Znajdź wielkość, której: a) 13% to 18,2 kg b)66 2\3% (66 i dwie trzecie) to 24 cm c) 33 1\3 (33 i jedna trzecia) to 18 dm d) 43% to 107,5 km e) 125% to 2h 30 min f) 175% to 3m 50cm g) 210% to 3h 30 min h) 1500% to 2 kg 75dag
Prosiłabym tym sposobem: np. 40% to 88 dag 1% - 88 : 40 = 2,2 100% - 2,2 * 100 = 220
Z góry dziękuje za rozwiązanie
WujekVit
A) (18,2kg*100%)/13% = 140kg b) (24cm*100%)/66 2/3% = 36cm c) (18dm*100%)/33 1/3% = 54 dm d) (107,5 km*100%)/43% = 250 km e) (150min*100%)/125% = 120min = 2h f) (350cm*100%)/175% = 200 cm = 2 m g) (210min*100%)/210% = 100min h) (275dag*100%)/1500% = 18 1/3 dag
b) (24cm*100%)/66 2/3% = 36cm
c) (18dm*100%)/33 1/3% = 54 dm
d) (107,5 km*100%)/43% = 250 km
e) (150min*100%)/125% = 120min = 2h
f) (350cm*100%)/175% = 200 cm = 2 m
g) (210min*100%)/210% = 100min
h) (275dag*100%)/1500% = 18 1/3 dag