 # Precalculus Notes

[Last Updated 22 January 2011]

## Roman Numerals

### Arabic Roman 1 I 2 II 3 III 4 IV 5 V 6 VI 7 VII 8 VIII 9 IX 10 X 11 XI 12 XII 13 XIII 14 XIV 15 XV 16 XVI 17 XVII 18 XVIII 19 XIX 20 XX 30 XXX 40 XL 50 L 60 LX 70 LXX 80 LXXX 90 XC 100 C 500 D 1000 M 1492 MCDXCII 1925 MCMXXV 1969 MCMLXIX 2009 MMIX

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### EXERCISES Convert each of the following to Roman or Arabic numerals: 1.1. 8 1.2. XVII 1.3. 3 1.4. X 1.5. 5 1.6. XIII 1.7. 20 1.8. XV 1.9. 9 1.10. XVI ANSWERS 1.1. VIII 1.2. 17 1.3. III 1.4. 10 1.5. V 1.6. 13 1.7. XX 1.8. 15 1.9. IX 1.10. 16

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### EXERCISES Convert each of the following to Roman or Arabic numerals: 2.1. 46 2.2. LXXXIX 2.3. 93 2.4. XXVIII 2.5. 76 2.6. LXII 2.7. 74 2.8. LX 2.9. 71 2.10. LV ANSWERS 2.1. XLVI 2.2. 89 2.3. XCIII 2.4. 28 2.5. LXXVI 2.6. 62 2.7. LXXIV 2.8. 60 2.9. LXXI 2.10. 55

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### EXERCISES Convert each of the following to Roman or Arabic numerals: 3.1. 1729 3.2. MDCVI 3.3. 1922 3.4. DXXXIX 3.5. 1155 3.6. MDXLVII 3.7. 666 3.8. DCXXV 3.9. 1539 3.10. MLIV ANSWERS 3.1. MDCCXXIX 3.2. 1606 3.3. MCMXXII 3.4. 809 3.5. MCLV 3.6. 1547 3.7. DCLXVI 3.8. 625 3.9. MDXXXIX 3.10. 1054

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## Prime and Composite Numbers

### PROBLEM Find the prime factors of the following numbers: (a) 33755 (b) 54321 (c) 48700 (d) 675 (e) 123456 (f) 347826 (g) 748647 (h) 450031 (i) 235550 (j) 76432 (k) 52744 (l) 11367 (m) 1958 (n) 1929 (o) 1800 (p) 46553 SOLUTION (a) 33755 = 5 · 6751 = 5 · 43 · 157 (b) 54321 = 3 · 18107 =3 · 19 · 953 (c) 48700 = 2 · 24350 = 22 · 12175 = 22 · 5 · 2435 = 22 · 52 · 487 (d) 675 = 5 · 135 = 52 · 27 = 52 · 33(e) 123456 = 2 · 61728 = 22 · 30864 = 23 · 15432 = 24 · 7716 = 25 · 3858 = 26 · 1929 = 26 · 3 · 643 (f) 347826 = 2 · 173913 = 2 · 3 · 57971 = 2 · 3 · 29 · 1999 (g) 748647 = 3 · 249549 = 32 · 83183 = 32 · 193 · 431 (h) 450031 = 37 · 12163 (i) 235550 = 2 · 117775 = 2 · 5 · 23555 = 2 · 52 · 4711 = 2 · 52 · 7 · 673 (j) 76432 = 2 · 38216 = 22 · 19108 = 23 · 9554 = 24 · 4777 = 24 · 17 · 281 (k) 52744 = 2 · 26372 = 22 · 13186 = 23 · 6593 = 23 · 19 · 347 (l) 11367 = 3 · 3789 = 32 · 1263 = 33 · 421 (m) 1958 = 2 · 979 = 2 · 11 · 89 (n) 1929 = 3 · 643 (o) 1800 = 2 · 900 = 22 · 450 = 23 · 225 = 23 · 5 · 45 = 23 · 52 · 9 = 23 · 32 · 52(p) 46553 = 13 · 3581

