Everyday Math Demystified, 2nd edition |
Stan Gibilisco |
Explanations for Quiz Answers in Chapter 1 |
1. If you arrange a lot of objects into a square array measuring a hundred wide by a hundred deep by one tall, then you'll have 100 x 100, or 10,000 objects. The correct choice is D. This fact holds true in the decimal numeration system (base ten). Here's a bit of "food for thought": Does this same notion work in other bases? For example, is it true that 100 x 100 = 10,000 in the octal system, or in the hexadecimal system, or in the binary system? Try these "extra credit" exercises and see for yourself! (Here's a hint: Convert all values to base-ten numerals, then multiply those, and then convert them back to expressions in the original base.) |
2. If you arrange a lot of objects into a cubical array measuring a hundred wide by a hundred deep by a hundred tall, then you'll end up with a total of 100 x 100 x 100, or 1,000,000 objects. The correct choice is B. For "extra credit," you can figure out whether or not this same principle works in the binary, octal, or hexadecimal systems. |
3. The Roman numeral M represents the decimal 1000, the Roman C represents the decimal 100, the Roman X represents the decimal 10, and the Roman I represents the decimal 1. When we write these four Roman symbols one after the other as MCXI, we get the decimal quantity 1000 + 100 + 10 + 1, or 1111. The correct choice is A. |
4. In an octal numeral, the digit farthest to the right represents a multiple of the decimal quantity 1. The next digit to the left represents a multiple of the decimal quantity 8. To the left of that, we have the multiple of the decimal quantity 8 x 8, or 64. To the left of that, we have the multiple of the decimal quantity 8 x 8 x 8, or 512. In the octal numeral 5103, the digit 5 occupies the place for the multiple of the decimal quantity 512, so it represents 5 x 512, or 2560, in decimal terms. The correct choice is A. |
5. We represent the decimal quantity 50 as the Roman numeral L, and we represent the decimal quantity 1 as the Roman numeral I. In order to denote the decimal numeral 52 in Roman form, we write LII, which stands for 50 + 1 + 1. The correct choice is C. |
6. In a decimal numeral, the digit farthest to the right indicates a multiple of 1, and the next digit to the left tells us a multiple of 10. To the left of that, we find a digit that gives us a multiple of 100. In the decimal numeral 335,427, the digit 4 lies in the spot that indicates a multiple of 100, so it represents 4 x 100, or 400. The answer is B. |
7. In the octal system, we have the digits 0, 1, 2, 3, 4, 5, 6, and 7 only. After 7, we encounter the octal quantity 10. When we see the digit 7 at the extreme right-hand end of an octal numeral and we want to increase the value by 1, we must replace the 7 with a digit 0, and then increase the digit immediately to its left by 1. According to this rule, the octal 70 follows the octal 67. Arithmetically, 67 + 1 = 70 in octal terms. The correct choice is A. |
8. Let's convert the Roman numeral to a decimal numeral, and then convert the decimal numeral to a binary numeral. The Roman symbol V represents the decimal quantity 5, and the Roman symbol I represents the decimal quantity 1. When we see the Roman numeral VIII, we can add the decimal values to get 5 + 1 + 1 + 1 = 8. In the binary system, the rightmost digit tells us a multiple of the decimal quantity 1; the next digit to the left gives us a multiple of the decimal quantity 2; going one more place to the left, we encounter a multiple of the decimal quantity 4; after that we get a multiple of the decimal quantity 8. When we write the binary numeral 1000, we get a decimal value of 8 + 0 + 0 + 0, which equals 8. The answer is B. |
9. The binary numeral 101010101 has nine digits. Let's track the decimal-equivalent place values from right to left:
We can see that the leftmost digit is 1, so we know that it represents the decimal quantity 1 x 256, or 256. The answer is C. |
10. The Roman numeration system has no symbol for zero. The correct choice is D. |