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Tentukan Basil pengurangan berbagai bentuk pecahan berikut! [tex]a. \: 6

Berikut ini adalah pertanyaan dari rezaganzz1432 pada mata pelajaran Matematika untuk jenjang Sekolah Dasar

Tentukan Basil pengurangan berbagai bentuk pecahan berikut!a. \: 6 \frac{1}{4} - 0.25 =
b. \: 65\% - \frac{1}{10} =
c. \: 3 \frac{4}{8} - 1.2 =
d. \: 8.8 - 3 \frac{1}{2} =
e. \: 3 \frac{1}{4} - 25\% =
f. \: 4.5 - 2 \frac{5}{8} =
g. \: 75\% - 0.055 =
h. \: 4 \frac{1}{8} - 2.375 =
i. \: 5 \frac{7}{8} - 15\% =
j. \: 9.55 - 120\% =
maaf kalo banyak , yang titik itu koma/desimal, tolong bantu jawab
​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawab:

\rm(a.)~~=\bf6\\\rm(b.)~~\bf\frac{11}{20}\\\rm(c.)~~=\bf2\frac{3}{10}\\\rm(d.)~~\bf5\frac{3}{10}\\\rm(e.)~~=\bf3\\\rm(f.)~~\bf1\frac{7}{8}\\\rm(g.)~~=\bf\frac{139}{200}\\\rm(h.)~~\bf 1\frac{3}{4}\\\rm(i.)~~=\bf5\frac{29}{40}\\\rm(j.)~~\bf8\frac{7}{20}

Penjelasan dengan langkah-langkah:

\rm(a.)~~6\frac{1}{4}-0,25=6\frac{1}{4}-\frac{25}{100}=6\frac{1}{4}-\frac{25\div25}{100\div25}\\=6\frac{1}{4}-\frac{1}{4}=6+(\frac{1}{4}-\frac{1}{4})=6+0=\bf6\\\\\rm(b.)~~65\%-\frac{1}{10}=\frac{65}{100}-\frac{1\times10}{10\times10}=\frac{65}{100}-\frac{10}{100}\\=\frac{65-10}{100}=\frac{55}{100}=\frac{55\div5}{100\div5}=\bf\frac{11}{20}

\rm(c.)~~3\frac{4}{8}-1,2=(3+\frac{4}{8})-(1+0,2)\\=(3+\frac{4\div4}{8\div4})-(1+\frac{2}{10})\\=(3+\frac{1}{2})-(1+\frac{2\div2}{10\div2})\\=(3+\frac{1}{2})-(1+\frac{1}{5})\\=(3+\frac{1}{2})-1-\frac{1}{5}\\=(3-1)+(\frac{1}{2}-\frac{1}{5})=2+(\frac{1}{2}-\frac{1}{5})\\=2+(\frac{5}{2\times5}-\frac{2}{5\times2})=2+(\frac{5}{10}-\frac{2}{10})\\=2+\frac{5-2}{10}=2+\frac{3}{10}=\bf2\frac{3}{10}

\rm(d.)~~8,8-3\frac{1}{2}=(8+0,8)-(3+\frac{1}{2})\\=(8+\frac{8}{10})-3-\frac{1}{2}=(8-3)+(\frac{8}{10}-\frac{1}{2})\\=5+(\frac{8}{10}-\frac{1}{2})=5+(\frac{8}{10}-\frac{5}{2\times5})\\=5+(\frac{8}{10}-\frac{5}{10})=5+\frac{8-5}{10}=5+\frac{3}{10}\\=\bf5\frac{3}{10}

\rm(e.)~~3\frac{1}{4}-25\%=3\frac{1}{4}-\frac{25}{100}=3\frac{1}{4}-\frac{25\div25}{100\div25}\\=3\frac{1}{4}-\frac{1}{4}=3+(\frac{1}{4}-\frac{1}{4})=3+0=\bf3

\rm(f.)~~4,5-2\frac{5}{8}=4\frac{5}{10}-2\frac{5}{8}=4\frac{5\div5\times4}{10\div5\times4}-2\frac{5}{8}\\=4\frac{1\times4}{2\times4}-2\frac{5}{8}=4\frac{4}{8}-2\frac{5}{8}=4+\frac{4}{8}-(2+\frac{5}{8})\\=4+\frac{4}{8}-2-\frac{5}{8}=(4-2)+(\frac{4}{8}-\frac{5}{8})\\=2+(\frac{4}{8}-\frac{5}{8})=2+\frac{4-5}{8}=2-\frac{5-4}{8}=2-\frac{1}{8}\\=(2-1)+(1-\frac{1}{8})=1+(\frac{8}{8}-\frac{1}{8})=1+\frac{8-1}{8}\\=1+\frac{7}{8}=\bf1\frac{7}{8}

\rm(g.)~~75\%-0,055=\frac{75}{100}-\frac{55}{1000}=\frac{750}{1000}-\frac{55}{1000}\\\\=\frac{750-55}{1000}=\frac{695}{1000}=\frac{695}{1000}=\bf\frac{139}{200}

\rm(h.)~~4\frac{1}{8}-2,375=(4+\frac{1}{8})-(2+0,375)\\=(4+\frac{1}{8})-(2+\frac{375}{1000})\\=~~~~~(4+\frac{1}{8})-(2+\frac{375\div125}{1000\div125})\\=(4+\frac{1}{8})-(2+\frac{3}{8})=(4+\frac{1}{8})-2-\frac{3}{8}\\=(4-2)+(\frac{1}{8}-\frac{3}{8})=2+\frac{1-3}{8}=2-\frac{3-1}{8}\\=2-\frac{2}{8}=2-\frac{2\div2}{8\div2}=2-\frac{1}{4}\\=(2-1)+(1-\frac{1}{4})=1+(1-\frac{1}{4})\\=1+(\frac{4}{4}-\frac{1}{4})=1+\frac{4-1}{4}=1+\frac{3}{4}\\=\bf 1\frac{3}{4}

(i.)~~5\frac{7}{8}-15\%=(5+\frac{7}{8})-\frac{15}{100}=5+(\frac{7}{8}-\frac{15}{100})\\=5+(\frac{7\times5}{8\times5}-\frac{15\div5\times2}{100\div5\times2})=5+(\frac{7\times5}{8\times5}-\frac{3\times2}{20\times2})\\=5+(\frac{35}{40}-\frac{6}{40})=5+\frac{35-6}{40}=5+\frac{29}{40}=\bf5\frac{29}{40}

(j.)~~9,55-120\%=(9+\frac{55}{100})-\frac{120}{100}\\=(9+\frac{55\div5}{100\div5})-\frac{100+20}{100}\\=(9+\frac{11}{20})-(\frac{100}{100}+\frac{20}{100})\\=(9+\frac{11}{20})-(1+\frac{20\div5}{100\div5})\\=(9+\frac{11}{20})-(1+\frac{4}{20})\\=(9+\frac{11}{20})-(1+\frac{20\div5}{100\div5})\\=(9+\frac{11}{20})-1-\frac{4}{20}\\=(9-1)+(\frac{11}{20}-\frac{4}{20})\\=8+(\frac{11}{20}-\frac{4}{20})\\=8+\frac{11-4}{20}=8+\frac{7}{20}\\=\bf8\frac{7}{20}

(xcvi)

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Last Update: Thu, 10 Nov 22
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