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## Fractions

### PROBLEM Add 1/5, 1/10, 1/15, 1/20, and 1/25. SOLUTION Define S = 1/5 + 1/10 + 1/15 + 1/20 + 1/25 Write the denominators as products of prime factors. S = 1/5 + 1/(2 · 5) + 1/(3 · 5) + 1/(22 · 5) + 1/52The lowest common denominator is 22 · 3 · 52 = 300. S = 60/300 + 30/300 + 20/300 + 15/300 + 12/300 = 137/300

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### PROBLEM Write each of the following fractions in the form 1/a + 1/b, where a and b are integers. (a) 1/2 (b) 1/12 (c) 1/25 (d) 1/37 SOLUTION (a) Assume that in general we can write a fraction 1/n in the form 1/(n+1) + 1/x, where x is an integer. Then 1/n = 1/(n+1) + 1/x => 1/x = 1/n - 1/(n+1) = (n+1) / [n(n+1)] - n / [n(n+1)] = 1 / [n(n+1)] => 1/n = 1/(n+1) + 1 / [n(n+1)] Thus, 1/2 = 1/(2+1) + 1/[2(2+1)] = 1/3 + 1/6 (b) 1/12 = 1/(12+1) + 1/[12(12+1)] = 1/13 + 1/156 (c) 1/25 = 1/(25+1) + 1/[25(25+1)] = 1/26 + 1/650 (d) 1/37 = 1/(37+1) + 1/[37(37+1)] = 1/38 + 1/1406

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## Real and Rational Numbers

### PROBLEM Express each of the following real numbers (where the underlined digits are repeated indefinitely) as a rational number: (a) 0.13(b) 0.123(c) 0.17(d) 0.414(e) 2.67479(f) 0.14583(g) 3.321(h) 3.14(i) 0.191(j) 0.4(k) 0.7144SOLUTION (a) Let x = 0.13. Then 10x = 1.3 (1) 100x = 13.3 (2) Subtract (1) from (2). 90x = 12 => x = 12/90 = 6/45 = 2/15 (b) Let x = 0.123 (1) Then 1000x = 123.123 (2) Subtract (1) from (2). 999x = 123 => x = 123/999 = 41/333 (c) Let x = 0.17 (1) Then 100x = 17.17 (2) Subtract (1) from (2). 99x = 17 => x = 17/99 (d) Let x = 0.414 (1) Then 1000x = 414.414 (2) Subtract (1) from (2). 999x = 414 => x = 414/999 = 138/333 = 46/111 (e) Let x = 2.67479 (1) Then 100000x = 267479.67479 (2) Subtract (1) from (2). 99999x = 267477 => x = 267477/99999 = 89159/33333 = (7)(47)(271) / (3)(41)(271) = (7)(47) / (3)(41) = 329/123 (f) Let x = 0.14583. Then 10000x = 1458.3 (1) 100000x = 14583.3 (2) Subtract (1) from (2). 90000x = 13125 => x = 13125/90000 = 2625/18000 = 525/3600 = 105/720 = 21/144 = 7/48 (g) Let x = 3.321 (1) Then 1000x = 3321.321 (2) Subtract (1) from (2). 999x = 3318 => x = 3318/999 = 1106/333 (h) Let x = 3.14 (1) Then 100x = 314.14 (2) Subtract (1) from (2). 99x = 311 => x = 311/99 (i) Let x = 0.191 (1) Then 1000x = 191.191 (2) Subtract (1) from (2). 999x = 191 => x = 191/999 (j) Let x = 0.4 (1) Then 10x = 4.4 (2) Subtract (1) from (2). 9x = 4 => x = 4/9 (k) Let x = 0.7144. Then 10x = 7.144 (1) 10000x = 7144.144 (2) Subtract (1) from (2). 9990x = 7137 => x = 7137/9990 = 793/1110

