Formulas, Functions and Sequences Drill Problem 1 In a certain sequence, the term an is given by the formula an = an-1 + 10. What is the positive difference between the 10th term and the 15th term? A) 5 B) 10 C) 25 D) 50 E) 100 Problem 2 In a certain sequence, the term an is given by the formula an = 10an-1. How many times greater is a10 than a8? A) 1 B) 3 C) 10 D) 30 E) 100 Problem 3 Billy decides how many dollars, d, to spend on a birthday present for his sister

using the formula d = 25 5365n m− where n is the number of days in the previous year (out

of 365) his sister was nice to him and m is the number of days she was mean to him. Assuming that Billy’s sister was either nice or mean to him every day of the previous year, how much to the nearest dollar will Billy spend on her present if she was nice to him 300 days last year? A) 16 B) 18 C) 20 D) 21 E) 25 Problem 4 Ashley and Beatrice received the same score on a physical fitness test. The scores for this test, t, are determined by the formula t = 3ps – 25m where s and p are the numbers

of sit-ups and push-ups the athlete can do in one minute and m is the number of minutes she takes to run a mile. Ashley did 10 sit-ups and push-ups and ran an 8-minute mile. Beatrice did half as many sit-ups and twice as many push-ups. If both girls received the same overall score, how many minutes did it take Beatrice to run the mile? A) 4 B) 8 C) 10 D) 16 E) 20 Problem 5 The expression a{}b is defined as a{}b = (a-b)(a+b). What is the value of (7{}6) {} (11{}11)? A) 0 B) 13 C) 169 D) 256 E) 311 Problem 6 If 5||10 = 5 and 1||(-2) = 1, which of the following could define the expression a||b? A) b-a

B) 2

3a b−

C) -ab/4

D) 15ba+

E) a+b+4 Problem 7 The height of a ball thrown straight up into the air can be determined by the formula h = -16t2 + vt + d, where t is the number of seconds since it was thrown, v is the initial speed of the throw (in feet per second), d is the height (in feet) at which the ball was released, and h is the height of ball t seconds after the throw. Two seconds after a ball is thrown, how high in the air is the ball if it was released at a height of 6 feet and a speed of 80 feet per second?

A) 96 feet B) 100 feet C) 102 feet D) 134 feet E) 230 feet Problem 8 At the car dealership where Tony has worked for the past 4 years, annual bonuses are determined by the formula b = y× (15c – 8d), where b is the employee’s bonus, c is the number of cars the employee sold that year, d is the number of sick days taken by the employee, and y is the number of years the employee has worked at the dealership. Tony sold 21 cars this year and received a bonus of $1,068. How many sick days did Tony take this year? A) 3 days B) 4 days C) 5 days D) 6 days E) 7 days Problem 9 If a#b = 2a b a− , where b ≥ 0, what is the value of (-4)#4? A) -36 B) -28 C) 12 D) 28 E) 36 Problem 10

The expression x$y is defined as 2x

y, where y ≠ 0. What is the value of 9$(6$2)?

A) 1/2 B) 9/2 C) 9/4 D) 18 E) 108

Problem 11 Amy deposited $1,000 into an account that earns 8% annual interest compounded every 6 months. Bob deposited $1,000 into an account that earns 8% annual interest compounded quarterly. If neither Amy nor Bob make any additional deposits or withdrawals, in 6 months how much more money will Bob have in his account than Amy? A) $40 B) $8 C) $4 D) $0.40 E) $0.04 Problem 12 The half-life of the carbon-14 isotope is 5,730 years. How many years must pass until a sample that starts out with 16,000 carbon-14 isotopes decays into a sample with 500 carbon-14 isotopes? The half-life of an isotope is the time required for half of a sample of the isotope to undergo radioactive decay. A) 180 years B) 1,146 years C) 5,730 years D) 28,650 years E) 183,360 years Problem 13

f(x) = 25

x− and g(x) = 3x – 2. If f(g(x)) = 1, what is the value of x?

A) -5/3 B) -1/3 C) 2/3 D) 1 E) 5/3 Problem 14 f(x) = 2x – 2 and g(x) = x/2 + 2. What is the value of 2f(g(2))?

