# Test Of Bs V Evening Course: Math305, Numerical Analysis 1, Chapter 1

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Fiza Zafar
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Quizzes Created: 1 | Total Attempts: 469
Questions: 15 | Attempts: 469  Settings  The contents of the test are from Chapter 1: Computer Arithmetic. There are 15 MCQs in the test. All MCQs are compulsory. The duration is 25 minutes.

• 1.

### Truncation error is caused by approximating

• A.

Irrational numbers

• B.

Fractions

• C.

Rational Numbers

• D.

Exact mathematical procedures or infinite process in finite.

D. Exact mathematical procedures or infinite process in finite.
Explanation
Truncation error occurs when an exact mathematical procedure or an infinite process is approximated or truncated to a finite number of steps. This can lead to inaccuracies in the final result, especially when dealing with irrational numbers or fractions. The error arises from the fact that the approximation cannot fully capture the precise value of the number or the exact steps of the process.

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• 2.

### True error is defined as

• A.

Present Approximation – Previous Approximation

• B.

True Value – Approximate Value

• C.

Abs (True Value – Approximate Value)

• D.

Abs (Present Approximation – Previous Approximation)

B. True Value – Approximate Value
Explanation
The true error is defined as the absolute difference between the true value and the approximate value. This means that it measures how far off the approximate value is from the true value. It is important to calculate and understand the true error in order to assess the accuracy of an approximation or measurement.

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• 3.

### The number 0.01850*10^3 has ________ significant digits

• A.

3

• B.

5

• C.

4

• D.

6

C. 4
Explanation
The number 0.01850*10^3 has 4 significant digits because all non-zero digits are considered significant, and the zeros between the non-zero digits are also significant. In this case, the digits 1, 8, 5, and 0 are all significant.

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• 4.

### A computer that represents only 4 significant digits with chopping would calculate 66.666*33.333 as

• A.

2220

• B.

2221

• C.

2221.17778

• D.

2222

B. 2221
Explanation
A computer that represents only 4 significant digits with chopping would calculate 66.666*33.333 as 2221. Chopping means that any digits beyond the desired number of significant digits are simply discarded, rather than rounded. In this case, the multiplication of 66.666 and 33.333 results in 2221. Therefore, the correct answer is 2221.

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• 5.

### In single precision, the number of bits available to store an integer are

• A.

31

• B.

23

• C.

24

• D.

63

A. 31
Explanation
In single precision, the number of bits available to store an integer is 31. Single precision is a data type used in computer programming to represent floating-point numbers with reduced precision compared to double precision. It consists of 32 bits, with 1 bit used for the sign, 8 bits for the exponent, and 23 bits for the significand (or mantissa). The remaining 31 bits are used to store the integer part of the number, allowing for a range of values to be represented.

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• 6.

### The result of rounding the number 1.5856 to two significant digits is:

• A.

1.59

• B.

1.5

• C.

1.6

• D.

1.58

C. 1.6
Explanation
When rounding a number to two significant digits, we look at the digit at the third significant digit. If it is 5 or greater, we round up the second significant digit. In this case, the third significant digit is 8, which is greater than 5. Therefore, we round up the second significant digit, which is 5, to 6. Hence, the result of rounding the number 1.5856 to two significant digits is 1.6.

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• 7.

### In single precision, the smallest positive number that can be stored is

• A.

2^(-127)

• B.

1.2*10^(-37)

• C.

0

• D.

2^(-126)

D. 2^(-126)
Explanation
In single precision, the smallest positive number that can be stored is 2^(-126). This is because single precision uses 32 bits, with 1 bit for the sign, 8 bits for the exponent, and 23 bits for the mantissa. The exponent is biased by 127, so the actual exponent value for the smallest positive number is -126. The mantissa is normalized, meaning the leading 1 is not stored, so the smallest positive number is 1.0 times 2^(-126), which is equal to 2^(-126).

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• 8.

### In double precision, the largest exponent corresponds to

• A.

255

• B.

2046

• C.

C-127

• D.

C-1023

B. 2046
Explanation
In double precision, the largest exponent corresponds to 2046. This is because the exponent in double precision is represented using bias notation, where the bias value is 1023. The largest exponent value that can be represented is obtained by subtracting the bias from the maximum possible exponent value, which is 2047. Therefore, 2047 - 1023 = 1024, which is the largest exponent value that can be represented. However, since the exponent also includes negative values, the largest positive exponent value is obtained by subtracting the bias from 2046, resulting in 1023, which corresponds to 2046.

