IanStewart
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.
Thank you
IanStewart !
Your explanations are clear and systematic!
I reviewed the weighted average concept over the past few days.
Could I have some follow-up questions?
1. Can the number of Model P television sold be 0 or 40?
I can understand why you say in your post that there are "41 conceivable situations," since the number of Model P can, theoretically, range from 0 to 40. But I am curious whether there is a constraint in this question that does not allow the number to be o or 40?
Quote:
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?
If we focus on the question itself, which says
"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," do we allow the possibility that all the 40 televisions sold are Model P or Model Q? In other words, the number of Model P can be 0 or 40, cannot it? (In this case, the only one type's price will be $141 itself.)
2. The features of "weighted-average" questions?
I did not find out that this question tests the weighted average concept but only thought that I could approach it algebraically. Reviewing this question and the stamp question, I think the two equations are similar:
4q-3X=564 (q is the price, X is the number of Model P sold)
15x+29y=440 (X is the number of one type of stamp sold and Y is the number of another type)
Hence, I am curious how we see that some questions test the weighted average and some do not. Is the key that if the question tests the concept of weighted average, it should
mention at least once the weighted average? (such as the average price, the average capacity or the average concentration)
Sometimes we are given the weighted average and try to figure out the ratio of the weightings of two sides, or sometimes we need to determine the absolute number of one side, or sometimes the direction is flipped.
IanStewart
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.
Thank you for your inspiring suggestions!