A) 0 B) 4 C) 6 D) 8 E) 12 Problem 15 An Olympic skater’s score, s, is calculated by the formula s = 10(3t – 5f)/m where t is the number of successful triple axels performed, f is the number of times the skater fell, and m is the length, in minutes, of the performance. If Tonya received a score of 6 for a 5 minute performance in which she had twice as many successful triple axels as she had falls, how many times did she fall? A) 2 B) 3 C) 4 D) 5 E) 6 Problem 16 Through savvy investments, John’s rich uncle exactly doubles his wealth every 8 years. John is now 33 years old, and at that age, the uncle had 13 million dollars ($13m). How much will John’s uncle be worth when John’s uncle retires at 65 years old? A) $26m B) $104m C) $130m D) $208m E) $260m Problem 17

a~b is defined as 5 baa−

− . What is the value of 1~((-1)~1)?

A) -9 B) -1 C) 1 D) 2 E) 5

Problem 18

An archer’s score is calculated by the formula 50 1010b a

s−+

where b is the number

of bull’s-eyes hit, a is the total number of arrows shot, and s is the time in seconds it took the archer to shoot. By how much would an archer who shot 10 arrows and hit all bull’s-eyes in 10 seconds beat an archer who shot twice as many arrows and hit half as many bull’s-eyes in 15 seconds? A) 2 B) 7 C) 10 D) 18 E) 20 Problem 19 Each term of a certain sequence is calculated by adding some constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence? A) 20 B) 15 C) 13 D) 12 E) 8 Problem 20

If a”b = 12 3a b−

and a%b = 3a – 2b, what is the value of 1%2 – 3”4 ?

A) -7/6 B) -1 C) -5/6 D) 2/3 E) 7/6 Problem 21 In a certain sequence, the term an is given by the formula an = an-1 + 5 where a1 = 1. What is the sum of the first 75 terms of this sequence?

A) 10150 B) 11375 C) 12500 D) 13950 E) 15375 Problem 22 In a certain sequence, the term an is given by the formula an = 2×an-1 where a1 = 1. What is the positive difference between the sum of the first ten terms of the sequence and the sum of the 11th and 12th terms of the same sequence? A) 1 B) 1024 C) 1025 D) 2048 E) 2049 Problem 23 An operation @ is defined by the equation a@b = (a – 1)(b – 2). If x@5 = 3@x, what is the value of x? A) -2 B) -1 C) 0 D) 1 E) 2 Problem 24 The wait time in hours, w, for a certain popular restaurant can be estimated by the

formula w = 10

n ks+ , where k is a constant, n is the number of parties waiting ahead of

you, and s is the size of your party. If a family of 4 has a wait time of 30 minutes when 2 other parties are ahead of it, how long would a family of 6 expect to wait if there are 8 parties ahead of it? A) 45.5 min B) 1 hr 15 min C) 1 hr 25 min D) 1 hr 45 min

E) 2 hrs Problem 25 A sequence is defined as an = 5×an-1 – 3 where a4 = 32. What is the first term of the sequence, a1? A) 1 B) 7 C) 16 D) 128 E) 157 Problem 26 A certain sequence is defined by the formula an = an-1 – 7. If a7 = 7 what is the value of a1? A) -49 B) -35 C) 0 D) 42 E) 49 Problem 27 If a!b = ab × b-a, which of the following is equal to (x!4) ÷ (4!x)? A) (4!x)(x!4) B) 4!(-x) C) (-4)!(-x) D) (x!4)2 E) (4!x)2 Problem 28 Monthly rent for units in a certain apartment building is determined by the

formula 25 10

5r tkf++

where k is a constant, r and t are the number of bedrooms and

bathrooms in the unit, respectively, and f is the floor number of the unit. A 2 bedroom, 2 bathroom unit on the first floor is going for $800/month. How much is the monthly rent on a 3 bedroom unit with 1 bathroom on the 3rd floor?

