# BSE Odisha 7th Class Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.3

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:
$$=\frac{8}{23}+\left(\frac{-5}{11}\right)=\frac{8 \times 11+(-5) \times 23}{23 \times 11}=\frac{88+(-115)}{253}=\frac{-27}{253}$$