Quick Sell Outlet sold a total of 40 televisions, each of

nice conceptual question and well explained by bunuel.

1 and 2 are clearly not sufficient.

1&2. if p = 120 then q = 90 then average cannot be 141 so its has to be p = 150 and q =120.

(1)+(2) If q=120, then p=90, so in this case both prices would be less than the average price ($141) which is not possible.



hi bunuel,
i don't understand this can you pls elaborate

If you only score 70 and 80 on tests in a class, is it possible for you to get an average test score of 90 in the class?

HI,

Even here when q=120 and p=150, the average still comes out to be less than 141. Hence how can we take this as a solution?

Let number of P televisions be m, and Q Televisions be n.

p*m + q*n/m+n = 141, m + n = 40
p*m + q*n = 5640
p = q - 30
q*(m+n) - 30*m = 5640. Now, when q = 120, m will be -ve which is not possible. So, only p = 120, q = 150

Average = \(\frac{{Total Sales}}{{No. of TVs sold}}\)

Let No. of Model P TVs be a
=> Model Q TVs = (40-a)

We are given

\(\frac{{ap + q(40-a)}}{40}\) = 141

We need to find the value of a.



S1) p = q-30
=> q(a+40) - a(30+q) = 5640
=> Not enough information to find the value of a.

Insufficient.

S2) p = $120 or q = $120
=> Clearly not enough information to find the value of a.

Insufficient.



p = q-30
if p = $120 then q = $150
if q = $120 then p = $90 => Not possible at the average is $141 [Ignore this case]

Inputting values of p & q in \(\frac{{ap + q(40-a)}}{40}\) = 141
=> 120a + 150(40-a)=5640
=> we can find the value of a.

Sufficient. The answer is C

Hi mate,

You're assuming that 1 unit of both TVs was sold.
if p = 120 and q = 150 then, for the average to be $141, more of model Q would need to be sold as the average is skewed towards 150.

Hope this helps.

Bunuel when I try solve questions like this, I always have the fear of missing out on that 1 crazy case where we would be able to solve the equation even with 2 unknowns.

In this particular problem for instance, we know the total money sold was 40*141 = 5640$. Now, how do I know that Statement 1 is not sufficient?

If I substitute p=q-30 into your equation, I get 40q-30x = 5640 -> And this is when fear comes because x has to be a non negative integer, and q is a price. Those constraints are the ones that make me fear I might be in front of a crazy question where there is actually only one combination of q and x that gives 5640. Is there any way to quickly prove/disprove that I am in front of such crazy case?

Lastly, related to the issue above... what is allowed as $ price on GMAT? is 0.00000121 allowed as a price? or do prices need to be of 0.00 form (given cents is the smallest currency)?

Thank you.

PS. ccooley VeritasKarishma I will be grateful if you could give input too. Thanks a bunch.

Yes, I understand your concern and when I looked at stmnt 1, I did ensure that there are multiple solutions possible. But I used a conceptual approach to eliminate (A) as the answer since I am too lazy to use a scratch pad.

141 is the average so I imagined a number line with 141 on it.

----------------------- (141) ------------------------

The actual price of P and Q has a difference of 30 so the distance between them on the number line will be 30

--------------p--------- (141) --------------q----------

Now since total number of TVs is 40, I think to myself, can I divide 40 into 2 parts such that their ratio is the distance between p and 141 and 141 and q? (Using scale method of averages) Yes.

40 = 20 + 20 (ratio 1:1)
So p could be 15 to the left of 141 and q could be 15 to the right of 141.

40 = 16 + 24 (ratio of 2:3)
So p could be 18 to the left of 141 and q could be 12 to the right of 141.

Also, even if the second case is hard to arrive at, I know that price can be in decimal. So I just split 40 = 10 + 30
So 30 needs to be split in the ratio 1:3 i.e. 7.50 and 22.50 which is fine.

Price would usually be upto 2 decimal places only except in cases in which you buy multiple items together such as you can buy a pack of 6 apples for $2.

Given: n = 40, Avg = 141, $p, $q, P Tvs, Q Tvs.
This is a weighted average question, we can use a number line to visualize as per VeritasKarishma method:

1) the Model P televisions sold for $30 less than the Model Q televisions

P______________________Q
($q-30)----($141)----------($q)

Not sufficient since we don't have values for the endpoints, can't determine ratio.


2) Either p=120 or q=120

P______________________Q
($120)-------($141)--------($q)

Q______________________P
($120)-------($141)--------($p)

We only have 1 endpoint. Not sufficient. Note that we don't know which is the smaller value.


3) From 1) $p=$q-30, so $p is the smaller value. From 2) $p OR $q = $120

If $q=120, this becomes impossible to do because we can't have a negative amount of TVs.

P_____51________-21_____Q
($90)-------($141)--------($120)

If $p=120:

P______21__________19____Q
($120)-------($141)--------($150)

So ratio of P Tv/Q Tv is 19/21 for a total of 40, with P=19 and Q=21. Sufficient.