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## Calculating Square Roots

### PROBLEM Calculate the square root of 4321 rounded off to two decimal places. SOLUTION Break up 4321 into groups of two digits. ---------------- | 43 21.00 00 00 Examine the first group of two digits, 43, and determine the largest integer which is less than or equal to its square root. The largest such integer is 6. Write 6 on top of the 43.    6 ---------------- | 43 21.00 00 00 Square the 6, write the result under the 43, subtract it from the 43 and bring down the next two digits, the 21.    6 ---------------- | 43 21.00 00 00 |-36 ----------------    7 21 Double the 6 and write the result with a "-" to the left of the "7 21".       6    ----------------    | 43 21.00 00 00    |-36    ---------------- 12-|  7 21 Determine how many times "12-", where "-" represents some digit between zero and nine, will go into 721 without exceeding it. Write this number, 5, at the top, above the 21, and where the "-" was.       6  5    ----------------    | 43 21.00 00 00    |-36    ---------------- 125|  7 21 Multiply 125 by 5 and write the result, 625, on the next line, subtract it, and bring down the next two digits.       6  5    ----------------    | 43 21.00 00 00    |-36    ---------------- 125|  7 21    |- 6 25    ----------------         96 00 Double the 65 and write the result, 130, with a "-" to the left of the "96 00".        6  5     ----------------     | 43 21.00 00 00     |-36     ----------------  125|  7 21     |- 6 25     ---------------- 130-|    96 00 Determine how many times 130-, where "-" is some digit between zero and nine, will go into 9600 without exceeding it. Write this number, 7, at the top, with a decimal point to the left of it, and where the "-" was.        6  5. 7     ----------------     | 43 21.00 00 00     |-36     ----------------  125|  7 21     |- 6 25     ---------------- 1307|    96 00 Multiply 1307 by 7 and write the result, 9149, on the next line, subtract it, and bring down the next two digits.        6  5. 7     ----------------     | 43 21.00 00 00     |-36     ----------------  125|  7 21     |- 6 25     ---------------- 1307|    96 00     |  - 91 49     ----------------           4 51 00 Double the 657 at the top and write the result, 1314, with a "-" to the left of the "4 51 00".        6  5. 7     ----------------     | 43 21.00 00 00     |-36     ----------------  125|  7 21     |- 6 25     ---------------- 1307|    96 00     |  - 91 49     ----------------   1314-|  4 51 00 Determine how many times 1314-, where "-" is some digit between zero and nine, will go into 45100 without exceeding it. Write this number, 3, at the top and where the "-" was.        6  5. 7  3     ----------------     | 43 21.00 00 00     |-36     ----------------  125|  7 21     |- 6 25     ---------------- 1307|    96 00     |  - 91 49     ----------------   13143|  4 51 00 Multiply 13143 by 3 and write the result, 39429, on the next line, subtract it, and bring down the next two digits.        6  5. 7  3     ----------------     | 43 21.00 00 00     |-36     ----------------  125|  7 21     |- 6 25     ---------------- 1307|    96 00     |  - 91 49     ----------------   13143|  4 51 00        |- 3 94 29        -------------             56 71 00 Double the 6573 at the top and write the result, 13146, with a "-" to the left of the "56 71 00".        6  5. 7  3     ----------------     | 43 21.00 00 00     |-36     ----------------  125|  7 21     |- 6 25     ---------------- 1307|    96 00     |  - 91 49     ----------------   13143|  4 51 00        |- 3 94 29        -------------    13146-|  56 71 00 Determine how many times 13146-, where "-" is some digit between zero and nine, will go into 567100 without exceeding it. Write this number, 4, at the top and where the "-" was.        6  5. 7  3  4     ----------------     | 43 21.00 00 00     |-36     ----------------  125|  7 21     |- 6 25     ---------------- 1307|    96 00     |  - 91 49     ----------------   13143|  4 51 00        |- 3 94 29        -------------    131464|  56 71 00 Multiply 131464 by 4 and write the result, 525856, on the next line and subtract it.        6  5. 7  3  4     ----------------     | 43 21.00 00 00     |-36     ----------------  125|  7 21     |- 6 25     ---------------- 1307|    96 00     |  - 91 49     ----------------   13143|  4 51 00        |- 3 94 29        -------------    131464|  56 71 00          |- 52 58 56            ---------              4 12 44 This confirms that 4 is the largest whole number of times 131464 goes into 567100. The square root of 4321, rounded off to two decimal places, is 65.73.

