SAT SUBJECT TEST MATH LEVEL 1

MISCELLANEOUS TOPICS

CHAPTER 18
Sequences

Exercises

1. The first term of sequence I is 10, and each term after the first term is 3 more than the preceding term. The first term of sequence II is 100, and each term after the first term is 3 less than the preceding term. If x is the 16th term of sequence I and y is the 16th term of sequence II, what is x + y ?

(A) 45

(B) 55

(C) 90

(D) 110

(E) 200

2. What is the 500th digit to the right of the decimal point when  is expressed as a decimal?

(A) 0

(B) 3

(C) 4

(D) 5

(E) 7

3. July 4, 2009 falls on a Saturday. What day of the week will it be on July 4, 2011? (Note: 2009, 2010, and 2011 are all regular years with 365 days.)

(A) Sunday

(B) Monday

(C) Tuesday

(D) Friday

(E) Saturday

4. The number of bacteria in a culture increases by 20% every 20 minutes. If there are 1,000 bacteria present at noon on a given day, to the nearest thousand, how many will be present at midnight of the same day?

(A) 9,000

(B) 43,000

(C) 174,000

(D) 591,000

(E) 709,000

5. A gum ball dispenser is filled with exactly 1,000 pieces of gum. The gum balls always come out in the following order: 1 red, 2 blue, 3 green, 4 yellow, and 5 white. After the fifth white, the pattern repeats, starting with 1 red, and so on. What is the color of the last gum ball to come out of the machine?

(A) red

(B) blue

(C) green

(D) yellow

(E) white

6. If the 21st term of an arithmetic sequence is 57 and the 89th term is 227, what is the 333rd term?

(A) 367

(B) 417

(C) 587

(D) 793

(E) 837

ANSWERS EXPLAINED

Answer Key

1. (D)

 

3. (B)

 

5. (D)

2. (A)

 

4. (E)

 

6. (E)

Solutions

Each of the problems in this set of exercises is typical of a question you could see on a Math 1 test. When you take the model tests in this book and, in particular, when you take the actual Math 1 test, if you get stuck on questions such as these, you do not have to leave them out—you can almost always answer them by using one or more of the strategies discussed in the “Tactics” chapter. The solutions given here do not depend on those strategies; they are the correct mathematical ones.

See Important Tactics for an explanation of the symbol ⇒, which is used in several answer explanations.

1. (D) Sequence I is an arithmetic sequence whose common difference is 3, and sequence II is an arithmetic sequence whose common difference is –3.

For sequence I: a16 = 10 + 15(3) = 10 + 45 = 55
For sequence II: a16 = 100 + 15(–3) = 100 –45 = 55

So x = y = 55, and x + y = 110.

 2. (A) Use your calculator to divide: 15 ÷ 37 = 0.405405405. So the question is equivalent to asking, “What is the 500th term in the repeating sequence 4, 0, 5, 4, 0, 5, 4, 0, 5, . . . ?” Since there are 3 terms in the repeating portion, divide 500 by 3:

500 ÷ 3 = 166.66  the quotient is 166
166 × 3 = 498  the remainder is 500 – 498 = 2

So the 500th term is the same as the second term: 0.

 3. (B) July 4, 2011 is exactly 2 years = 2 × 365 = 730 days after July 4, 2009. The days of the week form a repeating sequence in which 7 terms repeat.

730 ÷ 7 = 104.2857 . . .  the quotient is 104
104 × 7 = 728  the remainder is 730 – 728 = 2

So 730 days from Saturday will be the same day as 2 days from Saturday, namely Monday.

 4. (E) Every 20 minutes, the number of bacteria present is muliplied by a factor of 1.2. This creates a geometric sequence whose first term is 1,000 and whose common ratio is 1.2:

1,000, 1,200, 1,440, 1,728, . . .

The 12 hours from noon to midnight consist of 36 20-minute intervals. So the number of bacteria at midnight is

1,000 × (1.2)36 = 1,000(708.8) = 708,800  709,000.

 5. (D) Since the pattern repeats itself after every 15 gum balls, divide 1,000 by 15. The quotient is 66, and the remainder is 10. Therefore, the 1,000th gum ball is the same color as the 10th gum ball, which is yellow.

6. (E) Since an = a1 + (n – 1)d, we have

               57 = a21 = a1 + 20d
                            227 = a89 = a1 + 88d

Subtracting the first equation from the second yields

170 = 68d  d = 2.5

So 57 = a1 + 20(2.5) = a1 + 50  a1 = 7.
Finally, the 333rd term is

7 + 332(2.5) = 7 + 830 = 837