A

The only positive divisors of 19 are 1 and 19.
   Column A: 1 + 19 = 20.
   Column B: 1×19 = 19.

A

Since (a, b) is on the positive portion of the x-axis, a is positive and b = 0; so a + b is positive. Also, since (c, d) is on the negative portion of the y-axis, c is negative and d = 0; so c + d is negative. Column A is greater.

D

Column A: by the distributive law, 5(r + t) = 5r + 5t. Subtract 5r from each column, and compare 5t and t. They are equal if t = 0 and unequal otherwise. Neither column is always greater, and the two columns are not always equal (D).

A

Column A: there are 5 positive multiples of 5 less than 26: 5, 10, 15, 20, 25; their average is 15, the middle one [KEY FACT E51.
     Column B: there are 3 positive multiples of less than 26: 7, 14, 21; their average is 14. Column A is greater.

D

Since 400= 12×33+4, 100 months is 4 months more than 33 years. So 33 years from now it will again be June, and 4 months later it will be October. [See Section 14-P] **Look for a pattern. Since there are 12 months in a year, after every 12 months it will again be June; i.e., it will be June after 12, 24, 36, 48  120  360 months. So, 396 (33 × 12) months from now, it will again be June. Count 4 more months to October.

B

Use TACTIC 3 in Chapter 11: pick an easy-number. Since of the members to-use are boys, assume there are 9 members, 5 of whom are boys. Then the other 4 are girls, and the ratio of girls to boys is 4 to 5, or .

B

Since the surface area is 54, each of the six faces of the cube is a square whose area is 54 + 6 = 9. Then each edge is 3, and the volume is 3³ = 27.

C

ColumnB: a(b + a)-b(a + b)=
      ab + a²-ba-b² = a²-b².
     Column A: (a + b)(a -b) = a² -b².

D

If x and y represent the number of 5-cent stamps and 7-cent stamps, respectively, that Dalia used, then 5x + 7y = 75. There are infinitely many solutions to this equation, but there are only two solutions in which x and y are both positive integers: y = 10 and x = 1 or y = 5 and x = 8. Neither column is always greater, and the two columns are not always equal (D).

C

Draw a diagram, and on each small cube write the number of red faces it has. The cubes with three red faces are the eight comers. The cubes with no red faces are the "inside" ones that can't be seen. If you cut off the top and bottom rows, the front and back rows, and the left and right rows, you are left with a small 2-inch cube, none of whose faces is red. That 2-inch cube is made up of eight 1-inch cubes. The columns are equal (C).
   

B

Since , then Therefore, d -c = cd, which is positive. Then, d -c is positive, and so d ＞ c. Column B is greater.

C

The simplest solution is to realize that there is one palindrome between 100 and 109 (101), one between 390 and 399 (393), one between 880 and 889 (888), and in general, one out of every 10 numbers. So the probability is The answer is (C).
   **The more direct solution is to count the number of palindromes. Either systematically make a list and notice that there are 10 of them between 100 and 199, and 10 in each of the hundreds from the 100s to the 900s, for a total of 90; or use the counting principle: the first digit can be chosen in any of 9 ways, the second in any of 10 ways, and the third, since it must match the first, can be chosen in only 1 way (9 × 10 × 1 = 90). Since there are 900 three-digit numbers, the probability is

D

From the bottom graph, we can estimate the percentage distribution of total enrollment to be:
   Public 4-year 41%
   Private 4-year 21%
   Public 2-year 37%
    Private 2-year  1%
   Total public  78%
    Total private  22%
   78 -22 = 3.5, so there were 3.5 times as
   many students enrolled in public institutions as private ones.

E

In 1972, enrollment in private 4-year institutions was approximately 1,100,000 (22% of the total enrollment of 5,000,000). By 1995, the index for private 4-year institutions had increased from 80 to 120, a 50% increase.
   Therefore, the number of private 4-year students enrolled in 1995 was approximately 1,650,000 (50% more than the 1,100,000 students enrolled in 1972).

C

Let P = the price of the TV set. Then Jack paid 1.085(.90P), whereas Jill paid .90(1.085P). The columns are equal (C).
   **Use TACTIC 2, Chapter 12, and choose a convenient number: assume the TV cost $100. Jack paid $90 plus $7.65 tax (8.5% of $90) for a total of $97.65. Jill's cashier rang up $100 plus $8.50 tax and then deducted $10.85 (10% of $108.50) for a final cost of $97.65.