A) $825 B) $875 C) $900 D) $925 E) $1000 Problem 29 What is the sum of all the multiples of 3 between 250 and 350? A) 9000 B) 9900 C) 9990 D) 9999 E) 10002 Problem 30 Abby makes $8 an hour for the first 40 hours she works in a week (base pay) and $10 for each overtime hour after that. Becky makes $7 an hour base pay and $12 an hour overtime. Becky also makes a $25 bonus if she works any overtime hours. If Abby and Becky worked the same number of hours this week and received the same pay, how many hours did each work? A) 45 hours B) 45.5 hours C) 46 hours D) 47.5 hours E) 49 hours Problem 31 Town A has a population of 160,000 and is growing at a rate of 20% annually. Town B has a population of 80,000 and is growing at a rate of 50% annually. How many years will it be until Town B’s population is larger than that of Town A? A) 2 years B) 3 years C) 4 years D) 5 years E) 6 years

Problem 32 What is the ratio of the sum of the even positive integers between 1 and 100 (inclusive) and the sum of the odd positive integers between 100 and 150? A) 102 to 125 B) 50 to 51 C) 51 to 56 D) 202 to 251 E) 2 to 3 Problem 33

In a certain sequence, the term an is given by the formula 2 1

1

n nn

n

a aaa

− −

−

+= . If a1 =

4 and a4 = 3 what is the value of a2/a3? A) -2 B) -5/3 C) 1/2 D) 5/3 E) 2 Problem 34 A sequence is defined as an = an-1

2 – an-22 and an > 0. The first term, a1, is 2, and

the fourth term is equal to the first term multiplied by the sum of the second and third terms. What is the third term, a3? A) 0 B) 3 C) 5 D) 10 E) 16 Problem 35

The operator ? is defined by the following expression: a?b = 1 1a ba b+ +

− where

ab ≠ 0. What is the sum of the solutions to the equation x?2 = ?( 1)2

x − ?

A) -1 B) -0.75 C) -0.25 D) 0.25 E) 0.75 Problem 36 Dave plans to work 10 days a month at a convenience store for one year. His boss offered Dave a choice between two sets of wage payments. Dave could choose to be paid at the end of each work day, in which case the wage in dollars he makes for any given day of work is equal to the number of days he has worked total to that point. He could also choose to be paid monthly, in which case each month he would receive double what he received the previous month starting with $2 for the first month. Which plan should Dave choose, and how much more money will he make by choosing that plan as opposed to the other? A) Daily, $128 B) Daily, $1080 C) Monthly, $930 D) Monthly, $1080 E) Monthly, $2048 Problem 37 If a%b = 2 25 a b− and x%y = 25, where x and y are both positive integers, what

is the value of x + y? A) 16 B) 25 C) 28 D) 48 E) 49 Problem 38 Mike and Jim are both start driving east on the same road, but Jim starts 150 miles ahead of Mike. Mike decides to start out driving 10 miles per hour (mph) and double his speed every hour; Jim decides to start out driving 64 mph and cut his speed in half every hour. How long, in hours, until Mike passes Jim? A) 53/27 hours

B) 9/2 hours C) 62/13 hours D) 74/15 hours E) 16/3 hours Problem 39 In a certain sequence, each term beyond the second term is equal to the average of the previous two terms. If a1 and a3 are positive integers, which of the following is NOT a possible value of a5? A) -9/4 B) 0 C) 9/4 D) 75/8 E) 41/2 Problem 40 In a certain sequence, the term an is defined as the value of x that satisfies the equation 2 = (x/2) – an-1. If a6 = 156, what is the value of a2? A) 1 B) 6 C) 16 D) 26 E) 106

Problem 1 Solution: D This is an arithmetic sequence where the difference between successive terms is always 10. Regardless of the actual values of, for example, a10 and a11, the difference between the two is 10. The difference between a10 and a12 is 20. And the difference between a10 and a15 is 50. Problem 2 Solution: E This is a geometric sequence where the ratio of successive terms is always 10. From the formula, a9 = 10a8 and a10 = 10a9. Substitute for a9 in the second equation to give: a10 = 10(10a8) = 100a8. Problem 3 Solution: C Since the sister was either nice or mean every one of the 365 days last year, n+m must equal 365. Since n = 300, m = 365 – 300 = 65. Plug in these values for n and m to

