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

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New post 27 Dec 2013, 04:37
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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.

Data Sufficiency
Question: 7
Category: Algebra Simultaneous equations
Page: 153
Difficulty: 600


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Re: A certain dealership has a number of cars to be sold by its  [#permalink]

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New post 27 Dec 2013, 04:38
SOLUTION

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 --> \(# \ of \ cars=4s+23\) --> we need to know the number of salespeople (s) to answer the question. Not sufficient.

(2) If each of the salespeople sells 6 of the cars, 5 cars will remain unsold --> \(# \ of \ cars=6s+5\) --> we need to know the number of salespeople (s) to answer the question. Not sufficient.

(1)+(2) \(# \ of \ cars=4s+23=6s+5\) --> \(s=9\) --> \(# \ of \ cars=4s+23=4*9+23\). Sufficient.

Answer: C.
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Re: A certain dealership has a number of cars to be sold by its  [#permalink]

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New post 28 Dec 2013, 04:50
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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 s be the number of salespeople and cbe the number of cars sold.
To find: c

Statement 1:
\(c = 4s + 23\)
c has many values for different s.
Not sufficient.

Statement 2:
\(c = 6s + 5\)
c has many values for different s.
Not sufficient.

Combining Statements (1) and (2),
\(c = 4s +23\)
\(c= 6s + 5\)
Solving the above simultaneous equation, we get s = 9 and c = 59

Hence, answer is (C).
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Re: A certain dealership has a number of cars to be sold by its  [#permalink]

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New post 30 Dec 2013, 01:46
1
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
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A certain dealership has a number of cars to be sold by its sale  [#permalink]

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New post 10 Jan 2014, 10:09
TriColor wrote:
Q12:
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 sales 4 of the cars, 23 cars will remain unsold.
(2) If each of the salespeople sales 6 of the cars, 5 cars will remain unsold.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


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
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Re: A certain dealership has a number of cars to be sold by its  [#permalink]

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New post 12 Jun 2014, 09:55
1
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
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Re: A certain dealership has a number of cars to be sold by its  [#permalink]

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New post 05 Aug 2014, 08:25
1
Bunuel,

Are we supposed to take that the salespersons are same in both the statements?
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Re: A certain dealership has a number of cars to be sold by its  [#permalink]

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New post 12 Aug 2014, 03:12
2
sri30kanth wrote:
Bunuel,

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


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.
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A certain dealership has a number of cars to be sold by its  [#permalink]

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New post 06 Sep 2018, 05:51
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
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