1998 Budapest Lotto Open – Singles

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Singles
1998 Budapest Lotto Open
Final
Champion	Virginia Ruano Pascual
Runner-up	Silvia Farina
Score	6–4, 4–6, 6–3
Details
Draw	32
Seeds	8
Events
Singles Doubles
← 1997 · Budapest Lotto Open · 1999 →

Amanda Coetzer was the defending champion but did not compete that year.

Seventh-seeded Virginia Ruano Pascual won in the final 6–4, 4–6, 6–3 against Silvia Farina.

A champion seed is indicated in bold text while text in italics indicates the round in which that seed was eliminated.

	First round							Second round							Quarterfinals							Semifinals							Final
	1	S Testud	6	6
		M Maruska	3	4				1	S Testud	7	7
	Q	E Bes Ostariz	6	6				Q	E Bes Ostariz	6	6
		J Husárová	3	3											1	S Testud	7	6	4
		L Golarsa	0	6	7										6	F Li	5	7	6
		S Cecchini	6	2	5				L Golarsa	2	4
		W Probst	1	0				6	F Li	6	6
	5	F Li	6	6																		6	F Li	6	2	3
	3	S Farina	7	6																		3	S Farina	4	6	6
	LL	S Krivencheva	6	3				3	S Farina	6	6
		C Torrens Valero	6	6					C Torrens Valero	2	2
		P Suárez	3	2											3	S Farina	6	6
		A D-Balleret	2	4											8	M Sánchez Lorenzo	1	2
		L Ghirardi	6	6					L Ghirardi	4	6	3
	Q	R Sandu	4	6	4			8	M Sánchez Lorenzo	6	4	6
	8	M Sánchez Lorenzo	6	4	6																								3	Silvia Farina	4	6	3
	5	S Pitkowski	6	6																									7	V Ruano Pascual	6	4	6
	WC	Z Gubacsi	3	4				5	S Pitkowski	2	7	6
		K Studeníková	6	3	3				E Makarova	6	5	3
		E Makarova	4	6	6										5	S Pitkowski	5	6	6
		N Dechy	3	4											WC	J Halard-Decugis	7	3	2
	WC	J Halard-Decugis	6	6				WC	J Halard-Decugis	6	6
		G León García	6	6					G León García	3	0
	4	A-G Sidot	2	1																		5	S Pitkowski	6	1	5
	7	V Ruano Pascual	2	6	6																	7	V Ruano Pascual	2	6	7
	LL	S Načuk	6	1	2			7	V Ruano Pascual	6	4	7
	Q	T Garbin	4	6	1				F Lubiani	0	6	6
		F Lubiani	6	2	6										7	V Ruano Pascual	6	6
		E Gagliardi	3	5												R Kuti-Kis	3	0
		R Kuti-Kis	6	7					R Kuti-Kis	6	6
		C Cristea	6	6					C Cristea	3	2
	2	D Van Roost	4	0

• 1998 Budapest Lotto Open draw