1千克蘋果的價錢是a元,買6千克這樣的蘋果應(yīng)付________元,付出20元,應(yīng)找回________元(20>6a);當(dāng)a=1.6時,應(yīng)找回________元.
6a 20-6a 10.4
分析:(1)根據(jù)“單價×數(shù)量=總價”求出買6千克蘋果的總價;
(2)根據(jù)“所付的總錢數(shù)-應(yīng)付的錢數(shù)=應(yīng)找回的錢數(shù)”進(jìn)行解答即可;
(3)直接代入數(shù)據(jù)解答即可.
解答:(1)a×6=6a(元);
(2)20-6a(元);
(3)20-6×1.6,
=20-9.6,
=10.4(元),
答:買6千克應(yīng)付6a元,付出20元,應(yīng)找回20-6a元;當(dāng)a=1.6時,應(yīng)找回10.4元.
故答案為:6a,20-6a,10.4.
點評:此題考查了用字母表示數(shù),解題關(guān)鍵是先根據(jù)單價、數(shù)量和總價三者之間的關(guān)系求出買6千克蘋果的總價,進(jìn)而根據(jù)所付的總錢數(shù)、應(yīng)付的錢數(shù)和應(yīng)找回的錢數(shù)三者之間的關(guān)系解答.