Now the question tells us that CD is 2 units. So if we put C in between B and D, and 2 units away from D, we get something like this:
In other words, if we put D and E basically right next to each other, and B is 3.5 units away from E, then let’s say it’s 3.49 units from D. If C is 2 units from D, then B is 1.49 units from C, because
3.49 – 2 = 1.49.
Now, there are other possible answers here, so your work doesn’t have to look exactly like mine. What’s most important, I think, is not to get too thrown off by the fact that more than one outcome can be correct. For a question like this, we don’t have to figure out the entire range of possible solutions, as we might do in a math class in school. We just have to find one possible value and realize that there are other possible values out there.
Note that this question is fairly challenging for a lot of test-takers even though it only involves the most basic arithmetic (addition and subtraction) and the most basic geometry concepts (points on a line). This is the kind of thing you have to look out for on the SAT: simple concepts presented in strange ways.
Page 417, Question 13
This is yet another question that can’t really be set up with a typical algebraic formula. So let’s just think about it.
If the ratio of rainy days to sunny days is 3 to 2, then we can think about the days in terms of “blocks” of 5 days, because every 3 days of rain requires 2 days of sun, and 3 + 2 = 5.
So if we’re dealing with blocks of 5 days, and we have a 30-day month, then there can be 6 blocks of days in that month, because 6*5 = 30.
That means that there will be 6*3 rainy days, and 6*2 sunny days. So there will be 18 rainy days and 12 sunny days. (Just to be sure we haven’t made a mistake, at this point I’d probably quickly add 18 and 12 in my head to confirm that it’s 30, just like it needs to be.)
Now we have to be very careful at this point. The question is asking how many more rainy days there were than sunny days—the question is NOT asking how many rainy days there were, nor how many sunny days there were. It’s asking for the difference between those two numbers.
I’m sure a lot of people either answer 18 or 12 to this question, because they forget what it was asking, but the real answer is 6.
This is yet another example, then, of the critical role that reading comprehension and attention to detail will play in your SAT preparation.
Page 417, Question 14
Again, we have to be sure to read carefully. And we should probably try to avoid using traditional, formal series notation on this question. As is usually the case, this question will probably be a lot easier if we just think about it and figure it out using basic reasoning.
The gap between the 3rd term and the 6th term is exactly 3 terms (because 6 – 3 = 3). In terms of units, the same gap is 60 units, because 77 – 17 = 60.
So a difference of 3 terms corresponds to a difference of 60 units. That means each term is a difference of 20 units, because 60/3 = 20. (The fact that the actual arithmetic has been pretty easy so far is a good sign that we’re approaching the question correctly.)
We can confirm that each term is 10 more than the one before it by filling in the gaps in the original sequence: 17, 37, 57, 77.
So if each term is 20 units and the 6th term is 77, then we know the 7th term is 97 and the 8th term is 117.
Remember: read carefully, and don’t be afraid to think about basic math concepts in new ways. There’s a reason the College Board made the numbers fairly easy to work with here: they didn’t want you to have to use a formula. They wanted to give you the chance just to sit back for a second and think about the question, and count stuff out if you needed to. They could have asked for the thousandth term, or they could have made the difference between terms equal 13.4 instead of 20, or who knows what, but they didn’t. Instead, they made the arithmetic part fairly easy once we figured out what was going on.
Page 418, Question 15
In this question, the phrase “least value” indicates that there are at least two values that x might have. We need to keep that in mind as we work through the question.
We could choose to do this in a formal algebraic way. That would look like this:
|x – 3| = 1/2
x – 3 = 1/2 OR x – 3 = -1/2 (create the two absolute value possibilities)
x = (1/2) + 3 OR x = (-1/2) + 3 (isolate x)
x = 3.5 OR x = 2.5 (simplify the right-hand side)
Then we’d know that 2.5 is the correct answer, because it’s the lower of the two possible values.
Instead of the algebra, we could also just look at the expression in the question and realize that it’s basically just telling us that x is 1/2 a unit away from 3 on a number line. The two numbers that are 1/2 unit away from 3 are 2.5 and 3.5, and the smaller one of those is 2.5, so that’s the answer.