DITW = Do it their way
Trial and Error = pick an answer, play with it, see if it fits what they said in the problem…
Back Door = make up numbers for the variables in the problem, work out an answer that is based on the numbers you made up, then put those made up numbers back into the answer choices, ruling out any that don’t produce a matching answer.
More detailed explanations available in the book…
PRACTICE TEST #2, SECTION 3 – NO CALCULATOR
1. Yes, this is easy algebra. But even here, there is a minor shortcut: as you do your algebraic manipulations, keep in mind what they ask you to find. There is no reason to solve all the way for x here. After you subtract 6 from both sides, double both sides and then add 3. Or, starting from the beginning, you could double both sides of the equation and then subtract 9!
2. Trial and error. In fact, the first equation alone quickly lets you narrow down the options to B and C. B is the one that also works in the second equation.
4. A back door gift! In fact, try easy numbers: Say a=1, b=0…that alone does the trick.
5. Trial and error
6. Just know that slope = rise over run and then you can count boxes. The line on the left rises 2 and runs 5. Since the line on the right rises 4 it must run 10. Or do trial and error.
8. Trial and error – but start with D because you are looking for the largest possible answer.
D does not work – it makes the angle 45 degrees. But C works.
9. DITW – although you CAN avoid all of that with a really neat diagram. Try making your own graph. Draw it as carefully as you can. You will see that the intersection point has to be in the neighborhood of (-1, 4).
10. A funny thing happens when you try x = 0…
11. DITW — and review pages 221-222.
12. You need pretty nimble algebra skills to do this one their way. But lucky for you, this is a classic back door problem. I used F=4, N=6…got R=.4…then went to the answers…
14. Review percents and exponential decay (page 181) – then this is as straightforward as it can be.
15. Another classic back door: I used x=2, got 8/5 or 1.6…went to the answers…only D matched.
16. You just need to play around until you find one allowable combination. If you notice that $250 + $750 = $1000, you can quickly see that 3 of each type is one possibility. And one possibility is all you need.
17. The school way might be quickest here: distribute everything and group the terms, noting that you only care about the “linear” or “x” terms, not the constant term and not the x2 term. But still, this is DITW.
18. This shape again! You really need to immediately recognize that this is a pair of similar triangles. And the little one on top has lengths that are half of the corresponding lengths on the big triangle below.
19. DITW – but read up on radians (page 215). This is basically asking if you can recognize a 30-60-90 triangle and then convert the 30 degrees to radians.
20. When the lines have the same slope, there will be either no solutions or an infinite number of them (if they are actually the same line). 12 is 1/5 of 60 so a=2/5 and b=8/5 but their ratio is 2/8.
PRACTICE TEST #2, SECTION 4 – WITH CALCULATOR
1. You probably can do this by looking and thinking, but if you don’t see it, this is also a back door problem.
2. While this is a classic ratio problem, it can also be thought of as a probability scale-up problem. The probability of a bulb being inspected is 7/400. Then multiply that by the bigger sample size (20,000).
3. Trial and error works here if you don’t see the algebra.
4. I was going to say DITW – but their solution makes this harder than it needs to be. They told you that this is a direct proportion. So set up a proportion: 8/$120=20/x, cross multiply and divide.
7. The school way is to factor. But if you have a graphing calculator, you can graph it and see that the x-intercepts are at 2 and 4.
8. Trial and error works. But so does thinking! You’ve lost 2 points on 100 occasions and yet you still have 200 points left. Hmm…
10. No tricks — just a function question. Review pages 121-134 as needed. They are asking you to find the output g(3) and then use that as the input to the function.
11. DITW – but stay calm and don’t get distracted by irrelevant info. All that matters is the number of words, how many he can read per minute and the hours he will read each day.
12. Once again, this is a linear rate of change – so it’s b + mx…only this time they are using the letter y to represent the passing time instead of the final amount. This is a common way that the SAT likes to mess with you – using a different letter than the one you are used to.
14. They just want to see if you can read the graph! Start on the y-axis, go over to the graph and then down to the x-axis.
