If 2L is 3W, then 12L is 18W.
That means the area being covered can be expressed as 18W x 10L, or 180LW. Since LW is the size of each tile, that means we would need 180 tiles. So the answer is (E).
Notice that one of the wrong answers, (B), is 1/5 of the right answer. This makes sense, because a lot of people might accidentally determine the number of times that the big rectangle in the diagram could fit into the 12L x 10W region.
Page 703, Question 14
Most students I’ve worked with on this question have ultimately arrived at the right answer, but they often don’t see the easiest way to get there.
Most of them try to figure out the median slope by determining the slope of every single line segment in the question, and then trying to arrange them from least to greatest (which they usually do by converting them all to a common denominator). It’s quite time-consuming.
The much easier thing is to pay attention to the diagram and to think about what the concept of slope means in the first place. The slope of a line is a measurement of how slanted it is, in a manner of speaking. So the line segment with the median slope will be the one whose ‘slantiness’ puts it in the middle of all the other segments.
In other words, we can tell that line segment OC is the one with the median slope just by looking at the diagram. Now all we have to do is determine the slope of OC, and we’re done. Since OC starts at the origin and point C is at (4,3), that means it goes up 3 units and over 4, for a slope of 3/4. So (C) is right.
Notice that all the other answer choices are the slopes of the line segments, which allows test-takers to make all kinds of mistakes and still find a wrong answer that they’ll like. Remember to pay attention to details!
Page 705, Question 19
This question is one more excellent example of how we always have to be ready to apply basic math ideas in non-traditional ways. In this case, we’re asked to find equivalent proportions even though every proportion involves 4 variables and we’re never told the values of any of those variables. How is that possible?
In situations like this, it’s often helpful to fall back on the idea of trying to identify concepts related to the things that appear in the question. In this case, pretty much the only usable concept we have in front of us is the idea of cross-multiplying.
As it turns out, if we cross-multiply each answer choice, we’ll see that choice (A) gives us ac = bf, but all the other choices give us af = bc.
That means choice (A) is the one that’s not equivalent to the others.
So, in the end, we never learned the values of any of these variables, and we never did anything more complicated than cross-multiplying to answer the question. Many, many test-takers must have missed this question, but when I talk to students who’ve missed it I’ve never met one who didn’t know what cross-multiplying was. Remember that the challenge on the SAT Math section is to identify the basic concepts that will solve the problem.
Page 705, Question 20
Remember that these questions with roman numerals are typically going to be based on some kind of abstract property of the concepts in the question. If we can figure out what that property is, answering the question will usually be pretty straightforward.
For instance, the question tells us that roman numeral I will work out to ab - b. So we have to ask ourselves if there are any conditions in which that might be equal to zero.
That should raise the question, “when can subtracting one thing from another give us zero?”
The answer is that subtracting one value from another can only result in zero when the two values were equal to start with. This is a property of zero.
So ab - b can only equal zero if ab and b are equal to each other.
And that leads to another question: when can ab and b be equal? When can we multiply something by a and have it equal our original starting value, in this case b?
We can only multiply something by a without changing its value if a equals 1. This is a special property of the number 1.
So, in the end, ab - b can only equal zero when a equals 1, because if a is 1 then the expression ab - b is the same as the expression b - b, which must be zero.
So that means that the expressions in each of the other answer choices can only come out to be zero if the part of the expression before the box can possibly equal 1, like a could in roman numeral I.
For roman numeral III, then, it’s pretty clear that the value can work out to be zero if we just make a be 1 again.
But what about roman numeral II?
The (a + b) expression at the beginning of roman numeral II could work out to be 1 if a and b could be fractions or even if one could be negative. But the question tells us that a and b must be positive integers, which means the lowest possible value for a + b is 2. And that can’t work.