Odisha State Board BSE Odisha 7th Class Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.3 Textbook Exercise Questions and Answers.
BSE Odisha Class 7 Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.3
Question 1.
ପ୍ରଥମ ପରିମେୟ ସଂଖ୍ୟାରୁ ଦ୍ଵିତୀୟ ପରିମେୟ ସଂଖ୍ୟାକୁ ବିୟୋଗ କର ।
(କ) \(\frac{11}{2}, \frac{5}{4}\)
ସମାଧାନ:
\(=\frac{11 \times 2-5 \times 1}{4}=\frac{22-5}{4}=\frac{17}{4}=4 \frac{1}{4}\)
(ଖ) \(\frac{-3}{11}, \frac{7}{11}\)
ସମାଧାନ:
= \(\frac{-3}{11}+\frac{-7}{11}=\frac{(-3)+(-7)}{11}=\frac{-10}{11}\)
(ଗ) \(\frac{5}{4}, \frac{-4}{3}\)
ସମାଧାନ:
= \(\frac{5}{4}+\frac{4}{3}=\frac{5 \times 3+4 \times 4}{4 \times 3}=\frac{15+16}{12}=\frac{31}{12}=2 \frac{7}{12}\)
(ଘ) \(\frac{5}{42},\left(\frac{-6}{21}\right)\)
ସମାଧାନ:
= \(\frac{5}{42}+\frac{6}{21}=\frac{5+6 \times 2}{42}=\frac{5+12}{42}=\frac{17}{42}\)
Question 2.
ମାନ ନିର୍ଣ୍ଣୟ କର ।
(କ) \(\frac{6}{7}-\frac{-5}{7}\)
ସମାଧାନ:
= \(\frac{6}{7}+\frac{5}{7}=\frac{6+5}{7}=\frac{11}{7}=1 \frac{4}{7}\)
(ଖ) \(\frac{7}{24}-\frac{5}{36}\)
ସମାଧାନ:
= \(\frac{7}{24}+\left(\frac{-5}{36}\right)=\frac{7 \times 3+(-5) \times 2}{72}=\frac{21+(-10)}{72}=\frac{11}{72}\)
(ଗ) \(\frac{9}{10}-\frac{7}{-15}\)
ସମାଧାନ:
= \(\frac{9}{10}+\frac{7}{15}=\frac{9 \times 3+7 \times 2}{30}=\frac{27+14}{30}=\frac{41}{30}=1 \frac{11}{30}\)
(ଘ) \(\frac{8}{23}-\frac{5}{11}\)
ସମାଧାନ:
\(=\frac{8}{23}+\left(\frac{-5}{11}\right)=\frac{8 \times 11+(-5) \times 23}{23 \times 11}=\frac{88+(-115)}{253}=\frac{-27}{253}\)