Test 4 Section 3 - #9 (page 583)

This one is pretty tricky, but the 98th 1 will be followed by 98 0's, the 99th will be followed by 99 0's, and the 100th 1 will be followed by 100 0's. Then we'll be at the 101st 1, and we'll be done. The total number of 0's is 98+99+100 = 297. That's choice (D).

2010 PSAT - Saturday - Section 2 (Math)

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!

Test 3 Section 5 - #16 (page 530)

Well, according to the way the problem is set up, we're going to have one "1", two "2"'s, three "3"'s, etc.  We want to know when the first "12" will come.  Well, that's going to come after the eleventh "11".  Let's just count up how many terms will come first:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 66

So if 66 terms come before the first "12", the first "12" will be the 67th term.

Test 1 Section 7 - #16 (page 418)

This question is a pain. The thing you have to remember to make it easier is that W, X, Y, and Z all have to be integers between 0 and 9 for this to work. Let's look at the rules (and what they tell us):

X = W + Y + Z  -- This tells us X has to be the biggest number.
W = Y + 1 -- This one's obvious: W is bigger than Y, but only by 1.
Z = W - 5 -- This one tells us a LOT, because the lowest Z can be is 0.

Let's start putting things together. Because of the third rule, we know W is either 5, 6, 7, 8, or 9 (otherwise Z would be negative). But it can't be too big, because the second rule says W is only 1 bigger than Y, and the first rule says X equals the sum of all the other letters. Confused yet? If so, do a little guess and check with possibilities for W to see what happens to the other values.

Here's what I'm coming up with:


W = 5. Why?  Because if it was any bigger, like 6, Y would also be bigger, like 5 ( because of the second rule), so X would be bigger than 9 (by the first rule), which can't happen. Once you've got W, the rest sorta fall into place.

Y = 4.  Why?  Because W = 5 and the second rule tells us Y is one less.


Z = 0.  Why?  Because W = 5 and the second rule tells us W - 5 = Z.


X = 9.  Why?  Because the first rule tells us that X is the sum of all the others and 5 + 4 + 0 = 9

Our WXYZ number is 5940.

Test 1 Section 7 - #14 (page 417)

Set up a hangman-like set of blanks so that you can fill them in as you go:

__, __, __, __, __, __, __, __

Now, start to fill in what you know (the third and the sixth numbers):

__, __, 17, __, __, 77, __, __

When the question says the difference between two consecutive terms is always the same number, that means that we're counting up by that number (say we were counting by 2's, we'd have 2, 4, 6, 8, etc.).  What are we counting up by here?  How do we get from 17 to 77 in 3 steps?  We count up by 20's:

__, __, 17, 37, 57, 77, __, __

So all we need to do now is find what they asked for: the 8th term.  

__, __, 17, 37, 57, 77, 97, __

__, __, 17, 37, 57, 77, 97, 117

The eighth term is 117.

Test 7 Section 7 - #15 (page 788)

It would take us way too long to list all the numbers from 1-100 and check which are perfect squares.  Luckily we don't have to.  All we need to do is count all the ones that ARE, and then we know all the rest are NOT.  To do so, let's just list squares of integers until they get bigger than 100:

1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100
11² = 121 (Oops, too big!)

There are 100 numbers between 1 and 100 (inclusive), and we just counted 10 of them that ARE perfect squares. That means 90 are not.  Choice (E) is our answer.

Test 7 Section 7 - #13 (page 787)

Remember our conditions:

• One number is odd.
• One number is a multiple of 5.
• One number is the day of the month of Kyle's birthday.

The key is the part where they say that each number satisfies exactly one of the conditions.  That means, for example, if the number 15 were in a combination, we'd have to eliminate that choice, because 15 is both odd and a multiple of 5.  It satisfies 2 conditions, but the question said that's not allowed.  We can actually eliminate 3 choices just because of this: (B), (C), and (D) all have odd multiples of 5 (25, 15, and 15, respectively).

(A) 14-20-13
(B) 14-25-13
(C) 15-18-16
(D) 20-15-20(E) 34-30-21

So how do we know whether to pick (A) or (E)?  Well, if it were (E), our odd number would be 21, our multiple of 5 would be 30, and that'd mean our day of the month would have to be 34.  But no month has 34 days.  So we have to pick (A): 13 is odd, 20 is a multiple of 5, and 14 is a day of the month.

Test 7 Section 3 - #15 (page 773)

Honestly, when I'm faced with questions like this I just start listing and counting.  If you stay organized, you'll get this question right in not much time, and you won't have to worry about any equations.

I'm going to number the 4 experienced plumbers 1, 2, 3, and 4, and then the trainees 5, 6, 7, and 8.  Then I'm just going to follow the instructions in the problem (1 experienced plumber and 2 trainees) and see if I can recognize the pattern that emerges before I count up every combination.