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### PROBLEM Calculate the square root of p, rounded off to three decimal places. SOLUTION       1. 7  7  2  4    ----------------    |  3.14 15 92 65    |- 1    ----------------  27|  2 14    |- 1 89    ----------------   347|  25 15      |- 24 29      --------------     3542|  86 92         |- 70 84         -----------    35444|  16 08 65         |- 14 17 76         -----------             1 90 89 The square root of p, rounded off to three decimal places, is 1.772.

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## Algebra

### PROBLEM Twice the larger of two numbers is three more than five times the smaller, and the sum of four times the larger and three times the smaller is 71. What are the numbers? [from the movie Mean Girls (2004)] SOLUTION Let (x, y) = the (smaller, larger) number. Then 2y = 5x + 3 (1) and 4y + 3x = 71 (2) From (1), we get y = 5x/2 + 3/2 Subsituting this into (2), 4(5x/2 + 3/2) + 3x = 71 => 10x + 6 + 3x = 71 => 13x + 6 = 71 => 13x = 65 => x = 5 => y = (5/2)(5) + 3/2 = 14 The two numbers are 5 and 14.

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### PROBLEM Find an odd three-digit number whose digits add up to 12. The digits are all different and the difference between the first two digits equals the difference between the last two digits. [from the movie Mean Girls (2004)] SOLUTION Let (x, y, z) be the (units, tens, hundreds) digit. Then the required number satisfies the following four conditions: (1) x = 1, 3, 5, 7, or 9 (2) x + y + z = 12 (3) x ≠ y ≠ z ≠ x (4) z - y = y - x There are 65 numbers that satisfy condition (2): 129, 138, 147, 156, 165, 174, 183, 192 219, 228, 237, 246, 255, 264, 273, 282, 291 309, 318, 327, 336, 345, 354, 363, 372, 381 408, 417, 426, 435, 444, 453, 462, 471, 480 507, 516, 525, 534, 543, 552, 561, 570 606, 615, 624, 633, 642, 651, 660 705, 714, 723, 732, 741, 750 804, 813, 822, 831, 840 903, 912, 921, 930 31 of these also satisfy condition (1) and are odd: 129, 147, 165, 183 219, 237, 255, 273, 291 309, 327, 345, 363, 381 417, 435, 453, 471 507, 525, 543, 561 615, 633, 651 705, 723, 741 813, 831 903, 921 28 of these also satisfy condition (3) and have all-different digits: 129, 147, 165, 183 219, 237, 273, 291 309, 327, 345, 381 417, 435, 453, 471 507, 543, 561 615, 651 705, 723, 741 813, 831 903, 921 Four of these also satisfy condition (4): 147, 345, 543, 741

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### PROBLEM Three apples and two pears cost \$3.48. Four pears and one apple cost \$4.46. How much does one apple cost? How much does one pear cost? SOLUTION Let x = cost of one apple and y = cost of one pear. Then 3x + 2y = \$3.48 (1) 4y + x = \$4.46 (2) Multiply (2) by three. 12y + 3x = \$13.38 (3) Subtract (1) from (3). 10y = \$9.90 => y = \$0.99 Use (2) to solve for x in terms of y. x = \$4.46 - 4y = \$4.46 - (4)(\$0.99) = \$4.46 - \$3.96 = \$0.50 One apple costs \$0.50 or 50¢. One pear costs \$0.99 or 99¢.

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### PROBLEM Tommy has forty coins which are either quarters or dimes. The total value of the coins is \$7.75. How many quarters and how many dimes does Tommy have? SOLUTION Let x = number of quarters and y = number of dimes. Then 25x + 10y = 775 (1) x + y = 40 => y = 40 - x (2) Substituting (2) into (1), 25x + 10(40 - x) = 775 => 25x + 400 - 10x = 775 => 15x = 375 => x = 375/15 = 25 => y = 40 - 25 = 15 Tommy has 25 quarters and 15 dimes.

