Roman Numerals

Prime and Composite Numbers

Fractions

Real and Rational Numbers

Binary Numbers

Calculating Square Roots

Algebra

Law of Cosines

Synthetic Division

Quantitative Reasoning

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 |

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

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

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

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 = 2

(d) 675 = 5 · 135 = 5

(e) 123456 = 2 · 61728 = 2

(f) 347826 = 2 · 173913 = 2 · 3 · 57971 = 2 · 3 · 29 · 1999

(g) 748647 = 3 · 249549 = 3

(h) 450031 = 37 · 12163

(i) 235550 = 2 · 117775 = 2 · 5 · 23555 = 2 · 5

(j) 76432 = 2 · 38216 = 2

(k) 52744 = 2 · 26372 = 2

(l) 11367 = 3 · 3789 = 3

(m) 1958 = 2 · 979 = 2 · 11 · 89

(n) 1929 = 3 · 643

(o) 1800 = 2 · 900 = 2

(p) 46553 = 13 · 3581

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/(2

The lowest common denominator is 2

S = 60/300 + 30/300 + 20/300 + 15/300 + 12/300 = 137/300

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

Express each of the following real numbers (where the underlined digits are repeated indefinitely) as a rational number:

(a) 0.1

(b) 0.

(c) 0.

(d) 0.

(e) 2.

(f) 0.1458

(g) 3.

(h) 3.

(i) 0.

(j) 0.

(k) 0.7

SOLUTION

(a) Let x = 0.1

10x = 1.

100x = 13.

Subtract (1) from (2).

90x = 12 => x = 12/90 = 6/45 = 2/15

(b) Let

x = 0.

Then

1000x = 123.

Subtract (1) from (2).

999x = 123 => x = 123/999 = 41/333

(c) Let

x = 0.

Then

100x = 17.

Subtract (1) from (2).

99x = 17 => x = 17/99

(d) Let

x = 0.

Then

1000x = 414.

Subtract (1) from (2).

999x = 414 => x = 414/999 = 138/333 = 46/111

(e) Let

x = 2.

Then

100000x = 267479.

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 = 0.1458

10000x = 1458.

100000x = 14583.

Subtract (1) from (2).

90000x = 13125 => x = 13125/90000 = 2625/18000 = 525/3600 = 105/720 = 21/144 = 7/48

(g) Let

x = 3.

Then

1000x = 3321.

Subtract (1) from (2).

999x = 3318 => x = 3318/999 = 1106/333

(h) Let

x = 3.

Then

100x = 314.

Subtract (1) from (2).

99x = 311 => x = 311/99

(i) Let

x = 0.

Then

1000x = 191.

Subtract (1) from (2).

999x = 191 => x = 191/999

(j) Let

x = 0.

Then

10x = 4.

Subtract (1) from (2).

9x = 4 => x = 4/9

(k) Let x = 0.7

10x = 7.

10000x = 7144.

Subtract (1) from (2).

9990x = 7137 => x = 7137/9990 = 793/1110

Express each of the following decimal numbers as a binary number, accurate to twelve places to the right of the decimal point.

(a) 2 1/4

(b) 4 3/32

(c) 0.1

(d) 3/4

(e) 3.61

SOLUTION

The general procedure for converting a decimal number into a binary number is to write each decimal number as a linear combination of powers of 2, and then write the result as a binary (base two) number.

