# BSE Odisha 7th Class Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.5

Odisha State Board BSE Odisha 7th Class Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.5 Textbook Exercise Questions and Answers.

## BSE Odisha Class 7 Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.5

Question 1.
ନିମ୍ନଲିଖତ ସଂଖ୍ୟାମାନଙ୍କର ଗୁଣନାତ୍ମକ ବିଲୋମୀ ନିର୍ଣୟ କର ।

(କ) $$\frac{5}{9}$$
ସମାଧାନ:
$$\frac{9}{5}$$

(ଖ) $$\frac{-4}{3}$$
ସମାଧାନ:
$$\frac{-3}{4}$$

(ଗ) -2
ସମାଧାନ:
$$-\frac{1}{2}$$

(ଘ) 8
ସମାଧାନ:
$$\frac{1}{8}$$

(ଙ) 1 $$\frac{1}{2}$$
ସମାଧାନ:
1 $$\frac{1}{2}$$ = $$\frac{3}{2}$$ ∴ ଗୁଣନାତ୍ମକ ବିଲୋମୀ = $$\frac{3}{2}$$

(ଚ) $$\frac{11}{-12}$$
ସମାଧାନ:
$$\frac{-12}{11}$$

(ଛ) $$\frac{-2}{-19}$$
ସମାଧାନ:
$$\frac{-19}{-2}$$ ଚା $$\frac{19}{2}$$

(ଜ) -2 $$\frac{1}{7}$$
ସମାଧାନ:
-2 $$\frac{1}{7}$$ = $$\frac{-15}{7}$$ ∴ ଗୁଣନାତ୍ମକ ବିଲୋମୀ = $$\frac{7}{15}$$

Question 2.
ଭାଗଫଳ ଲେଖ ।

(କ) 3 ÷ $$\frac{4}{5}$$
ସମାଧାନ:
3 × ($$\frac{4}{5}$$ ର ବ୍ୟୁତ୍‌କ୍ରମ) =  3 × $$\frac{5}{4}$$ = $$\frac{15}{4}$$

(ଖ) $$\frac{-3}{5}$$ ÷ 2
ସମାଧାନ:
= $$\frac{-3}{5}$$ × (2 ର ବ୍ୟୁତ୍‌କ୍ରମ) = $$\frac{-3}{5}$$ × $$\frac{1}{2}$$ = –$$\frac{3}{10}$$

(ଗ) $$\frac{-4}{7}$$ ÷ 3
ସମାଧାନ:
$$\frac{-4}{7}$$ × (3 ର ବ୍ୟୁତ୍‌କ୍ରମ) = $$\frac{-4}{7}$$ × $$\frac{1}{3}$$ = –$$\frac{4}{21}$$

(ଘ) $$\frac{1}{5} \div \frac{6}{7}$$
ସମାଧାନ:
= $$\frac{1}{5}$$ × ($$\frac{6}{7}$$ ର ବ୍ୟୁତ୍‌କ୍ରମ) = $$\frac{1}{5}$$ × $$\frac{7}{6}$$ = $$\frac{7}{30}$$

(ଙ) $$\frac{-1}{8} \div \frac{3}{4}$$
ସମାଧାନ:
= $$\frac{-1}{8}$$ × ($$\frac{3}{4}$$ ର ବ୍ୟୁତ୍‌କ୍ରମ) = $$-\frac{1}{8} \times \frac{4}{3}=-\frac{4}{24}=-\frac{1}{6}$$

(ଚ) $$\frac{-7}{6} \div\left(\frac{-2}{3}\right)$$
ସମାଧାନ:
$$\frac{-7}{6}$$ × ($$\frac{-2}{3}$$ ର ବ୍ୟୁତ୍‌କ୍ରମ) = $$=-\frac{7}{6} \times\frac{-3}{2}=\frac{(-7) \times(-3)}{6 \times 2}=\frac{7}{4}$$

(ଛ) $$\frac{-5}{6} \div\left(\frac{-1}{4}\right)$$
ସମାଧାନ:
$$\frac{-5}{6}$$ × (-$$\frac{1}{4}$$ ର ବ୍ୟୁତ୍‌କ୍ରମ) = $$-\frac{5}{6} \times \frac{-4}{1}=\frac{(-5) \times(-4)}{6}=\frac{10}{3}$$

(ଜ) $$\frac{-3}{13} \div\left(\frac{-4}{65}\right)$$
ସମାଧାନ:
$$\frac{-3}{13}$$ × (-$$\frac{4}{65}$$ ର ବ୍ୟୁତ୍‌କ୍ରମ) = $$\left(-\frac{3}{13}\right) \times\left(-\frac{65}{4}\right)=\frac{(-3) \times(-65)}{13 \times 4}=\frac{3 \times 65}{13 \times 4}=\frac{3 \times 13 \times 5}{13 \times 4}=\frac{15}{4}=\frac{3}{4}$$