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### PROBLEM Jack can shovel the snow on someone's property in 5 hr. Joe can shovel the snow in 3 hr. (a) How long will it take Jack and Joe, working together, to shovel the snow? (b) What fraction of the entire job does each one do? (c) Assume that each is paid at the same hourly rate they would get if they did the job alone and that the total each would get paid if they worked alone is \$75. Who gets paid more if they work together? SOLUTION (a) The fraction of the job that (Jack, Joe) does per hour if working along is (1/5, 1/3). Let t be the time (in hours) it takes for both of them to do the job together. Then (1/5 + 1/3)t = 1 Multiply this equation by 15. (3 + 5)t = 15 => 8t = 15 => t = 15/8 hr = 1.875 hr (b) The fraction of the job that (Jack, Joe) does is [(1/5)(15/8), (1/3)(15/8)] = (3/8, 5/8). (c) (Jack, Joe) is paid [\$75/(5 hr), \$75/(3 hr)] = (\$15/hr, \$25/hr). The total (Jack, Joe) gets paid is [(\$15/hr)(15/8 hr), (\$25/hr)(15/8 hr)] = (\$225/8, \$375/8) = (\$28.125, \$46.875) => (\$28.13, \$46.88). So Joe gets paid more.

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## Law of Cosines

### PROBLEM A ranger in an observation tower can sight the north end of a lake 15 km away and the south end of the same lake 19 km away. The angle between these two lines of sight is 104°. How long is the lake? SOLUTION Draw triangle ABC as shown: south        north  end    a     end   C+----------+B     \        /      \      /      b\    /c        \  /         \/         A observation tower A is at the observation tower, B is at the north end of the lake, and C is at the south end. The distance from the tower to the north end is AB = c = 15 km The distance from the tower to the south end is AC = b = 19 km Angle A = 104°. Use the law of cosines to solve for a, the distance from the north end to the south end, which is the length of the lake. a = sqrt(b2 + c2 - 2bc cos A) = sqrt[(19 km)2 + (15 km)2 - (2)(19 km)(15 km) cos 104°] = 26.9 km

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## Synthetic Division

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### PROBLEM Use synthetic division to divide the given polynomial by the given binomial: (a) 2x3 + 3x2 - x - 2, x - 1 (b) 2x3 + 3x2 - x - 2, x + 2 (c) 3x4 + x3 - 2x2 + x - 1, x + 1 (d) x3 - 3x2 - x + 2, x - 4 (e) 2x3 - 3x2 + 9x - 4, x - 1/2 (f) 2x4 - 5x3 - 5x2 + 3x + 2, x + 1/2 (g) x5 - 4x + 2, x - 1 (h) 3x4 + 2x3 + x - 1, x + 2 (i) 4x3 + 3x - 3, x - 1/2 (j) 3x3 + 7x2 - 3, x - 2/3 [from Cannon, Lawrence O. and Joseph Elich 1994, Precalculus, Second Edition (New York: HarperCollins College Publishers), Develop Mastery exercises 3.2.19-28] SOLUTION (a) 2x3 + 3x2 - x - 2, x - 1 2  3  -1  -2  | 1    2   5   4  +--- ____________ 2  5   4   2 => 2x2 + 5x + 4 + 2 / (x - 1) (b) 2x3 + 3x2 - x - 2, x + 2 2  3  -1  -2 | -2   -4   2  -2 +---- ____________ 2 -1   1  -4 => 2x2 - x + 1 - 4 / (x + 2) (c) 3x4 + x3 - 2x2 + x - 1, x + 1 3  1  -2  1  -1 | -1   -3   2  0  -1 +---- _______________ 3 -2   0  1  -2 => 3x3 - 2x2 + 1 - 2 / (x + 1) (d) x3 - 3x2 - x + 2, x - 4 1  -3  -1   2 | 4     4   4  12 +--- _____________ 1   1   3  14 => x2 + x + 3 + 14 / (x - 4) (e) 2x3 - 3x2 + 9x - 4, x - 1/2 2  -3   9 -4 | 1/2     1  -1  4 +---- ____________ 2  -2   8  0 => 2x2 - 2x + 8 (f) 2x4 - 5x3 - 5x2 + 3x + 2, x + 1/2 2  -5  -5  3  2 | -1/2    -1   3  1 -2 +------ _______________ 2  -6  -2  4  0 => 2x3 - 6x2 - 2x + 4 (g) x5 - 4x + 2, x - 1 1  0  0  0  -4  2 | 1    1  1  1   1 -3 +--- _________________ 1  1  1  1  -3 -1 => x4 + x3 + x2 + x - 3 - 1 / (x - 1) (h) 3x4 + 2x3 + x - 1, x + 2 3  2  0   1  -1 | -2   -6  8 -16  30 +---- _______________ 3 -4  8 -15  29 => 3x3 - 4x2 + 8x - 15 + 29 / (x + 2) (i) 4x3 + 3x - 3, x - 1/2 4  0  3  -3 | 1/2    2  1   2 +----- ___________ 4  2  4  -1 => 4x2 + 2x + 4 - 1 / (x - 1/2) (j) 3x3 + 7x2 - 3, x - 2/3 3  7  0  -3 | 2/3    2  6   4 +----- ___________ 3  9  6   1 => 3x2 + 9x + 6 + 1 / (x - 2/3)

