// # pragma GCC target("avx2")
// # pragma GCC optimize("O3")
// # pragma GCC optimize("unroll-loops")
#ifdef harurun
#define debug(x) cerr<<#x<<": "<<x<<endl
#else
#define debug(x)
#endif
#include <bits/stdc++.h>
using namespace std;
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
using mint = modint998244353;
// using mint = modint1000000007;
// using mint = double;
#endif

//define
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<long long , long long>;
using vi = vector<int>;
using vll = vector<long long>;

#define rep(i,l,r) for (int i = (int)(l); i < (int)(r); i++)
#define rrep(i,l,r) for (int i = (int)(r-1); i >= (int)(l); i--)
#define len(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define elif else if
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
const int inf = 1e9;
const long long infl = 1LL<<60;
const int mod = 998244353;
ll pow(ll a, ll b, ll p){ ll ans = 1; while(b){ if(b & 1) (ans *= a) %= p; (a *= a) %= p; b /= 2; } return ans; }
template<class T, class U> bool chmin(T& a, const U& b){ if(a > T(b)){ a = b; return 1; } return 0; }
template<class T, class U> bool chmax(T& a, const U& b){ if(a < T(b)){ a = b; return 1; } return 0; }
template <class T>
using spq = priority_queue<T, vector<T>, greater<T>>;

template<class T>istream& operator>>(istream& i, vector<T>& v) {for(int j = 0; j < (int)(v).size(); j++) i >> v[j]; return i;}
struct IoSetup {
    IoSetup() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(15);
        cerr << fixed << setprecision(15);
    }
} iosetup;

using ld = long double;
void solve(){
    int n, x, y; cin >> n >> x >> y;
    int lim = 500 * 500 + 100;
    vector<int> dp(lim, -inf); dp[0] = 0;
    rep(i, 0, n){
        int w, v, c; cin >> w >> v >> c;
        chmin(c, x);
        int k = 1;
        while(c){
            int nk = min(k, c);
            int nw = w * nk;
            int nv = v * nk;
            rrep(s, 0, lim-nw-1){
                chmax(dp[s+nw], dp[s] + nv);
            }
            c -= nk;
            k *= 2;
        }
    } 
    ld ans = 0;
    rep(s, 1, lim){
        chmax(ans, (ld)(dp[s] + s / x * y) / (ld)s);
    }
    cout << ans << endl;
    return ;
}


int main(){
    int t;
    // cin >> t;
    t = 1;
    while(t--) solve();
    return 0;
}