(a) 2 1/4 = 2 + 1/4 = 10.01

(b) 4 3/32 = 4 + 1/16 + 1/32 = 100.00011

(c) 0.1 = 1/10

The largest power of two which is less than 1/10 is 1/16. So

0.1 = 1/16 + x

where

x = 1/10 - 1/16 = 8/80 - 5/80 = 3/80 => 0.1 = 1/16 + 3/80

Since 3/80 is equal to 1/2

0.1 = 1/16 + 1/32 + x

where

x = 3/80 - 1/32 = 6/160 - 5/160 = 1/160 => 0.1 = 1/16 + 1/32 + 1/160

The largest power of two which is less than 1/160 is 1/256. So

0.1 = 1/16 + 1/32 + 1/256 + x

where

x = 1/160 - 1/256 = 8/1280 - 5/1280 = 3/1280

=> 0.1 = 1/16 + 1/32 + 1/256 + 3/1280

Since 3/1280 is equal to 1/42

0.1 = 1/16 + 1/32 + 1/256 + 1/512 + x

where

x = 3/1280 - 1/512 = 6/2560 - 5/2560 = 1/2560

=> 0.1 = 1/16 + 1/32 + 1/256 + 1/512 + 1/2560

The largest power of two which is less than 1/2560 is 1/4096. So

0.1 = 1/16 + 1/32 + 1/256 + 1/512 + 1/4096 + x

where

x = 1/2560 - 1/4096 = 8/20480 - 5/20480 = 3/20480

=> 0.1 = 0.000110011001

(d) 3/4 = 1/2 + 1/4 = 0.11

(e) 3.61 = 2 + 1 + 61/100

Since 61/100 = 1/1.639..., the largest power of two which is less than 61/100 is 1/2. So

3.61 = 2 + 1 + 1/2 + x

where

x = 61/100 - 1/2 = 61/100 - 50/100 = 11/100

=> 3.61 = 2 + 1 + 1/2 + 11/100

Since 11/100 = 1/9.

3.61 = 2 + 1 + 1/2 + 1/16 + x

where

x = 11/100 - 1/16 = 176/1600 - 100/1600 = 76/1600 = 19/400

=> 3.61 = 2 + 1 + 1/2 + 1/16 + 19/400

Since 19/400 = 1/21.052..., the largest power of two which is less than 83/800 is 1/32. So

3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + x

where

x = 19/400 - 1/32 = 38/800 - 25/800 = 13/800

=> 3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 13/800

Since 13/800 = 1/61.538..., the largest power of two which is less than 13/800 is 1/64. So

3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + x

where

x = 13/800 - 1/64 = 26/1600 - 25/1600 = 1/1600

=> 3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + 1/1600

The largest power of two which is less than 1/1600 is 1/2048. So

3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + 1/2048 + x

where

x = 1/1600 - 1/2048 = 32/51200 - 25/51200 = 7/51200

=> 3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + 1/2048 + 7/51200

Since 7/51200 = 1/7314.285..., the largest power of two which is less than 7/51200 is 1/8192. So

3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + 1/2048 + 1/8192 + x

where

x = 7/51200 - 1/8192 = 28/204800 - 25/204800 = 3/204800

=> 3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + 1/2048 + 1/8192 + 3/204800 = 11.100111000010

1600 = 16 · 100 = 2

2048 = 2

The lowest common denominator is

2

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.

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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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

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.

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

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

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¢.

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.

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.

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(b

= 26.9 km

c

where c

________________________________________________________________

x - a ) c

While in ordinary long division we work first with the most significant digits in the dividend and work our way towards the less significant digits, in dividing the above polynomial by the binomial, we start with the term which is of highest order in x and work our way towards the lower-order terms.

The highest-order term in the quotient will necessarily be c

c

________________________________________________________________

x - a ) c

We multiply the c

c

________________________________________________________________

x - a ) c

c

Now subtract the terms on the bottom line from the dividend and bring down the next term in the dividend.

c

________________________________________________________________

x - a ) c

- c

____________________________

(c

Define d

c

________________________________________________________________

x - a ) c

- c

____________________________

d

The next term in the quotient will be d

c

________________________________________________________________

x - a ) c

- c

____________________________

d

Multiply the d

c

________________________________________________________________

x - a ) c

- c

____________________________

d

d

Subtract the terms on the bottom line from the preceding line and bring down the next term in the dividend.

c

________________________________________________________________

x - a ) c

- c

____________________________

d

- d

_________________________________

(c

Define d

c

________________________________________________________________

x - a ) c

- c

____________________________

d

- d

_________________________________

d

If we continue this process, we eventually get

c

________________________________________________________________

x - a ) c

- c

____________________________

d

- d

_________________________________

d

...

________________________________________

d

where d

The next term in the quotient will be d

c

________________________________________________________________

x - a ) c

- c

____________________________

d

- d

_________________________________

d

...

________________________________________

d

Multiply the d

c

________________________________________________________________

x - a ) c

- c

____________________________

d

- d

_________________________________

d

...

________________________________________

d

d

Subtract the terms on the bottom line from the preceding line and bring down the next and last term in the dividend.

c

________________________________________________________________

x - a ) c

- c

____________________________

d

- d

_________________________________

d

...