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## Quantitative Reasoning

### PROBLEM Consider three opaque boxes. One box contains all white balls, one all black balls, and one a mix of black and white balls. Each box is labeled, but the labels are all wrong. How many balls would you need to pull out to determine which box is which? [from Kirsner, Scott 2010, "Seeking Work? Be Prepared for Tests Aimed At Separating Stars from Duds," Boston Sunday Globe, G1] SOLUTION Suppose the boxes are labeled as shown. Then, because we are told that the labels are all wrong, the possible types of balls in each box are as written below the drawing of the box (B = black, W = white, B/W = black and white). +-----+  +-----+  +-----+ |  1  |  |  2  |  |  3  | |     |  |     |  |black| |black|  |white|  | and | |     |  |     |  |white| +-----+  +-----+  +-----+    W        B        B   B/W      B/W       W Thus, box 1 actually contains either all white or mixed black and white balls, box 2 actually contains either all black or mixed black and white balls, and box 3 actually contains either all black or all white balls. Suppose we draw one ball from box 3. If the ball is black, this means that box 3 contains all black balls, box 2 contains mixed black and white balls, and box 1 contains all white balls. If the ball drawn from box 3 is white, this means that box 3 contains all white balls, box 1 contains mixed black and white balls, and box 2 contains all black balls. Thus, by picking one ball from box 3, it is possible to determine what kind of balls is contained in all three of the boxes.

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### PROBLEM A car's odometer read 14632 miles on 26 October 2007, 16686 miles on 8 March 2008, and 20972 miles on 28 December 2008. Estimate the number of miles driven from 28 December 2007 to 28 December 2008 assuming that the miles driven each day is the same. SOLUTION From 26 October 2007 to 28 December 2007 is 63 days. 2008 is a leap year, so there are 29 days in February 2008. From 26 October 2007 to 8 March 2008 is therefore 134 days. On 28 December 2007, the odometer should have read about 14632 miles + (16686 miles - 14632 miles)(63 days) / (134 days) = 15597.69 miles. The number of miles driven from 28 December 2007 to 28 December 2008 was about 20972 miles - 15597.69 miles = 5374.313 miles => 5374 miles.

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### PROBLEM Gasoline is purchased for a car on the dates shown in the table below. Each time gas is purchased, the tank is filled up. The table shows the mileage on the car's odometer, the price paid, and the number of gallons of gasoline purchased. Date Odometer Reading (miles) Gasoline Purchased (gallons) Total Cost Cost per Gallon 2007 Apr 1 11640 9.553 \$25.40 \$2.659 2007 Apr 13 11778 12.983 \$35.69 \$2.749 2007 May 12 12040 15.255 \$44.84 \$2.939 From the data in the table, determine the miles per gallon and the cost per mile achieved by the car. SOLUTION From the data, the total number of miles driven during the period shown is 12040 miles - 11640 miles = 400 miles. The total gas used is 12.983 gal + 15.255 gal = 28.238 gal. The total cost of the gas used is \$35.69 + \$44.84 = \$80.53. The miles per gallon achieved by the car is (400 miles) / (28.238 gal) = 14.17 mpg. The cost per mile is \$80.53 / (400 miles) = \$0.2013 per mile. Note that we neglected the initial purchase of gas in our calculations because it was used to fill up the tank prior to the period during which the 400 miles were driven.

