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# Yesterday, an automobile dealership sold exactly 15 vehicles

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Yesterday, an automobile dealership sold exactly 15 vehicles  [#permalink]

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Updated on: 24 Dec 2013, 02:12
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Yesterday, an automobile dealership sold exactly 15 vehicles for a total of \$225,000. Did at least one of the vehicles sell for more than \$16,500?

(1) The median price for the 15 vehicles was \$13,000.

(2) The range for the price of the 15 vehicles was \$4,000.

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Originally posted by madhavsrinivas on 23 Dec 2013, 11:11.
Last edited by Bunuel on 24 Dec 2013, 02:12, edited 1 time in total.
Edited the question.
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Re: Yesterday, an automobile dealership sold exactly 15 vehicles  [#permalink]

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23 Dec 2013, 19:54
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madhavsrinivas wrote:
Yesterday, an automobile dealership sold exactly 15 vehicles for a total of \$225,000. Did at least one of the vehicles sell for more than \$16,500?

(1) The median price for the 15 vehicles was \$13,000.

(2) The range for the price of the 15 vehicles was \$4,000.

Dear madhavsrinivas
I'm happy to help.

First, here's a blog you may find helpful about mean and median:
http://magoosh.com/gmat/2012/common-gma ... tatistics/

In the prompt, we are told --- 15 vehicles sold for \$225,000, so the average price was \$225,000/15 = \$15,000. That's the average. That's important.

Statement #1: The median price for the 15 vehicles was \$13,000.
Very interesting. If they median was \$13K, then half the cars were at that price or below. For every car below \$13K, we need something above #17K to balance it out, so that the average stays at \$15K. If half the cars are below \$13K, and the highest cars are not above \$16,500, then it would be impossible to have an average of \$15K. This statement undoubted implies that at least one car is above \$17K, and therefore has to be above \$16,500. This gives a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.

Statement #2: The range for the price of the 15 vehicles was \$4,000.
Well, from the prompt, all we know is that the average is \$15K. It could be that five cars solve for \$13K, five for \$15K, and five for \$17K --- a range of \$4000, and this give a "yes" answer to the prompt question.
BUT, we could have three cars sold for \$15,000 - \$3200 = \$11,800, and twelve cars sold at \$15,000 + \$800 = \$15,800 --- still an average of \$15K, still a range of \$4000, but now a "no" answer to the prompt.
Based on this statement, we can construct an answer to the prompt either way. Nothing is definitive. This statement, alone and by itself, is not sufficient.

Answer = (A)

Does this make sense?
Mike
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Re: Yesterday, an automobile dealership sold exactly 15 vehicles  [#permalink]

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28 Sep 2016, 09:15
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There is an easy way to solve this kind of problems.
Mean=225,000\$/15=15,000\$.

(1) We know that the median is 13,000\$ , so we will MAXIMIZE the other numbers in order to see whether it is possible for a car to be under or equal to 16,500\$.

13,000\$ + 13,000\$ +...+ 13,000\$ (median) + 16,500\$ + 16,500\$ +...+ X (the term that we need to find out if it is over 16,5k.

Therefore, we have a total of 8*13,000\$ + 6*16,500\$ + X = 225,000\$
=> 104,000\$ + 99,000\$ + X = 225,000\$
=> X=225,000\$-203,000\$= 22,000\$ , So (1) is Sufficient.
X>16,500\$ when all the other numbers are MAXIMIZED, hence there is a X that is always over 16,500\$ in order for the statement to be true.

(2) The range is 4,000\$.
The numbers in the set could take different values in order for this statement to hold true.

For example, 7* 13,000\$ + 15,000\$ + 7* 17,000\$.
3*11,800\$ + 12*15,800\$.
So statement (2) is not sufficient.

Answer (A)
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Re: Yesterday, an automobile dealership sold exactly 15 vehicles  [#permalink]

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27 Mar 2014, 10:06
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mikemcgarry wrote:
madhavsrinivas wrote:
Yesterday, an automobile dealership sold exactly 15 vehicles for a total of \$225,000. Did at least one of the vehicles sell for more than \$16,500?

(1) The median price for the 15 vehicles was \$13,000.

(2) The range for the price of the 15 vehicles was \$4,000.

Dear madhavsrinivas
I'm happy to help.

