Selina solutions

# Question 1

Q1) Add the following fractions :

(i)1\frac{3}{4}\ and\ \frac{3}{8}

(ii)\frac{2}{5},2\frac{3}{15}\ and\ \frac{7}{10}

(iii)1\frac{7}{8}+1\frac{1}{2}+1\frac{3}{4}

(iv)3\frac{3}{4}+2\frac{1}{6}+1\frac{5}{8}

(v)2\frac{8}{9},\frac{11}{18}\ and\ 3\frac{5}{6}

(vi)3\frac{1}{8}+55\frac{1}{2}+\frac{5}{16}

Solution:

(i)1\frac{3}{4}\ and\ \frac{3}{8}

=\frac{7}{4}+\frac{3}{8}

=\frac{7\times2}{4\times2}+\frac{3}{8}\ \ \ \ (L.C.M\ of4,8=8)

=\frac{14}{8}+\frac{3}{8}=\frac{14+3}{8}=\frac{17}{8}=2\frac{1}{8}

(ii)\frac{2}{5},2\frac{3}{15}\ and\ \frac{7}{10}

=\frac{2}{5}+\frac{33}{15}+\frac{7}{10}

=\frac{2\times6}{5\times6}+\frac{33\times2}{15\times2}+\frac{7\times3}{10\times3} (L.C.M of 5,15 and 10 = 30)

=\frac{12}{30}+\frac{66}{30}+\frac{21}{30}=\frac{12+66+21}{30}=\frac{99}{30}=\frac{99\div3}{30\div3}

=\frac{33}{10}=3\frac{3}{10}

(iii)1\frac{7}{8}+1\frac{1}{2}+1\frac{3}{4}

=\frac{1\times8+7}{8}+\frac{1\times2+1}{2}+\frac{1\times4+3}{4}

=\frac{15}{8}+\frac{3}{2}+\frac{7}{4}=\frac{15\times1}{8\times1}+\frac{3\times4}{2\times4}+\frac{7\times2}{4\times2}

(L.C.M 8,2 and 4 is 8)

=\frac{41}{8}=5\frac{1}{8}

(iv)3\frac{3}{4}+2\frac{1}{6}+1\frac{5}{8}

=\frac{90}{24}+\frac{52}{24}+\frac{39}{24}=\frac{181}{24}=7\frac{13}{24}

(v)2\frac{8}{9},\frac{11}{18}\ and\ 3\frac{5}{6}

(L.C.M of 9,18 and 6 = 18)

=\frac{22}{3}=7\frac{1}{3}

(vi)3\frac{1}{8}+55\frac{1}{2}+\frac{5}{16}

=\frac{25}{8}+\frac{65}{12}+\frac{5}{16}

(L.C.M of 8,12 and 16 is 48)

=\frac{425}{48}=8\frac{41}{48}

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