Let's talk strategy here. Many explanations of Quantitative questions focus blindly on the math, but remember: the GMAT is a critical-thinking test. For those of you studying for the GMAT, you will want to internalize strategies that actually minimize the amount of math that needs to be done, making it easier to manage your time. The tactics I will show you here will be useful for numerous questions, not just this one. My solution is going to walk through not just what the answer is, but how to strategically think about it. Ready? Here is the full "GMAT Jujitsu" for this question:
The target of this question is "
total profit", and we know that profit is equal to the selling price minus the cost. Thus, the profit per unit for product P would be
\($_P = 10-8 = 2\), and the profit per unit for product Q would be
\($_Q = 13-9.5 = 3.5\). The total profit would therefore be (where \(P\) and \(Q\) represent the number of units of each product):
\(T=P($_P )+Q($_Q)\)\(T=P(2)+Q(3.5)\)Now that we have set up the basic equation, we need to pause in our discussion here. Many people spend too much time on Data Sufficiency questions because they think they need to get to the bitter end. The question asks “
Was the total profit on these items more than $2,000.00?” This is a
“Yes/No” question – a very common structure for Data Sufficiency problems. The fundamental trap for problems like these is to bait you into thinking that you actually need to solve for every value. You don’t. As soon as you have enough information to conclude whether you can come up with only one answer to the question, you can move on. For
“Yes/No” questions, if you can think of two situations (or two variable inputs) that are consistent with all of the problem’s constraints but come up with different answers to the question, you know a statement is insufficient. In my classes, I call this strategy
“Play Both Sides.”Let’s analyze each statement, and you will see what I mean. Statement #1 tells us that there are more units of \(P\) than there are units of \(Q\). Now, look at the shape of the equations. From the question stem, we know that
\(P+Q=834\). “834” is a bit of a messy number, so approximating this would make our analysis a lot easier. First, let’s assume that
\(P\) is roughly equal to
\(Q\), making both values around \(400\). Thus,
\(T=P(2)+Q(3.5)≈400(2)+400(3.5)≈800+1400≈2200\)So, it is possible to exceed \($2,000\), and we know we can answer
“Yes” to the question. The next question we ask ourselves is: Can we play both sides? Is it possible to have a solution where the total is
under \($2,000\)? If \(P\) dominated the relationship (in other words, if \(P\) were around \(800\), then we would have less than \($2,000\). After all,
\(800*2\) is only equal to
\($1,600\). We are not even close to the boundary of \($2,000\), so approximating is sufficient here. Since one scenario would clearly give us a combined profit of
less than \($2,000\) while the other would give a total profit of
more than \($2,000\), Statement #1 is insufficient by itself.
Statement #2 tells us that \(Q\) is “
at least 100 units.” We have
already shown by our analysis of Statement #1 that if \(Q\) is around \(400\), then the total will exceed \($2,000\). There is no reason to redo this. As we “
Play Both Sides”, we already know that there is a solution where \(Q\) is “
at least 100 units” that allows us to answer
[color=#00a651]“Yes”[/color] to the original question. So, we need to know if we can show that the total may be
less than \($2,000\). The phrase “
at least” is massive leverage. This is a minimum amount, so let’s see what a minimum amount would do – in other words, if
\(Q=100\). Again, approximating here is more enough. If
\(Q=100\), then
\(P≈700\). Thus:
\(T≈700(2)+100(3.5)≈1400+350≈1750\)Since we can also show that the total can be less than \($2,000\), we have “
played both sides” and Statement #2 is insufficient.
We now need to look at combining them to see if they are sufficient. But we have already
done all the analysis we need. The hypothetical situations we created for Statements #1 and #2 show two possibilities where (1)
P can be greater than Q and (2)
Q is at least 100. Since we can “
play both sides” of this
Yes/No question even when the statements are simultaneously true,
the answer is “E”.
Now, let’s look back at this problem from the perspective of strategy. This problem can teach us several patterns seen throughout the GMAT. First, minimize your math. You only need to do enough math to prove the answer to the question. With
Yes/No questions, this often means “
playing both sides” – coming up with two possible solutions that follow the rules of the problem, but give different answers to the question. Second, notice how
approximation can often cut through otherwise messy math. This saves time and energy, allowing you to focus on other, potentially harder, questions. And that is how you think like the GMAT.
_________________
Aaron PondVeritas Prep Teacher of the YearVisit me at https://www.veritasprep.com/gmat/aaron-pond/ if you would like to learn even more "GMAT Jujitsu"!