The following are TestTakers' suggestions for solving the problems on the October 16th, 2010 PSAT, Section 2 (a math section). Click "Read More" to view!

**1. Plug-in**here. Say

*n*= 2. Then "twice

*n*" will be 4. If

*k*is 3 more than that,

*k*is 7. Now just go through the answer choices to see which one yields 7 when 2 is plugged in for

*n*. Only (B) works.

**2. Backsolve**! You know one of these choices won't work, so just find which one won't by trying all of them. If (C) were the right answer, then

*x*= 0 wouldn't work. In other words, 2|0| - 3 < 7 would be false. But since -3

*is*less than 7, (C) is not our answer. The only choice that doesn't work is choice (A), because 2|5| - 3 is equal to 7, not less than 7. (A) is our answer.

**3.**

**Backsolve**again! When you see numbers in the answer choices, think backsolve, even if it's a geometry question! Start with (C): if

*x*= 18, then the angles go from

*x*, 2

*x*, 3

*x*, and 4

*x*to 18, 36, 54, and 72, respectively:

Since the line on the bottom is a straight line, those numbers need to add up to 180. Lo and behold, they do! So (C) is the correct answer.

(Of course, this problem could also be solved with algebra: 10

*x*= 180,*x*= 18)**4.**No fancy techniques here, just read the question very carefully. They want all the kids who are applying to College

*X*or College

*Y*(or both

*X*and

*Y*), but NOT College

*Z*. So cross off all the numbers that fall within the circle for

*Z*. You should be left with 11 applying to only

*X*, 10 applying to only

*Y*, and 9 applying to both

*X*and

*Y*. 11 + 10 + 9 = 30. Answer: (B).

**5.**We calculate averages by finding the sum of the numbers we want to average, and then dividing it by the number of numbers. Which is to say:

AVG = SUM / (# OF NUMBERS)

We know the average here, and we know the number of numbers, so let's put those into the equation to find the sum:

30 = SUM / 3

(30)(3) = SUM

90 = SUM

That's choice (E).

**6.**Again, nothing fancy here. Just count the dots that line up with the

*y*-coordinates we're looking at. There are 5 dots on the

*y*= 4 line, which is more than any other line. (B) is our answer.

**7.**We don't care about

*x*or

*y*individually here, we only care about the expression (

*x*+

*y*). This is a common theme on the SAT and PSAT: they treat whole expressions like (

*x*+

*y*) as something to be solved for. We start with:

(

*x*+*y*)^{3}= 64Take the cube root of each side:

(

*x*+*y*) = 4So if (x + y) = 4, then:

(

*x*+*y*) + 2= 4 + 2

= 6

(B) is our answer.

(You could also

**Backsolve**this. (C) says*x*+*y*+ 2 = 8, which would mean*x*+*y*= 6. But 6^{3}= 216, which is much bigger than 64, so we try (B) instead. If*x*+*y*+ 2 = 6, then*x*+*y*= 4, which we already know works from our work above.)**8.**Draw it, but be careful. In order to be sure you get this right, don't draw the streets in the order in which they're given. Start with Fern and Oak. The question says they're perpendicular:

The next thing they tell us is that Broad is parralel to Main, but we don't know anything about either of those streets yet, so ignore that for now. Ignore that for now and move on to the last thing they say: that Main is perpendicular to Fern. Since we've already drawn Fern, we can draw Main:

And now that we have Main, we can go back and draw Broad:

The question asks which streets are parallel.

II. Main and Oak

III. Oak and Broad

(E) is our answer.

**9.**Function (or operation) questions are all about following directions. They tell us that

*x*☆ = 2

*x*+ 1. That just means that whatever is to the left of the star gets multiplied by 2, and then 1 added to it. So, all of these are also true:

*y*☆ = 2

*y*+ 1

*a*☆ = 2

*a*+ 1

☃☆ = 2☃ + 1

What do you mean snowmen can't be math symbols? It's my blog, I'll do what I want.

Anyway, let's solve this. It works the same way with numbers as it does with variables or symbols:

(-4)☆ = 2(-4) + 1

(-4)☆ = -8 + 1

(-4)☆ = -7

Our answer is (B).

**10.**Just arithmetic here, but be careful so you don't make a silly mistake. The house price is $120,000. That's 2.5 times the annual income of the buyer, so the buyer's annual income is $120,000 / 2.5 = $48,000. If his annual income is $48,000, we divide that by 12 to find his monthly salary: $48,000 / 12 = $4000. Answer choice (D).

**11.**

**Backsolve**this! Start by checking if (C) will work. If

*x*= 5:

*x*

^{2}-

*y*

^{2}= 15

5

^{2}-*y*^{2}= 1525 -

*y*^{2}= 15*y*

^{2}= 10

Since the question stipulated that both x and y have to be integers and the square root of 10 isn't an integer, we know (C) isn't the right answer. Furthermore, we know we have to go higher because if we went lower,

*x*^{2}would be less than 15, which won't do us any good. So, let's try (D), which says*x*= 8:*x*

^{2}-

*y*

^{2}= 15

8

^{2}-*y*^{2}= 1564 -

*y*^{2}= 15*y*

^{2}= 49

*y*= 7

YES! This time

*y*works out to be an integer, so we're good. (D) is the correct answer.**12.**Note first that segments

*PQ*and

*PR*are BOTH radii (isn't the plural of "radius" weird?), so they're both equal to 4. That's very important, because it means our triangle is isosceles, and therefore angles

*Q*and

*R*have to be equal. But wait...if angles

*Q*and

*R*are the same, and angle

*P*is 60°...then...we've got an equilateral triangle! All the angles are 60°, so all the sides are 4, so the perimeter is 12. (B) is our answer.