First, here's a blog you may find helpful about mean and median:
http://magoosh.com/gmat/2012/common-gma ... tatistics/

In the prompt, we are told --- 15 vehicles sold for \$225,000, so the average price was \$225,000/15 = \$15,000. That's the average. That's important.

Statement #1: The median price for the 15 vehicles was \$13,000.
Very interesting. If they median was \$13K, then half the cars were at that price or below. For every car below \$13K, we need something above #17K to balance it out, so that the average stays at \$15K. If half the cars are below \$13K, and the highest cars are not above \$16,500, then it would be impossible to have an average of \$15K. This statement undoubted implies that at least one car is above \$17K, and therefore has to be above \$16,500. This gives a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.

Statement #2: The range for the price of the 15 vehicles was \$4,000.
Well, from the prompt, all we know is that the average is \$15K. It could be that five cars solve for \$13K, five for \$15K, and five for \$17K --- a range of \$4000, and this give a "yes" answer to the prompt question.
BUT, we could have three cars sold for \$15,000 - \$3200 = \$11,800, and twelve cars sold at \$15,000 + \$800 = \$15,800 --- still an average of \$15K, still a range of \$4000, but now a "no" answer to the prompt.
Based on this statement, we can construct an answer to the prompt either way. Nothing is definitive. This statement, alone and by itself, is not sufficient.

Answer = (A)

Does this make sense?
Mike

I did it in the following way. Mike please advice

So very well indeed average is 15.
At least 16,500 anyone?

Lets hit statement 1

1) Median is 13. So we have 7 terms that are at least 13, plus the median which is of course 13. Then we have 13*8 as a maximum value. Thus 225 - (13*8) = 121 to go. Now the other 7 cars must be 13 at most. Let's say that they are all equal to the median = 13. Then we will have 13*7=91 and 121-91=30. We are still missing 30. If we distribute those 30 equally among those we would have something like 4.2 more per car which will lead to a price of 13+4.2= 17>16.5 therefore this statement is sufficient

2) Range is 4k

Let's see so then we can assume that x and 1.4x account for the lowest and highest amount. If we have that 14 (1.4x) +x = 225 we can find that x=11. Therefore, x could be smaller than 17

Insufficient

Answer is thus A

Does it all make sense?

Cheers
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Re: Yesterday, an automobile dealership sold exactly 15 vehicles  [#permalink]

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27 Mar 2014, 10:53
1
jlgdr wrote:
I did it in the following way. Mike please advice

So very well indeed average is 15.
At least 16,500 anyone?

Lets hit statement 1

1) Median is 13. So we have 7 terms that are at least 13, plus the median which is of course 13. Then we have 13*8 as a maximum value. Thus 225 - (13*8) = 121 to go. Now the other 7 cars must be 13 at most. Let's say that they are all equal to the median = 13. Then we will have 13*7=91 and 121-91=30. We are still missing 30. If we distribute those 30 equally among those we would have something like 4.2 more per car which will lead to a price of 13+4.2= 17>16.5 therefore this statement is sufficient

2) Range is 4k

Let's see so then we can assume that x and 1.4x account for the lowest and highest amount. If we have that 14 (1.4x) +x = 225 we can find that x=11. Therefore, x could be smaller than 17

Insufficient

Answer is thus A
Does it all make sense?
Cheers
J

Dear jlgdr,
I'm happy to respond.

I really like your approach to Statement #1. I guess the only thing I would say it: clearly you & I can do all of that without a calculator, but I wonder if the average GMAT test taker would have the stomach for doing all that without a calculator. That was a number of calculations in a short space, and while you & I can do that, not everybody can. Other than that, your approach to #1 was excellent.

In #2, I have a criticism. Suppose x is the minimum, the lowest amount. Well, 1.4x would be a 40% increase from x, but that doesn't necessarily give us a range of \$4000. For example, if x = 11, then 1.4x = 1.4*11 = 15.4, a range of 4.4, or \$4400. If we are going to call x the minimum value, then in order to have a range of exactly 4000, we would have to use (x + 4000) --- or (x + 4), if we are dividing off the extraneous factor of a thousand. Then
Min = x
Max = x + 4
Assume we have one at the min and fourteen at the max, so
x + 14(x + 4) = 225
15x + 56 = 225
15x = 169
11 < x < 12, about 11.2
so, the max, (x + 4), would be less than 16.