16. This is just like our problem about bus-riders and kids who stay late (page 199). It is 7/25 and not 7/200 – we are only talking about the students who DID pass the exam.
17. Do not get distracted by chemistry. Seriously, all they are asking is can you find a number that is 20% less than 40. The quick way: 40×.8=32.
19. Median is middle number. And there are 600 students total. The median is the average of the 300th and 301st kid – but we won’t have to do that. Here’s why: starting with the top row of the data, 120 + 140 = 260. Next row: 80 + 110 = 190. And the running total is now 260 + 190 = 450. So somewhere in that second row, we went past that 300th and 301st kid!
20. This is another probability scale-up. Both schools have the same probability of 4 siblings: 10/300 or 1/30. But now you have to multiply by the size of the full population at each school. When you do that, you get 30 more kids at Washington.
21. A back door question. Try x = 150 and y = 155. It’s magic.
22. Yes, the fastest way to do this is to do the algebra. But once again, if that gives you trouble, the back door is open. Make up a P and and r, use your calculator to find I. Then put all of those numbers into each answer choice and rule out anything that doesn’t come out right. More magic.
23. I make my physics students do this kind of thing all the time. Look at the equation and ask yourself: what happens when we change the value of r? Since r is in the denominator, making r bigger makes the result smaller. And since r gets squared, whatever you do to r also gets squared. Now look at what they told you: observer A measured 16 times as much intensity. They used 16 there for a reason: 16 is 42. It must be that B is 4 times further away than A. So A is ¼ as far away as B. I know that this takes thinking and concentration, but it is easier than their way.
Wait! Still not convinced? You can also do this one by trial and error. But you have to be brave about making up numbers. Make up a P value. Then make up a distance for B. Then, to try answer choice A, make up the distance for A ¼ of what you made up for B. Plug them both into the formula and see if Observer A’s intensity comes out 16 times as big as B’s. If it does, you are done. If not, try another answer. But it does.
24. DITW – review page 168 for help with completing the squares.
25. Another back door! Just make up numbers so that a + b = 0 and then plot the points neatly on a graph. You will see a positive slope every time.
26. DITW – but really, all they are asking is do you know how to read a graph! Review pages 127-128 if you are shaky.
27. Just checking if you know that rate of change is also slope!
28. This is MUCH less work than you might think! The line they are asking for is downhill and steeper than 45 degrees. So of -3 and -1/3, the slope has to be -3. That narrows it down to A and B. But we also know that the line has a positive y-intercept. So much for A. It’s B.
29. While this one does have a clever, “math-y” solution, there are two less-clever solutions available. You can set ax2+b=3 and then use trial and error. You put in their a and b values and see if you get a positive value for x2.
Or, you can use your graphing calculator! Graph y=3and y=ax2+b using their a and b values one at a time. If the graphs intersect at two points, you are done.
31. I know there is an easier way, but for some students, trial and error is actually better. Try 10 years…you get 15 feet. Too small. Try 20 years. You get 30 feet. Too big. 15 years? 22.5 feet…getting close now…14 years? And we have a winner.
32. Again, if you need it, trial and error will work.
33. This is a perfect example of “The Case of the Missing Constant” (see page 126).
They are telling you that when x=3 is the input, you get f(3)=6 as the output. So put x=3 into the rule for the function, set the output equal to 6 and solve for b…
34. Use algebra if you like, but this time even the most devoted algebra students have to admit that playing with numbers is faster. You just need two numbers that add up to 250 and one is 40 more than the other. How about 150 and 100. Ooooh that’s almost it. OK, 145 and 105? Done.
35. And now we have yet another example of y = b + mx, dressed up in different letters. Jane must have started with $15 and must be depositing $18 per week.
37. DITW – but while this is certainly hard (and look where we are in the section!) it may be easier if you think of this as a function problem. They want you to find next year’s population as the output when you know this year’s population as the input. Then, to answer their question, you have to take the output from after one year and use it again as the input to find the next year after that one.
Once again, around 36 of the 58 had some alternate path available.