Secondary 5 Music Practice Paper: Intervals And Scales, Chords And Harmony, Rhythm And Tempo, Acoustics And Frequency, Score Analysis And Instrumentation, Aural Training And Examination Strategy (100 Marks)

Secondary 5 Music Practice Paper: Intervals and Scales, Chords and Harmony, Rhythm and Tempo, Acoustics and Frequency, Score Analysis and Instrumentation, Aural Training and Examination Strategy (100 marks)

2026-11-14

Music – Secondary 5 – Practice Paper

Secondary 5 Music practice paper with 24 parts and worked answers on equal-temperament frequencies, cents, scale degrees, transposing instruments, major, minor and

Paper details　Secondary 5 Music practice paper: 24 parts with worked answers on intervals, chords, tempo, acoustics, score analysis and aural skills.

Section A: Intervals and Scales

Equal temperament divides the octave into twelve equal steps with a frequency ratio of 2 to the power of one twelfth. Show the working.

#	Question	Answer and marking notes
1	Taking A4 = 440 Hz as the reference, find the frequencies of A5 and C5, correct to one decimal place.	Answer: A5 is 12 semitones above A4, so its frequency is 440 x 2^(12/12) = 880 Hz; C5 is 3 semitones above A4, so its frequency is 440 x 2^(3/12) = 523.25 Hz.
2	Find the number of semitones, the size in cents, and the frequency ratio of the interval from A4 to E5, correct to two decimal places.	Answer: A4 to E5 spans 7 semitones, a perfect fifth; cents = 7 x 100 = 700 cents; ratio = 2^(7/12) = 1.5, so E5 = 659.26 Hz.
3	Starting on C4 = 261.63 Hz, find the note name and frequency of the sixth degree of the major scale, correct to two decimal places, and state how many semitones it lies above C4.	Answer: the major scale steps are 2, 2, 1, 2, 2, so the sixth degree is A; it lies 9 semitones above C4, giving 261.63 x 2^(9/12) = 440 Hz, which matches the standard A4.
4	A B flat clarinet is a transposing instrument whose written pitch is a major second above its sounding pitch. If the score shows C5, find the sounding note name and frequency, correct to two decimal places, and the difference in cents.	Answer: the sounding note is B flat 4, frequency = 523.25 x 2^(-2/12) = 466.16 Hz; the difference is 1200 x log2(523.25 / 466.16) = 200 cents, one major second.

Section B: Chords and Harmony

Chords are built from a root with a third and a fifth; show the working and give frequencies in Hz.

#	Question	Answer and marking notes
5	Taking C4 = 261.63 Hz as the root, find the frequencies of E4 and G4 in the C major triad, correct to two decimal places.	Answer: a major triad adds 4 and 7 semitones to the root; E4 = 261.63 x 2^(4/12) = 329.63 Hz and G4 = 261.63 x 2^(7/12) = 392.0 Hz.
6	With A4 = 440 Hz, the third of the A minor triad is C5 = 523.25 Hz. Find how many semitones the third and the fifth lie above the root, and give the frequency of the fifth, correct to two decimal places.	Answer: a minor triad uses 3 semitones then 4 semitones; the third is 3 semitones above the root and the fifth is 7 semitones above it, so E5 = 440 x 2^(7/12) = 659.26 Hz.
7	With G3 = 196.00 Hz as the root, the G dominant seventh chord is G, B, D, F. Find the frequencies of B3, D4 and F4, correct to two decimal places, and state the semitone structure of a dominant seventh chord.	Answer: a dominant seventh chord uses 4, 3 and 3 semitones; B3 = 196.00 x 2^(4/12) = 246.94 Hz, D4 = 196.00 x 2^(7/12) = 293.66 Hz and F4 = 196.00 x 2^(10/12) = 349.23 Hz.
8	A perfect cadence in C major moves from the dominant G major chord to the tonic C major chord, with roots G3 = 196.00 Hz to C4 = 261.63 Hz. Find the frequency ratio and the difference in cents, and explain why the progression sounds stable.	Answer: the ratio is 261.63 / 196.00 = 1.33 and the difference is 1200 x log2(261.63 / 196.00) = 500 cents; the roots stand a perfect fifth apart, so their harmonics overlap strongly and the progression sounds stable.

Section C: Rhythm and Tempo

Tempo is given in beats per minute; give all durations in seconds and show the working.

#	Question	Answer and marking notes
9	A piece is played at 80 BPM in 4/4 time. Find the duration of one beat and of one bar in seconds, correct to two decimal places.	Answer: one beat = 60 / 80 = 0.75 s; a 4/4 bar has 4 beats, so a bar lasts 4 x 0.75 = 3.0 s.
10	A piece has 48 bars in 3/4 time at 120 BPM. Find the total playing time in minutes and seconds.	Answer: one beat = 60 / 120 = 0.5 s; a bar of 3 beats lasts 1.5 s; total = 48 x 1.5 = 72.0 s, that is 1 minutes 12 seconds.
11	A 120-bar piece in 4/4 is speeded up from 120 BPM to 144 BPM. Find the percentage increase in tempo, the new bar length and the new total playing time.	Answer: the increase is (144 – 120) / 120 x 100% = 20.0%; a bar lasts 4 x 60 / 144 = 1.67 s; total = 120 x 1.67 = 200.0 s, that is 3 minutes 20 seconds.
12	A 32-bar piece in 6/8 is counted in dotted crotchet beats at 132 BPM. Find the beat value, the bar length and the total time in seconds to one decimal place, and explain how 6/8 differs from 3/4.	Answer: one beat = 60 / 132 = 0.45 s; a 6/8 bar has two dotted crotchet beats = 0.91 s; total = 32 x 0.91 = 29.1 s. 6/8 is compound time with two beats each split into three quavers, while 3/4 is simple time with three separate crotchet beats.