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• 9.

### In single precision, the largest number that can be stored is

• A.

2^127

• B.

(2-2^(-24))*2^127

• C.

(2-2^(-52))*2^126

• D.

3.4*10^(38)

D. 3.4*10^(38)
Explanation
The given answer, 3.4*10^(38), is the largest number that can be stored in single precision. Single precision uses 32 bits to represent a number, with 1 bit for the sign, 8 bits for the exponent, and 23 bits for the significand. This allows for a range of values from approximately 1.18*10^(-38) to 3.4*10^(38), with a precision of about 7 decimal digits. Therefore, 3.4*10^(38) is the correct answer.

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• 10.

### In double precision, what is the smallest number that can be stored?

• A.

(2-2^(-52))*2^127

• B.

2^(-1022)

• C.

-1.8*10^(308)

• D.

(2-2^(-52))*2^1023

C. -1.8*10^(308)
Explanation
The given correct answer, -1.8*10^(308), represents the smallest number that can be stored in double precision. Double precision is a floating-point format that uses 64 bits to represent numbers, allowing for a greater range and precision compared to single precision. The smallest number that can be stored in double precision is determined by the exponent and the significand. In this case, the exponent is -308 and the significand is -1.8, resulting in the smallest possible value.

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• 11.

### Which of these numbers will overflow to infinity in 32 bit word length computer?

• A.

10^(40)

• B.

1.2*10^(-38)

• C.

3.4*10^(38)

• D.

2^(120)

A. 10^(40)
Explanation
The number 10^40 will overflow to infinity in a 32-bit word length computer because it is a very large number that exceeds the maximum value that can be represented in a 32-bit system. In a 32-bit system, the maximum value that can be represented is approximately 2^31 - 1, which is much smaller than 10^40. Therefore, when trying to store or calculate with 10^40 in a 32-bit system, it will result in an overflow and the value will be represented as infinity.

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• 12.

### Which of these numbers will underflow to zero in 64 bit word length computer?

• A.

2.2*10^(-308)

• B.

10^(-333)

• C.

1.8*10^(308)

• D.

2^(-1022)

B. 10^(-333)
Explanation
The number 10^(-333) will underflow to zero in a 64-bit word length computer because it is a very small number. In a computer, numbers are represented using a finite number of bits, and there is a limit to how small a number can be represented accurately. When a number is too small to be represented within the available range, it underflows and becomes zero. In this case, 10^(-333) is smaller than the smallest number that can be accurately represented in a 64-bit computer, so it will underflow to zero.

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• 13.

### Which of these numbers will overflow to negative infinity in a 32 bit word length computer?

• A.

7*10^(41)

• B.

-4*10^(39)

• C.

10^(-40)

• D.

-3.4*10^(38)

B. -4*10^(39)
Explanation
The given numbers are all written in scientific notation. In a 32-bit word length computer, the largest positive number that can be represented is approximately 3.4*10^(38). Any number larger than this will overflow, meaning it exceeds the maximum value that can be stored and will wrap around to the negative side of the number line. Among the given options, -4*10^(39) is the only number that is larger than the maximum positive value and will therefore overflow to negative infinity.

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• 14.

### What is the upper bound of relative round off error for a 64 bit word length computer?

• A.

2^(-23)

• B.

2^(-24)

• C.

2^(-52)

• D.

2^(-51)

C. 2^(-52)
Explanation
The upper bound of relative round off error for a 64 bit word length computer is 2^(-52). This is because a 64 bit computer can represent numbers with a precision of 53 bits, and the relative round off error is half of the least significant bit, which is 2^(-53). However, since the relative round off error is always positive, the upper bound is 2^(-52).

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• 15.

### How many significant digits will be lost in calculating y=x-sin(x) for x=1/15 with a calculator having 9 decimal digits of memory?

• A.

9

• B.

0

• C.

4

• D.

3

D. 3
Explanation
When calculating y=x-sin(x) for x=1/15, the calculator will only be able to store 9 decimal digits of memory. Since x=1/15 has an infinite number of decimal places, the calculator will round the number to fit within its memory capacity. This rounding will result in the loss of 3 significant digits. Therefore, 3 significant digits will be lost in the calculation.

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