I wrote a post a few weeks ago in which I discussed the importance of blending strategies on certain questions. It’s a mistake to pigeonhole a complex problem as one in which a single tool will be most effective. By test day you will have cultivated a veritable Swiss army knife of strategies and you want to be able to switch from one to another seamlessly.

This philosophy came to mind the other day when a student sent me the following official problem:

*For a certain art exhibit, a museum sold admission tickets to a group of 30 people every 5 minutes from 9:00 in the morning to 5:55 in the afternoon, inclusive. The price of a regular admission ticket was $10 and the price of a student ticket was $6. If on one day 3 times as man regular admissions tickets were sold as student tickets, was the total revenue from ticket sales that day?*

*A) 24,960*

*B) 25,920*

*C) 28,080*

*D) 28,500*

*E) 29,160*

Oh boy. There’s a lot going on here. So let’s start by simply finding the total number of tickets sold. We know that every 5 minutes, 30 tickets are sold. We know that there are twelve 5-minute increments each hour, so 12*30 = 360 tickets are sold each hour. We see that the museum will be open for a total of 9 hours, so a total of 9*360 = 3240 tickets are sold during that time.

We’ve got two different kinds of tickets – general and student. The general tickets were $10 and the student tickets were $6. And we know that 3 times as many general tickets were sold as student tickets. So the tickets were *overwhelmingly* for general admission. If they were all for general admissions tickets, we know that the revenue would have been 3240*10 = 32,400. Because 25% of the tickets were sold for $6, we know that the correct answer will be a bit below this value. If we were short on time, E would be a pretty reasonable guess.

But say we’ve achieved a level of mastery where we don’t need to guess. Hopefully, you recognized that if the ratio of general tickets to student tickets is 3:1, we’re dealing with a kind of weighted average, meaning we can use a number line to find the average overall ticket price, which will be much closer to $10 than to $6. So we know the average price is greater than $8, as this would be the average price if the same number of both kinds of tickets were sold. What about $9? On the number line, we’ll have the following: 6——–9—-10.

9 is three units away from 6 and one unit away from 10, thus yielding our desired 3:1 ratio. Now we know that the average price is $9 per ticket.

So all we have to do is calculate 3240 * 9, as 3240 tickets were sold for an average of $9 each, and we have our answer. That math isn’t too bad, but we can incorporate a couple more useful strategies to save some time. We know that 3000*9 = 27,000, so clearly 3240*9 is greater than 27,000. Now we can eliminate A and B from contention. Next, we can see that going from right to left, the first non-zero digit of 3240*9 will be 6, as 4*9 = 36. Among C, D, and E, the only answer choice that has a 6 in the tens place is E, which is our answer.

Takeaway: In a single question, we ended up doing a bit of estimation, using the answer choices, employing some rudimentary logic, and using the number line to simplify a weighted average. Just as important as what we did do, is what we *avoided* doing – a lot of grinding calculation.

We cannot emphasize this enough: the Quant section is not a math test. It’s an opportunity to demonstrate fluid thinking under pressure. So when you’re doing practice questions, work on employing every tool in your Swiss army knife of strategies. By the day of the test, the more fluidly you can switch from one tool to another, the better you’ll be able to handle even the most challenging problems.

**GMATPrep question courtesy of the Graduate Management Admissions Council.*

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*By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles by him here.*