B

Since A times ABA is a three-digit number, A has to be less than 4; but 1 times ABA is ABA, so A ≠ 1. Therefore, A = 2 or A = 3.
   
   Since there is no carrying, either 2 × B = C, a one-digit number, and B ＜ 5, or 3 × B = C, and B ＜ 3. In either case, column B is greater.

C

By the triangle inequality (KEY FACTS J12 and J13),
   · The third side must be less than 9 + 10 = 19. (Ⅲ is false.)
   · The third side must be greater than 10 -9 = 1. (Ⅰ is false.)
   · Any number between 1 and 19 could be the length of the third side. (Ⅱ is true.) The answer is C.

D

Check each choice. (A) If x = 5, = 2. (B) This one is more difficult; the only possibility if x = 1, in which case = 0. If you don't see that immediately, keep Choice B under consideration, and test the rest. (C) If x = 1, = 1. (D) For positive x,  is always greater than 1 but less than 2; cannot be an integer—that's it. If you didn't reason that out, check Choice E. is an integer if x = 6. You should have eliminated at least Choices, A, C, and E.

B

From the top graph, we see that among fourth-graders in 1996:
   25% did no homework;
   55% did less than 1 hour;
   5% did more than 2 hours.
   This accounts for 85% of the fourth-graders;
   the other 15% did between 1 and 2 hours of homework per day.

E

In 1984, approximately 540,000 eleventh-graders watched television 1 hour or less per day (27% of 2,000,000). By 1996, the number of eleventh-graders had increased by 10% to 2,200,000, but the percent of them who watched television 1 hour or less per day decreased to about 18%: 18% of 2,200,000 is 396,000. This is a decrease of 144,000, or approximately 150,000.

A

Since C = 2πr, then and area of circle =

A

Column A: Since the hypotenuse is 2, the length of each leg is , and the area is =1.
   Column B: Since the hypotenuse is 2 the shorter leg is 1, the longer leg is , and the area is
   
   Column A is greater.
   

C

B-A=
   (51 +52+53+...+99+ 100) -(1 +2+3+...+49+50)
   = (51-1)+(52-2)+(53-3)+...+(99-49)+(100-50)
   = 50 + 50 + 50 +...+ 50 + 50 = 50 × 50 = 2500.
   **If you know the formula, for adding up the first n positive integers, you can use it: A = = 25(51) = 1275. B is the sum of the integers from 1 to 100 minus the sum of the integers from 1 to 50:
   B = -1275 = 50(101) -1275 =
   5050 -1275 = 3875.
   Finally, B -A = 3875 -1275 = 2500.
   The columns are equal (C).

A

Since the area of each small circle is πr², the area of the white region is 3πr². Also, since the area of the large circle is πR², the shaded area is πR² -3πr² = π(R² -3r²). Since the areas of the white region and the shaded region are equal:
   3πr²=π(R²-3r²) 3r²=R²-3r²
   
   which is greater than 2. Column A is greater.

E

Since p pencils Cost c cents, each pencil costs cents. By dividing the number of cents we have by , we find out how many pencils we can buy. Since d dollars equals 100d cents, we divide 100d by , which is equivalent to multiplying 100d by
   You will probably prefer the alternate solution below.
   **Use TACTIC 2, Chapter 12. Assume 2 pen-cils cost 10 cents. So, pencils cost 5 cents each or 20 for one dollar. So, for 3 dollars, we can buy 60 pencils. Which of the choices equals 60 when p = 2, c = 10, and d = 3? Only

A

If a student earned a grade of g, she massed (100 -g) points. In adjusting the grades, the teacher decided to deduct only half that number: . So the student's new grade was
   
   Since this was done to each student's grade, the effect on the average was exactly the same. The new average was 50 + .

A

Use TACTIC 3, Chapter 11. Choose an appropriate number for the common perimeter. Any number will work; but since a triangle has 3 sides and a square has 4 sides, 12 is a good choice. Then, each side of the square is 3, and the area is 3² = 9. Each side of the equilateral[ triangle is 4, and the area is (KEY FACT J15). The ratio is