Lesson 4.5.4.7

4.5.4.7 Range, precision and normalisation of floating point Quiz: AQA Computer Science, Unit 5

20 questions

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Lesson 4.5.4.7, Range, precision and normalisation of floating point: 20 multiple choice questions for the AQA Computer Science (7517), Unit 5: Fundamentals of data representation, written with Revision Ninja.

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The 20 questions

  1. Which form generally gives a larger range for the same number of bits?

    • Fixed point
    • Floating point
    • Neither, since range depends only on the sign bit
    • Both give exactly the same range
  2. What does precision mean for a binary number representation?

    • The largest value that can be stored in the format, which is set by the number of exponent bits
    • How many significant bits of the value are stored accurately
    • The number of bits used for the sign, which is always one bit in every number format used
    • The speed at which numbers can be added together by the processor in a single clock cycle
  3. What is the purpose of normalising a floating point mantissa?

    • To remove the sign bit from the mantissa, so that every stored value is treated as positive
    • To make every exponent equal to zero, so that all values share a single common scale
    • To keep as many significant bits as possible by removing leading zeros
    • To convert the value into a fixed point number so that it can be compared with integers directly
  4. What form does a normalised positive binary mantissa take?

    • 0.0xxxx, with a 0 immediately after the binary point
    • 1.xxxx, with no binary point
    • 0.1xxxx, with a 1 immediately after the binary point
    • All zeros, to show a zero value
  5. What is an advantage of fixed point over floating point?

    • It normalises each value automatically
    • It gives greater precision for very large numbers
    • It gives a larger range for the same number of bits
    • Calculations are faster, since no exponent handling is needed
  6. What is the main disadvantage of fixed point form?

    • It requires an exponent for every value stored, which makes every calculation slower to complete
    • A limited range, since the position of the binary point cannot vary
    • It cannot represent any fractional values at all, so every stored number must be a whole number
    • It always needs more bits than floating point, because it must store a separate exponent for every value
  7. What mainly determines the range of a floating point number?

    • The number of sign bits only
    • The number of exponent bits
    • The number of mantissa bits only
    • The position of the binary point inside the mantissa
  8. Normalise the mantissa 0.0110 with exponent 3 in base 2. What is the normalised form?

    • 0.0011 with exponent 4
    • 0.0110 with exponent 3
    • 0.1100 with exponent 4
    • 0.1100 with exponent 2
  9. Normalise the mantissa 0.0001 with exponent 5 in base 2. What is the normalised form?

    • 0.1000 with exponent 2
    • 0.0001 with exponent 5
    • 0.0010 with exponent 3
    • 0.1000 with exponent 8
  10. Which statement about precision and range in floating point is correct?

    • Precision is set by the number of mantissa bits, and range by the number of exponent bits
    • Precision is set by the exponent bits, and range by the mantissa bits
    • Precision and range depend only on the sign bit
    • Fixed point always has greater precision than floating point with the same bits
  11. Using 8 bits in fixed point with 6 integer bits and 2 fractional bits, what is the largest unsigned value?

    • 255
    • 63.75
    • 16
    • 63
  12. A normalised mantissa is 0.1011 with exponent 4 in base 2. What is its decimal value?

    • 11
    • 2.75
    • 0.6875
    • 5.5
  13. Why is the leading bit of a normalised positive mantissa always 1?

    • The mantissa is shifted until its first significant bit sits just after the point, so that bit must be 1
    • The exponent is zero for every normalised number, so the leading bit is always the first bit of the exponent
    • The value zero is never stored in a mantissa, so the leading bit is always set to one for every value
    • The sign bit is always stored in that position, so the leading bit records whether the value is negative
  14. What is the decimal value of the fixed point number 0010.1100 with 4 integer bits and 4 fractional bits?

    • 2.75
    • 3.25
    • 2.5
    • 10.75
  15. A fixed point format has 4 integer bits and 2 fractional bits. What is the smallest positive step between values?

    • 0.5
    • 1
    • 0.25
    • 0.125
  16. Explain why normalisation improves precision in floating point.

    • Normalising makes a value exact, so rounding is never needed for any value stored in the mantissa
    • Normalisation sets every exponent to zero, so no bits are wasted in the exponent field of the format
    • Leading zeros carry no information, so shifting the mantissa keeps every stored bit significant
    • Normalisation adds extra bits to the mantissa for free, so the stored value is always more accurate
  17. A format with a fixed total number of bits gains more exponent bits by losing mantissa bits. What is the effect?

    • Both range and precision increase, since the total number of bits in the format has grown
    • Precision increases and range is unchanged, because the exponent field does not affect the scale
    • The range increases, but precision falls because fewer mantissa bits remain
    • Range falls and precision increases, because each mantissa bit covers a smaller span of values
  18. A normalised mantissa is 0.1101 with exponent -2 in base 2. What is its decimal value?

    • 0.8125
    • 0.40625
    • 3.25
    • 0.203125
  19. Why might a fixed point format be preferred in a speed-critical embedded task?

    • Calculations are simple integer-style operations with no exponent handling, so they run faster
    • Fixed point normalises every result automatically, which keeps the mantissa full of significant bits
    • Fixed point needs no bits for the fractional part, so every stored value is a whole number
    • Fixed point gives a larger range than floating point for the same bits, so it suits huge values
  20. Can the value 10^6 be stored in a 16-bit unsigned integer with no fractional bits?

    • No, since 10^6 exceeds the largest 16-bit unsigned value of 65535
    • Yes, since 16 bits can store up to 10^6
    • Yes, because fixed point ignores the number of bits
    • Yes, if the number is stored in two's complement with 16 bits

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