calculate d: d = 25 300 5 65 7500 325 7175365 365 365

⋅ − ⋅ −= = , which is approximately 19.7; $20

is the value to the nearest dollar. Problem 4 Solution: B First, calculate Ashley’s score: ta = 3×10×10 – 25×8 = 300 – 200 = 100. If Ashley and Beatrice scored the same score and Beatrice did half as many sit-ups (5) and twice as many push-ups (20): tb = 3×5×20 – 25m = 100. Solve for m: 300 – 25m = 100 and 25m = 200 so m = 8. Since p and s are multiplied together in the score formula, if we multiply s by 1/2 and p by 2, we keep the same value for 3ps. Beatrice will need the same mile time to achieve the same overall score as Ashley. Problem 5 Solution: C Following the rules of PEMDAS, we calculate the expressions within parentheses first: 7{}6 = (7-6)(7+6) = 1×13 = 13. 11{}11 = (11-11)(11+11) = 0. Substituting these back into the original expression: (13){}(0) = (13-0)(13+0) = 13×13 = 169.

Problem 6 Solution: B Plug in a = 5 and b = 10 to each of the expressions to determine which of them yield 5||10 = 5: (A) 10 – 5 = 5 YES

(B) 25

310− = 5 YES

(C) 5 104

− ⋅ = -12.5 NO

(D) 10 155+ = 5 YES

(E) 5 + 10 + 4 = 19 NO After using the first equation, (A), (B), and (D) remain as possible correct answer choices. Use the second equation to make further eliminations: 1||(-2) = 1. (A) -2 – 1 = -3 NO

(B) 21 ( )

32− − = 1 YES

(D) 2 151

− + = 13 NO

The answer is B. Problem 7 Solution: C The question presents a formula with a number of variables and values for all but one of those variables (h, the height of the ball, which is what we need to solve for). Solve for h by plugging in the values given for the other variables: t = 2 seconds, v = 80 feet/second, d = 6 feet. h = -16(2)2 + (80)(2) + 6 = 102 feet. Problem 8 Solution: D This question gives us a formula as well as values for all the variables in that formula but one (d, the number of sick days, which is what we need to solve for). Solve for d by plugging in the values given for the other variables: b = $1,068, y = 4 years, c = 21 cars. 1068 = 4 (15×21 – 8d); Simplify and divide both sides by 4: 267 = 315 – 8d; Isolate the variable: 8d = 315 – 267 = 48 and d = 6 days.

Problem 9 Solution: E The key to these “strange symbol” problems is not to be distracted by the unfamiliar notation but instead to identify the values of the variables (a and b) in the definition of the new symbol (#) and plug these values into the expression for the definition. In this problem a = (-4) and b = 4. Plugging in gives us: ( )24 4 ( 4)− − − = 16×2 + 4 = 36. Do not forget to keep the parentheses around the -4! Problem 10 Solution: B Order of operation rules (PEMDAS) stay the same even when we throw in new symbols. First, calculate the value of the expression in parentheses, 6$2. Plugging x = 6 and y = 2 into the definition for the $ operator yields 62/2 =36/2 = 18. Replace 6$2 with 18 in the original expression to give 9$18. Again, we use the definition of $ to solve. This time, x = 9 and y = 18 so 9$18 = 92/18 = 81/18 = 9/2. Problem 11 Solution: D Both Amy and Bob start with $1,000 and earn 8% interest annually; the difference is in how often this interest is compounded. Amy’s interest is compounded twice a year at 4% (8% / 2 times a year) each time. Bob’s interest is compounded four times a year at 2% (8% / 4 times a year) each time. After 6 months, Amy has $1,000×1.04 = $1,040.00 (1 interest payment at 4%) and Bob has $1,000× (1.02)2 = $1,040.40 (2 interest payments at 2%). The difference is $1,040.40 – $1,040.00 = $0.40. Problem 12 Solution: D After each half-life, the sample is left with half of the isotopes it started with in the previous period. After one half-life, the sample goes from 16,000 isotopes to 8,000. After two half-lives, it goes from 8,000 to 4,000. We can continue in this fashion to determine the total number of half-lives that have passed: 4,000 becomes 2,000 after 3 half-lives, 2,000 becomes 1,000 after 4 half-lives, 1,000 becomes 500 after 5 half-lives. The sample will have 500 isotopes after 5 half-lives or 5×5730 = 28,650 years. Algebraically, the equation is 500 = 16,000(1/2)t where t is the number of half-lives that need to pass. Solving yields: 1/32 = 1/2t and 32 = 2t or t = 5. Problem 13 Solution: B