Start with experienced plumber #1.  He can be accompanied by the following sets of trainees:

5, 6

5, 7

5, 8

6, 7

6, 8

7, 8

Remember, "5, 6" is the same as "6, 5" in this question, since in both cases trainees 5 and 6 will be accompanying experienced plumber 1.

OK, now what trainees can accompany experienced plumber #2?

5, 6

5, 7

5, 8

6, 7

6, 8

7, 8

The SAME ONES. Once we see the pattern, we don't need to finish listing. There are 6 possible combinations of trainees for each of 4 experienced plumbers. (6)(4) = 24. That's 24 possible combinations.

Test 2 Section 5 - #16 (page 468)

For questions like this, it's best to just start listing the numbers that are "tri-factorable" (by listing numbers that are made by multiplying 3 consecutive positive integers, starting with the lowest positive integers we can). It won't take us long:

1 x 2 x 3 = 6
2 x 3 x 4 = 24
3 x 4 x 5 = 60
4 x 5 x 6 = 120
5 x 6 x 7 = 210
6 x 7 x 8 = 336
7 x 8 x 9 = 504
8 x 9 x 10 = 720
9 x 10 x 11 = 990
10x11x12 = 1320  (Oops, too high!)

So how many were under 1000?   9 of them.

Test 5 Section 4 - #12 (page 654)

The median is just the middle number in a list. Since we're dealing with consecutive numbers, we can just list the 9 consecutive numbers that have 42 in the middle. If we want to be really careful, we can start in the middle and work outward until we have 9 numbers:

42

41, 42, 43

40, 41, 42, 43, 44

39, 40, 41, 42, 43, 44, 45

38, 39, 40, 41, 42, 43, 44, 45, 46

What's the largest number in our list?  46.

Test 4 Section 6 - #14 (page 597)

Note: In early editions of the Blue Book, the answer key is WRONG here.  Even though the answer key says it's 6, I assure you the answer is 5.

Even though they're describing a series here, treat it like a function: just follow the rules.  According to the rules laid out in the question, the the first 5 terms are going to be:

a, 3a, 9a, 27a, 81a

...add those up and you get:

a + 3a + 9a + 27a + 81a = 605
121a = 605

...solve for a:

a = 605/121 = 5.

Test 7 Section 7 - #15 (page 789)

They're asking for the number of integers from 1-100 that aren't the square of an integer, but it's going to be a lot easier and faster to count how many are the square of an integer*:

1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100 ...and we're done! (Any higher will be out of the range we care about.)

Count how many we've got. I'm counting 10.

There are 100 integers between 1 and 100 inclusive (inclusive just means "including the ends," so we are counting both 1 and 100), and only 10 of them are squares of integers. Which means 90 aren't. The answer is (E).

* This is kinda like a shaded region area problem in that way...it's easier to look for the opposite of what they ask for, and then subtract.

Test 7 Section 7 - #18 (page 789)

When they say they're looking for the "sum of consecutive integers from -22 to x, inclusive, is 72" they mean this:

-22 + -21 + -20 + ... + x = 72

That's the way your math teacher would write it, anyway. :)

What you need to recognize here is that for a while, you're going to be getting nothing but negative numbers for your sum.  -22 + -21 + -20 = -63, for example.  And it's only going to get worse when you add in -19.  You're going to have to go a long way before your numbers start to go positive again.  Think for a minute about when these numbers will first start getting positive.  Or, even better: ask yourself when your sum is going to be zero.

-22 + -21 + -20 + ... + 20 + 21 +22 = 0

You'll have zero when you add up every consecutive integer from -22 to 22; the positive and negative numbers will cancel each other out.  From there, we're cruising.  Watch the bolded bits.

-22 + -21 + -20 + ... + 20 + 21 +22 + 23 = 23
-22 + -21 + -20 + ... + 20 + 21 +22 + 23 + 24 = 47
-22 + -21 + -20 + ... + 20 + 21 +22 + 23 + 24 + 25 = 72

The last number we had to add was 25, so that's our x.  (B) is our answer.

Of course, the long way is just to get crackin' with the calculator (if your fingers are fast it won't take that long) and start adding things up.  When you get to 72, you're done.  No shame in doing it that way, especially if it's the first way you see, and you start on it right away. 

Test 2 Section 8 - #16 (page 486)

The key here is to plug in a few different numbers and look for a pattern. Start with n=2. If you have a 2x2 square, all 4 of the squares would lie along the boundary, so k=4. Now try a new number. If n=3, you'd have a 3x3 gameboard and only 8 of the squares would be along the boundary (the other square is smack in the middle), so k=8. Try one more number to detect a pattern. When n=4, you have a 4x4 gameboard, and 12 squares make up the border (the other 4 are in the middle), so k=12. So far we have k=4, k=8, and k=12... see the pattern? Multiples of 4. The only answer that's a multiple of 4 is (E).