CGPET 2013 Solved Paper

Let the frequency of first fork =n
then the frequency of last fork =3n
Since two successive forks given four beats.
Hence, we have
Frequency of first fork =n
Frequency of 2 nd fork =n+4
Frequency of 3rd fork =n+2×4
Frequency of 4 th fork =n+3×4
So, the frequency of N th fork =n+(N−1)4
But N th fork is the last fork
∴‌3n=n+(N−1)4
⇒2n=(N−1)4
‌ Here, ‌‌N=56
⇒2n=(56−1)×4=55×4=220
⇒n=‌

220

2

=110‌Hz