Substitute the expression for g(x) in for x in the expression for f(x): f(g(x)) = 2 ( ) 2 (3 2) 4 3

5 5 5g x x x− − − −

= = . Since f(g(x)) = 1, solve the equation 4 35

x− = 1; this

gives 4 – 3x = 5, so -3x = 1 and x = -1/3. Problem 14 Solution: D Follow the PEMDAS order and calculate the expressions inside parentheses first. g(2) = 2/2 + 2 = 3. Now the expression is 2f(3). f(3) = 2(3) – 2 = 4. Now the expression is 2×4 or 8. Problem 15 Solution: B If Tonya had twice as many triple axels as falls then t = 2f. Substituting that into the formula for the score gives s = 10(3(2f) – 5f)/m. Simplify: s = 10f/m. s = 6 and m = 5 so 6 = 10f/5 or f = 3. Problem 16 Solution: D Calculate how many 8 year periods are between 33 years old (when the uncle has $13m) and 65 (when he retires): (65-33)/8 = 4. The uncle’s wealth will double 4 times between 33 and retirement: $13m×24 = $13m×16 = $208m. As a check: the uncle has $13m at age 33, $26m at age 41, $52m at age 49, $104m at age 57, and $208m at age 65. Problem 17 Solution: B Follow PEMDAS order and start with the innermost parentheses, (-1)~1. In this

case a = -1 and b = 1: 5 1 41 1 1 4 31 1−

− − = − − = − + =− −

. So now we have 1~((-1)~1) =

1~3. Now a = 1 and b = 3: 5 31 1 2 11−

− = − = − .

Problem 18 Solution: D

Calculate each of the archers’ scores by plugging in the appropriate values of b, a,

and s. For the first archer, b = a = s = 10 and the score is 50 10 10 10 400 2010 10 20⋅ − ⋅

= =+

. For

the second archer, b is half of 10 or 5, a is double 10 or 20, and s = 15. The score for the

second archer is 50 5 10 20 50 210 15 25⋅ − ⋅

= =+

. The difference in scores is 20 – 2 = 18.

Problem 19 Solution: E Let the constant that we add to a term to get the next term be equal to k. Since term 2 = 27, term 3 = 27+k, term 4 = 27+2k, and term 5 = 27+3k. Since we know the value of term 5, 27+3k = 84; 3k = 57 and k = 19. To find the term that comes before the second term, we subtract k from the second term (since term 1 plus k equals term 2): 27 – 19 = 8 is the first term. Problem 20 Solution: C Substitute a = 1 and b = 2 into the definition of a%b: 3(1) – 2(2) = -1. Substitute

a = 3 and b = 4 into the definition of a”b: 1 1 12(3) 3(4) 6 6

= = −− −

. Subtracting gives (-1)

– (-1/6) = -1 + 1/6 = -5/6. Problem 21 Solution: D This is an arithmetic sequence. In other words, the difference between successive terms is always the same (in this case, that difference is equal to 5). Calculating the first few terms of the sequence gives 1, 6, 11, 16, 21, etc. We can rewrite this sequence as a direct function of the term number: an = 1 + 5(n-1). The 75th term of this sequence is a75 = 1 + 5(74) = 371. To find the sum of an arithmetic sequence, we take the average value of the terms times the number of terms, and the average is the midpoint between the first

and last terms. The average of the 1st and 75th terms is 1 371 1862

+= , and we know that

there are 75 terms. The sum of the first 75 terms then is 186×75 = 13950. Problem 22 Solution: E

