Демонстрационная контрольная КР-3 ГДЗ Контрольные Крайнева Виленкин 6 класс (Математика)

$$\frac{7}{16}=\frac{21}{48},\qquad \frac{11}{24}=\frac{22}{48},\qquad \frac{21}{48}<\frac{22}{48},$$

значит, $$\frac{7}{16}<\frac{11}{24}.$$

$$\frac{5}{11}=\frac{60}{132},\qquad \frac{5}{12}=\frac{55}{132},\qquad \frac{60}{132}>\frac{55}{132},$$

значит, $$\frac{5}{11}>\frac{5}{12}.$$

$$\frac{46}{45}=1\frac{1}{45}>1,\qquad 0{,}81<1,$$

значит, $$\frac{46}{45}>0{,}81.$$

$$\frac{17}{60}-\frac{7}{30}+\frac{1}{40}=\frac{34}{120}-\frac{28}{120}+\frac{3}{120}=\frac{9}{120}=\frac{3}{40}.$$

$$10-4\frac{6}{13}=9\frac{13}{13}-4\frac{6}{13}=5\frac{7}{13}.$$

$$1\frac{3}{4}+2\frac{5}{6}=1\frac{9}{12}+2\frac{10}{12}=3\frac{19}{12}=4\frac{7}{12}.$$

$$3\frac{4}{15}+5\frac{11}{12}=3\frac{16}{60}+5\frac{55}{60}=8\frac{71}{60}=9\frac{11}{60}.$$

$$6\frac{1}{6}-3\frac{4}{9}=6\frac{3}{18}-3\frac{8}{18}=5\frac{21}{18}-3\frac{8}{18}=2\frac{13}{18}.$$

$$3\frac{3}{8}+2\frac{5}{6}=3\frac{9}{24}+2\frac{20}{24}=5\frac{29}{24}=6\frac{5}{24}.$$

Планировалось собрать $$6\frac{5}{24}$$ т зерна. Из-за засухи собрали на $$1\frac{1}{4}=1\frac{6}{24}$$ т меньше:

$$6\frac{5}{24}-1\frac{6}{24}=5\frac{29}{24}-1\frac{6}{24}=4\frac{23}{24}.$$

$$a+3\frac{7}{9}=6\frac{5}{12}$$

$$a=6\frac{5}{12}-3\frac{7}{9}=6\frac{15}{36}-3\frac{28}{36}=5\frac{51}{36}-3\frac{28}{36}=2\frac{23}{36}.$$

$$2{,}75\cdot(c-5{,}03)=4{,}95$$

$$c-5{,}03=4{,}95:2{,}75=1{,}8$$

$$c=1{,}8+5{,}03=6{,}83.$$

Ответ: $$\frac{7}{16}<\frac{11}{24};\ \frac{5}{11}>\frac{5}{12};\ \frac{46}{45}>0{,}81;$$ $$\frac{3}{40};\ 5\frac{7}{13};\ 4\frac{7}{12};\ 9\frac{11}{60};\ 2\frac{13}{18};$$ $$4\frac{23}{24}\text{ т};\ a=2\frac{23}{36};\ c=6{,}83.$$