GraceSCKao wrote:
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?
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
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