This is a geometric sequence. In other words, the ratio of two successive terms is always the same (in this case that ratio is equal to 2). Calculating the first few terms of the series gives 1, 2, 4, 8, 16, etc. We can rewrite this sequence as a direct function of the term number: an = 2n-1. The sum of the first m-1 terms of this sequence is always equal to one less than the mth term. The sum of the first 10 terms of the sequence is 1 less than the 11th term (which is 210 = 1024) and the difference between the sum of the first 10 terms and the sum of the 11th and 12th terms is simply 1 plus the value of the 12th term or 1 + 211 = 1 + 2048 = 2049. Problem 23 Solution: B Use the definition of the new symbol to rewrite the equation x@5 = 3@x without the @ operator. For x@5, a = x and b = 5: x@5 = (x – 1)(5 – 2) = 3x – 3. For 3@x, a = 3 and b = x: 3@x = (3 – 1)(x – 2) = 2x – 4. Equating these two expressions gives us 3x – 3 = 2x – 4. Solving yields x = -1. Problem 24 Solution: B Start by solving for the constant, k. The problem says a family of 4 (s=4) has a wait time of 30 minutes (w=.5 hours – Don’t forget that w is in hours!) when 2 parties are

ahead of it (n=2). Plug these values into the formula: .5 = 2 410

k+ . Solve for k: 5 = 2 +

4k; 3 = 4k; k = 3/4. To solve for the wait time of the family of 6 with 8 parties ahead of

it, plug these values into the formula along with k = 3/4:

38 64

10w

⎛ ⎞+ ⎜ ⎟⎝ ⎠= = 1.25 hours.

The answer choices are denoted in hours and minutes. 0.25 hours is equal to .25×60 or 15 minutes. The answer is 1 hour 15 minutes. Problem 25 Solution: A The sequence is defined in such a way that given a term, we can use the definition to find the next term, but this process can also be reversed. The problem is giving us a term (a4) and asking for a previous term (a1). Manipulate the definition of the sequence

to make it easy to find the previous term in the sequence: 13

5n

naa −

+= .

43

3 32 3 75 5

aa + += = = . 3

23 7 3 2

5 5aa + +

= = = . 21

3 2 3 15 5

aa + += = = .

Problem 26 Solution: E This formula defines a sequence in which each term is 7 less than the previous term. In the problem, we are given the 7th term and asked to work backwards to find the first term. We know that a7 is 7 less than a6, making a6 equal to 7+7 or 14. To go from a7 to a1 will require 6 such steps of adding 7 to get the previous term: a1 = a7 + 6(7) = 7 + 42 = 49. Problem 27 Solution: D Compute the expressions for each of the terms: x!4 = x4×4-x and 4!x = 4x×x-4. Dividing the first by the second yields x4-(-4)×4-x - x = x8×4-2x or (x4×4-x)2 = (x!4)2. Alternatively, we could rewrite the definition of the ! operation as a!b = ab×1/ba = ab/ba. Recognize that a!b = 1/(b!a) and 4!x = 1/(x!4). (x!4)÷(4!x) = (x!4)×1/(4!x) = (x!4)× (x!4) = (x!4)2. Problem 28 Solution: A First, we need to solve for the constant k using the price information of the 2 bed,

2 bath unit (r = t = 2 and f = 1). 25 2 10 21 5

k ⋅ + ⋅+

= 800. Simplify and solve for k: k× (20/3)

= 800 and k = 120. Solve for the rent on the 3 bed, 1 bath on the 3rd floor (r = 3, t = 1,

and f = 3): 25 3 10 11203 5

⋅ + ⋅+

= 825.

Problem 29 Solution: B The first multiple of 3 in this range is 252 and the last is 348 (remember that to be a multiple of 3, the sum of a number’s digits must be a multiple of 3). The sum of an arithmetic sequence is the average value times the number of terms. The average value is the midpoint between 252 and 348, which is 300. To find the number of terms, take the high versus low term, divided by the size of the interval, and add 1. This yields (348-252)/3 + 1 = 96/3 + 1 = 32 + 1 = 33. The sum is then 300×33 = 9900. Problem 30 Solution: D