Does all this make sense, my friend?
Mike
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Re: Yesterday, an automobile dealership sold exactly 15 vehicles  [#permalink]

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25 Sep 2016, 09:13
This can be done in three steps with one calculation

All based around the idea that any set is made up of three sets (A) (MEDIAN) (B) where n in set A = n in set B and total n = 2n + 1

Answer:
225,000 – 13,000 = 212,000 / 14 is equal to a number > 15k
Multiply by 2X = a number > 30k
13,000 + Min average values above median = number > 30k
Min X > 16,500 therefore X must include 16,500 (Sufficient)

(2) is insufficient since it tells us nothing about the median or distribution of values
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Re: Yesterday, an automobile dealership sold exactly 15 vehicles  [#permalink]

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23 Feb 2017, 04:51
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PROMPT ANALYSIS
The total value of 15 cars is \$225000 ie.e the average price is \$15000.

SUPERSET
The answer to this question will be either YES or NO.

TRANSLATION
In order to find the value, we need:
Exact value of each car.
Any equation to solve for price of each car.
Any characteristics of of the set so that we can predict the range of prices.

STATEMENT ANALYSIS

St 1: We will try to negate with the help of the statement. In order to do that we will take all the values which can be evenly distributed and close to the average.

If we put all the prices in ascending order, the cost of 7th car will be \$13000. Let us assume that the value of all 8 cars be \$13000. That means the summation of all the values of 8 car is \$104000. The cost of rest of the 7 cars will be \$225000 - \$104000 = \$121000.Distributing it evenly again, we get the average price of rest of 7 cars nearly \$17528 which is greater than \$16500. Hence we can cay that at least 1 car will be there whose value is greater than \$16500. SUFFICIENT. Option b, c, e eliminated.

St 2: We take condition like 3 cars were sold for \$11800 and 12 cars sold for \$15800 each. For this, the range is \$4000 and average is \$15000 but the maximum price is less than \$16500.
We take another case wil 5 cars for \$13000, 5 cars for \$15000 and 5 cars for \$17000. In this case, it exceeds. Hence INSUFFICIENT. Option D eliminated

Option A
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Re: Yesterday, an automobile dealership sold exactly 15 vehicles  [#permalink]

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01 Jul 2017, 00:24
madhavsrinivas wrote:
Yesterday, an automobile dealership sold exactly 15 vehicles for a total of \$225,000. Did at least one of the vehicles sell for more than \$16,500?

(1) The median price for the 15 vehicles was \$13,000.

(2) The range for the price of the 15 vehicles was \$4,000.

Sum = \$225,000

(1) median = \$13,000

x1, x2, x3, x4, x5, x6, x7, 13,000, x8, x9, x10, x11, x12, x13, x14, x15
x1=x2=x3=x4=x5=x6=x7=13,000
8(13,000) + 7x = 225,000
x=17,286

Sufficient.
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Re: Yesterday, an automobile dealership sold exactly 15 vehicles  [#permalink]

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15 Aug 2018, 03:34
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To make the math easier, omit the last two 0's in every value:

madhavsrinivas wrote:
Yesterday, an automobile dealership sold exactly 15 vehicles for a total of \$2250. Did at least one of the vehicles sell for more than \$165?

(1) The median price for the 15 vehicles was \$130.

(2) The range for the prices of the 15 vehicles was \$40.

Statement 1:
Sum of the remaining 14 prices = 2250 - 130 = 2120.
Try to MINIMIZE the highest price so that none of the prices exceeds 165.
To minimize the highest price, we must MAXIMIZE the 7 lowest prices.
Since the median price = 130, the maximum sum for the 7 lowest prices = 7*130 = 910, with the result that the sum of the 7 highest prices = 2120 - 910 = 1210.
If each of the 7 highest prices = 165, then the sum of the 7 highest prices = 7*165 = 1155.
The sum in red is too small.
Implication:
For the sum of the 7 highest prices to be 1210, at least one of the prices must be GREATER THAN 165.
Since it is not possible to minimize the highest price so that none of the prices exceeds 165 -- implying that the highest price must be greater than 165 -- the answer to the question stem is YES.
SUFFICIENT.

