Home Chapter 7 Multiple Choice

# Multiple Choice

This activity contains 44 questions.

## Adding in binary, a decimal 26 + 27 will produce a sum of:

 111010 110110 110101 101011

## How many basic binary subtraction operations are possible?

 4 3 2 1

## When multiplying 13 × 11 in binary, what is the third partial product?

 1011 00000000 100000 100001

## When 1100010 is divided by 0101, what is the decimal remainder?

 2 3 4 6

 0010 0110 0011 1011 0011 1100 +0101 0101 +0001 1110 +0001 1111

 0111 1011    0100  0001    0101  1011 0111 1011    0101  1001    0101  1011 0111 0111    0101  1001    0101  1011 0111 0111    0100  0001    0101  1011

Subtract the following binary numbers.
 0101 1000 1010 0011 1101 1110 –0010 0011 –0011 1000 –0101 0111

 0011  0100    0110  1010    1000  0110 0011  0101    0110  1011    1000  0111 0011  0101    0110  1010    1000  0111 0011  0101    0110  1010    1000  0110

Multiply the following binary numbers.
 1010 1011 1001 ×0011 ×0111 ×1010

 0001  1110    0100  1101    0101  1011 0001  1110    0100  1100    0101  1010 0001  1110    0100  1101    0101  1010 0001  1101    0100  1101    0101  1010

## Divide the following binary numbers.

 0000  0010    0000  0010    1000  1111 0000  0010    0001  0010    0000  0100 0000  0011    0000  0010    0000  0100 0000  0010    0000  0010    0000  0100

## Solving –11 + (–2) will yield which two's-complement answer?

 1110 1101 1111 1001 1111 0011 1110 1001

## The two's-complement system is to be used to add the signed numbers 11110010 and 11110011. Determine, in decimal, the sign and value of each number and their sum.

 –14 and –13, –27 –113 and –114, 227 –27 and –13, 40 –11 and –16, –27

## Determine the two's-complement of each binary number.00110        00011        11101

 11001    11100    00010 00111    00010    00010 00110    00011    11101 11010    11101    00011

## Convert each of the signed decimal numbers to an 8-bit signed binary number (two's-complement).+7        –3        –12

 0000  0111    1111  1101    1111  0100 1000  0111    0111  1101    0111  0100 0000  0111    0000  0011    0000  1100 0000  0111    1000  0011    1000  1100

## Convert each of the following signed binary numbers (two's-complement) to a signed decimal number.00000101        11111100        11111000

 –5    +4    +8 +5    –4    –8 –5    +252    +248 +5    –252    –248

Convert each of the decimal numbers to two's-complement form and perform the addition in binary.

 0001  0100    0000  0101 0000  0110    0001  1001 0000  0110    0000  0101 1111  0110    1111  0101

## Perform subtraction on each of the following binary numbers by taking the two's-complement of the number being subtracted and then adding it to the first number.01001        0110000011        00111

 01100    10011 00110    00101 10110    10101 00111    00100

## Binary subtraction of a decimal 15 from 43 will utilize which two's complement?

 101011 110000 011100 110001

## An 8-bit register may provide storage for two's-complement codes within which decimal range?

 +128 to –128 –128 to +127 +128 to –127 +127 to –127

## Convert each of the decimal numbers to 8-bit two's-complement form and then perform subtraction by taking the two's-complement and adding.

 0001  0011 0000  1110 0010  1110 1110  0000

## The decimal value for E16 is:

 1210 1310 1410 1510

## Add the following hex numbers: 011016 + 1001016

 1012016 1002016 1112016 0012016

## Perform the following hex subtraction: ACE16 – 99916 =

 23516 13516 03516 33516

 3C 14 3B +25 +28 +DC

 60    3C    116 62    3C    118 61    3C    117 61    3D    117

 47 34 FA –25 –1C –2F

 22    18    CB 22    17    CB 22    19    CB 22    18    CC

## Which of the following is the primary advantage of using binary-coded decimal (BCD) instead of straight binary coding?

 Fewer bits are required to represent a decimal number with the BCD code. BCD codes are easily converted from decimal. the relative ease of converting to and from decimal BCD codes are easily converted to straight binary codes.

## Convert the decimal numbers 275 and 965 to binary-coded decimal (BCD) and add. Select the BCD code groups that reflect the final answer.

 1101 1110 1010 1110 1010 1110 0001 0010 0100 0000 0010 0011 0100 0000

 0110 0111 1001 0101 1000 1000

 0000  1011    0000  1111    0001  0001 0001  0001    0001  0101    0001  0001 0000  1011    0000  1111    0001  0111 0001  0001    0001  0101    0001  0111

## What are the two types of basic adder circuits?

 sum and carry half-adder and full-adder asynchronous and synchronous one- and two's-complement

## The summing outputs of a half- or full-adder are designated by which Greek symbol?

 omega theta lambda sigma

## How many inputs must a full-adder have?

 2 3 4 5

## A half-adder circuit would normally be used each time a carry input is required in an added circuit.

 True False

## The binary adder circuit is designed to add _____ binary numbers at the same time.

 2 4 6 8

 True False

 two single bits and one carry bit two 2-bit binary numbers two 4-bit binary numbers two 2-bit numbers and one carry bit

## What logic function is the sum output of a half-adder?

 AND exclusive-OR exclusive-NOR NAND

 Yes No

## The carry propagation delay in 4-bit full-adder circuits:

 is cumulative for each stage and limits the speed at which arithmetic operations are performed is normally not a consideration because the delays are usually in the nanosecond range decreases in direct ratio to the total number of full-adder stages increases in direct ratio to the total number of full-adder stages, but is not a factor in limiting the speed of arithmetic operations

 determine sign and magnitude reduce propagation delay add a 1 to complemented inputs increase ripple delay

## Why is a fast-look-ahead carry circuit used in the 7483 4-bit full-adder?

 to decrease the cost to make it smaller to slow down the circuit to speed up the circuit

## The BCD addition of 910 and 710 will give initial code groups of 1001 + 0111. Addition of these groups generates a carry to the next higher position. The correct solution to this problem would be to:

 ignore the lowest order code group because 0000 is a valid code group and prefix the carry with three zeros add 0110 to both code groups to validate the carry from the lowest order code group disregard the carry and add 0110 to the lowest order code group add 0110 to the lowest order code group because a carry was generated and then prefix the carry with three zeros

## How many BCD adders would be required to add the numbers 97310 + 3910?

 3 4 5 6

## An input to the mode pin of an arithmetic/logic unit (ALU) determines if the function will be:

 one's-complemented arithmetic or logic positive or negative with or without carry

## The selector inputs to an arithmetic/logic unit (ALU) determine the:

 selection of the IC arithmetic or logic function data word selection clock frequency to be used