has a number of cars to be sold by its salespeople. How many cars are to be sold?

(1) If each of the salespeople sells 4 of the cars, 23 cars will remain unsold

(2) If each of the salespeople sells 6 of the cars, 5 cars will remain unsold

"The" in both statements refer to the salespeople mentioned in the stem.

A certain dealership has a number of cars to be sold by its salespeople. How many cars are to be sold?

(1) If each of the salespeople sells 4 of the cars, 23 cars will remain unsold.
(2) If each of the salespeople sells 6 of the cars, 5 cars will remain unsold.

Let there n be number of cars and K be the no. of sales people.

Using St 1, we can say that total no.of cars ie. n = 4K+23
We cannot find n using this information. Hence A and D ruled out

Using st2,we get that n =6K+5. Again using this equation, we cannot find n so B is ruled out

Combining both the statements, we get 4K+23= 6K+5 or 2K=18 or K =9 (No. of Sales people) and n =59.

Ans C

Just look at the remainder: 23 and 5 --> So the change is 23 - 5 = 18, and the change in the #of sold cars is 2 (from 4 to 6)
18/2 = 9 Salesperson just plugin this number and the answer is:
1) 9*4 + 23 = 59
2) 9*6 +5 = 59

Bunuel,

Are we supposed to take that the salespersons are same in both the statements?

Solution:

Question Stem Analysis:

We need to determine the number of cars to be sold by the salespeople in a car dealership. We can let the number of cars to be sold be c and the number of salespeople be s. We need to determine the value of c.

Statement One Alone:

We see that c = 4s + 23. However, we have one equation and two variables, and we can’t determine the value of c. Statement one alone is not sufficient.

Statement Two Alone:

We see that c = 6s + 5. However, we have one equation and two variables, so we can’t determine the value of c. Statement two alone is not sufficient.

Statements One and Two Together:

With the two statements, we have two linear equations and two variables. Note that neither equation is dependent on the other, which means that one equation is not a linear multiple of the other., Thus, we can determine the value of c (and also s). Both statements together are sufficient.

Answer: C

They don't really give you an equation in the stem, all we need to find is T, given that X(i)*S(j) = Cars to be sold, where i,j denote the observations: numbers of cars for observation (i) sold per seller (j). In other words, how many cars each person sells (we don't know that everyone sells equally many cars).

1) This gives us x(i) = 4 for all observations x(i). So we have 4*S = T - 23 <-- this many cars have been sold. Clearly insufficient, we have 2 variables.
2) This gives us x(i) = 6, for all x(i).. So we have 6*S = T - 5.. Also insufficient, for the same reason as 1.

1 + 2. First, note that the information in 1 and 2 is NOT the same, one is not a multiple of the other. So we have 2 equations, 2 unknowns, and thus we can solve for T.

Answer: C

Hi,

Alternate but lengthy approach
Let number of cars be n , and numbers of sales persons be S

then

From Stmt 1: n= 4S+23

From Stmt 2: n= 6S+5

Combining 1 and 2 we have

From Stmt 2 :\(\frac{n}{6}\)reminder 5
or \(\frac{4S+23}{6}\)

Note that we will have a cylicity of 3 for reminders here,this is what i mean,
For s=1 n= 27 and \(\frac{27}{6}\) reminder will be 3
For s=2 n= 31 and \(\frac{31}{6}\) reminder will be 1
For s=3 n= 35 and \(\frac{35}{6}\) reminder will be 5
For s=4 n= 39 and \(\frac{39}{6}\) reminder will be 3


we get reminder 5 when n=35 or 47 or 59

Now lets check at which value does it give reminder of 23 when divided by 4.

But when \(\frac{35}{6}\) we have 5 sales person and 5 cars remaining but when \(\frac{35}{4}\) when we have 5 sales person each selling 4 cars we will have 15 cars remaining
But when \(\frac{47}{6}\) we have 7 sales person and 5 cars remaining but when \(\frac{47}{4}\) when we have 7 sales person each selling 4 cars we will have 19 cars remaining
But when \(\frac{59}{6}\) we have 9 sales person and 5 cars remaining but when \(\frac{59}{4}\) when we have 9 sales person each selling 4 cars we will have 23 cars remaining Bingo

So Answer is C
Probus

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