Statement 2:
Case 1: highest price = 160, lowest price = 120, the remaining 13 prices = (2250 - 160 - 120)/13 ≈ 151
In this case, none of the prices exceeds 165, so the answer to the question stem is NO.
Case 2: highest price = 170, lowest price = 130, the remaining 13 prices = (2250 - 170 - 130)/13 = 150
In this case, the highest price exceeds 165, so the answer to the question stem is YES.
INSUFFICIENT.

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Re: Yesterday, an automobile dealership sold exactly 15 vehicles  [#permalink]

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19 Aug 2019, 19:01
Official Explanation from Veritas

In the question stem, you learn that the total price for the 15 vehicles was \$225,000, so the average price per vehicle was \$15,000. You must then analyze the statements to decide if enough information is given to prove that at least one of the cars was sold for more than \$16,500.

Statement (1) tells you that the median price was \$13,000, guaranteeing that 7 of the vehicles sold for \$13,000 or less, 7 for \$13,000 or more, and one for exactly \$13,000 in the middle. At this point, you should harness the Min/Max approach taught in the Word Problems book. Since \$13,000 must be in the middle, you should maximize the lower priced vehicles all at \$13,000 (remember that if the lowest 8 are all \$13,000 then the median is still \$13,000) giving 8 vehicles at \$13,000 accounting for \$104,000 of the \$225,000. That leaves \$121,000 that must be accounted for by the remaining 7 highest priced vehicles. If you divide 121 by 7 (or just estimate) you see clearly that the average price of those 7 vehicles is slightly above \$17,000. This guarantees that at least one vehicle is above \$16,500 because if all 7 vehicles were below \$16,500 then it would be impossible to meet that total of \$121,000. It is important to note that if any of the lower priced vehicles were less than \$13,000 (which of course they could be), then the higher priced vehicles would have to account for even more money and the average price required on those highest priced vehicles would be even greater (this is why you wanted to maximize the lower priced vehicles). Statement (1) is thus sufficient.

In the second statement, you are given the range but this information combined with the average given in the question stem is not sufficient. For instance you could have 11 vehicles at \$16,000, 1 at \$13,000, and 3 at \$12,000 giving the proper range and average with no vehicles over \$16,500. Or you could easily make more than one be over \$16,500 with the same range – for instance 5 vehicles at \$17,000, 5 vehicles at \$13,000, and 5 vehicles at \$15,000 giving the proper range and average with multiple vehicles over \$16,500. Statement (2) is thus not sufficient. Answer is (A).
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Yesterday, an automobile dealership sold exactly 15 vehicles  [#permalink]

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19 Aug 2019, 19:21
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madhavsrinivas wrote:
Yesterday, an automobile dealership sold exactly 15 vehicles for a total of \$225,000. Did at least one of the vehicles sell for more than \$16,500?

(1) The median price for the 15 vehicles was \$13,000.

(2) The range for the price of the 15 vehicles was \$4,000.

Given: Yesterday, an automobile dealership sold exactly 15 vehicles for a total of \$225,000.
Average price of 15 vehicles = \$225,000/15=\$15,000

Asked: Did at least one of the vehicles sell for more than \$16,500?

(1) The median price for the 15 vehicles was \$13,000.
Median price is the price of 8th vehicle when the prices are arranged in increasing order.
8 vehicle have price less than or equal to \$13,000 or a sum of less than or equal \$13,000*8=\$104,000.
Remaining 7 vehicles will have total price greater than or equal \$121,000 averaging \$17,285>\$16,500. Since for minimum price of vehicle, all remaining 7 have to be of the same price equal to average.
SUFFICIENT

(2) The range for the price of the 15 vehicles was \$4,000.
Average price of 15 vehicles = >15,000
Range of prices =\$4,000
To minimise maximum price, let us assume that all 14 vehicles have equal price p and 1 vehicle have price p-4000
Total price of 15 vehicles = 14p+p-4000=225000
15p=229000
p=15,266<\$16,500
To maximise maximum price, let us assume that 14 vehicles have price p and 1 vehicle has price p+4000
14p+p+4000=225000
p=221000/15= \$14,733
p+4000=\$18,733>\$16,500
NOT SUFFICIENT

IMO A

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Yesterday, an automobile dealership sold exactly 15 vehicles   [#permalink] 19 Aug 2019, 19:21
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