Since their base pay for the first 40 hours is different, we know that each had to work overtime in order to equalize their pay for the week. Assuming she worked overtime, Abby’s pay is 8(40)+10(h-40) where h is the number of hours worked. Becky’s pay is 7(40)+12(h-40)+25. For the pay to be equal for the same h, we get: 8(40)+10(h-40) = 7(40)+12(h-40)+25 or 10h – 80 = 12h – 175. 2h = 95 and h = 47.5. Problem 31 Solution: C The inequality to be solved in this problem is 160,000× (1.2)t < 80,000× (1.5)t. However, given that the values in the answer choices are relatively small, it is probably faster to set up a table and calculate the population of each town after every year. After 1 year: A has 160,000(1.2) = 192,000 people and B has 80,000(1.5) = 120,000 people. After 2 years: A = 192,000(1.2) = 230,400 and B = 120,000(1.5) = 180,000. After 3 years: A = 230,400(1.2) = 276,480 and B = 180,000(1.5) = 270,000. After 4 years: A = 276,480(1.2) = 331,776 and B = 270,000(1.5) = 405,000. You can probably stop after calculating the populations for after 3 years if you recognize that the populations are almost equal after 3 years and B will certainly pass A in one more year given its much higher growth rate. Problem 32 Solution: A First, calculate the sum of the even integers between 1 and 100 (2, 4, 6… 98, 100). We can think of this list as 50 even integers to be summed or, more usefully, we can think of it as 25 integer pairs each of which sum to 102 (2 + 100, 4 + 98, …, 50 + 52). The sum of these 25 pairs is simply 25×102. This is an alternate approach to the usual formula that tells us to compute the average value times the number of terms. Next, calculate the sum of the odd integers between 100 and 150 (101, 103, 105… 147, 149). Like before, we can turn this list into pairs, though we need to be careful because there are an odd number of integers in this list and the middle number (125) will not get a pair: (101 + 149, 103 + 147… 123 + 127). There are 12 pairs that each add to 250 and one leftover integer, 125. The sum is 12×250 + 125 or 12.5×250. The ratio be asked for in the question is (25×102) to (12.5×250). We can divide both sides by 25 to yield the ratio 102 to 125. Problem 33 Solution: E There are two equations that we can write using the formula that defines the sequence and that uses either the two terms we know (a1 and a4) or the two whose ratio we are looking for (a2 and a3). Those two equations are the formulas for a3 (which uses

a1 and a2) and for a4 (which uses a2 and a3). This gives us: 1 2 23

2 2

4a a aaa a+ +

= = and

2 34

3

a aaa+

= or 2 3

3

3 a aa+

= . Plug in our expression for a3 in terms of a2 into the second

equation:

22

2

2

2

4

3 4

aaaa

a

++

=+

. Multiply both sides by 2

2

4 aa+ to get 2 2

22 2

4 43 a aaa a+ +

= + ,

and mutiply by a2 to get 22 2 212 3 4a a a+ = + + . This gives us a2

2 – 2a2 – 8 = 0 or (a2 – 4)(a2 + 2) = 0 so a2 = 4 or -2. If a2 = 4, a3 = (4+4)/4 = 2 and a2/a3 = 4/2 = 2. If a2 = -2, a3 = (4-2)/(-2) = -1 and a2/a3 = (-2)/(-1) = 2. In either case, the ratio is 2. Problem 34 Solution: C The problem gives two ways to calculate the fourth term: (1) the definition of the sequence tells us that a4 = a3

2 – a22 and (2) we are told that a4 = a1(a2 + a3) = 2(a2 + a3).

Setting these two equal gives a32 – a2

2 = 2(a2 + a3). Factor the left side: (a3 + a2)(a3 – a2) = 2(a2 + a3). Since an > 0, we know that (a3 + a2) does not equal 0 and we can divide both sides by it: a3 – a2 = 2 and a3 = a2 + 2. Using the definition of a3, we know a3 = a2

2 – a1

2 = a22 – 4. Substituting for a3 yields: a2 + 2 = a2

2 – 4 and a22 – a2 – 6 = 0. Factor

and solve: (a2 – 3)(a2 + 2) = 0; a2 = 3 or -2. an must be positive, so a2 = 3 and a3 = a2 + 2 = 3 + 2 = 5. Problem 35 Solution: D

Use the definition of ? to rewrite the equation: 1 2 1 1 1 1 12 2 1

x xx x

⎛ ⎞+ + + − +− = −⎜ ⎟−⎝ ⎠

.

Simplifying yields: 1 3 1 12 2

x xx x+ +

− = . Let z = 1xx+ . Substitute z into the equation:

z – 3/2 = (1/2)z or z = 3. To solve 1xx+ = 3, take two cases:

(1) 1 0xx+

> In this case, x+1 and x have the same sign; in other words x > 0 or x < -1.

1xx+ = 3 or x = .5.