Section D: Acoustics and Frequency

Use the cent formula 1200 x log2(frequency ratio) and the sound intensity level 10 x log10(I / I0) with I0 = 1e-12 W per square metre.

#	Question	Answer and marking notes
13	Standard A4 is 440 Hz and a tuning fork gives 442 Hz. Find the difference in cents to one decimal place and state whether the fork is sharp or flat.	Answer: the difference is 1200 x log2(442 / 440) = 7.9 cents; the fork is about 7.9 cents sharp because 442 Hz is above 440 Hz, close enough by ear but producing slow beats in ensemble.
14	A string has a fundamental frequency of 220 Hz. Find the frequency of the third harmonic and the ratio of its wavelength to the fundamental wavelength, correct to two decimal places.	Answer: the third harmonic is 220 x 3 = 660 Hz; wavelength is inversely proportional to frequency, so the ratio is 1 / 3 = 0.33, one third of the fundamental wavelength.
15	A sound has an intensity of 1e-4 W per square metre. Find its sound intensity level in decibels, and the increase in decibels when the intensity doubles.	Answer: the level is 10 x log10(1e-4 / 1e-12) = 10 x log10(1e8) = 80 dB; when the intensity doubles it becomes 10 x log10(2e8) = 83.0 dB, an increase of about 3 dB for every doubling.
16	A hall has a reverberation time of 1.8 seconds and the music is played at 100 BPM in 4/4. Find the reverb length in beats to one decimal place and explain how excessive reverb affects fast passages.	Answer: one beat lasts 60 / 100 = 0.6 s, so the reverb spans 1.8 / 0.6 = 3.0 beats. A reverb of about 3 beats smears the notes of fast passages together and reduces clarity.

Section E: Score Analysis and Instrumentation

Support the analysis with bar counts and percentages, giving percentages to one decimal place.

#	Question	Answer and marking notes
17	A 96-bar piece falls into A (32 bars), B (40 bars) and A prime (24 bars). Find the percentage of each section and name the form.	Answer: A = 32 / 96 = 33.3 percent, B = 40 / 96 = 41.7 percent and A prime = 24 / 96 = 25.0 percent, totalling 100 percent. Section A returns after B, so the form is ternary, ABA.
18	A full score has 24 staves: 8 woodwind, 6 brass and 10 strings. Find the percentage of string staves and the number of non-string staves.	Answer: checking 8 + 6 + 10 = 24 staves; strings are 10 / 24 x 100 percent = 41.7 percent; non-string staves = 24 – 10 = 14.
19	Each system holds 4 bars and each page holds 2 systems. How many systems and pages are needed for a 96-bar piece?	Answer: systems = 96 / 4 = 24; pages = 24 / 2 = 12, an exact division needing no filler page.
20	Dynamic marks run from ppp, pp, p, mp, mf, f, ff to fff, eight levels in all. Count the levels from pp to ff inclusive and state the median level of that span.	Answer: pp, p, mp, mf, f and ff give 6 levels; between pp and ff there are 4 interior levels, so the midpoint falls between mf and f, making mf to f the central dynamic for planning the whole piece.

Section F: Aural Training and Examination Strategy

Answer aural questions with cents, semitone counts and time-signature grouping, and show the marks calculation.

#	Question	Answer and marking notes
21	Three intervals are heard with widths of 200, 700 and 1200 cents. Name each interval and give its semitone count.	Answer: 200 cents is 2 semitones, a major second; 700 cents is 7 semitones, a perfect fifth; 1200 cents is 12 semitones, an octave; dividing cents by 100 gives the semitone count.
22	A melody has 6 quavers per bar, clearly grouped as two groups of 3. Give the most likely time signature and state the number of main beats per bar and quavers per beat.	Answer: 6/8 is most likely; the 6 quavers split as 3 + 3, giving 2 main beats per bar with 3 quavers each, whereas 3/4 would group them as 2 + 2 + 2.
23	Give the semitone counts of a major sixth and a perfect fourth, and compute their sum and difference.	Answer: a major sixth spans 9 semitones and a perfect fourth 5; the sum is 9 + 5 = 14 semitones and the difference is 9 – 5 = 4.
24	Internal assessment weights listening at 40 percent and performing at 60 percent. A candidate scores 70 percent in listening and 80 percent in performing; find the overall percentage.	Answer: the total is 40 x 70 percent plus 60 x 80 percent = 28.0 + 48.0 = 76.0 percent.

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