Done algebraically:
1) the Model P televisions sold for $30 less than the Model Q televisions

\(\frac{P}{Q}= \frac{$q - $141}{$141 - $(q-30)}\) ---> Can't determine P/Q ratio because we don't have a value for $q.

2) Either p=120 or q=120

\(\frac{P}{Q}= \frac{$q - $141}{$141 - $120}\) OR \(\frac{Q}{P}= \frac{$p - $141}{$141 - $120}\) ---> Can't determine P/Q ratio because we don't have a value for $p.

3)

If $q=120 then \(\frac{P}{Q}= \frac{$120 - $141}{$141 - $90}\) ---> \(\frac{P}{Q}= \frac{-$19}{$51}\) ---> Not possible to have negative amount of TVs.

If $p=120 then \(\frac{P}{Q}= \frac{$150 - 141}{141 - 120}\) ---> \(\frac{P}{Q}= \frac{$21}{$19}\) --> So amount of P TVs is 19.

Solution:

Statement One Alone:

The Model P televisions sold for $30 less than the Model Q televisions

Suppose 20 Model P televisions were sold for $126 each and suppose 20 Model Q televisions were sold for $156 each. Then, the average selling price of 40 televisions is (20*126 + 20*156)/40 = (2,520 + 3,120)/40 = 5,640/40 = $141. In this scenario, 20 of the 40 televisions were Model P.

Suppose, on the other hand, that 28 Model P televisions were sold for $132 and 12 Model Q televisions were sold for $162. Then, the average selling price of the 40 televisions is (28*132 + 12*162)/40 = (3,696 + 1,944)/40 = 5,640/40 = $141. In this scenario, 28 of the 40 televisions were Model P.

Statement one alone is not sufficient. Eliminate answer choices A and D.

Statement Two Alone:

Either p=120 or q=120

Suppose 10 model P televisions were sold for $120 and 30 model Q televisions were sold for $148. Then, the average selling price of a television was (10*120 + 30*148)/40 = (1,200 + 4,440)/40 = 5,640/40 = $141. In this scenario, 10 model P televisions were sold.

Suppose, on the other hand, that 30 model P televisions were sold for $120 and 10 model Q televisions were sold for $204. Then, the average selling price of a television was (30*120 + 10*204)/40 = (3,600 + 2,040)/40 = 5,640/40 = $141. In this scenario, 30 model P televisions were sold.

Statement two alone is not sufficient. Eliminate answer choice B.

Statements One and Two Together:

If Model Q television sells for $120, then a Model P television must sell for 120 - 30 = $90 since statement one tells us that a Model P television sells for $30 less than a Model Q television. Then, the average selling price of a television will be between 90 and 120, which is inconsistent with the given fact that the average selling price of a television is $141. Thus, it must be true that a Model P television sells for $120 and a Model Q television sells for 120 + 30 = $150.

If we let n be the number of Model P televisions sold, then the number of Model Q televisions sold must be 40 - n. We can create the following equation:

(120*n + 150*(40 - n))/40 = 141

120n + 6,000 - 150n = 141*40

-30n = 5,640 - 6,000

-30n = -360

n = 12

So, the number of Model P televisions sold was 12. Statements one and two together are sufficient.

Answer: C

Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions?

Average selling price = Total selling price/ total no of Tv's sold.

Let the number of Model P television be X and no of Model Q television be Y.

141 = (p* X + q* Y)/ 40

141 *40= p* X + q* Y

Y = 40 - X
141 *40= p* X + q*(40 - X)

141 *40 = (p-q)* X + q * 40
We need to find the value of X.
1) the Model P televisions sold for $30 less than the Model Q televisions

p = q - 30
p-q = -30
141 *40= -30* X + q * 40
Since we have 2 variable and 1 equation, Statement 1 alone is insufficient to find the value of X.

2) Either p=120 or q=120
Statement 2 is insufficient as we dont have exact value of p and q.

Combining 2 statements:
Case 1: if p = 120, then q= 150
30 X = 150*40 - 141*40 = 9*40
X = 360/30 = 12

Case 2 : if q = 120 , p= 90
30X = q *40 - 141*40
Since q is 120 which is less than 141, X will be a negative value, which is not possible.

So we can confirm that the value of x is 12 after combining 2 statements.
Option C is the answer.

Thanks,
Clifin J Francis,
GMAT SME

Hi Bunuel

One basic query, when we are referring to plural televisions in statement-01 "The Model P televisions sold for $30 less than the Model Q televisions"
Why did you not write this equation in term of overall difference: (40-x)*q - p*x = 30

Please guide, Is my interpretation incorrect?

