Suppose G is a graph (of order n) obtained by adding k chords to Cn:x1x2…xnx1(x1x2…xn are all the venices of G). We assume that all the chords are drawn in side the bounded region of Cn. Let f(k) be the number of distinct cycles of G

let m(k) =min {f(k) : for all possible such G }and let M(k)=max{f(k):for all possible such G}. Yap H. P. and Teo S. K. raised the following problem: Is it true that m(k)=(k+1)(k+2)/2 and M(k)=2k+k? For each m satisfying m (k)≤m≤M(k)

can we find a graph G by adding k chords to Cn so that G has f(k)=m cycles?We prove that if F is a path of Cn

then G-E(F) has at most one cycle which contains all chords ;G has at most two cycles which contains all chords. We apply it to obtain that(1) M(k)<2k+1 -2 for k≥4. (2)m (k) = (k+ 1 ) (k + 2)/2 for k≤n - 3. (3)m (k)a (k+ 1 ) (k + 2)/2 +k - 1 > (k+ 1) (k+2)/2 for k>n - 3≥1. (4) M(k)≥2k+K(k+ 1)/2>2K+k for 2≤k≤[n/2] (5). For 4≤k≤n-3

f(k)=(k+1) (k+ 2)/2 or f(k)≥(k+ 1 ) (k+2)/2+ (k-2). and the answer to the problem of Yap H. P. and Teo S. K. is negative.