This question is probing your understanding of “tuplets” - irregular divisions
of the beat in which a group of notes has to be played faster or slower than their
notated value. The most common tuplet is the triplet (3 ) in which the three notes
need to be played in the time normally taken to play two of the notes - e.g. a triplet
of quavers must be played in the time taken to play a pair of quavers. But any number
of notes can be grouped into a “tuplet” and follow the table of rules
Tuplets played to the time of
3 2 of smallest note duration
in tuplet
5,6,7 4 of smallest note duration in
tuplet
9,10,11 8 of smallest note duration in
tuplet
2 3 of smallest note duration
in tuplet (compound time)
4 3 of smallest note duration
in tuplet (compound time)
In the first bar we have two semi-quaver triplet groups, each played to the time
taken by 2 semi-quavers
( i.e a quaver each). The first bar therefore has a total of 3 quavers. The second
bar also has three quaver beats - the first quintuplet (5) of demi-semi-quavers
is played in the time of 4 demi-semi-quavers - a quaver.
The first two bars therefore have a time signature of 3/8
In bar 3, a triplet is indicated over a crotchet quaver pair - tuplets can be
written over notes of different duration ! But from the rules table above, working
to the smallest note duration of the quaver in triplet, the triplet has to be played
in the time normally taken up by two quavers - i.e. a crotchet. So the first two
triplets take up two crotchets. The sextuplet (6) of semi-quavers has to be played
in the time normally taken up by 4 semi-quavers - i.e. another crotchet. So the
third bar is made up of note groups and rests totalling 5 crotchets and has a time
signature of 5/4
