Rozwiązania zadań z matematyki
Przygotowanie do kartkówki dla klasy 6
Zadanie 8
- a) $0,3 \cdot 0,7 + 0,2^2 = 0,21 + 0,04 = \mathbf{0,25}$
- b) $4,5+3,6-0,4+0,05 = 8,1 – 0,4 + 0,05 = \mathbf{7,75}$
- c) $0,54:0,9:(1-0,7) = 0,6 : 0,3 = \mathbf{2}$
- d) $1\frac{5}{6}+\frac{2}{3} = 1\frac{5}{6} + \frac{4}{6} = 1\frac{9}{6} = \mathbf{2\frac{1}{2}}$
- e) $4\frac{2}{5}\cdot1\frac{1}{2} = \frac{22}{5} \cdot \frac{3}{2} = \frac{66}{10} = \mathbf{6\frac{3}{5}}$
- f) $2\frac{3}{5}-\frac{11}{20} = 2\frac{12}{20} – \frac{11}{20} = \mathbf{2\frac{1}{20}}$
- g) $6\frac{1}{2}:1\frac{3}{10} = \frac{13}{2} : \frac{13}{10} = \frac{13}{2} \cdot \frac{10}{13} = \mathbf{5}$
Zadanie 2
- a) $\frac{19}{25}+0,4 = 0,76 + 0,4 = \mathbf{1,16}$
- b) $3\frac{1}{6}+1,4 = 3\frac{1}{6} + 1\frac{2}{5} = 3\frac{5}{30} + 1\frac{12}{30} = \mathbf{4\frac{17}{30}}$
- c) $3,7-1\frac{4}{25} = 3,7 – 1,16 = \mathbf{2,54}$
- d) $4\frac{1}{7}-2,5 = 4\frac{1}{7} – 2\frac{1}{2} = 3\frac{16}{14} – 2\frac{7}{14} = \mathbf{1\frac{9}{14}}$
- e) $\frac{3}{7}\cdot0,28 = \frac{3}{7} \cdot \frac{28}{100} = \frac{12}{100} = \mathbf{0,12}$
- f) $2,5\cdot1\frac{3}{5} = \frac{5}{2} \cdot \frac{8}{5} = \mathbf{4}$
- g) $4,9:1\frac{2}{5} = 4,9 : 1,4 = \mathbf{3,5}$
- h) $2\frac{1}{3}:2,1 = \frac{7}{3} : \frac{21}{10} = \frac{7}{3} \cdot \frac{10}{21} = \frac{10}{9} = \mathbf{1\frac{1}{9}}$
Zadanie 3
- a) $2\frac{4}{9}+4\frac{3}{8} = 2\frac{32}{72} + 4\frac{27}{72} = \mathbf{6\frac{59}{72}}$
- b) $0,5+3,8 = \mathbf{4,3}$
- c) $8\frac{3}{5}-3\frac{2}{3} = 7\frac{24}{15} – 3\frac{10}{15} = \mathbf{4\frac{14}{15}}$
- d) $5,3-3,7 = \mathbf{1,6}$
- e) $3\frac{3}{7}\cdot2\frac{4}{5} = \frac{24}{7} \cdot \frac{14}{5} = \frac{48}{5} = \mathbf{9\frac{3}{5}}$
- f) $0,4\cdot1,3 = \mathbf{0,52}$
- g) $3\frac{1}{4}:8\frac{2}{3} = \frac{13}{4} : \frac{26}{3} = \frac{13}{4} \cdot \frac{3}{26} = \mathbf{\frac{3}{8}}$
- h) $4,2:0,06 = 420 : 6 = \mathbf{70}$