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### PROBLEM Jack, a taxpayer, is in the 35% income tax bracket. (a) If his annual income in 2007 was \$75,000, and his income tax was deducted from his paycheck throughout the year at the rate corresponding to his tax bracket, how much income tax did he pay in 2007? (b) Jack paid \$20,000 in property tax in 2007. If this amount is deductible for income tax purposes (i.e., it counts as negative income), how much of an income tax refund should he get when he files his income tax forms? (c) If the \$20,000 in property tax is not deductible for income tax purposes but is taxed at 28% instead of 35%, how much of an income tax refund should he get? SOLUTION (a) Jack paid (0.35)(\$75,000) = \$26,250 income tax in 2007. (b) The income tax that Jack should pay for 2007 is (0.35)(\$75,000 - \$20,000) = \$19,250. Since \$26,250 was deducted from his paycheck, he should get a \$26,250 - \$19,250 = \$7,000 refund. (c) The income tax that Jack should pay for 2007 is (0.35)(\$55,000) + (0.28)(\$20,000) = \$24,850. Since \$26,250 was deducted from his paycheck, he should get a \$26,250 - \$24,850 = \$1,400 refund.

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### PROBLEM Joan, a taxpayer, earned \$85,000 in 2007. (a) Joan's income tax was deducted from her paycheck throughout the year at the rate corresponding to her tax bracket. If Joan is in the 35% tax bracket, how much income tax was deducted from her paycheck? (b) Joan paid \$10,000 in property tax in 2007. If this amount is taxed at 28% instead of 35%, how much of a refund should Joan get when she files her income tax forms? (c) How would your answer to part (b) change if Joan was in the 25% tax bracket? SOLUTION (a) (0.35)(\$85,000) = \$29,750 was deducted from Joan's paycheck. (b) Joan should pay (0.35)(\$75,000) + (0.28)(\$10,000) = \$29,050 income tax for 2007, so she should get a refund of \$29,750 - \$29,050 = \$700. (c) If Joan was in the 25% tax bracket, (0.25)(\$85,000) = \$21,250 would have been deducted from her paycheck. The amount of income tax she should actually pay is (0.25)(\$75,000) + (0.28)(\$10,000) = \$21,550. So she would owe \$21,550 - \$21,250 = \$300 in additional income tax when she files her income tax forms.

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### PROBLEM If 0 < st < 1, then which of the following can be true? (A) s < -1 and t > 0 (B) s < -1 and t < -1 (C) s > -1 and t < -1 (D) s > 1 and t < -1 (E) s > 1 and t > 1 [from Educational Testing Service advertisement in U.S. News and World Report, May 2009, p. 69] SOLUTION The question is asking us to select the choice which doesn't necessarily contradict the condition 0 < st < 1. We examine each choice. (A) If s < -1 and t > 0, st must be negative, which contradicts 0 < st < 1. (B) if s < -1 and t < -1, st must be greater than 1, which contradicts 0 < st < 1. (C) If s > -1 and t < -1, possible choices for s and t are s = - 1/4 and t = - 3/2. This yields st = (- 1/4)(- 3/2) = 3/8 = 0.375, which satisfies the condition 0 < st < 1. Thus, (C) is not necessarily a contradiction, and (C) can be true. (D) If s > 1 and t < -1, st < -1, which contradicts 0 < st < 1. (E) If s > 1 and t > 1, st > 1, which contradicts 0 < st < 1. Thus, the only choice which doesn't necessarily contradict 0 < st < 1 is (C).

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### PROBLEM If u > t, r > q, s > t, and t > r, which of the following must be true? I. u > s II. s > q III. u > r (A) I only (B) II only (C) III only (D) I and II (E) II and III [from Educational Testing Service advertisement in U.S. News and World Report, May 2009, p. 69] SOLUTION Combining the inequalities, we have s, u > t > r > q. Thus, s > q and u > r, but u might not be greater than s. It follows that II and III are true, and the answer is (E).

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