Think of it as a weighted average question where the middle value is 141. The first condition tells you there is a gap of 30 between the two outside values but you don't know how that is nested around 141. Condition 2 tells you that one of the outside values is 120, but you don't know the other one. Together if one of the values is 120 then the other one must be 150, not 90, for combined average to be 141. From there you can use the weighted average method described by Dan the GMAT Man to get your ratio

I have a query?
Is it possible to solve this question through the weighted average/ mixtures method?
Please help with the methodology, if it's possible?

Hi experts IanStewart

I ran into this question in my practice exam, and although I got it correctly, I spent more than three minutes because I, worried that the first statement could be sufficient itself, thought carefully and slowly about it.

In school we were taught that one equation cannot solve two variables, but in the GMAT world, if the variables are positive or non-negative integers, sometimes one equation can provide an unique solution of the two variables. An example would be the first statement of this question https://reurl.cc/9GW6MO.

After I read this television question, I wrote an equation:
px+q(40−x)=141*40 (X is the number of Model P televisions sold.)

With the first statement, the equation evolved into:
(q-30)X+q(40-X)=5640

Although it is one equation containing two variables, I could not kill (A) and moved on, because I was not sure that it must be insufficient. So, I simplified this equation and got:
qX-30X+q40-qX=5640 => q40-30X=5640 => 4q-3X=564

X must be an integer, but q is not necessarily an integer since it represents a price.
Is this the main reason the first statement is not sufficient?

Or, there is no absolute rule regarding this issue. Sometimes an equation is insufficient even though we know that the two variables are both integers, and we just need to test cases in each question to confirm?

For this question, if x=1, 4q=567, q=567/4, which is a strange price but not impossible. If x=2, 4q=570, q=570/4. So there is no unique value for x.

I hope to compare this statement with another one (whose link is attached above.)

With the first statement, we can get an equation:
15x+29y=440

X and Y must be integers, and after due check, we can know that there is only one solution of X and Y.
Hence, the first statement is sufficient.

Could you share your approach for this type of questions containing "mixed variables" and "integer constraints"?
Thank you.

This television question is just a weighted average problem, and I would never consider solving a weighted average problem algebraically. As you correctly point out, if you produce an algebraic equation, it may not be obvious if it has one solution or more than one, especially when we know certain quantities must be integers. Instead we can solve weighed averages visually (the method goes by the technical name 'alligation', and you can find it explained in detail in several places, e.g. in my Word Problems book). Here, the overall average price $141 must be somewhere in between the price of tv P and the price of tv Q. Using only Statement 1, if $141 is exactly halfway between the two prices:

--------126-----------141-----------156----------

then we must have sold an equal number of tv's of each type. If, on the other hand, the $141 average is closer to one tv's average price, then sales of that tv must have predominated, so if we had this situation:

---------135----141---------------------------165-----

we would have sold many more tv's at $135 than at $165 (and by alligation principles, we can very quickly work out that 80% of the tv's were sold at $135 in this case, because the distances on the number line above to the middle number $141 are in a 6 to 24, or 1 to 4 ratio, so 1/5 of the tv's were one type, 4/5 the other).

So Statement 1 is almost immediately insufficient -- we have 41 conceivable situations (selling 0, 1, 2, up to 40 tv's of type P), and in each situation we'll be able to draw a different number line like the ones above, so all are possible. But because we have only a finite number of situations, only a finite number of solutions exist -- we can't have a number line with 140 and 170 at either end, for example, because then we'd need to sell a fractional number of televisions of each type. In many of the 41 possible situations, the prices turn out not to be integers (if, say, one tv is type Q, the remaining 39 are type P, it turns out television P sells for $140.25), but prices don't need to be integers, so that turns out not to be an issue.

Once we use both Statements, because $141 must be between the two prices, the prices can only be $120 and $150, and then if you know alligation it will be instantly obvious the question has only one answer.

In general, you should never be counting equations and counting unknowns to decide the answer to a GMAT DS question -- you'll actually get more questions wrong than right doing that. It is true that when we have two unknowns in two equations that represent lines, then if those lines aren't parallel, there will be exactly one intersection point of those two lines, and thus exactly one solution for each unknown. But there are so many other possible situations -- we might have two identical lines, or non-linear equations, or more than two unknowns, or constraints on the unknowns (e.g. that they must be integers), or we might need to solve for some arithmetic combination of unknowns (e.g. for x +y or x/y, not for x and y alone) -- and in all of those situations, you cannot reliably count equations and unknowns to tell if you have exactly one solution. Those 'exceptions' turn out not to be exceptional on the GMAT -- they are very common on the test.

As for how to tell when a linear equation with positive integer unknowns (e.g. the equation you get from Statement 1 of the $0.29 stamps question) has one solution or more than one solution (either situation is possible, depending on the equation), that's too lengthy a topic to cover here, but that's usually a Number Theory problem, and my Number Theory book discusses it in detail. The best approach depends on the specific numbers in the equation.

edit: I didn't read the earlier replies before posting, but Karishma has also explained this question in the best way, using a number line, so you can also see her solution to read more about how to use that approach here