(2) 1 0xx+

< In this case, x+1 and x have opposite signs; in other words -1 < x < 0.

1xx+ = -3 or x = -.25. The sum of the solutions is 0.5 + (-0.25) = 0.25.

Problem 36 Solution: C Calculate the total amounts paid out by each plan. In the daily plan, Dave receives $1 for day 1, $2 for day 2, …, and $120 for his last day of work (10 days a month × 12 months). To find the total amount paid under this plan, we need to find the sum of the integers from 1 to 120. We can either use the formula for the sum of the

integers from 1 to n 2

2n n⎛ ⎞+

⎜ ⎟⎝ ⎠

or we can recognize that this sequence consists of 60 pairs

of numbers that sum to 121 (1+120, 2+119, etc.). Either way, the sum comes out to 60×121 = 7260. Under the monthly plan, David makes $2 in month 1, $4 in month 2, $8 in month 3, …, and $212 or $4,096 for the last month. Recognize that the sum of this sequence up to the nth term is always 2 less than the value of the n+1 term (2+4 is 2 less than 8, 2+4+8 is 2 less than 16, etc.). The sum of 2+4+8+…4096 is equal to 213 – 2 = 8192 – 2 = 8190. Dave should select the monthly plan which will give him $8190 – $7260 = $930 more than the daily plan. Problem 37 Solution: B

Simplify the equation 2 25 25yx =− to give 2 2x y− = 25 which means either

x2 – y2 = 25 or y2 – x2 = 25, depending on which of the two variables is larger. Because we are looking for the sum x+y though, it doesn’t matter which is larger and we can solve either equation (you can verify this yourself by solving both equations). So we can focus on the equation x2 – y2 = 25 or x2 – y2 = 52. Rearranging gives 52 + y2 = x2 which is the form of the Pythagorean Theorem. x and y must be integers, so we are looking for a Pythagorean triple that has 5 as the length of one of the legs of the right triangle. The only such triple is 5, 12, 13, so x+y = 12+13 = 25. Problem 38 Solution: C At the start, Jim is 150 miles ahead. After 1 hour, Mike has driven 10 miles and Jim has driven 64. Jim is now 150 + 64 – 10 = 204 miles ahead. At this point, Mike doubles his speed to 20 mph and Jim halves his speed to 32 mph. After the second hour goes by, Jim is 204 + 32 – 20 = 216 miles ahead. After 3 hours, Jim is 216 + 16 – 40 =

192 miles ahead. After 4 hours, Jim is 192 + 8 – 80 = 120 miles ahead. During the fifth hour, Mike will be traveling at 160 mph and Jim at 4 mph, so Mike is closing the gap with Jim at a speed of 160 – 4 or 156 mph. It will take 120/156 or 10/13 hours for Mike to close the 120 mile gap that existed at the beginning of the fifth hour. The total time it took for Mike to pass Jim is 4 + 10/13 = 62/13 hours. Problem 39 Solution: D Since a1 and a3 are integers, a2 must also be an integer: a3 = (a1 + a2)/2 or INT = (INT + a2)/2 so 2(INT) = INT + a2 and a2 = 2(INT) – INT which is itself an integer. a4 will be the average of two integers. If a2 + a3 is even, a4 will be an integer. If a2 + a3 is odd, a4 will be a decimal ending in .5. If a4 is an integer, a5 can be an integer or can be a decimal ending in .5. If a4 is a decimal ending in .5, a5 must be a decimal ending in .25 or .75. a5 cannot be a decimal ending in .375 such as 75/8 = 9.375. Note that a5 can be negative: Even if a1 and a3 are positive, that does not rule out the possibility that a2 (and subsequent terms) could be negative. Problem 40 Solution: B First solve for x in the equation to reveal the recursive formula for calculating a_n: 4 = x – 2(a_(n-1)) and x = 4 + 2(a_(n-1)) or a_n = 4 + 2(a_(n-1)). Since we have a_6 and want to calculate a previous term, a_4, it may be useful to rewrite the equation in a form that allows us to solve for the previous term: a_(n-1) = (a_n – 4)/2. a_5 = (156-4)/2 = 76. a_4 = (76-4)/2 = 36. a_3 = (36-4)/2 = 16. And a_2 = (16-4)/2 = 6.