**13. Plug-In (carefully)!**We know these two fractions need to equal each other, so let's say

*x*= 20, and

*w*= 15. That way both fractions will equal 5, a nice easy number to work with:

which is the same as saying 5 = 5, which is true. Now all we need to do is plug our values into the next expression to get our answer!

Done. Answer: (A).

**14. Backsolve!**The pattern is generated in this problem by multiplying each term after the first by the same number,

*k*. Our job is to find

*k*. We do that by trying the answer choices we're given until one works. Since we don't know the first term but we do know the second, we find the first (

*a*) by dividing 15 by our answer choice (if this is confusing or wordy, I'll clarify in a moment). Then, we'll multiply 15 by our answer choice to get

*c*. Then we'll multiply our new

*c*by 15 again to see if our end result is 135. If it is, we've found our answer! Let's make a table:

Answer choice (k) | a (15 / k) | 15 (given) | c (15 x k) | 135 (c x k) |

(C) 5 | 3 | 15 | 75 | 375 (too big, go smaller) |

(B) 4.5 | 3.333... | 15 | 67.5 | 303.75 (better, but still too big) |

(A) 3 | 5 | 15 | 45 | 135 (perfect!) |

So, (A) is our answer.

**15.**To solve this one, break the 30 integers into a few groups. First, we know 18 of them are multiples of 5. That leaves 12 that aren't multiples of 5. Now we have to break up the 18 multiples of 5 into the 10 odd ones, which leaves 8 even ones. The breakdown looks something like this:

The question wants to know how many even integers at most

*could*there be, so we assume every integer we don't know for a fact is odd to be even. That means the 12 non-multiples of 5, for our purposes, are all even. If that's true, then we'd have 20 even integers. 20 is the most we could have, since we know 10 are odd for sure. Answer: (C).**16.**On this graph, steeper slopes correspond to greater rates of population growth. Since we can look at both the slopes of actual population growth (the solid lines) and the theoretical constant rate (the dashed line), all we need to do is compare the slopes of the 20-year intervals to the slope of the dashed line and count how many of them have slopes that go up and to the right more steeply:

**1880 - 1900: steeper than dashed line**

**1900 - 1920: steeper than dashed line**

1920 - 1940: less steep than dashed line

**1940 - 1960: steeper than dashed line**

1960 - 1980: less steep than dashed line

That's 3 intervals that are steeper than the dashed line, so our answer is (C).

**17.**To say triangles are similar is to say that all their angles are the same. A pair of equilateral triangles

*must*be similar, therefore, because all their angles are 60°.

That's not true about isosceles, or right triangles, though. You could have, for example, an isosceles that has angles of 20°, 20°, and 140°, and another with angles of 30°, 30°, and 120°. Both of those would be isosceles, but they would not be similar to each other. Since the question says "

*must*be similar," you can eliminate a choice if you can come up with a counterexample.Likewise, it's easy to imagine two right triangles that aren't similar. Take the two special ones you're given at the beginning of each math section on the SAT and PSAT: the 45°-45°-90° and the 30°-60°-90°. Again, both right triangles, but not similar.

Therefore, our answer is (A).

**18.**To make 0.30608, we basically want to break the following:

into this:

0.30608 = 0.3 + 0.006 + 0.00008

In order to find

*j*,*k*, and*n*, just count the number of spaces the decimal place would have to move to make those decimals. (For example, the decimal place would have to move one place to make the 3 into 0.3). So*j*= 1,*k*= 3, and*n*= 5. 1 + 3 + 5 = 9. Answer choice (D).**19.**Plot this! They talk about

*g*(

*x*) here, but they don't show it, so we need to draw it right on the page. If you use a graphing calculator this is pretty easy, but it's not much harder to just draw by hand. Remember the basic slope-intercept form of a line:

*y*=

*mx*+

*b*. (

*m*is your slope,

*b*is your

*y*-intercept.)

We've got:

So our slope is 3/5, and our

*y*-intercept is zero. That's all we need to know to draw this line. Slope is rise / run, or [change in*y*] / [change in*x*]. Since one of our points is (0, 0) our other has to be a change of 5 along the*x*-axis, and a change of 3 along the*y*-axis. That's (5, 3). Connect those two points, and you've got your line, which clearly intersects*f*(*x*) twice -- at around (2.5, 1.5) and around (5, 3). They ask for how many values of*x*does*f*(*x*) =*g*(*x*), which is just a fancy way of asking how many times the two intersect, so our answer is (C).**20. PLUG-IN!**A #20 plug-in! Hopefully you're only attempting this after working carefully and making sure you got all the earlier questions right because you know this is worth no more points than any other question, but what a gift! It's still a #20, though, so even with plug-in it's going to be a bit tricky. Here's how you do it.

Let's say our profit from the sale of tickets was $90 (that's

*z*). Then let's say our expenses totaled $10 (that's*x*). I chose those numbers because now the total amount of money we took in from ticket sales (profit plus expenses) was $100. That's a nice easy number to work with.Now set your ticket price (

*y*). Let's make it easy and say tickets were $5 each, which means we had**20**people attend the dance to make up our $100 in total ticket sales.To review:

*x*= 10

*y*= 5

*z*= 90

The question asks us to find an expression for how many tickets were sold, so all we're doing is plugging these numbers into all the answer choices to see which one gives us

**20**, since we figured out above that's how many tickets we sold. Only (E) works.