We study the disproportionate version of the classical cake‐cutting problem: how efficiently can we divide a cake, here 0,1, among n⩾2 agents with different demands α1,α2,⋯,αn summing to 1? When all the agents have equal demands of α1=α2=⋯=αn=1/n, it is well known that there exists a fair division with n−1 cuts, and this is optimal. For arbitrary demands on the other hand, folklore arguments from algebraic topology show that O(nlogn) cuts suffice, and this has been the state of the art for decades. Here, we improve the state of affairs in two ways: we prove that disproportionate division may always be achieved with 3n−4 cuts, and also give an effective algorithm to construct such a division. We additionally offer a topological conjecture that implies that 2n−2 cuts suffice in general, which would be optimal.