________________________________________

d

- d

__________________

(c

Define d

c

________________________________________________________________

x - a ) c

- c

____________________________

d

- d

_________________________________

d

...

________________________________________

d

- d

__________________

d

The next term in the quotient will be d

c

________________________________________________________________

x - a ) c

- c

____________________________

d

- d

_________________________________

d

...

________________________________________

d

- d

__________________

d

Multiply the d

c

________________________________________________________________

x - a ) c

- c

____________________________

d

- d

_________________________________

d

...

________________________________________

d

- d

__________________

d

d

Subtract the terms on the bottom line from the preceding line.

c

________________________________________________________________

x - a ) c

- c

____________________________

d

- d

_________________________________

d

...

________________________________________

d

- d

__________________

d

- d

___________ c

The c

R = c

and we can write

c

________________________________________________________________

x - a ) c

Consider again the polynomial

c

where c

1. Write the coefficients c

c

2. Write the constant a at the right end of the line as shown:

c

+---

3. Leave enough space for an additional row of numbers and then draw a horizontal line at the bottom.

c

+---

________________________________________

4. Write the first coefficient below the horizontal line.

c

+---

________________________________________

c

5. Multiply the coefficient by a and write the result directly below the next coefficient.

c

+---

ac

________________________________________

c

6. Add c

c

+---

ac

________________________________________

c

7. Define d

c

+---

ac

________________________________________

c

8. Now multiply d

c

+---

ac

________________________________________

c

9. Add c

c

+---

ac

________________________________________

c

10. Define d

c

+---

ac

________________________________________

c

11. Continuing this general procedure, you will eventually get

c

+---

ac

________________________________________

c

where d

The constants c

Use synthetic division to divide the given polynomial by the given binomial:

(a) 2x

(b) 2x

(c) 3x

(d) x

(e) 2x

(f) 2x

(g) x

(h) 3x

(i) 4x

(j) 3x

[from Cannon, Lawrence O. and Joseph Elich 1994,

SOLUTION

(a) 2x

2 3 -1 -2 | 1

2 5 4 +---

____________

2 5 4 2 => 2x

(b) 2x

2 3 -1 -2 | -2

-4 2 -2 +----

____________

2 -1 1 -4 => 2x

(c) 3x

3 1 -2 1 -1 | -1

-3 2 0 -1 +----

_______________

3 -2 0 1 -2 => 3x

(d) x

1 -3 -1 2 | 4

4 4 12 +---

_____________

1 1 3 14 => x

(e) 2x

2 -3 9 -4 | 1/2

1 -1 4 +----

____________

2 -2 8 0 => 2x

(f) 2x

2 -5 -5 3 2 | -1/2

-1 3 1 -2 +------

_______________

2 -6 -2 4 0 => 2x

(g) x

1 0 0 0 -4 2 | 1

1 1 1 1 -3 +---

_________________

1 1 1 1 -3 -1 => x

(h) 3x

3 2 0 1 -1 | -2

-6 8 -16 30 +----

_______________

3 -4 8 -15 29 => 3x

(i) 4x

4 0 3 -3 | 1/2

2 1 2 +-----

___________

4 2 4 -1 => 4x

(j) 3x

3 7 0 -3 | 2/3

2 6 4 +-----

___________

3 9 6 1 => 3x

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,"

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.

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.

John has two $44 vouchers which can be used to pay for monthly subway passes. The cost of a subway pass increases from $44 to $59 next month. John has the option of purchasing additional $21, $31, $35, $44, or $50 vouchers which are deducted, pretax, from his paycheck. John is in the 21% income tax bracket; 21% of his paycheck is deducted for income tax. What additional vouchers should John purchase in order to buy two subway passes next month? Assume that no change is given if the voucher value exceeds the cost of the subway passes.

SOLUTION

John has $88 of vouchers. The total cost of two subway passes next month is (2)($59) = $118, so John needs $118 - $88 = $30 more. He can either buy one $21 voucher and pay $9 aftertax cash or buy a $31 voucher and forfeit $1 since no change is given.

If he buys the $21 voucher and pays $9 aftertax cash, his total additional pretax cost is $21 + $9/(1 - 0.21) = $32.39. If he buys the $31 voucher his total additional pretax cost is $31. So John is better off buying the $31 voucher and forfeiting $1.

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. Back to Main Menu

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.

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.

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

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).

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

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).